Author Topic: Problem set 2.6 Q4  (Read 733 times)

Jingxuan Zhang

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Problem set 2.6 Q4
« on: February 08, 2018, 06:42:42 AM »
So now the boundary condition amounts to an ODE with initial condition $\psi(0)=\phi(0)/2$? And then do we just leave the integral as it is after variation of parameter?
I am little diffident about the change of variable in the integral I made, so should I get
$$\psi(x-ct)=e^{\alpha(ct-x)}\int_0^{ct-x} e^{\alpha y}(\phi'(y)+\alpha \phi(y)) \,dy +e^{\alpha(ct-x)}\phi(0)/2 ?$$
It seems to me that I can be simplified further but I don't know how.

EDIT: This is solved in today's tutorial. So in fact it simplifies a little bit. But how about the integration constant, the initial condition for this ODE? On textbook it says (in our case) $\psi(0)=\phi(0)/2$, so then should I still have the term $e^{\alpha(ct-x)}\phi(0)/2$?
« Last Edit: February 08, 2018, 03:56:48 PM by Jingxuan Zhang »

Victor Ivrii

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Re: Problem set 2.6 Q4
« Reply #1 on: February 10, 2018, 03:46:13 PM »
Integration constant should be selected to make $u$ continuous at the origin and thus along $x=ct$.