problem solving examples of quadratic equation

20 Quadratic Equation Examples with Answers

Quadratic equations have the form $latex ax^2+bx+c$. Depending on the type of quadratic equation we have, we can use various methods to solve it. Some of the most important methods are methods for incomplete quadratic equations, the factoring method, the method of completing the square, and the quadratic formula.

Here, we will look at a brief summary of solving quadratic equations. Then, we will look at 20 quadratic equation examples with answers to master the various methods of solving these typesof equations.

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Learning to solve quadratic equations with examples.

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Summary of quadratic equations

20 quadratic equation examples with answers.

Recall that quadratic equations are equations in which the variables have a maximum power of 2. These equations have the general form $latex ax^2+bx+c=0$. For example, the equations $latex 4x^2+x+2=0$ and $latex 2x^2-2x-3=0$ are quadratic equations.

There are several methods that we can use to solve quadratic equations depending on the type of equation we have. The most common methods are by factoring, completing the square, and using the quadratic formula.

We can solve incomplete quadratic equations of the form $latex ax^2+c=0$ by completely isolating x ². Then we can take the square root of both sides of the equation.

To solve incomplete quadratic equations of the form $latex ax^2+bx=0$, we have to factor x from both terms. Then, we can form an equation with each factor and solve them.

When we have complete quadratic equations of the form $latex ax^2+bx+c=0$, we can use factorization and write the equation in the form $latex (x+p)(x+q)=0$ which will allow us to find its roots easily.

Finally, when it is not possible to solve a quadratic equation with factorization, we can use the general quadratic formula:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

You can learn or review the methods for solving quadratic equations by visiting our article: Solving Quadratic Equations – Methods and Examples .

The following 20 quadratic equation examples have their respective solutions using different methods. Try to solve the problems yourself before looking at the solution.

Find the solutions to the equation $latex x^2-25=0$.

This equation is an incomplete quadratic equation that does not have the bx term. We can solve this equation by isolating the x ² term and taking the square root of both sides of the equation:

$latex x^2-25=0$

$latex x^2=25$

Taking the square root of both sides, we have:

$latex x=\pm \sqrt{25}$

$latex x=\pm 5$

The solutions to the equation are $latex x=5$ and $latex x=-5$.

What are the solutions to the equation $latex x^2-4x=0$?

This is an incomplete quadratic equation that does not have the c term. To solve this equation, we need to factor x and then form an equation with each factor:

$latex x^2-4x=0$

$latex x(x-4)=0$

Forming an equation with each factor, we have:

$latex x=0~~$ or $latex ~~x-4=0$

$latex x=0~~$ or $latex ~~x=4$

The solutions of the equation are $latex x=0$ and $latex x=4$.

Solve the equation $latex x^2+5x+6=0$.

We can solve this equation using the factoring method. For this, we look for two numbers that when multiplied are equal to 6 and when added are equal to 5.

The two numbers we are looking for are 2 and 3. Therefore, we have:

$latex x^2+5x+6=0$

$latex (x+2)(x+3)=0$

Now, we form an equation with each factor and solve:

$latex x+2=0~~$ or $latex ~~x+3=0$

$latex x=-2~~$ or $latex ~~x=-3$

The solutions to the equation are $latex x=-2$ and $latex x=-3$.

What are the roots to the equation $latex x^2-6x-7=0$?

We can solve this equation by factoring. For this, we look for two numbers, which when multiplied are equal to -7 and when added are equal to -6.

The numbers we are looking for are -7 and 1. Therefore, we have:

$latex x^2-6x-7=0$

$latex (x-7)(x+1)=0$

$latex x-7=0~~$ or $latex ~~x+1=0$

$latex x=7~~$ or $latex ~~x=-1$

The solutions to the equation are $latex x=7$ and $latex x=-1$.

Solve the equation $latex 2x^2-32=0$.

This equation is an incomplete quadratic equation of the form $latex ax^2+c=0$. We can solve this equation by solving for x ² and taking the square root of both sides:

$latex 2x^2-32=0$

$latex 2x^2=32$

$latex x^2=16$

$latex x=\pm \sqrt{16}$

$latex x=\pm 4$

The solutions of the equation are $latex x=4$ and $latex x=-4$.

Find the roots to the equation $latex 4x^2+8x=0$.

This equation is an incomplete quadratic equation of the form $latex ax^2+bx=0$. To solve this equation, we can factor 4 x from both terms and then form an equation with each factor:

$latex 4x^2+8x=0$

$latex 4x(x+2)=0$

$latex 4x=0~~$ or $latex ~~x+2=0$

$latex x=0~~$ or $latex ~~x=-2$

The solutions to the equation are $latex x=0$ and $latex x=-2$.

Find the solutions to the equation $latex x^2+4x-6=0$ using the method of completing the square.

To complete the square, we take the coefficient b , divide it by 2, and square it. Then, we have:

$$\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2$$

Now, we add and subtract that value to the quadratic equation:

$$x^2+4x-6=x^2+4x+2^2-2^2-6$$

Now, we can complete the square and simplify:

$latex = (x+2)^2-4-6$

$latex = (x+2)^2-10$

Thus, we have the equation:

$latex (x+2)^2=10$

⇒ $latex x+2=\sqrt{10}$

Solving, we have:

⇒ $latex x=-2\pm \sqrt{10}$

Find the solutions of the equation $latex x^2-8x+4=0$ to two decimal places.

We can identify the coefficients $latex a=1$, $latex b=-8$, and $latex c=4$. Using these values in the quadratic formula, we have:

$$x=\frac{-(-8)\pm \sqrt{( -8)^2-4(1)(4)}}{2(1)}$$

$$=\frac{8\pm \sqrt{64-16}}{2}$$

$$=\frac{8\pm \sqrt{48}}{2}$$

$$x=7.46 \text{ or } 0.54$$

The solutions are $latex x=7.46$ and $latex x=0.54$.

Solve the equation $latex 2x^2+8x-10=0$ using the method of completing the square.

We can divide the entire equation by 2 to make the coefficient of the quadratic term equal to 1:

⇒ $latex x^2+4x-5=0$

Now, we take the coefficient b , divide it by 2 and square it. Therefore, we have:

Adding and subtracting that value to the quadratic expression, we have:

$$x^2+4x-5=x^2+4x+2^2-2^2-5$$

Completing the square and simplifying, we have:

$latex = (x+2)^2-4-5$

$latex = (x+2)^2-9$

Now, we write the equation like this:

⇒ $latex (x+2)^2=9$

And we take the square root of both sides:

⇒ $latex x+2=\pm 3$

⇒ $latex x=1$ or $latex x=-5$

Use the quadratic formula to solve the equation $latex x^2-10x+25=0$.

We can identify the coefficients $latex a=1$, $latex b=-10$, and $latex c=25$. Using them in the general quadratic formula, we have:

$$x=\frac{-(-10)\pm \sqrt{( -10)^2-4(1)(25)}}{2(1)}$$

$$=\frac{10\pm \sqrt{100-100}}{2}$$

$$=\frac{10\pm \sqrt{0}}{2}$$

$$=\frac{10}{2}$$

$latex x=5$

In this case, we have a single repeated root $latex x=5$.

Find the roots of the equation $latex 4x^2+5=2x^2+20$.

First, we need to simplify this equation and write it in the form $latex ax^2+bx+c=0$:

$latex 4x^2+5=2x^2+20$

$latex 4x^2-2x^2+5-20=0$

$latex 2x^2-15=0$

Now, we can see that it is an incomplete quadratic equation that does not have the bx term. Therefore, we can solve it by solving for x ² and taking the square root of both sides:

$latex 2x^2=15$

$latex x^2=\frac{15}{2}$

$latex x=\pm \sqrt{\frac{15}{2}}$

Solve the equation $latex 5x^2+5x=2x^2+10x$.

To solve the equation, we have to start by writing it in the form $latex ax^2+bx+c=0$. Therefore, we have:

$latex 5x^2+5x=2x^2+10x$

$latex 5x^2-2x^2+5x-10x=0$

$latex 3x^2-5x=0$

We see that it is an incomplete equation that does not have the term c . Thus, we can solve it by factoring x :

$latex x(3x-5)=0$

$latex x=0~~$ or $latex ~~3x-5=0$

$latex x=0~~$ or $latex ~~x=\frac{5}{3}$

Solve the equation $latex 3x^2+5x-4=x^2-2x$ using the general quadratic formula. Express the solutions to two decimal places.

To use the general formula, we have to start by writing the equation in the form $latex ax^2+bx+c=0$:

$latex 3x^2+5x-4=x^2-2x$

$latex 2x^2+7x-4=0$

Now, we have the coefficients $latex a=2$, $latex b=7$, and $latex c=-4$. Therefore, using these values in the quadratic formula, we have:

$$x=\frac{-(7)\pm \sqrt{( 7)^2-4(2)(-4)}}{2(2)}$$

$$=\frac{-7\pm \sqrt{49+32}}{4}$$

$$=\frac{-7\pm \sqrt{81}}{2}$$

$$=\frac{-7\pm 9}{2}$$

$$=-8 \text{ or }1$$

The solutions of the equation are $latex x=-8$ and $latex x=1$.

Solve the following equation $$(3x+1)(2x-1)-(x+2)^2=5$$

To solve this equation, we need to expand the parentheses and simplify to the form $latex ax^2+bx+c=0$. Therefore, we have:

$$(3x+1)(2x-1)-(x+2)^2=5$$

$$6x^2-x-1-(x^2+4x+4)=5$$

$latex 5x^2-5x-5=5$

$latex 5x^2-5x-10=0$

Now, we can solve by factorization:

$latex 5(x^2-x-2)=0$

$latex 5(x+1)(x-2)=0$

$latex x+1=0~~$ or $latex ~~x-2=0$

$latex x=-1~~$ or $latex ~~x=2$

Use the method of completing the square to solve the equation $latex -x^2+3x+1=-2x^2+6x$.

We have to start by writing the equation in the form $latex ax^2+bx+c=0$:

$latex -x^2+3x+1=-2x^2+6x$

$latex x^2-3x+1=0$

Now, we see that the coefficient b in this equation is equal to -3. Therefore, we have:

$$\left(\frac{b}{2}\right)^2=\left(\frac{-3}{2}\right)^2$$

Adding and subtracting this value to the quadratic equation, we have:

$$x^2-3x+1=x^2-2x+\left(\frac{-3}{2}\right)^2-\left(\frac{-3}{2}\right)^2+1$$

$latex = (x-\frac{3}{2})^2-\left(\frac{-3}{2}\right)^2+1$

$latex = (x-\frac{3}{2})^2-\frac{5}{4}$

Now, we write the equation as follows:

$latex ⇒ (x-\frac{3}{2})^2=\frac{5}{4}$

⇒ $latex x-\frac{3}{2}=\sqrt{\frac{5}{4}}$

⇒ $latex x-\frac{3}{2}=\frac{\sqrt{5}}{2}$

⇒ $latex x=\frac{3}{2}\pm \frac{\sqrt{5}}{2}$

Prove that the equation $latex 5x^2+4x+10=0$ has no real solutions using the general formula.

We can use the values $latex a=5$, $latex b=4$, and $latex c=10$ in the quadratic formula:

$$x=\frac{-(4)\pm \sqrt{( 4)^2-4(5)(10)}}{2(5)}$$

$$=\frac{-5\pm \sqrt{16-200}}{10}$$

$$=\frac{-5\pm \sqrt{-184}}{10}$$

We can see that we got a negative number inside the square root. $latex \sqrt{-184}$ is not a real number, so the equation has no real roots.

→ Quadratic Equations Calculator

Solve the following equation $$\frac{4}{x-1}+\frac{3}{x}=3$$

This equation does not appear to be quadratic at first glance. However, we can multiply it by $latex x(x-1)$ to eliminate the fractions, and we have:

$$\frac{4}{x-1}+\frac{3}{x}=3$$

$$4x+3(x-1)=3x(x-1)$$

$$4x+3x-3=3x^2-3x$$

$latex 3x^2-10x+3=0$

Now, we can factor this equation to solve it:

$latex (3x-1)(x-3)=0$

$latex 3x-1=0~~$ or $latex ~~x-3=0$

$latex x=\frac{1}{3}~~$ or $latex ~~x=3$

Find the solutions to the following equation $$\frac{2x+1}{x+5}=\frac{3x-1}{x+7}$$

To simplify fractions, we can cross multiply to get:

$$ (2x+1)(x+7)=(3x-1)(x+5)$$

Expanding and simplifying, we have:

$$ 2x^2+15x+7=3x^2+14x-5 $$

$latex x^2-x-12=0$

Factoring and solving, we have:

$latex (x+3)(x-4)=0$

$latex x+3=0~~$ or $latex ~~x-4=0$

$latex x=-3~~$ or $latex ~~x=4$

Find two numbers such that their sum equals 17 and their product equals 60.

To solve this problem, we can form equations using the information in the statement. We use the letters X (smaller number) and Y (larger number) to represent the numbers:

$latex X+Y=17~~[1]$

$latex XY=60~~[2]$

Writing equation 1 as $latex Y=17-X$ and substituting it into the second equation, we have:

$latex X(17-X)=60$

We can expand and write it in the form $latex ax^2+bx+c=0$:

$latex 17X-X^2=60$

$latex X^2-17X+60=0$

Now, we can solve the equation by factoring:

$latex (X-12)(X-5)=0$

$latex X-12=0~~$ or $latex ~~X-5=0$

$latex X=12~~$ or $latex ~~X=5$

Therefore, we have:

• If $latex X=5$, we have $latex Y=17-5=12$. This solution is the correct one because X<Y.
• If $latex X=12$, we have $latex Y=17-12=5$

The numbers are 12 and 5.

If the area of a rectangle is 78 square units and its longest side is 7 units longer than its shortest side, what are the lengths of the sides?

To solve this problem, we have to use the given information to form equations. Let’s represent the shorter side with x . This means that the longest side is equal to x +7.

Now considering that the area of a rectangle is found by multiplying the lengths of its sides, we have:

$latex x(x+7)=78$

Expanding and writing the equation in the form $latex ax^2+bx+c=0$, we have:

$latex x^2+7x-78=0$

Solving by factorization, we have:

$latex (x+13)(x-6)=0$

$latex x+13=0~~$ or $latex ~~x-6=0$

$latex x=-13~~$ or $latex ~~x=6$

Since we can’t have negative lengths, we have $latex x=6$, so the lengths are 6 and 13.

Interested in learning more about quadratic equations? Take a look at these pages:

• Quadratic Equations Calculator
• Solving Quadratic Equations – Methods and Examples
• How to Solve Quadratic Equations? Step-by-Step
• 10 Quadratic Equations Word Problems

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Quadratic Equations

An example of a Quadratic Equation :

The function can make nice curves like this one:

The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x 2 ).

It is also called an "Equation of Degree 2" (because of the "2" on the x )

Standard Form

The Standard Form of a Quadratic Equation looks like this:

• a , b and c are known values. a can't be 0
• x is the variable or unknown (we don't know it yet)

Here are some examples:

Have a Play With It

Play with the Quadratic Equation Explorer so you can see:

• the function's graph, and
• the solutions (called "roots").

Hidden Quadratic Equations!

As we saw before, the Standard Form of a Quadratic Equation is

But sometimes a quadratic equation does not look like that!

For example:

How To Solve Them?

The " solutions " to the Quadratic Equation are where it is equal to zero .

They are also called " roots ", or sometimes " zeros "

There are usually 2 solutions (as shown in this graph).

And there are a few different ways to find the solutions:

Just plug in the values of a, b and c, and do the calculations.

We will look at this method in more detail now.

About the Quadratic Formula

First of all what is that plus/minus thing that looks like ± ?

The ± means there are TWO answers:

x = −b + √(b 2 − 4ac) 2a

x = −b − √(b 2 − 4ac) 2a

Here is an example with two answers:

But it does not always work out like that!

• Imagine if the curve "just touches" the x-axis.
• Or imagine the curve is so high it doesn't even cross the x-axis!

This is where the "Discriminant" helps us ...

Discriminant

Do you see b 2 − 4ac in the formula above? It is called the Discriminant , because it can "discriminate" between the possible types of answer:

• when b 2 − 4ac is positive, we get two Real solutions
• when it is zero we get just ONE real solution (both answers are the same)
• when it is negative we get a pair of Complex solutions

Complex solutions? Let's talk about them after we see how to use the formula.

Using the Quadratic Formula

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x 2 + 6x + 1 = 0

Answer: x = −0.2 or x = −1

Let's check the answers:

Remembering The Formula

A kind reader suggested singing it to "Pop Goes the Weasel":

Try singing it a few times and it will get stuck in your head!

Or you can remember this story:

x = −b ± √(b 2 − 4ac) 2a

"A negative boy was thinking yes or no about going to a party, at the party he talked to a square boy but not to the 4 awesome chicks. It was all over at 2 am. "

Complex Solutions?

