This is a question from 'Introduction to Partial Differential Equations' by Peter J. Olver.
The Poisson-Dabroux equation is $$u_{tt}-u_{xx}-\frac{2}{x}u_{x}=0$$
Solve the initial value problem $u(0,x)=0 , u_{t}(0,x)=g(x)$ where $g(x)=g(-x)$. Hint: Set $w=xu$
I used the hint to get the following
$$ w_{xx}= xu_{xx}+2u_{x}$$
Using the given equation to get $$xu_{tt}-xu_{xx}-2u_{x}=0$$
Using the w values obtained before and plugging them into the above equation gives:
Using D'Alembert's formula for a general solution:
$$w(t,x)=\int_{x-t}^{x+t} s g(s) ds$$ Now this is where I'm stuck. I tried using integration by parts but I ended up with a more complicated integral. Also, I reckon the fact that $g$ is even is used somewhere in the calculation of this integral but I can't figure out where either. The solution to this question is not in the solution manual either. Any help appreciated. Thank you!


You're already done. There's nothing wrong with having the solution as an integral. There's no indication that you need to solve it

$$ u(x,t) = \frac{1}{x} \int_{x-t}^{x+t}sg(s) ds $$

  • $\begingroup$ Thank you! I assumed I had to solve it as I hadn't used the fact that g is even anywhere. Is there any way of solving it without knowing g? $\endgroup$ – ro262 May 28 '18 at 12:35

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