2
$\begingroup$

I'm trying to solve this Cauchy problem, which appears to be rather basic, but it seems like I'm missing something.

The problem is as follows:

$ u_x+u_t+u=0 \\ 0<\alpha*t<x<\infty \\ \alpha >0 , \ \alpha\neq1 \\ u(\alpha*t,t)=1, t>0\\u(x,0)=0,x>0 $

What I did was using the following sets of equations and initial conditions: $ \frac {dx}1=\frac {dt}1=\frac {du}{0-1*u}=ds \\ \Gamma_{s=0}:\{x=\tau, t=\tau/\alpha, u_0=1 \} $

After some integration and algebra, I reached the answer $ u(x,t)=e^{-\frac12(x+t+(t-x)(\frac{\alpha+1}{\alpha-1}))} $, but I can't seem to confirm whether this was the right way to do this, or whether the answer is correct, especially considering the fact I haven't used the initial condition $u(x,0)=0,x>0 $ anywhere.

Any help on how to approach this sort of problems? I've read various solutions of Cauchy problems, but couldn't find any that had two initial conditions rather than an initial curve of sorts.

$\endgroup$
  • 1
    $\begingroup$ Integrate the first equality to get $$x(t) = t + c_{1}$$ then integrate the second to get $$\ln u = -t + f(c_{1}) \implies u = f(c_{1}) e^{-t} = f(x-t)e^{-t}$$ Now you can apply your initial condition. $\endgroup$ – Mattos Jul 13 '18 at 0:54
  • $\begingroup$ It's kind of hard to grasp why the added constant after the second integration is necessarily a function of the first added constant. Anyway, the point now is I have to find a function $ f(x-t) $ such that $ f(\alpha*t-t)=e^t $ and$ f(x-0)=0$? $\endgroup$ – Tom Lyden Jul 14 '18 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.