When the Discriminant (the value b 2 − 4ac ) is negative we get a pair of Complex solutions ... what does that mean?

It means our answer will include Imaginary Numbers . Wow!

Example: Solve 5x 2 + 2x + 1 = 0

Answer: x = −0.2 ± 0.4 i

The graph does not cross the x-axis. That is why we ended up with complex numbers.

In a way it is easier: we don't need more calculation, we leave it as −0.2 ± 0.4 i .

Example: Solve x 2 − 4x + 6.25 = 0

Answer: x = 2 ± 1.5 i

BUT an upside-down mirror image of our equation does cross the x-axis at 2 ± 1.5 (note: missing the i ).

Just an interesting fact for you!

• Quadratic Equation in Standard Form: ax 2 + bx + c = 0
• Quadratic Equations can be factored
• Quadratic Formula: x = −b ± √(b 2 − 4ac) 2a
• positive, there are 2 real solutions
• zero, there is one real solution
• negative, there are 2 complex solutions

Solving Quadratics by the Quadratic Formula

How to solve quadratic equations using the quadratic formula.

There are times when we are stuck solving a quadratic equation of the form [latex]a{x^2} + bx + c = 0[/latex] because the trinomial on the left side can’t be factored out easily. It doesn’t mean that the quadratic equation has no solution. At this point, we need to call upon the straightforward approach of the quadratic formula to find the solutions of the quadratic equation or put simply, determine the values of [latex]x[/latex] that can satisfy the equation.

In order use the quadratic formula, the quadratic equation that we are solving must be converted into the “standard form”, otherwise, all subsequent steps will not work. The goal is to transform the quadratic equation such that the quadratic expression is isolated on one side of the equation while the opposite side only contains the number zero, [latex]0[/latex].

Take a look at the diagram below.

In this convenient format, the numerical values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are easily identified! Upon knowing those values, we can now substitute them into the quadratic formula then solve for the values of [latex]x[/latex].

• The Quadratic Formula

• Where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are the coefficients of an arbitrary quadratic equation in the standard form, [latex]a{x^2} + bx + c = 0[/latex].

Slow down if you need to. Be careful with every step while simplifying the expressions. This is where common mistakes usually happen because students tend to “relax” which results to errors that could have been prevented, such as in the addition, subtraction, multiplication and/or division of real numbers.

Examples of How to Solve Quadratic Equations by the Quadratic Formula

Example 1 : Solve the quadratic equation below using the Quadratic Formula.

By inspection, it’s obvious that the quadratic equation is in the standard form since the right side is just zero while the rest of the terms stay on the left side. In other words, we have something like this

This is great! What we need to do is simply identify the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] then substitute into the quadratic formula.

That’s it! Make it a habit to always check the solved values of [latex]x[/latex] back into the original equation to verify.

Example 2 : Solve the quadratic equation below using the Quadratic Formula.

This quadratic equation is absolutely not in the form that we want because the right side is NOT zero. I need to eliminate that [latex]7[/latex] on the right side by subtracting both sides by [latex]7[/latex]. That takes care of our problem. After doing so, solve for [latex]x[/latex] as usual.

The final answers are [latex]{x_1} = 1[/latex] and [latex]{x_2} = – {2 \over 3}[/latex].

Example 3 : Solve the quadratic equation below using the Quadratic Formula.

This quadratic equation looks like a “mess”. I have variable [latex]x[/latex]’s and constants on both sides of the equation. If we are faced with something like this, always stick to what we know. Yes, it’s all about the Standard Form. We have to force the right side to be equal to zero. We can do just that in two steps.

I will first subtract both sides by [latex]5x[/latex], and followed by the addition of [latex]8[/latex].

Values we need:

[latex]a = – 1[/latex], [latex]b = – \,8[/latex], and [latex]c = 2[/latex]

Example 4 : Solve the quadratic equation below using the Quadratic Formula.

Well, if you think that Example [latex]3[/latex] is a “mess” then this must be even “messier”. However, you’ll soon realize that they are really very similar.

We first need to perform some cleanup by converting this quadratic equation into standard form. Sounds familiar? Trust me, this problem is not as bad as it looks, as long as we know what to do.

Just to remind you, we want something like this

Therefore, we must do whatever it takes to make the right side of the equation equal to zero. Since we have three terms on the right side, it follows that three steps are required to make it zero.

The solution below starts by adding both sides by [latex]3{x^2}[/latex], followed by subtraction of [latex]3x[/latex], and finally the addition of [latex]5[/latex]. Done!

After making the right side equal to zero, the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are easy to identify. Plug those values into the quadratic formula, and simplify to get the final answers!

Example 5 : Solve the quadratic equation below using the Quadratic Formula.

First, we need to rewrite the given quadratic equation in Standard Form, [latex]a{x^2} + bx + c = 0[/latex].

• Eliminate the [latex]{x^2}[/latex] term on the right side.

• Eliminate the [latex]x[/latex] term on the right side.

• Eliminate the constant on the right side.

After getting the correct standard form in the previous step, it’s now time to plug the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] into the quadratic formula to solve for [latex]x[/latex].

• From the converted standard form, extract the required values.

[latex]a = 1[/latex], [latex]b = – \,4[/latex], and [latex]c = – \,14[/latex]

• Then evaluate these values into the quadratic formula.

You might also like these tutorials:

• Quadratic Formula Practice Problems with Answers
• Solving Quadratic Equations by Square Root Method
• Solving Quadratic Equations by Factoring Method
• Solving Quadratic Equations by Completing the Square

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Course: Algebra 1   >   Unit 14

The quadratic formula, understanding the quadratic formula.

• Worked example: quadratic formula (example 2)
• Worked example: quadratic formula (negative coefficients)
• Quadratic formula
• Using the quadratic formula: number of solutions
• Number of solutions of quadratic equations
• Quadratic formula review
• Discriminant review

Worked example

• a ‍   is the coefficient in front of x 2 ‍   , so here a = 1 ‍   (note that a ‍   can’t equal 0 ‍   -- the x 2 ‍   is what makes it a quadratic).
• b ‍   is the coefficient in front of the x ‍   , so here b = 4 ‍   .
• c ‍   is the constant, or the term without any x ‍   next to it, so here c = − 21 ‍   .

What does the solution tell us?

Second worked example, tips when using the quadratic formula.

• Be careful that the equation is arranged in the right form: a x 2 + b x + c = 0 ‍   or it won’t work!
• Make sure you take the square root of the whole ( b 2 − 4 a c ) ‍   , and that 2 a ‍   is the denominator of everything above it
• Watch your negatives: b 2 ‍   can’t be negative, so if b ‍   starts as negative, make sure it changes to a positive since the square of a negative or a positive is a positive
• Keep the + / − ‍   and always be on the look out for TWO solutions
• If you use a calculator, the answer might be rounded to a certain number of decimal places. If asked for the exact answer (as usually happens) and the square roots can’t be easily simplified, keep the square roots in the answer, e.g. 2 − 10 2 ‍   and 2 + 10 2 ‍  
• Practice using the quadratic formula .
• Watch Sal do an example:
• Prove the quadratic formula:

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Last modified on August 3rd, 2023

Quadratic Equation Word Problems

Here, we will solve different types of quadratic equation-based word problems. Use the appropriate method to solve them:

• By Completing the Square
• By Factoring
• By Quadratic Formula
• By graphing

For each process, follow the following typical steps:

• Make the equation
• Solve for the unknown variable using the appropriate method
• Interpret the result

The product of two consecutive integers is 462. Find the numbers?

Let the numbers be x and x + 1 According to the problem, x(x + 1) = 483 => x 2 + x – 483 = 0 => x 2 + 22x – 21x – 483 = 0 => x(x + 22) – 21(x + 22) = 0 => (x + 22)(x – 21) = 0 => x + 22 = 0 or x – 21 = 0 => x = {-22, 21} Thus, the two consecutive numbers are 21 and 22.

The product of two consecutive positive odd integers is 1 less than four times their sum. What are the two positive integers.

Let the numbers be n and n + 2 According to the problem, => n(n + 2) = 4[n + (n + 2)] – 1 => n 2 + 2n = 4[2n + 2] – 1 => n 2 + 2n = 8n + 7 => n 2 – 6n – 7 = 0 => n 2 -7n + n – 7 = 0 => n(n – 7) + 1(n – 7) = 0 => (n – 7) (n – 1) = 0 => n – 7 = 0 or n – 1 = 0 => n = {7, 1} If n = 7, then n + 2 = 9 If n = 1, then n + 1 = 2 Since 1 and 2 are not possible. The two numbers are 7 and 9

A projectile is launched vertically upwards with an initial velocity of 64 ft/s from a height of 96 feet tower. If height after t seconds is reprented by h(t) = -16t 2 + 64t + 96. Find the maximum height the projectile reaches. Also, find the time it takes to reach the highest point.

Since the graph of the given function is a parabola, it opens downward because the leading coefficient is negative. Thus, to get the maximum height, we have to find the vertex of this parabola. Given the function is in the standard form h(t) = a 2 x + bx + c, the formula to calculate the vertex is: Vertex (h, k) = ${\left\{ \left( \dfrac{-b}{2a}\right) ,h\left( -\dfrac{b}{2a}\right) \right\}}$ => ${\dfrac{-b}{2a}=\dfrac{-64}{2\times \left( -16\right) }}$ = 2 seconds Thus, the time the projectile takes to reach the highest point is 2 seconds ${h\left( \dfrac{-b}{2a}\right)}$ = h(2) = -16(2) 2 – 64(2) + 80 = 144 feet Thus, the maximum height the projectile reaches is 144 feet

The difference between the squares of two consecutive even integers is 68. Find the numbers.

Let the numbers be x and x + 2 According to the problem, (x + 2) 2 – x 2 = 68 => x 2 + 4x + 4 – x 2 = 68 => 4x + 4 = 68 => 4x = 68 – 4 => 4x = 64 => x = 16 Thus the two numbers are 16 and 18

The length of a rectangle is 5 units more than twice the number. The width is 4 unit less than the same number. Given the area of the rectangle is 15 sq. units, find the length and breadth of the rectangle.

Let the number be x Thus, Length = 2x + 5 Breadth = x – 4 According to the problem, (2x + 5)(x – 4) = 15 => 2x 2 – 8x + 5x – 20 – 15 = 0 => 2x 2 – 3x – 35 = 0 => 2x 2 – 10x + 7x – 35 = 0 => 2x(x – 5) + 7(x – 5) = 0 => (x – 5)(2x + 7) = 0 => x – 5 = 0 or 2x + 7 = 0 => x = {5, -7/2} Since we cannot have a negative measurement in mensuration, the number is 5 inches. Now, Length = 2x + 5 = 2(5) + 5 = 15 inches Breadth = x – 4 = 15 – 4 = 11 inches

A rectangular garden is 50 cm long and 34 cm wide, surrounded by a uniform boundary. Find the width of the boundary if the total area is 540 cm².

Given, Length of the garden = 50 cm Width of the garden = 34 cm Let the uniform width of the boundary be = x cm According to the problem, (50 + 2x)(34 + 2x) – 50 × 34 = 540 => 4x 2 + 168x – 540 = 0 => x 2 + 42x – 135 = 0 Since, this quadratic equation is in the standard form ax 2 + bx + c, we will use the quadratic formula, here a = 1, b = 42, c = -135 x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$ => ${\dfrac{-42\pm \sqrt{\left( 42\right) ^{2}-4\times 1\times \left( -135\right) }}{2\times 1}}$ => ${\dfrac{-42\pm \sqrt{1764+540}}{2}}$ => ${\dfrac{-42\pm \sqrt{2304}}{2}}$ => ${\dfrac{-42\pm 48}{2}}$ => ${\dfrac{-42+48}{2}}$ and ${\dfrac{-42-48}{2}}$ => x = {-45, 3} Since we cannot have a negative measurement in mensuration the width of the boundary is 3 cm

The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

Let the length of the other two sides be x and x + 4 According to the problem, (x + 4) 2 + x 2 = 20 2 => x 2 + 8x + 16 + x 2 = 400 => 2x 2 + 8x + 16 = 400 => 2x 2 + 8x – 384 = 0 => x 2 + 4x – 192 = 0 => x 2 + 16x – 12x – 192 = 0 => x(x + 16) – 12(x + 16) = 0 => (x + 16)(x – 12) = 0 => x + 16 = 0 and x – 12 = 0 => x = {-16, 12} Since we cannot have a negative measurement in mensuration, the lengths of the sides are 12 and 16

Jennifer jumped off a cliff into the swimming pool. The function h can express her height as a function of time (t) = -16t 2 +16t + 480, where t is the time in seconds and h is the height in feet. a) How long did it take for Jennifer to attain a maximum length. b) What was the highest point that Jennifer reached. c) Calculate the time when Jennifer hit the water?

Comparing the given function with the given function f(x) = ax 2 + bx + c, here a = -16, b = 16, c = 480 a) Finding the vertex will give us the time taken by Jennifer to reach her maximum height  x = ${-\dfrac{b}{2a}}$ = ${\dfrac{-16}{2\left( -16\right) }}$ = 0.5 seconds Thus Jennifer took 0.5 seconds to reach her maximum height b) Putting the value of the vertex by substitution in the function, we get ${h\left( \dfrac{1}{2}\right) =-16\left( \dfrac{1}{2}\right) ^{2}+16\left( \dfrac{1}{2}\right) +480}$ => ${-16\left( \dfrac{1}{4}\right) +8+480}$ => 484 feet Thus the highest point that Jennifer reached was 484 feet c) When Jennifer hit the water, her height was 0 Thus, by substituting the value of the height in the function, we get -16t 2 +16t + 480 = 0 => -16(t 2 + t – 30) = 0 => t 2 + t – 30 = 0 => t 2 + 6t – 5t – 30 = 0 => t(t + 6) – 5(t + 6) = 0 => (t + 6)(t – 5) = 0 => t + 6 = 0 or t – 5 = 0 => x = {-6, 5} Since time cannot have any negative value, the time taken by Jennifer to hit the water is 5 seconds.

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Quadratic Equation

Quadratic equations are second-degree algebraic expressions and are of the form ax 2 + bx + c = 0. The term "quadratic" comes from the Latin word "quadratus" meaning square, which refers to the fact that the variable x is squared in the equation. In other words, a quadratic equation is an “equation of degree 2.” There are many scenarios where a quadratic equation is used. Did you know that when a rocket is launched, its path is described by a quadratic equation? Further, a quadratic equation has numerous applications in physics, engineering, astronomy, etc.

Quadratic equations have maximum of two solutions, which can be real or complex numbers. These two solutions (values of x) are also called the roots of the quadratic equations and are designated as (α, β). We shall learn more about the roots of a quadratic equation in the below content.

What is Quadratic Equation?

A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is ax 2 + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term. The important condition for an equation to be a quadratic equation is the coefficient of x 2 is a non-zero term (a ≠ 0). For writing a quadratic equation in standard form, the x 2 term is written first, followed by the x term, and finally, the constant term is written.

Further, in real math problems the quadratic equations are presented in different forms: (x - 1)(x + 2) = 0, -x 2 = -3x + 1, 5x(x + 3) = 12x, x 3 = x(x 2 + x - 3). All of these equations need to be transformed into standard form of the quadratic equation before performing further operations.

Roots of a Quadratic Equation

The roots of a quadratic equation are the two values of x, which are obtained by solving the quadratic equation. These roots of the quadratic equation are also called the zeros of the equation. For example, the roots of the equation x 2 - 3x - 4 = 0 are x = -1 and x = 4 because each of them satisfies the equation. i.e.,

• At x = -1, (-1) 2 - 3(-1) - 4 = 1 + 3 - 4 = 0
• At x = 4, (4) 2 - 3(4) - 4 = 16 - 12 - 4 = 0

There are various methods to find the roots of a quadratic equation. The usage of the quadratic formula is one of them.

Quadratic Formula

Quadratic formula is the simplest way to find the roots of a quadratic equation . There are certain quadratic equations that cannot be easily factorized, and here we can conveniently use this quadratic formula to find the roots in the quickest possible way. The two roots in the quadratic formula are presented as a single expression. The positive sign and the negative sign can be alternatively used to obtain the two distinct roots of the equation.

Quadratic Formula: The roots of a quadratic equation ax 2 + bx + c = 0 are given by x = [-b ± √(b 2 - 4ac)]/2a.

This formula is also known as the Sridharacharya formula .

Example: Let us find the roots of the same equation that was mentioned in the earlier section x 2 - 3x - 4 = 0 using the quadratic formula.

a = 1, b = -3, and c = -4.

x = [-b ± √(b 2 - 4ac)]/2a = [-(-3) ± √((-3) 2 - 4(1)(-4))]/2(1) = [3 ± √25] / 2 = [3 ± 5] / 2 = (3 + 5)/2 or (3 - 5)/2 = 8/2 or -2/2 = 4 or -1 are the roots.

