Inviscid Burgers' Equation and Implicit Solutions

Question

Consider the Burgers' equation:

$u_{t}+uu_{x}=0 \quad u(x,0)=x^{3}$

Find a solution to the PDE in terms of a parameter $x_{0}$ which is implicitly related to $(x,t)$. Show that this parameter is unique for given $(x,t)$. Moreover, sketch the characteristics and the profiles $u(x,0),u(x,1),u(x,3)$ of the solution.

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I have found a solution $u(x,t)=x_{0}^{3}$ for $x_{0}$ being implicitly related to $(x,t)$ (where in the characteristics $x$ is a function of $t$) as follows:

$x(t)=x_{0}^{3}t+x_{0}$

I've shown that the parameter is unique by contradiction (if we have two different parameters we get a negative time value). However, I don't know how to precisely sketch the characteristics and the profiles of the solution.

The characteristics will be of the form of lines with slope $x_{0}^{3}$ as seen from the equation of $x(t)$, and with increasing slope (since $x^{3}$ is an increasing function in $x$) which implies that the characteristics will span the whole plane and will be parallel to each other. However, I don't know how to make this more precise. Moreover, I'm lost in sketching the profiles since I don't have a clear intuition of what characteristics are. We have the characteristic $u(x,0)$ given, yet I don't know how the profiles $u(x,1), u(x,3)$ would look like. What would be a way to get a better intuition on this?

Thanks for the help!

Answer 1

This isn't the answer to the question, but a too long comment to be put in the comments section.

So, you will be able to compare the exact result below with the sketches of characteristics and profiles that you will obtain without fully solving the PDE.

$$u_{t}+uu_{x}=0 \quad u(x,0)=x^{3}$$

FIRST PART : Solving $\quad u_t+uu_x=0\quad $without taking account of the initial condition.

Set of characteristic differential equations : $\quad \frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}$

A first equation of characteristic curve comes from $du=0\quad\to\quad u=c_1$

A second equation of characteristic curve comes from $\quad \frac{dt}{1}=\frac{dx}{c_1} \quad\to\quad x-c_1t=c_2$

General solution on the form of an implicit equation : $$\Phi\left(c_1\:,\:c_2\right)=\Phi\left(u\:,\:x-ut\right)=0$$ where $\Phi$ is any differentiable function of two variables. An equivalent form is : $$u=F\left(x-ut\right)$$ where $F$ is any differentiable function of one variable.

SECOND PART :

Initial condition : $u(x,0)=x^3=F\left(x-0\right) \quad\to\quad F(X)=X^3$

Instead of $x$, we use a dummy variable $X$ in order to avoid confusion between the case $X=x$ if $t=0$ and $X=(x-ut)$ if $t\neq 0$.

Thus, the particular solution according to the given initial condition is : $$u=(x-ut)^3$$ This is expressed on the form of an implicit equation.