2
$\begingroup$

I have the PDE listed below and I am not quite sure how to solve it.

$\frac{\partial^2 g}{\partial y^2} - \frac{\partial^2 g}{\partial z^2} - 2 \frac{\partial g}{\partial x}=0$

I have tried separation of variables, however, it does not seem to work. I have been reading about the method of characteristics but it mostly applies to first order PDE. I am unfamiliar with that approach and I am not sure how I may apply it to this problem.

I am more interested in the process of how to solve this problem than the solution but I would gladly accept the solution as well so I can work it out for myself.

I appreciate any assistance.

$\endgroup$
  • $\begingroup$ The easiest way is taking a guess as to the structure of $g$, and trying to see if that structure hosts a solution. Other methods are ... not fun. Check out this link. It might give you some ideas as to where to start. $\endgroup$ – Don Thousand Mar 24 at 22:57
1
$\begingroup$

Hint

The method "Separation of Variables" works. Define$$g(x,y,z)=f_1(x)f_2(y)f_3(z)$$therefore $$f_1(x)f_2''(y)f_3(z)-f_1(x)f_2(y)f_3''(z)-2f_1'(x)f_2(y)f_3(z)=0$$which yields to$${f_2''(y)\over f_2(y)}-{f_3''(z)\over f_3(z)}-2{f_1'(x)\over f_1(x)}=0$$and hence $${f_2''(y)\over f_2(y)}=k_1\\{f_3''(z)\over f_3(z)}=k_2\\{f_1'(x)\over f_1(x)}={k_1-k_2\over 2}$$

$\endgroup$
  • $\begingroup$ Excellent. Thank you. $\endgroup$ – dsmalenb Mar 24 at 23:05
  • $\begingroup$ Your welcome. Good luck! $\endgroup$ – Mostafa Ayaz Mar 24 at 23:08

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.