Proof of Quadratic Formula

Consider an arbitrary quadratic equation: ax 2 + bx + c = 0, a ≠ 0

To determine the roots of this equation, we proceed as follows:

ax 2 + bx = -c ⇒ x 2 + bx/a = -c/a

Now, we express the left-hand side as a perfect square , by introducing a new term (b/2a) 2 on both sides:

• x 2 + bx/a + (b/2a) 2 = -c/a + (b/2a) 2

The left-hand side is now a perfect square:

(x + b/2a) 2 = -c/a + b 2 /4a 2 ⇒ (x + b/2a) 2 = (b 2 - 4ac)/4a 2

This is good for us, because now we can take square roots to obtain:

x + b/2a = ±√(b 2 - 4ac)/2a

x = (-b ± √(b 2 - 4ac))/2a

Thus, by completing the squares, we were able to isolate x and obtain the two roots of the equation.

Nature of Roots of the Quadratic Equation

The roots of a quadratic equation are usually represented to by the symbols alpha (α), and beta (β). Here we shall learn more about how to find the nature of roots of a quadratic equation without actually finding the roots of the equation.

The nature of roots of a quadratic equation can be found without actually finding the roots (α, β) of the equation. This is possible by taking the discriminant value, which is part of the formula to solve the quadratic equation. The value b 2 - 4ac is called the discriminant of a quadratic equation and is designated as 'D'. Based on the discriminant value the nature of the roots of the quadratic equation can be predicted.

Discriminant: D = b 2 - 4ac

• D > 0, the roots are real and distinct
• D = 0, the roots are real and equal.
• D < 0, the roots do not exist or the roots are imaginary .

Now, check out the formulas to find the sum and the product of the roots of the equation.

Sum and Product of Roots of Quadratic Equation

The coefficient of x 2 , x term, and the constant term of the quadratic equation ax 2 + bx + c = 0 are useful in determining the sum and product of the roots of the quadratic equation. The sum and product of the roots of a quadratic equation can be directly calculated from the equation, without actually finding the roots of the quadratic equation. For a quadratic equation ax 2 + bx + c = 0, the sum and product of the roots are as follows.

• Sum of the Roots: α + β = -b/a = - Coefficient of x/ Coefficient of x 2
• Product of the Roots: αβ = c/a = Constant term/ Coefficient of x 2

Writing Quadratic Equations Using Roots

The quadratic equation can also be formed for the given roots of the equation. If α, β, are the roots of the quadratic equation, then the quadratic equation is as follows.

x 2 - (α + β)x + αβ = 0

Example: What is the quadratic equation whose roots are 4 and -1?

Solution: It is given that α = 4 and β = -1. The corresponding quadratic equation is found by:

x 2 - (α + β)x + αβ = 0 x 2 - (α + β)x + αβ = 0 x 2 - (4 - 1)x + (4)(-1) = 0 x 2 - 3x - 4 = 0

Formulas Related to Quadratic Equations

The following list of important formulas is helpful to solve quadratic equations.

• The quadratic equation in its standard form is ax 2 + bx + c = 0
• For D > 0 the roots are real and distinct.
• For D = 0 the roots are real and equal.
• For D < 0 the real roots do not exist, or the roots are imaginary.
• The formula to find the roots of the quadratic equation is x = [-b ± √(b 2 - 4ac)]/2a.
• The sum of the roots of a quadratic equation is α + β = -b/a.
• The product of the Root of the quadratic equation is αβ = c/a.
• The quadratic equation whose roots are α, β, is x 2 - (α + β)x + αβ = 0.
• The condition for the quadratic equations a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0 having the same roots is (a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2 .
• When a > 0, the quadratic expression f(x) = ax 2 + bx + c has a minimum value at x = -b/2a.
• When a < 0, the quadratic expression f(x) = ax 2 + bx + c has a maximum value at x = -b/2a.
• The domain of any quadratic function is the set of all real numbers.

Methods to Solve Quadratic Equations

A quadratic equation can be solved to obtain two values of x or the two roots of the equation. There are four different methods to find the roots of the quadratic equation. The four methods of solving the quadratic equations are as follows.

• Factorizing of Quadratic Equation
• Using quadratic formula (which we have seen already)

Method of Completing the Square

• Graphing Method to Find the Roots

Let us look in detail at each of the above methods to understand how to use these methods, their applications, and their uses.

Solving Quadratic Equations by Factorization

Factorization of quadratic equation follows a sequence of steps. For a general form of the quadratic equation ax 2 + bx + c = 0, we need to first split the middle term into two terms, such that the product of the terms is equal to the constant term. Further, we can take the common terms from the available term, to finally obtain the required factors as follows:

• x 2 + (a + b)x + ab = 0
• x 2 + ax + bx + ab = 0
• x(x + a) + b(x + a)
• (x + a)(x + b) = 0

Here is an example to understand the factorization process.

• x 2 + 5x + 6 = 0
• x 2 + 2x + 3x + 6 = 0
• x(x + 2) + 3(x + 2) = 0
• (x + 2)(x + 3) = 0

Thus the two obtained factors of the quadratic equation are (x + 2) and (x + 3). To find its roots, just set each factor to zero and solve for x. i.e., x + 2 = 0 and x + 3 = 0 which gives x = -2 and x = -3. Thus, x = -2 and x = -3 are the roots of x 2 + 5x + 6 = 0.

Further, there is another important method of solving a quadratic equation. The method of completing the square for a quadratic equation is also useful to find the roots of the equation.

The method of completing the square in a quadratic equation is to algebraically square and simplify, to obtain the required roots of the equation. Consider a quadratic equation ax 2 + bx + c = 0, a ≠ 0. To determine the roots of this equation, we simplify it as follows:

• ax 2 + bx + c = 0
• ax 2 + bx = -c
• x 2 + bx/a = -c/a

Now, we express the left-hand side as a perfect square, by introducing a new term (b/2a) 2 on both sides:

• (x + b/2a) 2 = -c/a + b 2 /4a 2
• (x + b/2a) 2 = (b 2 - 4ac)/4a 2
• x + b/2a = + √(b 2 - 4ac)/2a
• x = - b/2a + √(b 2 - 4ac)/2a
• x = [-b ± √(b 2 - 4ac)]/2a

Here the '+' sign gives one root and the '-' sign gives another root of the quadratic equation. Generally, this detailed method is avoided, and only the quadratic formula is used to obtain the required roots.

Graphing a Quadratic Equation

The point(s) where the graph cuts the horizontal x-axis (typically the x-intercepts ) is the solution of the quadratic equation. These points can also be algebraically obtained by equalizing the y value to 0 in the function y = ax 2 + bx + c and solving for x.

Quadratic Equations Having Common Roots

Consider two quadratic equations having common roots a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0. Let us solve these two equations to find the conditions for which these equations have a common root. The two equations are solved for x 2 and x respectively.

(x 2 )(b 1 c 2 - b 2 c 1 ) = (-x)/(a 1 c 2 - a 2 c 1 ) = 1/(a 1 b 2 - a 2 b 1 )

x 2 = (b 1 c 2 - b 2 c 1 ) / (a 1 b 2 - a 2 b 1 )

x = (a 2 c 1 - a 1 c 2 ) / (a 1 b 2 - a 2 b 1 )

Hence, by simplifying the above two expressions we have the following condition for the two equations having the common root.

(a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2

Maximum and Minimum Value of Quadratic Expression

The maximum and minimum values of the quadratic expressions are of further help to find the range of the quadratic expression: The range of the quadratic expressions also depends on the value of a. For positive values of a( a > 0), the range is [ F(-b/2a), ∞), and for negative values of a ( a < 0), the range is (-∞, F(-b/2a)].

• For a > 0, Range: [ f(-b/2a), ∞)
• For a < 0, Range: (-∞, f(-b/2a)]

Note that the domain of a quadratic function is the set of all real numbers, i.e., (-∞, ∞).

Tips and Tricks on Quadratic Equation:

Some of the below-given tips and tricks on quadratic equations are helpful to more easily solve quadratic equations.

• The quadratic equations are generally solved through factorization. But in instances when it cannot be solved by factorization, the quadratic formula is used.
• The roots of a quadratic equation are also called the zeroes of the equation.
• For quadratic equations having negative discriminant values, the roots are represented by complex numbers.
• The sum and product of the roots of a quadratic equation can be used to find higher algebraic expressions involving these roots.

☛Related Topics:

• Roots Calculator
• Quadratic Factoring Calculator
• Roots of Quadratic Equation Calculator

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Quadratic Equations Examples

Example 1: James is a fitness enthusiast and goes for a jog every morning. The park where he jogs is rectangular in shape and measures 12m × 8m. An environmentalist group plans to revamp the park and decides to build a pathway surrounding the park. This would increase the total area to 140 sq m. What will be the width of the pathway?

Let’s denote the width of the pathway by x.

Then, the length and breadth of the outer rectangle is (12+2x) m and (8+2x) m.

Given that, area = 140

(12 + 2x)(8 + 2x) = 140

2(6 + x) 2(4 + x) = 140

(6 + x)(4 + x) = 35

24 + 6x + 4x + x 2 = 35

x 2 + 10x -11 = 0

x 2 + 11x - x - 11 = 0

x(x + 11) - 1(x + 11) = 0

(x + 11)(x - 1) = 0

(x + 11) =0 and (x - 1) = 0

x = -11 and x = 1

Since length can’t be negative, we take x = 1.

Answer: Therefore the width of the pathway is 1 m.

Example 2: Rita throws a ball upwards from a platform that is 20m above the ground. The height of the ball from the ground at a time 't', is denoted by 'h'. Suppose h = -4t 2 + 16t + 20. Find the maximum height attained by the ball.

We can rearrange the terms of the quadratic equation

h = -4t 2 + 16t + 20

in such a way that it is easy to find the maximum value of this equation.

= -4t 2 + 16t + 20

= -4(t 2 - 4t - 5)

= -4((t - 2) 2 - 9)

= -4(t - 2) 2 + 36

We should keep the value of (t - 2) 2 minimum in order to find the maximum value of h.

So, the minimum value (t - 2) 2 can take is 0.

Answer: Therefore the maximum height attained is 36m.

Example 3: Find the quadratic equation having the roots 5 and 8 respectively.

The quadratic equation having the roots α, β, is x 2 - (α + β)x + αβ = 0.

Given α = 5, and β = 8.

Therefore the quadratic equation is:

x 2 - (5 + 8)x + 5×8 = 0

x 2 - 13x + 40 = 0

Answer: Hence the required quadratic equation is x 2 - 13x + 40 = 0

Example 4: The quad equation 2x 2 + 9x + 7 = 0 has roots α, β. Find the quadratic equation having the roots 1/α, and 1/β.

The quadratic equation having roots that are reciprocal to the roots of the equation ax 2 + bx + c = 0, is cx 2 + bx + a = 0.

The given quadratic equation is 2x 2 + 9x + 7 = 0.

Hence the required equation having reciprocal roots is 7x 2 + 9x + 2 = 0.

Method 2: From the given equation,

α + β = -9/2 and α β = 2/7.

The new equation should have its roots to be 1/α and 1/β.

Their sum = 1/α + 1/β = (α + β) / α β = -9/7 Their product = 1/α β = 2/7 Thus, the required equation is, x 2 - (1/α + 1/β)x + 1/α β = 0 x 2 - (-9/7)x + 2/7 = 0 Multiplying both sides by 7, 7x 2 + 9x + 2 = 0

Answer: Therefore the equation is 7x 2 + 9x + 2 = 0.

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Practice Questions on Quadratic Equation

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FAQs on Quadratic Equation

What is the definition of a quadratic equation.

A quadratic equation in math is a second-degree equation of the form ax 2 + bx + c = 0. Here a and b are the coefficients, c is the constant term, and x is the variable. Since the variable x is of the second degree, there are two roots or answers for this quadratic equation. The roots of the quadratic equation can be found by either solving by factorizing or through the use of the quadratic formula.

What is the Quadratic Formula?

The quadratic equation formula to solve the equation ax 2 + bx + c = 0 is x = [-b ± √(b 2 - 4ac)]/2a. Here we obtain the two values of x, by applying the plus and minus symbols in this formula. Hence the two possible values of x are [-b + √(b 2 - 4ac)]/2a, and [-b - √(b 2 - 4ac)]/2a.

How do You Solve a Quadratic Equation?

There are several methods to solve quadratic equations, but the most common ones are factoring, using the quadratic formula, and completing the square.

• Factoring involves finding two numbers that multiply to equal the constant term, c, and add up to the coefficient of x, b.
• The quadratic formula is used when factoring is not possible, and it is given by x = [-b ± √(b 2 - 4ac)]/2a.
• Completing the square involves rewriting the quadratic equation in a different form that allows you to easily solve for x.

What is Determinant in Quadratic Formula?

The value b 2 - 4ac is called the discriminant and is designated as D. The discriminant is part of the quadratic formula. The discriminants help us to find the nature of the roots of the quadratic equation, without actually finding the roots of the quadratic equation.

What are Some Real-Life Applications of Quadratic Equations?

Quadratic equations are used to find the zeroes of the parabola and its axis of symmetry . There are many real-world applications of quadratic equations.

• They can be used in running time problems to evaluate the speed, distance or time while traveling by car, train or plane.
• Quadratic equations describe the relationship between quantity and the price of a commodity.
• Similarly, demand and cost calculations are also considered quadratic equation problems.
• It can also be noted that a satellite dish or a reflecting telescope has a shape that is defined by a quadratic equation.

How are Quadratic Equations Different From Linear Equations?

A linear degree is an equation of a single degree and one variable, and a quadratic equation is an equation in two degrees and a single variable. A linear equation is of the the form ax + b = 0 and a quadratic equation is of the form ax 2 + bx + c = 0. A linear equation has a single root and a quadratic equation has two roots or two answers. Also, a quadratic equation is a product of two linear equations.

What Are the 4 Ways To Solve A Quadratic Equation?

The four ways of solving a quadratic equation are as follows.

• Factorizing method
• Roots of Quadratic Equation Formula Method
• Method of Completing Squares
• Graphing Method

How to Solve a Quadratic Equation by Completing the Square?

The quadratic equation is solved by the method of completing the square and it uses the formula (a + b)^2 = a 2 + 2ab + b 2 (or) (a - b)^2 = a 2 - 2ab + b 2 .

How to Find the Value of the Discriminant?

The value of the discriminant in a quadratic equation can be found from the variables and constant terms of the standard form of the quadratic equation ax 2 + bx + c = 0. The value of the discriminant is D = b 2 - 4ac, and it helps to predict the nature of roots of the quadratic equation, without actually finding the roots of the equation.

How Do You Solve Quadratic Equations With Graphing?

The quadratic equation can be solved similarly to a linear equal by graphing. Let us take the quadratic equation ax 2 + bx + c = 0 as y = ax 2 + bx + c . Here we take the set of values of x and y and plot the graph. The two points where this graph meets the x-axis, are the solutions of this quadratic equation.

How Important Is the Discriminant of a Quadratic Equation?

The discriminant is very much needed to easily find the nature of the roots of the quadratic equation. Without the discriminant, finding the nature of the roots of the equation is a long process, as we first need to solve the equation to find both the roots. Hence the discriminant is an important and needed quantity, which helps to easily find the nature of the roots of the quadratic equation.

Where Can I Find Quadratic Equation Solver?

To get the quadratic equation solver, click here . Here, we can enter the values of a, b, and c for the quadratic equation ax 2 + bx + c = 0, then it will give you the roots along with a step-by-step procedure.

What is the Use of Discriminants in Quadratic Formula?

The discriminant (D = b 2 - 4ac) is useful to predict the nature of the roots of the quadratic equation. For D > 0, the roots are real and distinct, for D = 0 the roots are real and equal, and for D < 0, the roots do not exist or the roots are imaginary complex numbers . With the help of this discriminant and with the least calculations, we can find the nature of the roots of the quadratic equation.

How do you Solve a Quadratic Equation without Using the Quadratic Formula?

There are two alternative methods to the quadratic formula. One method is to solve the quadratic equation through factorization, and another method is by completing the squares. In total there are three methods to find the roots of a quadratic equation.

How to Derive Quadratic Formula?

The algebra formula (a + b) 2 = a 2 + 2ab + b 2 is used to solve the quadratic equation and derive the quadratic formula. This algebraic formula is used to manipulate the quadratic equation and derive the quadratic formula to find the roots of the equation.

• The Quadratic Formula

What it is, what it does, and how to use it

What is the Quadratic Formula?

The quadratic formula is:

What does this formula tell us?

The quadratic formula calculates the solutions of any quadratic equation.

What is a quadratic equation?

A quadratic equation is an equation that can be written as ax ² + bx + c where a ≠ 0 . In other words, a quadratic equation must have a squared term as its highest power.

Examples of quadratic equations

$$ y = 5x^2 + 2x + 5 \\ y = 11x^2 + 22 \\ y = x^2 - 4x +5 \\ y = -x^2 + + 5 $$

Non Examples

$$ y = 11x + 22 \\ y = x^3 -x^2 +5x +5 \\ y = 2x^3 -4x^2 \\ y = -x^4 + 5 $$

Ok, but what is a 'solution'?

