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Consider the initial-value problem for Burgers' Equation $$\left\{ \begin{array}{ll} u_{t} + (\frac{u^{2}}{2})_{x}=0 & \mathbb{R} \times (0, \infty) \\ u=g & \text{on } \mathbb{R} \times {t=0} \end{array} \right.$$

with initial data $$g(x)= \begin{cases} 1, & x\leq 0 ,\\ 1-x, & 0\leq x \leq 1 ,\\ 0, & x \geq 1.\end{cases}$$

This is an example out of Evans PDE Text (Ex. 1 p. 139) and I am trying to understand how we can find $u(x,t)$ from here, it just says in the book and thus $$u(x,t) = \begin{cases} 1, & x\leq t ,\\ \tfrac{1-x}{1-t}, & t \leq x \leq 1,\\ 0, & x \geq 1.\end{cases} \quad (0 \leq t \leq 1)$$

What is the general way to write down this form for $u(x,t)$? How does the author find $u(x,t)$ for this problem? The book omits any details.

EditPiAf
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James.Welch
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2 Answers2

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$$u_t+u\:u_x=0$$ Characteristic system of DEs : $\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}$

First characteristic curve : $du=0 \quad\to\quad u=c_1$

Second characteristic curve : $\frac{dt}{1}=\frac{dx}{c_1}\quad\to\quad x-c_1 t=c_2$

Equation of general solution : $\Phi\left( u\;,\:x-u\:t \right)=0$ for any differentiable function $\Phi$.

Equivalent expression of the general solution expressed on implicit form : $$u=F(x-ut)\qquad \text{any differentiable function }F.$$

Initial conditions : $u(x,0)=F(x-0\:u)=F(x)\qquad$ $\begin{cases} u(x,0)=1 & x \leq 0 \\ u(x,0)=1-x & 0\leq x\leq 1 \\ u(x,0)=0 & 1\leq x \end{cases}$

This determines the function $F(X)$ , any dummy variable $X$ : $\quad\begin{cases} F(X)=1 & X \leq 0 \\ F(X)=1-X & 0\leq X\leq 1 \\ F(X)=0 & 1\leq X \end{cases}$

With $X=x-u\:t \qquad \begin{cases} u=1 & x-u\:t \leq 0 \\ u=1-(x-u\:t) & 0\leq x-u\:t\leq 1 \\ u=0 & 1\leq x-u\:t \end{cases}$

Case $u=1-(x-u\:t) \quad\to\quad u=\frac{1-x}{1-t}\quad$ if $\quad 0\leq x-\frac{1-x}{1-t}\:t\leq 1 \qquad 0\leq \frac{x-t}{1-t}\leq 1$

$$ \begin{cases} u(x,t)=1 & x \leq t \\ u(x,t)=\frac{x-t}{1-t} & 0\leq \frac{x-t}{1-t}\leq 1 \\ u(x,t)=0 & 1\leq x \end{cases}$$

JJacquelin
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Alternatively, let's apply the method of characteristics in parameter-dependent form:

  • $\frac{d u}{d s} = 0$, letting $u(0)=g(x_0)$ we know $u=g(x_0)$;
  • $\frac{d t}{d s} = 0$, letting $t(0)=0$ we know $t=s$;
  • $\frac{d x}{d s} = u$, letting $x(0)=x_0$ we know $x=x_0+ut$;

This solution can be rewritten in implicit form: $$ u = g(x-ut) . $$ The expression of $g$ gives the piecewise solution

  • $u=1$ if $x-ut \leq 0,\quad$ i.e. $\quad u=1$ if $x\leq t$;
  • $u=1-x+ut$ if $0\leq x-ut \leq 1,\quad$ i.e. $\quad u=\frac{1-x}{1-t}$ if $t\leq x \leq 1$;
  • $u=0$ if $x-ut \geq 1,\quad$ i.e. $\quad u=0$ if $x\geq 1$.

Hence the proposed solution for small times $t<1$ is retrieved. See also this post for an illustration.

EditPiAf
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