Solution of Burgers' equation

Question

Can you help me to solve this problem and explain the method used?

$u\equiv u(x,t)$ with $x\in\mathbb{R}$

$$ u_t+\left( \frac{1}{2}u^2\right)_x=0 $$ with initial data

$$u_0(x)=u(x,0)= \begin{cases} 1-x^2, &\text{for } x\in [-1,1] \\\\ 0, &\text{else} \end{cases} $$

I tried to find $u$ using $u=u_0(x-ut)$ but I don't get the solution. In particular with the boundary conditions.

Thanks for your help.

EDIT: Solving with the characteristics method I find this solution $$u(x,t)=1-\left( \frac{1\pm \sqrt{(1-4\chi t (x-\chi t)}}{2\chi t} \right)^2$$ where $\chi\equiv\chi_{[-1,1](x)}$ Solving with Upwind scheme for Burgers's equation the solution does not agree with the numerical one.

Answer 1

I will be using method of characteristics (which is described in Wikipedia article for Burger's euqation in detail).

Writing down the equation:

$$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0$$

This has the form of a full time derivative for some function:

$$\frac{du}{dt}=\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x} \frac{dx}{dt}$$

Which gives us the following system of ODEs:

$$\frac{du}{dt}=0 \\ \frac{dx}{dt}=u$$

The solutions have the form:

$$u=u_0 \\ x=x_0+u_0 t$$

This seems surprising, but actually this means that for any $x$:

$$u \left(x+u(x,0)t,t \right)=u(x,0) \tag{1}$$

Now we want to get an explicit solution using our initial conditions.

• First, consider the simple case $|x|>1$, which means $u(x,0)=0$:

$$u(x,t)=u(x,0)=0 \tag{2}$$

• Now, the case $|x| \leq 1$, which means $u(x,0)=1-x^2$:

Let's make a substitution:

$$x+(1-x^2)t=y$$

We need to find $x(y,t)$ now to substitute on the right hand side of $(1)$:

$$t x^2-x+y-t=0$$

$$x=\frac{1}{2t} (1 \pm \sqrt{1+4t(t-y)})$$

Which makes our solution in this range to be:

$$u(y,t)=1-\frac{1}{4t^2} (1 \pm \sqrt{1+4t(t-y)})^2$$

Or, simply renaming the variable again, we have:

$$u(x,t)=1-\frac{1}{4t^2} (1 \pm \sqrt{1+4t(t-x)})^2 \tag{3}$$

You can directly check (by taking the derivatives) that the original equation is satisfied by this function.

If we want the limit at $t \to 0$ to exist, we need to choose the "$-$" sign in (3).

---

One thing we need to account for is the range $|x| \leq 1$ which separates the first solution from the second one. Not sure if we just can set the same condition on $y$ or not.

Let's consider the condition on $x$:

$$\left| \frac{1}{2t} (1 - \sqrt{1+4t(t-y)}) \right| \leq 1$$

Assume $t>0$:

$$\left| 1 - \sqrt{1+4t(t-y)} \right| \leq 2t$$

The solution is quite interesting:

$$-1 \leq y \leq 1, \qquad t \leq \frac12$$

$$-1 \leq y \leq \frac{4t^2+1}{4t}, \qquad t > \frac12$$

The latter case contradicts the initial condition on $x$ which we used to get our first, zero solution.

When plotting the solution in Mathematica for different times we see a 'crashing wave', which initially, doesn't go beyond $x=1$.

However, for larger times, it does go beyond the initial boundary:

Answer 2

As you have a right-moving phase left of a stationary phase, you will get a shock front that separates both. At the points $(t,y_s(t))$ of the shock front, the equation $y_s(t)=x+tu_0(x)$ has two solutions, $x_1(t)\in(-1,1)$ and $x_2(t)=y(t)\ge 1$. As $$ y=x+t(1-x^2)\iff 4ty=1+4t^2-(2tx-1)^2 $$ the maximum of $y$ for fixed $t$ and $x\in[-1,1]$ is

• at $x=1$ for $t<\frac12$ with $y_{max}=1$ and
• at $x=\frac1{2t}$ with $y_{max}=\frac1{4t}+t$ after that.

This means that the phases "start colliding" at $t=\frac12$.

It is known that the shock front moves at the mean of the speeds of the phases, in this case at $$ y_s'(t)=\frac12u_0(x_1(t)),~~ y_s(\frac12)=1, $$ as $u_0(x_2(t))=0$. On the other hand we know that $$ y_s'(t)=x_1'(t)+u_0(x_1(t))+tu_0'(x_1(t))x_1'(t) $$ which allows to translate the differential equation for $y_s$ into a differential equation for $x_1$, $$ -\frac12(1-x_1^2)=(1-2tx_1)x_1',~~ x_1(\frac12)=1 \\~\\ (1-x_1^2)\frac{dt}{dx_1}=4x_1t-2,~~ t(1)=\frac12 $$ which can be solved as linear ODE $$ \frac{d}{dx}\left(t(x)(1-x^2)^2\right)=t'(x)(1-x^2)^2-4t(x)x(1-x^2)=-2(1-x^2) \\~\\ t(x)=\frac{-2(3x-x^3)+C}{3(1-x^2)^2}. $$ To get a finite value at $t(1)$ we get necessarily $C=4$ so that $$ t(x)=\frac{2(2+x)}{3(1+x)^2} $$ indeed satisfies the initial condition. Solving the quadratic equation for $x_1(t)\in[-1,1]$ gves \begin{align} &&9t^2(1+x)^2&=2(3t(1+x))+6t \\& \iff&(3t(1+x)-1)^2&=1+6t \\& \implies& x_1(t)&=\frac{1+\sqrt{1+6t}}{3t}-1=\frac{1-3t+\sqrt{1+6t}}{3t} \\& \implies& y_s(t)&=\frac{1-3t+\sqrt{1+6t}}{3t}+t\left(1-\frac{2+9t^2+2(1-3t)\sqrt{1+6t}}{9t^2}\right) \\&& &=\frac{1-3t+\sqrt{1+6t}}{3t}-\frac{2+2(1-3t)\sqrt{1+6t}}{9t} \\&& &=\frac{1-9t+(1+6t)\sqrt{1+6t}}{9t} \end{align}

python script for the picture

tf = 3.0
for x in np.arange(-1.5,tf,0.05):
    u = 0; ts = tf
    if abs(x)<1: u = 1-x**2; ts = min( tf, 2*(2+x)/(3*(1+x)**2))
    plt.plot([x, x+ts*u], [0, ts], 'b',lw=0.5)
    if ts < tf: ys = x+ts*u; plt.plot( [ys,ys], [0,ts], 'b', lw = 0.5)

t = np.arange(0.5,tf+1e-3,0.05);
ys = (1-9*t+(1+6*t)**1.5)/(9*t)
plt.plot(ys,t,'r',lw=2.5); 
plt.xlabel('x'); plt.ylabel('t'); 
plt.title('method of lines with shock front'); 
plt.grid(); plt.show()