Well a solution can be thought in two ways:

Example of the quadratic formula to solve an equation

Use the formula to solve theQuadratic Equation: $$ y = x^2 + 2x + 1 $$ .

Just substitute a,b, and c into the general formula:

$$ a = 1 \\ b = 2 \\ c = 1 $$

Below is a picture representing the graph of y = x² + 2x + 1 and its solution.

Quadratic Formula Song

A catchy way to remember the quadratic formula is this song (pop goes the weasel).

Practice Problems

Use the quadratic formula to find the solutions to the following equation: y = x² − 2x + 1 and its solution.

Use the quadratic formula to find the solutions to the following equation: y = x² − x − 2 and its solution.

In this quadratic equation, y = x² − x − 2 and its solution:

Use the quadratic formula to find the solutions to the following equation: y = x² − 1 and its solution.

In this quadratic equation, y = x² − 1 and its solution:

Calculate the solutions of the the quadratic equation below by using the quadratic formula : y = x² + 2x − 3 and its solution.

In this quadratic equation, y = x² + 2x − 3 and its solution:

Below is a picture of the graph of the quadratic equation and its two solutions.

Use the quadratic formula to find the solutions to the following equation: y = x² + 4x − 5 and its solution.

In this quadratic equation, y = x² + 4x − 5 and its solution:

Use the quadratic formula to find the solutions to the following equation: y = x² − 4x + 5 and its solution.

In this quadratic equation, y = x² − 4x + 5 and its solution:

Below is a picture of this quadratic's graph.

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What is a Quadratic Equation?

How to use the quadratic formula, how to solve quadratic equations, solving quadratic equations by graphing, practice questions, quadratic formula – explanation & examples.

In this article, we will learn how to solve quadratic equations using two methods, namely the quadratic formula and the graphical method . Before we can dive into this topic, let’s recall what a quadratic equation is.

A quadratic equation in mathematics is defined as a polynomial of second degree whose standard form is ax 2 + bx + c = 0, where a, b and c are numerical coefficients and a ≠ 0.

The term second degree means that at least one term in the equation is raised to the power of two. In a quadratic equation, the variable x is an unknown value, for which we need to find the solution.

Examples of quadratic equations are: 6x² + 11x – 35 = 0, 2x² – 4x – 2 = 0, 2x² – 64 = 0, x² – 16 = 0, x² – 7x = 0, 2x² + 8x = 0 etc.  From these examples, you can note that, some quadratic equations lack the term “c” and “bx.”

Suppose ax 2 + bx + c = 0 is our standard quadratic equation. We can derive the quadratic formula by completing the square as shown below.

Isolate the term c to right side of the equation

ax 2 + bx = -c

Divide each term by a.

x 2 + bx/a = -c/a

Express as a perfect square x 2 + b​x/a + (b​/2a) 2 = – c/a ​+ (b/2a) 2

(x + b/2a) 2 = (-4ac+b 2 )/4a 2

(x + b/2a) = ±√ (-4ac + b 2 )/2a

x = – b/2a ±√ (b 2 – 4ac)/2a

x = [- b ±√ (b 2 – 4ac)]/2a………. (This is the quadratic formula)

The presence of the plus (+) and minus (-) in the quadratic formula implies that there are two solutions, such as:

x 1 = (-b + √b2 – 4ac)/2a

x 2 = (-b – √b2 – 4ac)/2a

The above two values of x are known as roots of the quadratic equation. The roots of a quadratic equation depend on the nature of the discriminant. The discriminant is part of the quadratic formula in the form of b  2 – 4 ac. A quadratic equation has two different real roots of the discriminant.

Let’s solve a few examples of problems using the quadratic formula.

Use the quadratic formula to find the roots of x 2 -5x+6 = 0.

Comparing the equation with the general form ax 2 + bx + c = 0 gives,

a = 1, b = -5 and c = 6

b 2 – 4ac = (-5)2 – 4×1×6 = 1

Substitute the values in the quadratic formula

x 1 = (-b + √b2-4ac)/2a

⇒ (5 + 1)/2

x 2 = (-b – √b2-4ac)/2a

⇒ (5 – 1)/2

Solve the quadratic equation below using quadratic formula:

3x 2  + 6x + 2 = 0

Comparing the problem with the general form of quadratic equation ax 2  + bx + c = 0 gives,

a = 3, b = 6 and c = 2

x = [- b ± √ (b 2 – 4ac)]/2a

⇒ [- 6 ± √ (6 2 – 4* 3* 2)]/2*3

⇒ [- 6 ± √ (36- 24)]/6

⇒ [- 6 ± √ (12)]/6

x 1 = (-6 + 2√3)/6

⇒ -(2/3) √3

x 2 = (-6– 2√3)/6

⇒ -(4/3) √3

Solve 5x 2 + 6x + 1 = 0

Comparing with the quadratic equation, we get,

a = 5, b = 6, c = 1

Now apply the quadratic formula:

x = −b ± √ (b 2 − 4ac)2a

Substitute the values of a, b and c

⇒ x = −6 ± √ (6 2 − 4×5×1)2×5

⇒ x = −6 ± √ (36 − 20)10

⇒ x = −6 ± √ (16)10

⇒ x = −6 ± 410

⇒ x = − 0.2, −1

Solve 5x 2 + 2x + 1 = 0

The coefficients are;

a = 5, b = 2, c = 1

In this case, the discriminant is negative:

b 2 − 4ac = 2 2 − 4×5×1

Now apply the quadratic formula;

x = (−2 ± √ −16)/10

⇒√ (−16) = 4

Where i is the imaginary number √−1

⇒x = (−2 ± 4i)/10

Therefore, x = −0.2 ± 0.4i

Solve x 2 − 4x + 6.25 = 0

According to the standard form of a quadratic equation ax 2  + bx + c = 0, we can observe that;

a = 1, b = −4, c = 6.25

Determine the discriminants.

b 2 − 4ac = (−4) 2 – 4 × 1 × 6.25

= −9 ………………. (negative discriminant)

⇒ x = −(−4) ± √ (−9)/2

⇒ √ (−9) = 3i; where i is the imaginary number √−1

⇒ x = (4 ± 3i)/2

Hence, x = 2 ± 1.5i

How to Graph a Quadratic Equation?

To graph a quadratic equation, here are the steps to follow:

• Given a quadratic equation, rewrite the equation by equating it to y or f(x)
• Choose arbitrary values of x and y to plot the curve
• Now graph the function.
• Read the roots where the curve crosses or touches the x-axis.

Graphing is another method of solving quadratic equations. The solution of the equation is obtained by reading the x-intercepts of the graph.

There are three possibilities when solving quadratic equations by graphical method:

• An equation has one root or solution if the x-intercept of the graph is 1.
• An equation with two roots has 2 x -intercepts
• If there is no x – intercepts, then an equation has no real solutions.

Let’s graph a few examples of quadratic equations.  In these examples, we have drawn our graphs using graphing software, but for you to understand this lesson very well, draw your graphs manually.

Solve the equation x 2  + x – 3 = 0 by graphical method

Our arbitrary values are shown in the table below:

The x- intercepts are x =  1.3 and  x =  –2.3. Therefore, the roots of the quadratic equation are x = 1.3 and x = –2.3

Solve the equation 6x – 9 – x 2  = 0.

Choose arbitrary values of x.

The curve touches the x-axis at x = 3. Therefore, 6 x  – 9 –  x 2  = 0 has one solution(x = 3).

Solve the equation x 2  + 4x + 8 = 0 by graphical method.

In this example, the curve does not touch or cross the x -axis. Therefore, the quadratic equation x 2  + 4x + 8 = 0 does not have any real roots.

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• Mathematics

How to Solve a Quadratic Equation: A Step-by-Step Guide

Last Updated: May 27, 2024 Fact Checked

Factoring the Equation

Using the quadratic formula, completing the square, practice problems and answers, expert q&a.

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,431,452 times.

A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. [1] X Research source There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.

Quadradic Formula for Solving Equations

• Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4. [3] X Research source

• 3x = -1 ..... by subtracting
• 3x/3 = -1/3 ..... by dividing
• x = -1/3 ..... simplified
• x = 4 ..... by subtracting
• x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.

• So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.

• 4x 2 - 5x - 13 = x 2 -5
• 4x 2 - x 2 - 5x - 13 +5 = 0
• 3x 2 - 5x - 8 = 0

• {-b +/-√ (b 2 - 4ac)}/2
• {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3)

• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
• {5 +/-√(25 + 96)}/6
• {5 +/-√(121)}/6

• (5 + 11)/6 = 16/6
• (5-11)/6 = -6/6

• x = (-1, 8/3)

• 2x 2 - 9 = 12x =
• In this equation, the a term is 2, the b term is -12, and the c term is -9.

• 2x 2 - 12x - 9 = 0
• 2x 2 - 12x = 9

• 2x 2 /2 - 12x/2 = 9/2 =
• x 2 - 6x = 9/2

• -6/2 = -3 =
• (-3) 2 = 9 =
• x 2 - 6x + 9 = 9/2 + 9

• x = 3 + 3(√6)/2
• x = 3 - 3(√6)/2)

• If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example: [14] X Research source Thanks Helpful 0 Not Helpful 0
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 3 Not Helpful 0
• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. Thanks Helpful 1 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
• ↑ https://www.mathsisfun.com/algebra/factoring-quadratics.html
• ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
• ↑ https://www.cuemath.com/algebra/quadratic-equations/
• ↑ https://www.purplemath.com/modules/solvquad4.htm
• ↑ http://www.purplemath.com/modules/quadform.htm
• ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
• ↑ http://www.mathsisfun.com/algebra/completing-square.html
• ↑ https://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm

About This Article

To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down! Did this summary help you? Yes No

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How to Solve Quadratic Equations

How to Solve Quadratic Equations? There are basically three methods to solve quadratic equations. They are:

• Using Quadratic formula
• Factoring the quadratic equation
• Completing the square

A quadratic equation is an equation that has the highest degree equal to two. The standard form of the quadratic equation is ax 2 + bx + c = 0, where a, b, c are constants and a ≠ b ≠ 0. Here, x is an unknown variable for which we need to find the solution. Let us learn here how to solve quadratic equations.

Solving Quadratic Equations – Using Quadratic Formula

The quadratic formula is used to find solutions of quadratic equations. If ax 2 + bx + c = 0, then solution can be evaluated using the formula given below;

Thus, the formula will result in two solutions here.

Here, b 2 – 4ac is the discriminant (D).

D = b 2 – 4ac

• If D = 0, the two roots of quadratic equation are real and equal
• If D > 0, the roots are real and unequal
• If D < 0, the roots are not real, i.e. imaginary

Example: Solve x 2 – 5x + 6 = 0.

Solution: Given,

x 2 – 5x + 6 = 0

a = 1, b = -5, c = 6

By the quadratic formula, we know;

b 2 – 4ac = (-5) 2 – 4 × 1 × 6 = 25 – 24 = 1 > 0

Thus, the roots are real.

x = [-b ± √(b 2 – 4ac)]/ 2a

= [-(-5) ± √1]/ 2(1)

= [5 ± 1]/ 2

i.e. x = (5 + 1)/2 and x = (5 – 1)/2

x = 6/2, x = 4/2

Therefore, the solution of x 2 – 5x + 6 = 0 is 3 or 2.

Solving Quadratic Equations – By Factorisation

We can write the quadratic equation as a product of factors having degree less than or equal to two. This method of solving quadratic equations is called factoring the quadratic equation.

Let us learn by an example.

Example: Solve 6m 2 – 7m + 2 = 0 by factoring method.

Solution: 6m 2 – 4m – 3m + 2 = 0

⟹ 2m(3m – 2) – 1(3m – 2) = 0

⟹ (3m – 2) (2m – 1) = 0

⟹ 3m – 2 = 0 or 2m – 1 = 0

⟹ 3m = 2 or 2m = 1

Therefore, the solutions of the given equation are:

m = ⅔ or m = ½

Solving Quadratic Equation – Completing Square

To solve the quadratic equation using completing the square method, follow the below given steps.

• First make sure the equation is in the standard form: ax 2 + bx + c = 0
• Now, divide the whole equation by a, such that the coefficient of x 2 is 1.
• Write the equation with a constant term on the Right side of equation
• Add the square of half of coefficient of x on both sides and complete the square
• Write the left side equation as a square term and solve

Let us understand with the help of an example.

Example: Solve 4x 2 + x = 3 by completing the square method.

4x 2 + x = 3

Divide the whole equation by 4.

x 2 + x/4 = ¾

Coefficient of x is ¼

Half of ¼ = ⅛

Square of ⅛ = (⅛) 2

Add (⅛) 2 on both sides of the equation.

x 2 + x/4 + (⅛) 2 = ¾ – (⅛) 2

x 2 + x/4 + 1/64 = ¾ + 1/64

(x + ⅛) 2 = (48+1)/64 = 49/64

Taking square root on both sides, we get;

x+1/8 = √(49/64) = ±7/8

x = ⅞ – ⅛ = 6/8 = ⅜

x = – ⅞ – ⅛ = -8/8 = -1

Hence, the solution of quadratic equation is x = ⅜ or x = -1.

Practice Questions

Solve the following quadratic equations.

• 2s 2 + 5s = 3
• 9x 2 – 6x – 2 = 0
• 3x 2 – 11x – 4 = 0
• 2x 2 – 12x – 9 = 0

Frequently Asked Questions on Solving Quadratic Equations

What are the methods to solve quadratic equations, what is the formula to solve quadratic equations.

The quadratic formula is given by: x = [b±√(b²-4ac))]/(2a)

What is the discriminant of the quadratic formula?

The discriminant of quadratic formula is b2 – 4ac. It helps to determine the nature of the roots.

When is the quadratic equation has imaginary roots?

When discriminant of the quadratic equation is less than zero, then the roots are imaginary or non-real.

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Solving Quadratic Equations

Many word problems Involving unknown quantities can be translated for solving quadratic equations

Methods of solving quadratic equations are discussed here in the following steps.

Step I: Denote the unknown quantities by x, y etc.

Step II: use the conditions of the problem to establish in unknown quantities.

Step III: Use the equations to establish one quadratic equation in one unknown.

Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.

Now we will learn how to frame the equations from word problem:

1.  The product of two consecutive integers is 132. Frame an equation for the statement. What is the degree of the equation?

Method I: Using only one unknown

Let the two consecutive integers be x and x + 1

Form the equation, the product of x and x + 1 is 132.

Therefore, x(x + 1) = 132

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

This is the equation of the statement, x denoting the smaller integer.

Method II: Using more than one unknown

Let the consecutive integers be x and y, x being the smaller integer.

As consecutive integers differ by 1, y - x = 1 ........................................... (i)

Again, from the question, the product of x and y is 132.

So, xy = 132 ........................................... (ii)

From (i), y = 1 + x.

Putting y = 1 + x in (ii),

x(1 + x) = 132

Solving the quadratic equation, we get the value of x. Then the value of y can be determined by substituting the value of x in y = 1 + x.

2. The length of a rectangle is greater than its breadth by 3m. If its area be 10 sq. m, find the perimeter.

Suppose, the breadth of the rectangle = x m.

Therefore, length of the rectangle = (x + 3) m.

So, area = (x + 3)x sq. m

Hence, by the condition of the problem

(x + 3)x = 10

⟹ x\(^{2}\) + 3x - 10 = 0

⟹ (x + 5)(x - 2) = 0

So, x = -5,2

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore x = 2

Hence, breadth = 2 m

and length = 5 m

Therefore, Perimeter = 2(2 + 5) m = 14 m.

x = -5 does not satisfy the conditions of the problem length or breadth can never be negative. Such a root is called an extraneous root. In solving a problem, each root of the quadratic equation is to be verified whether it satisfies the conditions of the given problem. An extraneous root is to be rejected.

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations 

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring

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Trigonometric Equations - Multiple-Choice WS/Practice (20 problems + solutions)

Subject: Mathematics

Age range: 16+

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Last updated

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This is an engaging multiple-choice practice on solving trigonometric equations(basic, multiple angles, of quadratic type, such that can be solved by factoring). There are 5 pages, containing a total of 20 problems each having five optional answers. The first 8 problems are finding all the solutions of each equation given, the next 12 problems are determining the sum of the solutions of an equation on a closed interval. Students have an empty field below each problem where to write down their solution.

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Error Estimates and Adaptivity of the Space-Time Discontinuous Galerkin Method for Solving the Richards Equation

• Open access
• Published: 20 August 2024
• Volume 101 , article number  11 , ( 2024 )

Cite this article

You have full access to this open access article

• Vít Dolejší   ORCID: orcid.org/0000-0001-6356-934X 1 ,
• Hyun-Geun Shin 1 &
• Miloslav Vlasák 2  

We present a higher-order space-time adaptive method for the numerical solution of the Richards equation that describes a flow motion through variably saturated media. The discretization is based on the space-time discontinuous Galerkin method, which provides high stability and accuracy and can naturally handle varying meshes. We derive reliable and efficient a posteriori error estimates in the residual-based norm. The estimates use well-balanced spatial and temporal flux reconstructions which are constructed locally over space-time elements or space-time patches. The accuracy of the estimates is verified by numerical experiments. Moreover, we develop the hp -adaptive method and demonstrate its efficiency and usefulness on a practically relevant example.

