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I managed to understand the method of characteristics to get the solution of the transport equation. However, I am getting stucked finding the final solution for the Burgers' equation $u_t + u u_x = 0$ given the following initial solution

$$ u_0(x) = \begin{cases} 0, & x < -1 \\ 1+x, & -1 < x < 0 \\ 1-x, & 0 < x < 1 \\ 0, & 1 < x \\ \end{cases} $$

The solution I get is of the form

$$u(x,t) = u(x-u_0(x)t)$$

and according to my calculations this gives

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

which does not matches with the correct answer in my notes. Can anyone explain me how to do this correctly?

EditPiAf
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BRabbit27
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1 Answers1

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The peak at height $u=1$ travels with speed equal to $1$, so you should have $u(x,t)=1$ when $x=t$, and since the inviscid Burgers equation has the property that a piecewise linear wave profile stays piecewise linear (as can be seen by looking at the characteristic curves), the wave profile $u(x,t)$ for a fixed $t \in (0,1)$ (before shock formation) is simply given by connecting the point $(x,u)=(-1,0)$ with a straight line to $(x,u)=(t,1)$, and from there to $(x,u)=(1,0)$ with a straight line. (And $u=0$ outside the interval $-1<x<1$, of course.)

ViktorStein
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Hans Lundmark
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