2

I am just learning the method of characteristics. Suppose I want to solve $$xu_t - tu_x = u \quad u(x,0) = h(x)$$

I write $$\dot{t}(s) = x\\ \dot{x}(s) = -t \\ \dot{z}(s) = z$$

If $t(0) = 0$ and $x(0) = x_0$, we have (I think) $$t(s) = x_0\sin(s)\\ x(s) = x_0\cos(s)\\ z(s) = h(x_0)e^s$$

Where do I go from here? First I'd like to get rid of the $s$, then I'd like to get rid of the $x_0$, but I'm not seeing how to do that.

dfeuer
  • 9,069
Eric Auld
  • 28,127
  • The case is surprisingly subtle. Sure this is the one? – Did Dec 17 '13 at 11:37
  • @Did Yes, this is what the instructor wrote, although he could have made a mistake, of course. We only ask that there be a local solution (it may not exist for all time). Does that help? Also, we are only interested in ${(x,t)\mid t>0}$. – Eric Auld Dec 17 '13 at 15:46

3 Answers3

1

The idea of the method of characteristics is to compute $u$ along some curve $s\mapsto (x(s),t(s))$, here $(x(s),t(s))=(x_0\cos s,x_0\sin s)$ for some $x_0$. You showed that, for every solution $u$ defined on the circle centered at $(0,0)$ with radius $|x_0|$, the function $z:s\mapsto u(x(s),t(s))$ is such that $z(s)=h(x_0)\mathrm e^s$. Thus, $u(x_0\cos s,x_0\sin s)= h(x_0)\mathrm e^s$ for every $x_0$ and $s$. Solving $(x,t)=(x_0\cos s,x_0\sin s)$ yields $x_0=\sqrt{x^2+t^2}$ and $s=\arg(x,t)$ hence one would get $$u(x,t)=h\left(\sqrt{x^2+t^2}\right)\mathrm e^{\arg(x,t)}.$$ The trouble, however, is that this does not define a regular function $u$ on the whole plane. For example, following the circle centered at $(0,0)$ with radius $\sqrt{x^2+t^2}$ anticlockwise during one period, one obtains by continuity that $u(x,t)=u(x,t)\mathrm e^{2\pi}$ for every $(x,t)\ne(0,0)$, which is absurd unless $u=0$. A second problem is that, following semi-circles centered at $(0,0)$, one should have $u(-x,0)=u(x,0)\mathrm e^{\pi}$ and $u(-x,0)=u(x,0)\mathrm e^{-\pi}$ for every $x\ne0$, which is absurd as well.

To sum up:

  • If $h(-x)=\mathrm e^\pi h(x)$ for every $x\gt0$, then the function $u$ exists and is given by the formula above, on every slit plane $\{(x,t)\mid\theta\lt\arg(x,t)\lt2\pi+\theta\}$ such that $-\pi\lt\theta\lt0$.
  • If $h(-x)=\mathrm e^{-\pi} h(x)$ for every $x\gt0$, then the function $u$ exists and is given by the formula above, on every slit plane $\{(x,t)\mid\theta\lt\arg(x,t)\lt2\pi+\theta\}$ such that $0\lt\theta\lt\pi$.
  • Otherwise, the function $u$ exists and is given by the formula above, either on the slit plane $\{(x,t)\mid-\pi\lt\arg(x,t)\lt\pi\}$, or on the slit plane $\{(x,t)\mid0\lt\arg(x,t)\lt2\pi\}$.
Did
  • 279,727
1

Clearly, the domain of a solution of $xu_t-tu_x=u$, can not contain any circle centered at the origin. But this is the only restriction!

Take $\Omega=\mathbb R^2\smallsetminus C$, where $C$ is a curve starting from the origin and going to infinity, assume also that $C$ does not intersect to positive $x$-axis. Let $h=h(x), x\in(0,\infty)$ be a $C^1$ function. Then using the ideas of the previously exposed solution, you can obtain a unique solution $u=u(x,t)$ of the PDE defined in $\Omega$, such that $u(x,0)=h(x)$.

Yiorgos S. Smyrlis
  • 83,933
0

$xu_t-tu_x=u$

$\dfrac{u_t}{t}-\dfrac{u_x}{x}=\dfrac{u}{xt}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=\dfrac{1}{t}$ , letting $t(0)=0$ , we have $t^2=2s$

$\dfrac{dx}{ds}=-\dfrac{1}{x}$ , letting $x(0)=x_0$ , we have $x^2=-2s+x_0^2=-t^2+x_0^2$

$\dfrac{du}{ds}=\dfrac{u}{xt}=\dfrac{u}{\sqrt{2s(x_0^2-2s)}}$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)e^{\tan^{-1}\frac{\sqrt{2s}}{\sqrt{x_0^2-2s}}}=f(\sqrt{x^2+t^2})e^{\tan^{-1}\frac{t}{x}}$

$u(x,0)=h(x)$ :

$f(x)=h(x)$

$\therefore u(x,t)=h(\sqrt{x^2+t^2})e^{\tan^{-1}\frac{t}{x}}$

doraemonpaul
  • 16,178
  • 3
  • 31
  • 75