Avoid common mistakes on your manuscript.

1 Introduction

Fluid flows in variably saturated porous media are usually described by the Richards equation [ 33 ], which is expressed in the form

where \(\partial _t\) denotes the derivative with respect to time, \(\nabla \cdot \) and \(\nabla \) are the divergence and gradient operators, respectively, \(\psi \) is the sought pressure head (= normalized pressure), z is the vertical coordinate, \(\theta \) is the water content function, \({\textbf{K}}\) is the hydraulic conductivity tensor and g is the source term. In addition, the active pore volume \(\vartheta \) is related to \(\theta \) by the following relation

where \(S_s,\theta _s\ge 0\) are material parameters. The hydraulic conductivity satisfies \({\textbf{K}}(\psi )= {\textbf{K}}_s {{{\mathcal {K}}}}_r(\psi )\) , where \({\textbf{K}}_s\) is the saturated conductivity tensor, and \({{{\mathcal {K}}}}_r\in [0,1]\) is the relative saturation. The functions \(\theta \) and \({{{\mathcal {K}}}}_r\) are given by constitutive relations, e.g., by van Genuchten’s law [ 27 ] and by Mualem’s law [ 31 ], respectively.

The Richards equation belongs to the nonlinear parabolic problems, and it can degenerate, in particular \({\textbf{K}}\rightarrow 0\) or \(\tfrac{{\mathrm d}{\vartheta }}{{\mathrm d}\psi }\rightarrow 0\) . Due to the degeneracy, the numerical solution is challenging, and various techniques have been developed for its solution in the last decades, see [ 25 ] for a survey.

In [ 14 ], we presented the adaptive space-time discontinuous Galerkin (STDG) method for the numerical solution of ( 1 ). This technique is based on a piecewise polynomial discontinuous approximation with respect to both the spatial and temporal coordinates. The resulting scheme is sufficiently stable, provides high accuracy, and is suitable for the hp -mesh adaptation. This is an important property, since the weak solution of the Richards equation is (only) piecewise regular and exhibits singularities along the material interfaces and the unsaturated/saturated zone (when \(\psi \approx 0\) ). Therefore, an adaptive method that allows different meshes at different time levels, can achieve an accurate approximation with a relatively small number of degrees of freedom.

The numerical experiments presented in [ 14 ] showed the potential of the adaptive STDG method. However, the mesh adaptation used is based on interpolation error estimates that do not guarantee an upper error bound. The aim of this work is to overcome this bottleneck, derive a posteriori error estimates, and use them in the hp -mesh adaptation framework.

A posteriori error estimates for the numerical solution of the Richards equation have been treated in many papers for different numerical methods. We mention the finite volume framework with multistep time discretization in [ 5 ], the mixed finite element method in [ 6 ], the two-point finite volume discretization in [ 8 ], the lowest-order discretization on polytopal meshes in [ 38 ], finite element techniques in [ 30 ] and the references cited therein.

Guaranteed error estimates without unknown constants are usually obtained by measuring the error in a dual norm of the residual. Introducing reconstructed fluxes from the space \(H^1(\textrm{div},{\varOmega })\) , the upper bound can then be obtained directly. In [ 18 ], we developed this approach to the higher-order STDG method for nonlinear parabolic problems, where the temporal discontinuities were treated by temporal flux reconstructions considering the time jumps.

In this paper, we extend the approach [ 18 ] to the Richards equation ( 1 ). Although the definition of the temporal and spatial flux reconstructions as well as the derivation of the upper bounds is straightforward, the proof of the lower bound (efficiency) is rather tricky since the term \(\theta (\psi )\) in the time derivative is not a polynomial function for a polynomial \(\psi \) . In contrary to [ 18 ], the proof of efficiency requires the additional oscillatory data terms. We construct spatial fluxes by solving local Neumann problems defined on space-time patches that generalize the approach from [ 22 ]. Moreover, we provide numerical experiments verifying derived error estimates. Compared to [ 18 ], the resulting effectivity indices are much closer to one. This is the first novelty of this paper.

Secondly, we deal with the errors arising due to iterative solution of nonlinear algebraic systems. We introduce a cheap stopping criterion for iterative solvers and justify it by numerical experiments. Thirdly, we introduce a space-time adaptive algorithm that employs the anisotropic hp -mesh adaptation technique [ 15 ]. The algorithm admits local adaptation of size and shape of mesh elements and the local adaptation of degrees of polynomial approximation with respect to space. However, the size of the time step can vary globally, and the degree of polynomial approximation with respect to time is fixed. Using the equidistribution principle, the algorithm gives an approximate solution with the error estimate under the given tolerance. The performance of the adaptive algorithm is demonstrated numerically, including a practically relevant example.

The rest of the paper is organized as follows. In Sect.  2 , we introduce the problem considered, its STDG discretization is briefly described in Sect.  3 . The main theoretical results are derived in Sect.  4 , where the upper and lower bounds are proved. Two possible spatial reconstructions are discussed in Sect.  5 together with the stopping criteria of iterative solvers. The numerical verification of the error estimates is given in Sect.  6 . Furthermore, we present the resulting hp -mesh adaptation algorithm in Sect.  7 and demonstrate its performance by numerical examples. Finally, we conclude with some remarks in Sect.  8 .

2 Problem Formulation

Let \({\varOmega }\subset {\mathbb {R}}^d\) ( \(d=2,3\) ) be the domain occupied by a porous medium and \(T>0\) the physical time to be reached. For simplicity, we assume that \({\varOmega }\) is polygonal. By \({\varGamma }:=\partial {\varOmega }\) , we denote the boundary of \({\varOmega }\) which consists of two disjoint parts: the Dirichlet boundary \({\varGamma _\textrm{D}}\) and the Neumann boundary \({\varGamma _\textrm{N}}\) . We write the Richards equation ( 1 ) in a different form, which is more suitable for the analysis. We seek a function \(u= u(x,t):{\varOmega }\times (0,T)\rightarrow {\mathbb {R}}\) , which represents a hydraulic head (with the physical unit \(\textrm{L}\) ). The quantity \(u\) is related to the pressure head \(\psi \) by \(u= \psi + z\) . The Richards equation ( 1 ) reads

where \(g:{\varOmega }\times (0,T)\rightarrow {\mathbb {R}}\) represents a source term if g is positive or a sink term if g is negative, \({\vartheta }:{\mathbb {R}}\rightarrow {\mathbb {R}}\) denotes the dimensionless active pore volume, and \({\textbf{K}}:{\mathbb {R}}\rightarrow {\mathbb {R}}^{d\times d}\) is the hydraulic conductivity with the physical unit \(\textrm{L}\cdot \textrm{T}^{-1}\) (L = length, T = time). Moreover, \(u_D\) is a trace of a function \(u^*\in L^2(0,T;H^1({\varOmega }))\) on \({\varGamma _\textrm{D}}\times (0,T)\) , \(g_N\in L^2(0,T; L^2({\varGamma _\textrm{N}}))\) and \(u_0\in L^2({\varOmega })\) . We note that with respect to ( 1 ), we should write \({\vartheta }= {\vartheta }(u-z)\) and \({\textbf{K}}= {\textbf{K}}(\theta (u-z))\) , however, we skip this notation for simplicity. We assume that the function \({\vartheta }(u)\) is non-negative, non-decreasing and Lipschitz continuous. Moreover, the tensor \({\textbf{K}}(u)\) is symmetric, positively semi-definite, and Lipschitz continuous.

In order to introduce the weak solution, we set \(H(\textrm{div},\varOmega )=\{v\in L^2({\varOmega })^d:\nabla \cdot v\in L^2({\varOmega })\}\) and define the spaces

where \({\vartheta }^\prime (u)=\partial _t{\vartheta }(u) = \frac{{\textrm{d}} {\vartheta }}{{\textrm{d}}u} \partial _tu\) denotes the time derivative (in the weak sense). Obviously, if \(v\in Y\) then \({\vartheta }(v)\in C([0,T],L^2({\varOmega }))\) . In order to shorten the notation, we set the physical flux

Definition 1

We say that \(u\in Y\) is the weak solution of ( 3 ) if \(u-u^* \in V\) and

where \(\big ({u},{v}\big )_{{\varOmega }}:=\int _{\varOmega }u v {\,{\mathrm d}x}\) and \(\big ({u},{v}\big )_{{\varGamma _\textrm{N}}}:=\int _{\varGamma _\textrm{N}}u v {\,{\mathrm d}S}\) .

The existence and uniqueness of the Richards equation is studied in [ 2 ], see also the later works [ 3 , 28 ].

3 Space-time discretization

We briefly describe the discretization of ( 6 ) by the space-time discontinuous Galerkin (STDG) method, for more details, see [ 13 , 14 ]. Let \(0=t_0<t_1<\ldots <t_r=T\) be a partition of the time interval (0,  T ) and set \(I_m=(t_{m-1},t_m)\) and \(\tau _m=t_m-t_{m-1}\) . For each \(m=0,\dots ,r\) , we consider a simplicial mesh \({{\mathcal {T}}_h^m}\) covering \({\overline{{\varOmega }}}\) . For simplicity, we assume that \({{\mathcal {T}}_h^m}\) , \(m=0,\dots ,r\) are conforming, i.e., neighbouring elements share an entire edge or face. However, this assumption can be relaxed by the technique from [ 12 ].

For each element \(K\in {{\mathcal {T}}_h^m}\) , we denote by \({\partial K}\) its boundary, \({n_K}\) its unit outer normal and \({h_K}=\text{ diam }(K)\) its diameter. In order to shorten the notation, we write \({{\partial K}_{\!N}}:={\partial K}\cap {\varGamma _\textrm{N}}\) . By the generic symbol \(\gamma \) , we denote an edge ( \(d=2\) ) or a face ( \(d=3\) ) of \(K\in {{\mathcal {T}}_h^m}\) and \(h_{\gamma }\) denotes its diameter. In the following, we speak only about edges, but we mean faces for \(d=3\) . We assume that

\({{\mathcal {T}}_h^m}\) , \(m=0,\dots , r\) are shape regular , i.e., \({h_K}/\rho _K\le C\) for all \(K\in {{\mathcal {T}}_h}\) , where \(\rho _K\) is the radius of the largest d -dimensional ball inscribed in K and constant C does not depend on \({{\mathcal {T}}_h^m}\) for \(h\in (0,h_0)\) , \(m=0,\dots ,r\) .

\({{\mathcal {T}}_h^m}\) , \(m=0,\dots , r\) are locally quasi-uniform , i.e., \({h_K}\le C h_{K^\prime }\) for any pair of two neighbouring elements K and \(K^\prime \) , where the constant C does not depend on \(h\in (0,h_0)\) , \(m=0,\dots ,r\) .

Let \(p_K\ge 1\) be an integer denoting the degree of polynomial approximation on \(K\in {{\mathcal {T}}_h^m}\) , \(m=0,\dots ,r\) and \(P_{p_K}(K)\) be the corresponding space of polynomial functions on K . Let

denote the spaces of discontinuous piecewise polynomial functions on \({{\mathcal {T}}_h^m}\) with possibly varying polynomial approximation degrees. Furthermore, we consider the space of space-time discontinuous piecewise polynomial functions

where \(q\ge 0\) denotes the time polynomial approximation degree and \(P_{q}(I_m,{S_{hp,m}})\) is the Bochner space, i.e., \(v\in P_{q}(I_m,{S_{hp,m}})\) can be written as \(v(x,t)=\sum _{j=0}^q t^j\,v_j(x)\) , \(v_j\in {S_{hp,m}}\) , \(j=0,\dots ,q\) .

For \(v\in {{S_{hp}^{\tau q}}}\) , we define the one-sided limits and time jumps by

where \(u_0\) is the initial condition. In the following, we use the notation

where M is either element \(K\in {{\mathcal {T}}_h^m}\) or its (part of) boundary \({\partial K}\) . The corresponding norms are denoted by \({\left\| \cdot \right\| }_{M} \) and \({\left\| \cdot \right\| }_{M,m} \) , respectively. By \(\sum _{K,m}=\sum _{m=1}^r\sum _{K\in {{\mathcal {T}}_h^m}}\) , we denote the sum over all space-time elements \(K\times I_m\) , where \(K\in {{\mathcal {T}}_h^m}\) and \(m=1,\ldots ,r\) .

Moreover, we define the jumps and mean values of \(v\in {S_{hp,m}}\) on edges \(\gamma \subset {\partial K},\ K\in {{\mathcal {T}}_h^m}\) by

where \(v^{\scriptscriptstyle (+)} \) and \(v^{\scriptscriptstyle (-)} \) denote the traces of v on \({\partial K}\) from interior and exterior of K , respectively, and \(u_D\) comes from the Dirichlet boundary condition. For vector-valued \(v\in [{S_{hp,m}}]^d\) , we set \([{v}] = (v^{\scriptscriptstyle (+)} -v^{\scriptscriptstyle (-)} )\cdot {n_K}\) for \(\gamma \in {\varOmega }\) and similarly for boundary edges.

For each space-time element \(K\times I_m\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=1,\ldots ,r\) , we define the forms

where \(\alpha >0\) is a sufficiently large penalization parameter ( \(\alpha \sim p_K^2/h_K\) ) and \(\beta \in \{0,\tfrac{1}{2},1\}\) corresponds to the choice of the variants of the interior penalty discretization (SIPG with \(\beta =0\) , IIPG with \(\beta =1/2\) and NIPG with \(\beta =1\) ), see, e.g., [ 13 , Chapter 2].

We introduce the space-time discontinuous Galerkin discretization of ( 3 ).

Definition 2

The function \({u_h^{\tau }}\in {{S_{hp}^{\tau q}}}\) is called the approximate solution of ( 6 ) obtained by the space-time discontinuous Galerkin method (STDGM), if

with form \({A_{K,m}}\) given by ( 12 ) and \(\{\cdot \}\) defined by ( 9 ).

We note that \({u_h^{\tau }}\) is discontinuous with respect to time at \(t_m,\ m=1,\dots ,r-1\) . The solution between \(I_{m-1}\) and \(I_m\) is stuck together by the “time-penalty” term \(\big ({ \big \{{{\vartheta }(u)}\big \}_{m-1}},{v^{m-1}_+}\big )_{K}\) which also makes sense for u and v belonging to different finite element spaces.

Finally, we derive some identities that will be used later. Let \({{\mathcal {F}}_h^m}\) denote the set of all interior edges \(\gamma \not \subset {\varGamma }\) of mesh \({{\mathcal {T}}_h^m}\) and \({{\mathcal {F}}_D^m}\) the set of boundary edges of \({{\mathcal {T}}_h^m}\) lying on \({\varGamma _\textrm{D}}\) . Then, the identity

holds for a piecewise smooth vector-valued function w and a piecewise smooth scalar function z .

Using identity ( 15 ) and the following obvious formulas valid for interior edges \(\left\langle {\left\langle {{\textbf{K}}(u) \nabla u}\right\rangle }\right\rangle =\left\langle {{\textbf{K}}(u) \nabla u}\right\rangle \) , \(\left\langle {\alpha [{u}]}\right\rangle =\alpha [{u}]\) , \([{\left\langle {{\textbf{K}}(u) \nabla u}\right\rangle }]=0\) , \([{\alpha [{u}]}]=0\) , we gain

Consequently, from ( 12 ) and ( 16 ), we obtain the identity

4 A Posteriori Error Analysis

4.1 error measures.

In order to proceed to the derivation of error estimators, we define the spaces of piecewise continuous functions with respect to time by

Obviously, \(Y^0\subset Y\subset {Y^\tau }\subset X\) and \({{S_{hp}^{\tau q}}} \subset {Y^\tau }\) . Moreover, we have the following result.

Let \(u\in Y^0\) be the weak solution of ( 6 ). Then it satisfies

with \({a_{K,m}}\) given by ( 12 ) and the time jump \(\{\cdot \}_{m-1}\) defined by ( 9 ). Moreover, there exists a unique solution \(u\in Y^\tau \) such that \(u-u^*\in V^\tau \) and satisfies ( 19 ).

