2
$\begingroup$

Consider the nonlinear wave problem $$ \begin{cases} u_t + (x+u)u_x = 0\\ u(x,0) = x \end{cases} $$

a) Write down Monge's equations in the for $du/d\tau$...,etc, and solve them.

b) Plot the characteristics in the tx plane

So, I honestly don't know how to do this. I know the process when we have an equation of the form $u_t + c(u)u_x = 0$, but not when that function also includes x?

All I can confidently do is write Monge's equations for the problem: $$ \begin{cases} \dfrac{dt}{d\tau} = 1,\\ \dfrac{dx}{d\tau} = x+u,\\ \dfrac{du}{d\tau}=0 \end{cases} $$

Which gives that

$$ \begin{cases} u = c_1 \text{ (const. w.r.t } \tau)\text{ and } \\\dfrac{dx}{dt} = x + u \end{cases} $$

But I don't know how to go about solving this and I'm struggling to find resources on this.

If anyone could explain what I need to do here or point me to some resources on this it would be much appreciated.

I tried to follow this answer, as it seemed like a similar problem, however the solution I got was $$ u(x,y) = \frac{x(1-t)}{1+t}, $$ but this doesn't work when substituted into the PDE so I must have done something wrong or perhaps that method is not what I want to be doing?

$\endgroup$
2
  • 2
    $\begingroup$ $$u = c_{1} \implies x' = x + c_{1} \implies x = - c_{1} + c_{2} e^{t} = - u + c_{2} e^{t} \implies c_{2} = (x + u)e^{-t}$$ which gives $$u = f((x + u)e^{-t})$$ and you can check satisfies the PDE. Now apply the initial condition. $\endgroup$ Commented Mar 2 at 12:48
  • $\begingroup$ @MatthewCassell Okay, I got solution $u(x,t) = \frac{x}{2e^t - 1}$, which works. So how do I sketch the characteristics? Am I right that the equation is $x = c_2e^t - u$? If so, what values can I place for $c_2$ and $u$ in order to sketch these? $\endgroup$
    – spooleey
    Commented Mar 2 at 13:48

1 Answer 1

2
+50
$\begingroup$

Regarding the PDE

$$ \begin{cases} u_t + (x+u)u_x = 0\\ u(x,0) = x \end{cases} $$

we have

$$ \frac{dt}{1}=\frac{dx}{x+u}=\frac{du}{0} $$

and solving,

$$ \cases{ u = c_1\\ e^{t + c_2} = x+c_1 \Rightarrow e^t c_3 = x + c_1 } $$

but $c_1 = u$ and $c_3 = G(u)$, so

$$ e^tG(u) = x+u $$

Now $G(\cdot)$ is determined according to the boundary conditions:

$$ e^0G(u(0,x))=x+u(0,x)\Rightarrow G(x) = 2x $$

then we follow with

$$ e^t(2 u)=x+u\Rightarrow u(t,x) = \frac{x}{2e^t-1} $$

Along the characteristic curves, $u(x,t) = c_0$ so we have

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .