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I am trying to solve Burgers equation using the method of characteristics: $$\partial_tu+u \partial_x u=0$$ with the following initial condition: $$\begin{cases} 1 & x<0 \\ 1-x & 0\leq x \leq 1 \\ 0 & x>1 \end{cases}$$ Then using the Method of characteristics I find that: $$X(t)=u(t,X(t))t+x$$ And $$u(t,X(t))=u(0,x)$$ And so for example if $x\in [0,1]$ then $u(t,X(t))=1-x=1-(X(t)-u(t,X(t)) t)$

which means that $u(t,X(t))=\frac{1-X(t)}{1-t}$

The problem is that I can't find the explicit cases

EditPiAf
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BrianTag
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1 Answers1

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According to the equation $X(t) = u(0,x)t + x$ in OP,

  • if $u(0,x)=1$, then $X = t+x$ with $x<0$, i.e. $X < t$.

  • if $u(0,x)=1-x$, then $X = u(0,x)t+ x$ with $0\leq x\leq 1$. Thus, $$u = u\big(0,X - u t\big) = 1-X + u t\, ,$$ so that $u(t,X(t)) = \tfrac{1-X}{1-t}$ for $t\leq X \leq 1$.

  • if $u(0,x)=0$, then $X = x$ with $x>1$, i.e. $X>1$.

Finally, for $0\leq t <1$, the solution reads $$ u(t,x) = \left\lbrace \begin{aligned} & 1 &&\text{if}\quad x<t \\ & \tfrac{1-x}{1-t} &&\text{if}\quad t\leq x\leq 1\\ & 0 &&\text{if}\quad x>1 \end{aligned}\right. $$

See also this post for a book reference (Evans, Ex. 1 p. 139).

EditPiAf
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