The proof follows directly by comparing formulas ( 19 )–( 20 ) with ( 6 ) and the fact that \(\big ({\{{\vartheta }(u)\}_{m-1}},{v^{m-1}_+}\big )_{K}=0\) for \(u\in Y^0\) . For the proof of uniqueness, we employ the fact that \(C_0^\infty (\varOmega )\) is dense in \(L^2(\varOmega )\) , i.e., there exists a sequence \(\{v_\varepsilon \}\subset C_0^\infty (\varOmega )\) for any \(v\in L^2(\varOmega )\) such that \(\Vert v_\varepsilon -v\Vert \rightarrow 0\) as \(\varepsilon \rightarrow 0\) , cf. [ 34 , Theorem 3.14]. We apply \(v=v_{s,\varepsilon _1}(x)v_{t,\varepsilon _2}(t)\) in ( 19 ), where the spatial component \(v_{s,\varepsilon _1}\in \{v\in H^1(\varOmega ):v|_{{\varGamma _\textrm{D}}}=0\}\) tends to \(\{{\vartheta }(u)\}_{m-1}\) as \(\varepsilon _1\rightarrow 0\) and the time component \(v_{t,\varepsilon _2}\) is given as 0 outside the interval \((t_{m-1},t_{m-1}+\varepsilon _2)\) and \(v_{t,\varepsilon _2}=1-(t-t_{m-1})/\varepsilon _2\) on \((t_{m-1},t_{m-1}+\varepsilon _2)\) , i.e., \(v_{t,\varepsilon _2}(t)\) tends to 0 as \(\varepsilon _2\rightarrow 0\) . Therefore, all the terms containing time integrals in ( 19 ) tend to 0 when \(\varepsilon _2\) tends to 0. Since \(v^{m-1}_+=v_{s,\varepsilon _1}\) , the remaining jump term tends to \(\Vert \{{\vartheta }(u)\}_{m-1}\Vert ^2\) as \(\varepsilon _1\) tends to 0. From this it follows that \(\{{\vartheta }(u)\}_{m-1}=0\) . Then it is possible to see that any solution of ( 19 ) satisfies the original weak formulation ( 6 ). Since the weak problem ( 6 ) has a unique solution, cf. [ 2 ], the extended problem ( 19 ) has a unique solution as well. \(\square \)

In virtue of [ 11 , § 2.3.1], we define a parameter \({d_{K,m}}\) associated with the space-time element \(K\times I_m\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=1,\dots ,r\) . The parameter \({d_{K,m}}\) represents a user-dependent weight, typically with physical units \((\textrm{T}\,\textrm{L})^{1/2}\) so that the error measure has the same physical unit as the energy norm. In this paper, we use two choices

where \({\left\| \cdot \right\| }_{m,\infty } := {\left\| \cdot \right\| }_{L^\infty ({\varOmega }\times I_m)} \) . We note that the following error analysis is independent of the choice of \({d_{K,m}}\) . Moreover, we define the norm in the space \({V^\tau }\) (cf. ( 18 )) by

In virtue of ( 19 ), we introduce the error measure as a dual norm of the residual

where \({b_{K,m}}\) is given by ( 20 ). The residual \({{\mathcal {R}}}(v)\) represents a natural error measure for \(u-v\in {V^\tau }\) , cf. [ 11 , Remark 2.3]. In Sect.  4 , we estimate \({{\mathcal {R}}}({u_h^{\tau }})\) for \({u_h^{\tau }}\) being the solution of ( 13 ).

Since the approximate solution \({u_h^{\tau }}\) belongs to the space of discontinuous function \({{S_{hp}^{\tau q}}}\not \subset {V^\tau }\) , we introduce the second building block measuring the nonconformity of the solution in spatial variables. Therefore, similarly to [ 18 ], we define the form

where \( C_{K,m,{\textbf{K}},\alpha }=\alpha ^2 +{\left\| {\textbf{K}}({u_h^{\tau }})\right\| }_{L^\infty (K\times I_m)}^{2} \) . The scaling factors are chosen such that \({{\mathcal {J}}}(v)^{1/2}\) has the same physical unit as \({{\mathcal {R}}}({u_h^{\tau }})\) .

We note that \({{\mathcal {J}}}(v)\) measures also the violation of the Dirichlet boundary condition since \({{\mathcal {J}}}(v)\) contains the term \({\left\| v-u_D\right\| }_{{\partial K}\cap {\varGamma _\textrm{D}},m} \) , cf. ( 11 ).

The final error measure is then defined by

where \({{\mathcal {R}}}({u_h^{\tau }})\) is given by ( 23 ) and \({{\mathcal {J}}}({u_h^{\tau }})\) by ( 24 ).

The error measure \({{\mathcal {E}}}({u_h^{\tau }})=0\) if and only if \({u_h^{\tau }}=u\) is the weak solution given by ( 6 ).

Obviously, if \({u_h^{\tau }}=u\) , then \({{\mathcal {J}}}({u_h^{\tau }})=0\) and \({{\mathcal {R}}}({u_h^{\tau }})=0\) due to ( 19 ). On the other hand, if \({{\mathcal {J}}}({u_h^{\tau }})=0\) , then \({u_h^{\tau }}\in {Y^\tau }\) and \({u_h^{\tau }}-u^*\in {V^\tau }\) . Moreover, \({{\mathcal {R}}}({u_h^{\tau }})=0\) and the uniqueness of ( 19 ) imply that \({u_h^{\tau }}\) is the weak solution ( 6 ). \(\square \)

4.2 Temporal and Spatial Flux Reconstructions

Similarly as in [ 18 ], we define a temporal reconstruction \({R_h^{\tau }} = {R_h^{\tau }}(x,t)\) as a continuous function with respect to time that mimics \({\vartheta }({u_h^{\tau }})\) , \({u_h^{\tau }}\in {{S_{hp}^{\tau q}}}\) . Let \(r_m\in P_{q+1}(I_m)\) be the right Radau polynomial on \(I_m\) , i.e., \(r_m(t_{m-1})=1\) and \(r_m(t_m)=0\) , and \(r_m\) is orthogonal to \(P_{q-1}(I_m)\) with respect to the \(L^2(I_m)\) inner product. Then we set

where \(\big \{{\cdot }\big \}\) is given by ( 9 ). The temporal flux reconstruction \({R_h^{\tau }}(x,t)\) is continuous in time, namely \({R_h^{\tau }}\in H^1(0,T, L^2({\varOmega }))\) and it satisfies the initial condition due to

Moreover, by the integration by parts and the properties \(r_m(t_{m-1})=1\) , \(r_m(t_m)=0\) , we obtain

which together with definition ( 26 ) implies

Finally, using the orthogonality of \(r_m\) to \(P_{q-1}(I_m)\) , we obtain from ( 28 ), the formula

Consequently, if \({u_h^{\tau }}\) is the approximate solution given by ( 13 ), then it satisfies

Obviously, the reconstruction \({R_h^{\tau }}\) is local and explicit, so its computation is fast and easy to implement.

The spatial flux reconstruction needs to define a function \({\sigma _h^{\tau }}\in L^2(0,T,H(\textrm{div},\varOmega ))\) which mimics the flux \({\sigma }({u_h^{\tau }}, \nabla {u_h^{\tau }}) = {\textbf{K}}({u_h^{\tau }})\nabla {u_h^{\tau }}\) , cf. ( 5 ). In particular, \({\sigma _h^{\tau }}|_{K\times I_m} \in P_q(I_m, {\textrm{RTN}}_{p}(K))\) where

is the Raviart-Thomas-Nedelec finite elements, cf. [ 7 ] for more details. We assume that the reconstructed flux \({\sigma _h^{\tau }}\) has to be equilibrated with the temporal flux \({R_h^{\tau }}\)

and with the Neumann boundary condition

In Sect.  5 we present two possible constructions of \({\sigma _h^{\tau }}\) including the choice of the spatial polynomial degree p in ( 32 ).

4.3 Auxiliary Results

In the forthcoming numerical analysis, we need several technical tools. We will employ the scaled space-time Poincarè inequality , cf. [ 11 , Lemma 2.2]: Let \(\varphi _{K,m}\in P_0(K\times I_m)\) be the \(L^2\) -orthogonal projection of \(\varphi \in H^1(K\times I_m)\) onto a constant in each space-time element \(K \times I_m\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=0,\dots ,r\) . Then,

where \(C_{\textrm{P}}\) is the Poincarè constant equal to \(1/\pi \) for simplicial elements and the last equality follows from ( 22 ).

Moreover, we introduce the space-time trace inequality

Let \(\varphi _{\gamma ,m}\in P_0(\gamma \times I_m)\) be the \(L^2\) -orthogonal projection of \(\varphi \in H^1(K\times I_m)\) onto a constant on each \(\gamma \times I_m\) , where \(\gamma \subset {\partial K}\) is an edge of \(K\in {{\mathcal {T}}_h^m}\) . Then there exists a constant \(C_{\textrm{T}}>0\) such that

where \(C_{\textrm{T}}= \max (c_T,C_{\textrm{P}})\) , \(C_{\textrm{P}}\) is from ( 35 ) and \(c_T>0\) is the constant from the (space) trace inequality.

The proof is straightforward, we present it for completeness. Let \(\varphi \in H^1(K \times I_m)\) and, for all \(t\in I_m\) , set \({{\tilde{\varphi }}}(t):= |\gamma |^{-1} \int _{\gamma } \varphi (x,t){\,{\mathrm d}S}\) . Observing that \((\varphi -{{\tilde{\varphi }}})\) and \(({{\tilde{\varphi }}}-\varphi _{\gamma ,m})\) are \(L^2(\gamma \times I_m)\) -orthogonal, we have

Using the standard trace inequality (e.g., [ 21 , Lemma 3.32]), we have

where \(c_T>0\) is a constant whose values can be set relatively precisely, see the discussion in [ 37 , Section 4.6]. Hence, integrating the square of ( 38 ) over \(I_m\) and using the fact that \(h_{\gamma }\le {h_K}\) , \(\gamma \subset {h_K}\) , we estimate the first term on the right-hand side of ( 37 ) as

Using the fact that \(\varphi _{\gamma ,m} = \tau _m^{-1} \int _{I_m} {{\tilde{\varphi }}}(t){\,{\mathrm d}t}\) , the one-dimensional Poincarè inequality on \(I_n\) and the Cauchy–Schwarz inequality yield

Collecting bounds ( 37 ), ( 39 ), ( 40 ) and the definition of the norm ( 22 ) yields ( 36 ). \(\square \)

4.4 Reliability

We presented the upper bound of \({{\mathcal {R}}}({u_h^{\tau }})\) , cf. ( 23 ).

Let \(u\in Y\) be the weak solution of ( 6 ) and \({u_h^{\tau }}\in {{S_{hp}^{\tau q}}}\) be the approximate solution given by ( 13 ). Let \({R_h^{\tau }}\in H^1(0,T, L^2({\varOmega }))\) be the temporal reconstruction given by ( 26 ) and \({{\sigma _h^{\tau }}}\in L^2(0,T,H(\textrm{div},\varOmega ))\) be the spatial reconstruction satisfying ( 33 ). Then

where \(C_{\textrm{P}}\) is the constant from Poincarè inequality ( 35 ), \(C_{\textrm{T}}\) is the constant from the trace inequality ( 36 ) and the estimators \(\eta _{R,K,m}\) , \(\eta _{S,K,m}\) , \(\eta _{T,K,m}\) , and \(\eta _{N,K,m}\) are given by

The proof of Theorem  1 can be found in [ 19 ] for the case of the homogeneous Dirichlet boundary condition. For completeness, we present its modification including mixed Dirichlet-Neumann boundary conditions.

Starting from ( 20 ), adding the terms \(\pm \big ({{R_h^{\tau }}},{v}\big )_{K,m}\) and \(\pm \big ({\nabla \cdot {\sigma _h^{\tau }}},{v}\big )_{K,m}\) , and using the integration by parts, we obtain

The terms \(\xi _i\) , \(i=1,\dots ,4\) are estimated separately.

Let \({v_{K,m}}\in P_0(K\times I_m)\) be the piecewise constant projection of \(v\in {V^\tau }\) given by the identity \( \big ({{v_{K,m}}},{1}\big )_{K,m} = \big ({v},{1}\big )_{K,m} \) . Using the Cauchy–Schwarz inequality, assumption ( 33 ), the Poincarè inequality ( 35 ), and ( 22 ), we have

Furthermore, by the Cauchy–Schwarz inequality and ( 22 ), we obtain

The use of ( 29 ), and a similar manipulations as in ( 45 ), give

Hence, estimates ( 45 )–( 46 ), the Cauchy inequality and ( 22 ) imply

Furthermore, let \({v_{\gamma ,m}}\in P_0(\gamma \times I_m)\) , \(\gamma \subset {{\partial K}_{\!N}}\) be the \(L^2\) -orthogonal projection from Lemma  3 . Then using assumption ( 34 ), the Cauchy inequality and the space-time trace inequality ( 36 ), we have

The particular estimates ( 44 ), ( 47 ), and ( 48 ), together with the discrete Cauchy–Schwarz inequality, imply ( 41 ). \(\square \)

Obviously, if \({\partial K}\cap {\varGamma _\textrm{N}}\ne \emptyset \) , then \(\eta _{N,K,m}=0\) .

4.5 Efficiency

The aim is to show that the local individual error estimators \(\eta _{R,K,m}\) , \(\eta _{S,K,m}\) and \(\eta _{T,K,m}\) from ( 41 )–(42) are locally efficient, i.e., they provide local lower bounds to the error measure up to a generic constant \(C>0\) which is independent of u , \({u_h^{\tau }}\) , h and \(\tau \) , but may depend on data problems and the degrees of polynomial approximation p and q . A dependence of the estimate up to this generic constant we will denote by \(\lesssim \) .

In order to derive the local variants of the error measure, we denote by \({\omega _K}\) the set of elements sharing at least a vertex with \(K\in {{\mathcal {T}}_h^m}\) , i.e.,

Moreover, we define the functional sub-spaces \(V_{D,m} = \{ v \in {V^\tau }: \text{ supp }\,(v) \subset \overline{ D \times I_m } \}\) for any set \(D\subset {\varOmega }\) (cf. ( 18 )) and the corresponding error measures (cf. ( 23 ))

Obviously, the definition of \(V_{D,m}\) and \({{\mathcal {R}}}_{D,m}({u_h^{\tau }})\) together with the shape regularity implies

Moreover, for each space-time element \(K\times I_m\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=1,\dots ,r\) , we introduce the \(L^2(K\times I_m)\) -projection of the non-polynomial functions, namely

Finally, for each vertex a of the mesh \({{\mathcal {T}}_h^m}\) , we denote by \({\omega _a}\) a patch of elements \(K\in {{\mathcal {T}}_h^m}\) that share this vertex. By \(p_a= \max _{K\in {\omega _a}} p_K\) we denote the maximal polynomial degree on \({\omega _a}\) . Then, for each a of \(K\in {{\mathcal {T}}_h^m}\) , we define a vector-valued function \({\overline{\sigma }}_a = {\overline{\sigma }}_a({u_h^{\tau }}, \nabla {u_h^{\tau }}) \in P_q(I_m, {\textrm{RTN}}_{p_a}(K))\) (cf. ( 32 )) by

where \(\left\langle {\cdot }\right\rangle \) denotes the mean value on \(\gamma \subset {\partial K}\) and \(\psi _a\) is a continuous piecewise linear function such that \(\psi _a(a)=1\) and it vanishes at the other vertices of K . Finally, we set \({\overline{\sigma }}|_{K\times I_m}=\sum _{{a\in K}}{\overline{\sigma }}_a\) .

The proof of the local efficiency of the error estimates presented is based on the choice of a suitable test function in ( 23 ). We set

where \(\chi _K(x)\) is the standard bubble function on K , \(\varPhi _m(t)\) is the Legendre polynomial of degree \(q+1\) on \(I_m\) (and vanishing outside) and \(P_h\big (\big \{{{\vartheta }({u_h^{\tau }})}\big \}_{m-1}\big )\in P_{p_K}(K)\) is the \(L^2(K)\) -projection weighted by \(\chi _K(x)\) given by

We note that

in general, compare with ( 52 ).

Using the inverse inequality, the polynomial function w given by ( 54 ) can be estimated as

Similarly as in [ 11 ] or [ 18 ], we introduce the oscillation terms

The goal is to prove the lower bounds of the proposed error estimates, namely to estimate \(\eta _{T,K,m}\) , \(\eta _{R,K,m}\) and \(\eta _{S,K,m}\) by \({{\mathcal {R}}}_{K,m}({u_h^{\tau }})\) and the oscillation terms ( 58 ), \(K\in {{\mathcal {T}}_h}\) , \(m=1,\dots ,r\) .

Let \(\eta _{T,K,m}\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=1,\dots , r\) be the error estimates given by (42), then

where \({{\mathcal {R}}}_{K,m}\) are the local error measures defined by ( 49 )–( 50 ) and the oscillation terms \(\eta _{G,K,m}\) , \(\eta _{{\vartheta },K,m}\) and \(\eta _{{\vartheta }^\prime ,K,m}\) are given by ( 58 ).

