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Solve the initial value problem $$u_t+3uu_x=0 ,\quad u(0,x)=\left\{\begin{matrix} 2 & x<1\\ 0& x>1 \end{matrix}\right.$$

My Idea:

given that $u_t+3uu_x=0$

Now $\frac{dt}{1}=\frac{dx}{3u}=\frac{du}{0}$

$\frac{dt}{1}=\frac{du}{0} \rightarrow u=c_1$

$\frac{dt}{1}=\frac{dx}{u} \Rightarrow \frac{dt}{1}=\frac{dx}{3c_1} \Rightarrow 3c_1t-x=c_2$

any one can help me from here

EditPiAf
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Inverse Problem
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2 Answers2

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Let us assume that you want to solve $u_t + 3uu_x = 0$, with the proposed initial data $$ u(x,0) = \left\lbrace \begin{aligned} &2 &&\text{if}\quad x < 1 \, ,\\ &0 &&\text{if}\quad x > 1 \, .\\ \end{aligned} \right. $$ We follow the steps in this post. According to the Lax entropy condition, the solution to this Riemann problem is a shock wave (characteristic curves intersect in the $x$-$t$ plane), which speed $s$ is given by the Rankine-Hugoniot condition $$ s = \frac{3}{2}\left(0 + 2\right) = 3 \, . $$ One obtains the following solution: $$ u(x,t) = \left\lbrace \begin{aligned} &2 &&\text{if}\quad x < 1+st \, ,\\ &0 &&\text{if}\quad x > 1+st \, .\\ \end{aligned} \right. $$

EditPiAf
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Use the method of characteristics:

I assume you want to solve $$u_t+kuu_x=0.$$

Method of charactersitics results in: $$\dfrac{dt}{1}=\dfrac{dx}{ku}=\dfrac{du}{0}.$$

Using $$\dfrac{dt}{1}=\dfrac{du}{0}$$

gives $u=c_1$.

Using $$\dfrac{dt}{1}=\dfrac{dx}{ku}=\dfrac{dx}{kc_1}$$

$$\implies kc_1t=x+c_2 \implies c_2=kc_1t-x.$$

We know that

$$u=c_1=F(c_2)=F(kc_1t-x)=F(ktu-x)$$

in the last step I used $u=c_1$.

MrYouMath
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    You are also not consistent. The characteristics you use are for $1\cdot u_t+ku\cdot u_x=0\cdot u$. – Lutz Lehmann Jan 22 '18 at 11:44
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    Your original PDE should have $x$ and $t$ switched, after that it's right. – Ian Jan 22 '18 at 12:24
  • @MrYouMath.. i think the final solution should be in $x,t$ – Inverse Problem Jan 22 '18 at 14:32
  • @LutzL & Ian Thank you for pointing this out. – MrYouMath Jan 22 '18 at 17:15
  • @suresh you have to simplify the equation by using the conditions that you have given. But the general solution of your PDE should look like the solution that I have written. – MrYouMath Jan 22 '18 at 17:19
  • @suresh I am curious about how one deduces the solution from the proposed characteristics (recall that the method of characteristics is useless in the vicinity of shocks!). You may get nonunique solutions or non-entropic shocks (see my answer for details). – EditPiAf Jan 24 '18 at 09:33