Sketching base characteristics of first order pde

Question

$$xu_{x} - uu_{t} = t$$

with intial data: $u(1,t) = t, −\infty < t < \infty$

So I want to sketch the base characteristics and I'm not too sure what I'm doing.

characteristic equations: $\frac{\partial{x}}{\partial{\tau}}=x$, $\frac{\partial{t}}{\partial{\tau}}=-u$, $\frac{\partial{u}}{\partial{\tau}}=t$

initial conditions:

$x(0)=1$, $t(0)=\xi$, $u(0)=\xi$

$x=e^\tau$ so $\tau=lnx$

Notice that:

$\frac{\partial^2{u}}{\partial{\tau^2}}=\frac{\partial{t}}{\partial{\tau}}=-u$,

This gives me the second order ODE

$\frac{\partial^2{u}}{\partial{\tau^2}}+u=0$

Solving I get:

\begin{align} u&=c_{1}\cos(\tau) + c_{2}\sin(\tau)\\ t&=c_{1}\sin(\tau) - c_{2}\cos(\tau) \end{align}

From here I don't know what I'm doing?

Answer 1

You correctly got

$$ u_{\tau\tau} = -u $$

The initial conditions are

$$ u(0) = \xi, \quad u_\tau(0) =t(0)= \xi $$

which gives

$$ u = \xi(\cos(\tau) + \sin(\tau)) $$

and

$$ t = \xi(\cos(\tau) - \sin(\tau)) $$

Cancel out $\xi$ to get

$$ u = \frac{\cos(\tau)+\sin(\tau)}{\cos(\tau)-\sin(\tau)}t = \frac{\cos(\ln x)+\sin(\ln x)}{\cos(\ln x)-\sin(\ln x)}t $$

Answer 2

$$xu_{x} - uu_{t} = t$$ $$\frac{dx}{x}=\frac{dt}{-u}=\frac{du}{t}=d\tau$$ A first characteristic equation comes from $\frac{dt}{-u}=\frac{du}{t}$ : $$u^2+t^2=c_1$$ $u=\pm\sqrt{c_1-t^2}$

A second characteristic equation comes from $\frac{dx}{x}=\frac{dt}{-u}=\frac{dt}{\mp\sqrt{c_1-t^2}}$ :

$\ln|x|\pm\sin^{-1}\left(\frac{t}{\sqrt{c_1}}\right)=c_2$

$$\ln|x|\pm\sin^{-1}\left(\frac{t}{\sqrt{u^2+t^2}}\right)=c_2$$ The general solution of the ODE on the form of implicit equation $c_2=F(c_1)$ is : $$\ln|x|\pm\sin^{-1}\left(\frac{t}{\sqrt{u^2+t^2}}\right)=F(u^2+t^2)$$ where $F$ is an arbitrary function, to be determined according to the specified condition.

CONDITION : $u(1,t)=t$ .

$\ln|1|\pm\sin^{-1}\left(\frac{t}{\sqrt{t^2+t^2}}\right)=F(t^2+t^2)$

$\pm\sin^{-1}\left(\frac{t}{\sqrt{2t^2}}\right)=F(2t^2) = \pm\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)=\pm\frac{\pi}{4}$

Thus $F$ is a constant function. We put it into the above general solution : $$\ln|x|\pm\sin^{-1}\left(\frac{t}{\sqrt{u^2+t^2}}\right)=\pm\frac{\pi}{4}$$ Solving for u :

$\frac{t}{\sqrt{u^2+t^2}}=\pm\sin\left(\pm\frac{\pi}{4}-\ln|x|\right)$

Note that squaring introduces extra solutions which will be rejected latter.

$u^2= \frac{t^2}{\sin^2\left(-\ln|x|\pm\frac{\pi}{4}\right)}-t^2= t^2\cot^2\left(-\ln|x|\pm\frac{\pi}{4}\right)$

$u=\pm t\cot\left(-\ln|x|\pm\frac{\pi}{4}\right)$

One have to check in putting into the PDE and the condition, which determines the signs both $+$.

The solution of the PDE according to the specified condition is :

$$u(x,t)= t\cot\left(-\ln|x|+\frac{\pi}{4}\right)$$