We start the proof by the putting function w from ( 54 ) as the test function in ( 50 ), i.e.

since \(\text{ supp } (w) = K\times I_m\) , cf. ( 54 ). Then, using ( 20 ) and the fact that w vanishes on \({\partial K}\) , we have

The functions \(\overline{\vartheta ^\prime ({u_h^{\tau }})}\) , \( {\overline{g}}\) and \({\sigma _h^{\tau }}\) are polynomials of degree q in time whereas w and \(\nabla w\) are the (Legendre) polynomial of degree \((q+1)\) in time, cf. ( 54 ). Due to the \(L^2(I_m)\) -orthogonality of the Legendre polynomials, we have \(\xi _1=0\) , since

Moreover, using inequality ( 57 ), relations ( 54 )-( 55 ) and the equivalence of norms on finite dimensional spaces,

Furthermore, let \(w_{K,m} = \frac{1}{K\times I_m} \int _{K\times I_m} w {\,{\mathrm d}x}{\,{\mathrm d}t}\) be the mean value of w on the space-time element \(K\times I_m\) . Due to ( 52 ), the Cauchy–Schwarz inequality and ( 35 ), we have

Similarly, the Cauchy–Schwarz inequality and ( 22 ) imply

Collecting ( 61 )–( 66 ), we have

Moreover, using ( 42c ), ( 26 ), integration by parts, the boundedness of the Radau polynomials, the triangle inequality and ( 58 ), we have

Hence, ( 67 ) and ( 68 )

which proves the theorem. \(\square \)

Let \(\eta _{S,K,m}\) and \(\eta _{R,K,m}\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=1,\dots , r\) be the error estimates given by (42), then

where \({{\mathcal {R}}}_{{\omega _K},m}\) is the local error measures defined by ( 49 )–( 50 ) and the oscillation terms \(\eta _{G,K,m}\) , \(\eta _{{\vartheta },K,m}\) and \(\eta _{{\vartheta }^\prime ,K,m}\) are given by ( 58 ).

The proof is in principle identical with the proof [ 18 , Lemmas 7-9], we present the main step for completeness. Let \({\overline{g}}\) and \({\overline{\sigma }}\) be the projection given by ( 52 ) and ( 53 ). Using the triangle inequality, the inverse inequality and ( 58 ), we obtain

The first term on the right-hand side of ( 72 ) can be estimated as in [ 36 , Theorem 4.10] by

where the resulting oscillation terms are estimated with the aid ( 58 ). Moreover, the last term on the right-hand side of ( 72 ) together with ( 42b ) and assumption ( 58 ), reads

which proves ( 70 ).

The proof of ( 71 ) is based on the decomposition

While the second term on the right-hand side of ( 75 ) can be estimated by assumption ( 58 ), the estimate of the first term is somewhat more technical. It depends on the flux reconstruction used. For the flux reconstruction in Sect.  5.2 , the proof is identical to the proof of [ 18 , Lemma 9], which mimics the stationary variant [ 24 , Theorem 3.12]. On the other hand, using the flux reconstruction from Sect.  5.1 , it is possible to apply the technique from [ 11 , Lemma 7.5], where the final relation has to be integrated over \(I_m\) . \(\square \)

5 Spatial Flux Reconstructions and Stopping Criteria

We present two ways of reconstructing the spatial flux \({\sigma _h^{\tau }}\in L^2(0,T,H(\textrm{div},\varOmega ))\) that satisfies the assumptions ( 33 )–( 34 ). The first one, proposed in [ 19 ] for the case of homogeneous Dirichlet boundary condition, is defined by the volume and edge momenta of the Raviart-Thomas-Nedelec (RTN) elements, cf. [ 7 ], and is easy to compute. The second approach is based on the solution of local Neumann problems on patches associated with each vertex of the mesh. This idea comes from, e.g., [ 24 ], its space-time variant was proposed in [ 18 ] for nonlinear convection-diffusion equations. Finally, in Sect.  5.3 , we discuss the errors arising from the solution of algebraic systems and introduce a stopping criterion for the appropriate iterative solver.

5.1 Element-Wise Variant

We denote by \(p_{K,\max }\) the maximum polynomial degree over the element K and its neighbours that share the entire edge with K and \(p_{\gamma ,\max }\) the maximum polynomial degree on neighbouring elements having a common edge \(\gamma \) . Let \({\textrm{RTN}}_{p_{K,\max }}(K)\) be the space of RTN finite elements of order \(p_{K,\max }\) for element \(K\in {{\mathcal {T}}_h^m}\) , cf. ( 32 ), and \({u_h^{\tau }}\in {{S_{hp}^{\tau q}}}\) be the approximate solution. The spatial reconstruction \({\sigma _h^{\tau }}\) is defined element-wise: for each \(K\in {{\mathcal {T}}_h^m}\) , find \({{\sigma _h^{\tau }}}|_{K\times I_m}\in P_q(I_m,{\textrm{RTN}}_{p_{K,\max }}(K))\) with \({{\sigma _h^{\tau }}\cdot n}|_{\gamma \times I_m}\in P_q(I_m,P_{p_{\gamma ,\max }}(\gamma ))\) such that

The edge momenta in ( 76 ) are uniquely defined and since \(p_{\gamma ,\max }\le p_{K,\max }\) , \({\sigma _h^{\tau }}\) in ( 76 ) is well defined as well. Here, the numerical flux \(\left\langle {{\textbf{K}}({u_h^{\tau }}) \nabla {u_h^{\tau }}}\right\rangle \cdot {n} - \alpha [{{u_h^{\tau }}}]\cdot {n}\) is conservative on interior edges, which implies that \({\sigma _h^{\tau }}\cdot {n}\) are the same on each interior edge \(\gamma \) and therefore the resulting reconstruction \({{\sigma _h^{\tau }}}\in L^2(0,T,H(\textrm{div},\varOmega ))\) globally.

Obviously, the first relation in ( 76 ) with \(p_K\le p_{\gamma ,\max }\) directly implies assumption ( 34 ). Moreover, using the Green theorem, ( 76 ), ( 12 ), ( 31 ) and \(p_K\le p_{\gamma ,\max }\le p_{K,\max }\) , we obtain

which justifies the assumption ( 33 ).

5.2 Patch-Wise Variant

For each vertex a of the mesh \({{\mathcal {T}}_h^m}\) , we denote by \({\omega _a}\) a patch of elements \(K\in {{\mathcal {T}}_h^m}\) sharing this vertex. By \(p_a= \max _{K\in {\omega _a}} p_K\) we denote the maximal polynomial degree on \({\omega _a}\) . Let \(P_{p_a}^{*}({\omega _a})\) be the space of piecewise polynomial discontinuous functions of degree \(p_a\) on \({\omega _a}\) with mean value zero for \(a \notin \partial {\varOmega }\) . We define the space

We set the local problems on patches \({\omega _a}\) for all vertices a : find \({\sigma _h^{\tau }}\in P_q(I_m, W^N_{{\textrm{RTN}},p_a}({\omega _a}))\) and \(r^\tau _a \in P_q(I_m, P_{p_a}^{*}({\omega _a}))\) such that

and \(\ell _{m,\gamma }: {S_{hp,m}}\rightarrow [{S_{h0,m}}]^d\) is the lifting operator defined by

Then the final reconstructed flux is obtained by summing up \({\sigma }^\tau _a\) on each element that contains vertex a , i.e.,

The assumption ( 34 ) follows directly from ( 78 ) and \(p_K\le p_a\) . Inserting the hat function \(\psi _a v\) for \(a\not \in \partial \varOmega \) and \(v\in P_q(I_m)\) in ( 17 ), using ( 5 ), ( 82 ) and omitting the zero terms, we have

Applying ( 13 ) and ( 31 ), we gain for \(a\not \in \partial \varOmega \) and \(v\in P_q(I_m)\)

From this it follows that the second relation in ( 79 ) holds element-wise, i.e.

Then ( 33 ) follows from

and from \(p_K\le p_a\) .

5.3 Stopping Criteria for Iterative Solvers

The space-time discretization ( 13 ) leads to a system of nonlinear algebraic equations for each time level \(m=1,\dots ,r\) . These systems have to be solved iteratively by a suitable solver, e.g., the Picard method, the Newton method or their variants. Therefore, it is necessary to set a suitable stopping criterion for the iterative solvers so that, on the one hand, the algebraic errors do not affect the quality of the approximate solution and, on the other hand, an excessive number of algebraic iterations is avoided.

However, the error estimates presented in Sect.  4 do not take into account errors arising from the inaccurate solution of these systems. Indeed, the aforementioned reconstructions fulfill assumption ( 33 ) only if the systems given by ( 13 ) are solved exactly. The full a posteriori error analysis including algebraic errors has been treated, e.g., in [ 8 , 23 , 29 ]. These error estimators are based on additional flux reconstructions that need to be evaluated at each iteration, and therefore, the overall computational time is increased.

To speed up the computations and control the algebraic errors, we adopt the technique of [ 17 ]. This approach offers (i) the measurement of algebraic errors by a quantity similar to the error measure ( 23 ), (ii) the setting of the stopping criterion for iterative solvers with one parameter corresponding to the relative error, and (iii) a fast evaluation of the required quantities.

For each \(m=1,\dots ,r\) , we define the estimators (cf. ( 23 ))

where the norm \({\left\| \cdot \right\| }_{{V^\tau }} \) is given by ( 22 ),

The space \({S_{hp+1}^{\tau q+1}}\) is an enrichment space of \({S_{hp}^{\tau q}}\) by polynomials of the space degree \(p_K+1\) and the time degree \(q+1\) for each \(K\times I_m\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=1,\dots ,r\) . Finally, we define the global in time quantities

Obviously, if \({u_h^{\tau }}\) fulfills ( 13 ) exactly, then \({\eta _{\textrm{alg}}^m}({u_h^{\tau }})=0\) for all \(m=0,\dots ,r\) . Moreover, if \({u_h^{\tau }}\) is the weak solution ( 6 ) then \({\eta _{\textrm{spa}}^m}({u_h^{\tau }})=0\) for all \(m=0,\dots ,r\) . Comparing ( 88 ) with ( 23 ), the quantity \({\eta _{\textrm{spa}}}({u_h^{\tau }})\) exhibits a variant of the error measure \({{\mathcal {R}}}({u_h^{\tau }})\) . Nevertheless, \({\eta _{\textrm{spa}}}({u_h^{\tau }})\) is neither lower nor upper bound of \({{\mathcal {R}}}({u_h^{\tau }})\) since \({S_{hp+1}^{\tau q+1}}\not \subset {V^\tau }\) and \({V^\tau }\not \subset {S_{hp+1}^{\tau q+1}}\) .

The quantities ( 88 ) can be evaluated very fast since the suprema (maxima) are the sum of the suprema (maxima) for all space-time elements \(K\times I_m\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=1,\dots ,r\) , which are computable separately, cf. [ 17 ] for details. Hence, we prescribe the stopping criterion for the corresponding iterative solver as

where \(c_A\in (0,1)\) is the user-dependent constant. The justification of this approach and the influence of algebraic errors on the error estimates are studied numerically in Sect.  6.1.1 .

6 Numerical Experiments

We present numerical experiments that justify the a posteriori error estimates ( 41 )–(42). Since the error measure ( 23 ) is the dual norm of the residual, it is not possible to evaluate the error even if the exact solution is known. Therefore, similarly to [ 18 ], we approximate the error by solving the dual problem given for each time interval \(I_m,\ m=1,\dots ,r\) : Find \( \psi _m \in {Y^{\tau }_{m}}=L^2(I_m,H^1({\varOmega })) \) ,

where (cf. ( 21a )–( 22 ))

Then we have \({{\mathcal {R}}}({u_h^{\tau }})^2 = \sum ^r_{m=1}{\left\| \psi \right\| }_{{Y^{\tau }_{m}}}^{2} \) . We solve ( 92 ) for each \(m=1,\dots ,r\) by linear conforming finite element on a global refinement of the space-time mesh \({{\mathcal {T}}_h^m}\times I_m\) which is proportional to the space and time polynomial approximation degrees. We denote this quantity by \(\widetilde{{{\mathcal {R}}}}({u_h^{\tau }})\) . The second error contribution \({{\mathcal {J}}}\) given by ( 24 ) is computable, so the total error \({{\mathcal {E}}}\) (cf. ( 25 )) is approximated by \(\widetilde{{{\mathcal {E}}}}({u_h^{\tau }}):= \left( \widetilde{{{\mathcal {R}}}}({u_h^{\tau }})^2+{{\mathcal {J}}}({u_h^{\tau }})\right) ^{1/2}\) .

Sometimes, this approximate evaluation of the (exact) error is not sufficiently accurate for fine grids and high polynomial approximation degrees. In this case, very fine global refinement is required and then the resulting algebraic systems are too large to be solved in a reasonable time.

All numerical experiments were carried out using the patch-wise reconstruction from Sect.  5.2 using the in-house code ADGFEM [ 10 ]. The arising nonlinear algebraic systems are solved iteratively by a Newton-like method, we refer to [ 14 ] for details. Each Newton-line iteration leads to a linear algebraic system that is solved by GMRES method with block ILU(0) preconditioner.

6.1 Barenblatt Problems

We consider two nonlinear variants of ( 3 ) following from the Barenblatt problem [ 4 ] where the analytical solution exists. The first variant reads

where the analytical solution is

Using the substitution \(v:=u^{1/m}\) , we have the second variant

having the solution

For both problems (( 94 )–( 95 ) and ( 96 )–( 97 )), the computational domain is \(\varOmega = (-6, 6)^2\) and the Dirichlet boundary condition is prescribed on all boundaries by ( 95 ) or ( 97 ). The final time is \(T=1\) .

We carried out computation using a sequence of uniform triangular grids (having 288, 1152, 4608 and 18432 triangles) with several combinations of polynomial approximation degrees with respect to space ( p ) and time ( q ). The time step has been chosen constant \(\tau =0.01\) . Besides the error quantities ( \(\widetilde{{{\mathcal {R}}}}({u_h^{\tau }})\) and \({{\mathcal {J}}}({u_h^{\tau }})\) ) and its estimators \(\eta \) , \(\eta _{R}:=\sum _{K,m}\eta _{R,K,m}\) , \(\eta _{S}:=\sum _{K,m}\eta _{S,K,m}\) and \(\eta _{T}:=\sum _{K,m}\eta _{T,K,m}\) , we evaluate the effectivity indices

In addition, we present the experimental order of convergence (EoC) of the errors and the estimators for each pair of successive meshes.

Tables  1 – 4 show the results for both Barenblatt problems (( 94 )–( 95 ) with \(m=0.25\) and ( 96 )–( 97 ) with \(m=2\) ) with two variants of the scaling parameter \({d_{K,m}}\) , \(K\in {{\mathcal {T}}_h^m}\) , \(m=1,\dots ,r\) given by ( 21a ) and ( 21b ). The quantity \(\#\textrm{DoF}\) represents the number of degrees of freedom in the space, that is, \(\#\textrm{DoF}=\dim {S_{hp,m}}\) , \(m=1,\dots ,r\) . We observe a good correspondence between \(\widetilde{{{\mathcal {R}}}}({u_h^{\tau }})\) and \(\eta \) , the effectivity index \(i_{\textrm{eff}}\) varies between 1 and 2.5 for all tested values of p and q and both variants of \({d_{K,m}}\) (( 21a ) and ( 21b )).

Finally, we note that the experimental orders of convergence EoC in Tables  1 – 4 ) of the error \(\widetilde{{{\mathcal {R}}}}({u_h^{\tau }})\) and its estimate \(\eta \) are \(O(h^p)\) for the choice ( 21b ) of the scaling parameter \({d_{K,m}}\) but only \(O(h^{p-1})\) for the choice ( 21a ). This follows from the fact that \(\tau _m \ll h_K\) for the computations of the Barenblatt problem and then the dominant part of \({d_{K,m}}\) is \(\tau _m^{-2} T {\left\| \tfrac{{\mathrm d}{\vartheta }}{{\mathrm d}u}\right\| }_{K,m,\infty } \) , cf. ( 21a ), which implies that \({d_{K,m}}= O(h^0)\) (the time step is the same for all computations). The dominant part of the error estimator is \(\eta _{S,K,m}\) , hence if \({\left\| {\sigma _h^{\tau }}- {\sigma }({u_h^{\tau }}, \nabla {u_h^{\tau }}) \right\| }_{K,m} = O(h^{p})\) then \(\eta _{S,K,m}= O(h^{p-1})\) , cf. ( 42b ). Nevertheless, comparing the pairs of Tables  1 – 2 and Tables  3 – 4 , we found that the effectivity indexes are practically independent of the choice of \({d_{K,m}}\) .

6.1.1 Justification of the Algebraic Stopping Criterion ( 91 )

We present the numerical study of the stopping criterion ( 91 ) which is used in the iterative solution of algebraic systems given by ( 13 ). We consider again the Barenblatt problem ( 94 )–( 95 ) with \(m=0.25\) and ( 96 )–( 97 ) with \(m=2\) . The user-dependent constant \(c_A\) in ( 91 ) has been chosen as \(10^{-1}\) , \(10^{-2}\) , \(10^{-3}\) and \(10^{-4}\) . Tables  5 and 6 show the estimators \(\eta \) , \({{\mathcal {J}}}({u_h^{\tau }})\) , \({\eta _{\textrm{alg}}}\) and \({\eta _{\textrm{alg}}}\) , cf. ( 90 ), for selected meshes and polynomial approximation degrees and the scaling parameter \({d_{K,m}}\) chosen by ( 21a ).

Additionally, we present the total number of steps of the Newton-like solver \({N_{\textrm{non}}}\) , the total number of GMRES iterations \({N_{\textrm{lin}}}\) and the computational time in seconds. The computational time has only an informative character.

We observe that the error estimators \(\eta \) , \({{\mathcal {J}}}({u_h^{\tau }})\) and also \({\eta _{\textrm{spa}}}\) converge to the limit values for decreasing \(c_A\) in ( 91 ) which mimic the case when the algebraic errors are negligible. Moreover, the relative differences between the actual values \(\eta \) and \({{\mathcal {J}}}({u_h^{\tau }})\) and their limits correspond more or less to the value of \(c_A\) . Obviously, smaller values of \(c_A\) cause prolongation of the computational time, due to a higher number of iterations, with a negligible effect on accuracy. Thus, the choice \(c_A=10^{-2}\) seems to be optimal in order to balance accuracy and efficiency.

The presented numerical experiments indicate that the estimator \({\eta _{\textrm{spa}}}({u_h^{\tau }})\) gives an upper bound of \({{\mathcal {R}}}({u_h^{\tau }})\) , however, this observation is not supported by the theory. The quantity \({\eta _{\textrm{spa}}}({u_h^{\tau }})\) is used only in the stopping criterion ( 91 ).

6.2 Tracy Problem

Tracy problem represents a standard benchmark, where the analytical solutions of the Richards equation are available [ 35 ]. We consider the Gardners constitutive relations [ 26 ]

where \(\psi = u - z\) is the pressure head, z is the vertical coordinate and the material parameters \({\textbf{K}}_s=1.2{\mathbb {I}}\) , \(\theta _s=0.5\) , \(\theta _r=0.0\) , and \(\alpha =0.1\) are the isotropic conductivity, saturated water content, residual water content, and the soil index parameter related to the pore-size distribution, respectively.

The computational domain is \({\varOmega }=(0,1)^2\) , the initial condition is set \(u = u_r: = -10\) in \({\varOmega }\) where \(u_r\) corresponds to the hydraulic head when the porous medium is dry. On the top part of the boundary \({\varGamma }_1:=\{(x, z),\ x\in (0,1),\ z= 1\}\) , we prescribe the boundary condition

and on the rest of boundary \({\varGamma }\) we set \(u = u_r\) . We note that this benchmark poses an inconsistency between the initial and boundary conditions on \({\varGamma }_1\) . Hence, the most challenging part is the computation close to \(t=0\) . In order to avoid the singularity at \(t=0\) , we investigate the error only on the interval \(t\in [1.0 \times 10^{-5}, 1.1 \times 10^{-4}]\) with the fixed time step \(\tau \) is \(1.0 \times 10^{-6}\) .

We perform a computation using a sequence of uniform triangular grids with several combinations of polynomial approximation degrees and the choice ( 21b ), the results are shown in Table  7 . We observe reasonable values of the effectivity indices except for the finest grids and the higher degrees of polynomial approximation, where the effectivity indices \(i_{\textrm{eff}}\) are below 1. Based on the values of EoC, we suppose that \(i_{\textrm{eff}}\) below 1 is not caused by the failure of the error estimator but due to an inaccurate approximation \(\widetilde{{{\mathcal {R}}}}({u_h^{\tau }})\) of the exact error; see Remark  3 .

7 Mesh Adaptive Algorithm

We introduce the mesh adaptive algorithm which is based on the a posteriori error estimates \(\eta \) , cf. ( 41 ). Let \(\delta >0\) be the given tolerance, the goal of the algorithm is to define the sequence of time steps \(\tau _m\) , meshes \({{\mathcal {T}}_h^m}\) and spaces \({S_{hp,m}}\) , \(m=1,\dots ,r\) such that the corresponding approximate solution \({u_h^{\tau }}\in {{S_{hp}^{\tau q}}}\) given by ( 13 ) satisfies the condition

Another possibility is to require \(\left( \eta ^2 + {{\mathcal {J}}}({u_h^{\tau }})\right) ^{1/2} \le \delta \) , then the following considerations have to be modified appropriately.

The mesh adaptation strategy is built on the equi-distribution principle, namely the sequences \(\{\tau _m,\, {{\mathcal {T}}_h^m},\, {S_{hp,m}}\}_{m=1}^r\) should be generated such that

where \(\eta _m:= (\sum _{K\in {{\mathcal {T}}_h^m}} \eta _{K,m}^2 )^{1/2}\) is the error estimate corresponding to the time interval \(I_m\) , \(m=1,\dots ,r\) and \(\#{{\mathcal {T}}_h^m}\) denotes the number of elements of \({{\mathcal {T}}_h^m}\) . Obviously, if all the conditions in ( 102 ) are valid, then the criterion ( 101 ) is achieved.

Space-time mesh adaptive algorithm.

Based on ( 101 )–( 102 ), we introduce the abstract Algorithm 1. The size of \(\tau _m\) , \(m=1,\dots ,r\) (step  8 of the algorithm) are chosen to equilibrate estimates of the spatial and temporal reconstruction, \(\eta _{S,m}:= (\sum _{K\in {{\mathcal {T}}_h^m}} (\eta _{S,K,m})^2)^{1/2}\) and \(\eta _{T,m}:= (\sum _{K\in {{\mathcal {T}}_h^m}} (\eta _{T,K,m})^2)^{1/2}\) , cf. (42). Particularly, we set the new time step according to the formula

where \(c_F\in (0,1)\) is the security factor and \(q\ge 0\) is the polynomial degree with respect to time. Therefore, \(q+1\) corresponds to the temporal order of convergence.

The construction of the new mesh (step  11 in Algorithm 1) is based on the modification of the anisotropic hp -mesh adaptation method from [ 15 , 20 ]. Having the actual mesh \({{\mathcal {T}}_h^m}\) , for each \(K\in {{\mathcal {T}}_h^m}\) we set the new volume of K according the formula

where \(\delta _{K,m}\) is the local tolerance from ( 102b ), | K | is the volume of | K | and \(\varLambda :{\mathbb {R}}^+\rightarrow {\mathbb {R}}+\) is a suitable increasing function such that \(\varLambda (1) = 1\) . For particular variants of \(\varLambda \) , we refer to [ 15 , 20 ].

When the new volume of mesh elements is established by ( 104 ), the new shape of K and a new polynomial approximation degree \(p_K\) are optimized by minimizing the interpolation error. This optimization is done locally for each mesh element. In one adaptation level, we admit the increase or decrease of \(p_K\) by one. Setting the new area, shape, and polynomial approximation degree for each element of the current mesh, we define the continuous mesh model [ 16 ] and carry out a remeshing using the code ANGENER [ 9 ].

The generated meshes are completely non-nested and non-matching, hence the evaluation of the time-penalty term (cf. Remark  1 ) is delicate. We refer to [ 20 ] where this aspect is described in detail and numerically verified. The presented numerical analysis takes into account the errors arising from the re-meshing in the temporal reconstruction \({R_h^{\tau }}\) , which contains term \( \big \{{{\vartheta }({u_h^{\tau }})}\big \}_{m-1}\) , cf. ( 26 ). The following numerical experiments show that the error estimator is under the control also after each remeshing.

7.1 Barenblatt Problem

We apply Algorithm 1 to the Barenblatt problem ( 96 ) with \(m=2\) . Table  8 shows the error estimators obtained by adaptive computation for three different tolerances \(\delta \) . Compared with the error estimators from Table  4 , we observe that the adaptive computations achieve significantly smaller error estimates using a significantly smaller number of degrees of freedom. We note that we are not able to present the quantity \(\widetilde{{{\mathcal {R}}}}\) (cf. ( 92 )–( 93 )) approximating the error since the finite element code used for the evaluation of \(\widetilde{{{\mathcal {R}}}}\) supports only uniform grids.

Figure  1 shows the performance of Algorithm 1, where each dot corresponds to one time step \(m=1,\dots ,r\) . We plot the values of the accumulated estimators \({\overline{\eta }}_m =\sum _{i=1}^m \eta _i\) for all \(m=1,\dots ,r\) . The red nodes correspond to all computed time steps, including the rejected ones (steps  11 – 12 of Algorithm 1) whereas the blue nodes mark only the accepted time steps. The rejected time steps indicate the re-meshing. Moreover, we plot the “accumulated” tolerance \(\delta (t_m/T)^{1/2}\) , cf. ( 101 ) and ( 102a ). We observe that the resulting estimator \(\eta \) at \(t=T\) is below the tolerance \(\delta \) by a factor of approximately 2.5 since conditions ( 102 ) are stronger than ( 101 ).

Barenblatt problem, ( 96 )–( 97 ), \(m=2\) , performance of Algorithm 1, accumulated error estimator \({\overline{\eta }}_m\) and the “accumulated” tolerance \(\delta (t_m/T)^{1/2}\) for \(m=1,\dots ,r\)

Figure  2 , left, shows the hp -mesh obtained by Algorithm 1 at the final time \(T=1\) , each triangle is highlighted by a color corresponding to the polynomial degree used \(p_K\) , \(K\in {{\mathcal {T}}_h^m}\) . We observe a strong anisotropic refinement about the circular singularity of the solution when \(u\rightarrow 0^+\) , see the analytical formula ( 97 ). Outside of this circle, large triangles with the smallest polynomial degree ( \(p=1\) ) are generated. On the other hand, due to the regularity of the solution in the interior of the circle, the polynomial degrees \(p=2\) or \(p=3\) are generated.

Barenblatt problem, hp -mesh obtained by Algorithm 1 (left) and the error estimators \(\eta _{K,m}\) , \(K\in {{\mathcal {T}}_h^m}\) at \(T=1\)

Moreover, Fig.  2 , right, shows the error estimator \(\eta _{K,m}\) , \(K\in {{\mathcal {T}}_h^m}\) at \(T=1\) . The elements in the exterior of the circle have small values of \(\eta _{K,m}\approx 10^{-17}\) – \(10^{-14}\) due to a constant solution and negligible errors. On the other hand, the values of \(\eta _{K,m}\) for the rest of elements \(K\in {{\mathcal {T}}_h^m}\) are in the range \(10^{-13}\) – \(10^{-11}\) due to the equidistant principle used.

7.2 Single Ring Infiltration

We deal with the numerical solution of the single ring infiltration experiment, which is frequently used for the identification of saturated hydraulic conductivity, cf. [ 32 , 39 ] for example. We consider the Richards equation ( 3 ) where the active pore volume \(\vartheta \) is given by ( 2 ), the water content function \(\theta \) is given by the van Genuchten’s law [ 27 ] and the conductivity \({\textbf{K}}(u) ={\textbf{K}}_s {{{\mathcal {K}}}}_r(u)\) is given by the Mualem function [ 31 ], namely

where \(\psi = u - z\) is the pressure head, z is the vertical coordinate and the material parameters \({\textbf{K}}_s=0.048\,{\mathbb {I}}\,\,\mathrm {m\cdot hours^{-1}}\) , \(\theta _s=0.55\) , \(\theta _r=0.0\) , \(\alpha =0.8\,\mathrm {m^{-1}}\) , \(n=1.2\) , \(m=1/6\) and \(S_s=10^{-3}\,\mathrm {m^{-1}}\) (cf. ( 2 )).

The computational domain together with the boundary parts is sketched in Fig.  3 a. On the boundary part \(\varGamma _D\) we set the Dirichlet boundary condition \(u= 1.05\,\textrm{m}\) , and on \({\varGamma _\textrm{N}}={\varGamma }{\setminus }{\varGamma _\textrm{D}}\) we consider the homogeneous Neumann boundary condition. The smaller “magenta” vertical lines starting at \({\varGamma _\textrm{D}}\) belong to \({\varGamma _\textrm{N}}\) . At \(t=0\) , a dry medium with \(u= \psi + z =-2\,\textrm{m}\) is prescribed. We carried out the computation until the physical time \(T=2\,\textrm{hours}\) . The inconsistency of the initial and boundary condition on \({\varGamma _\textrm{D}}\) makes the computation quite difficult for \(t\approx 0\) .

Single ring infiltration problem

Figure  3 b verifies the conservativity of the adaptive method. We plot the quantities

where F ( t ) is the total flux of the water through the boundary \({\varGamma }\) till time t and \(\varDelta V(t)\) is the changes of the water content in the domain between times 0 and t . From equation ( 3 ) and the Stokes theorem, we have the conservation law \(F(t) = \varDelta V(t)\) for all \(t\in [0,T]\) . Therefore, we also show the relative difference between these quantities \(|F(t) - \varDelta V(t)|/ \varDelta V(t)\) for \(t>0\) in Fig.  3 b the vertical label on the right. We observe that, except for the time close to zero, where the inconsistency between initial and boundary conditions is problematic, the relative difference is at the level of several percent.

Furthermore, Fig.  4 shows the accumulated estimators \({\overline{\eta }}_m =\sum _{i=1}^m \eta _i\) for time levels \(t_m\) , \(m=1,\dots ,m\) . The red nodes correspond to all computed time steps, including the rejected steps whereas the blue line connects only the accepted time steps. The rejected time steps are followed by the remeshing which is carried out namely for small t . We observe that the elimination of the rejected time steps causes that the errors arising from the remeshing do not essentially affect the total error estimate \(\eta \) .

Single ring infiltration, performance of Algorithm 1, accumulated error estimator \({\overline{\eta }}_m\) with respect to \(t_m\) , \(m=1,\dots ,r\)

Moreover, Fig.  5 shows the hp -meshes, the hydraulic head and the error estimator \(\eta _{K,m}\) , \(K\in {{\mathcal {T}}_h^m}\) at selected time levels obtained from Algorithm 1 with \(\delta =5.0 \times 10^{-3}\) . We observe the mesh adaptation namely at the (not sharp) interface between the saturated and non-saturated medium and also in the vicinity of the domain singularities. The error estimators \(\eta _{K,m}\) , \(K\in {{\mathcal {T}}_h^m}\) indicate an equi-distribution of the error.

Single ring infiltration, hydraulic head (top ), the corresponding hp -meshes obtained by Algorithm 1 (center) and the error estimators \(\eta _{K,m}\) , \(K\in {{\mathcal {T}}_h^m}\) (bottom) at \(t=0.4\) , \(t=0.8\) and \(t=2\) \(\,\textrm{hours}\) (from left to right)

8 Conclusion

We derived reliable and efficient a posteriori error estimates in the residual-based norm for the Richards equation discretized by the space-time discontinuous Galerkin method. The numerical verification indicates the effectivity indexes between 1 and 2.5 for the tested examples. Moreover, we introduced the hp -mesh adaptive method handling varying non-nested and non-matching meshes and demonstrated its efficiency for simple test benchmark and its applicability for the numerical solution of the single ring infiltration experiment.

It will be possible to generalize the presented approach to genuinely space-time hp -adaptive method, where the (local) polynomial order q in time is varied as well. However, the question is of potential benefit. Based on our experience, the setting \(q=1\) gives sufficiently accurate approximation for the majority of tested problems.

On the other hand, the choice \(q=0\) would be sufficient only in subdomains of \({\varOmega }\) where the solution is almost constant in time. Therefore, we suppose that the benefit of local varying of polynomial order in time will be low.

Although the presented numerical examples are two-dimensional, it would be possible to apply the presented error estimates and mesh adaptation to three-dimensional problems as well. We refer, e.g., to [ 1 ] and the references therein, where the anisotropic mesh adaptation techniques are developed for time-dependent 3D problems.

Data Availability

No datasets were generated or analysed during the current study.

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Open access publishing supported by the National Technical Library in Prague. This work has been supported by the Czech Science Foundation Grant No. 20-01074 S (V.D.), the Charles University grant SVV-2023-260711, and the Grant Agency of Charles University Project No. 28122 (H.S.), European Development Fund-Project “Center for Advanced Aplied Science” No. CZ.02.1.01/0.0/0.0./16 019/0000778 (M.V.). V.D. acknowledges the membership in the Nečas Center for Mathematical Modeling ncmm.karlin.mff.cuni.cz.

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Dolejší, V., Shin, HG. & Vlasák, M. Error Estimates and Adaptivity of the Space-Time Discontinuous Galerkin Method for Solving the Richards Equation. J Sci Comput 101 , 11 (2024). https://doi.org/10.1007/s10915-024-02650-x

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Received : 15 January 2024

Revised : 28 May 2024

Accepted : 28 July 2024

Published : 20 August 2024

DOI : https://doi.org/10.1007/s10915-024-02650-x

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• Space-time discontinuous Galerkin method
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