$$u''=w^2 u$$ on 0

Question

$$u''=w^2 u$$ on 0<y<H with $$u(0,w)=F_1(w)$$$$u(H,w)=F_2(w)$$

richyw · Answer

$$u=A(w)e^{wy}+B(w)e^{-wy}$$

richyw · Answer

I have literally no idea how my book gets the solution. $$u(w,y)=F_2(w)\frac{\sinh wy}{\sinh wH}+F_1(w) \frac{\sinh (w(H-y))}{\sinh wH}$$

accessdenied · Answer

I believe if you start from u(y, w) = A(w) e^(wy) + B(w) e^(-wy)
and substitute in the boundary conditions into that, you could "eventually" get that through a lengthy process of solving for A(w) and B(w) in terms of F1(w) and F2(w).  I feel like there should be an easier way in this case, but this is the only one I know of at this time.  :x

accessdenied · Answer

If you wanted to go that route which takes a lot of careful simplifications, it goes like this:
   Find equation for u(0, w) = A(w) e^(w(0)) + B(w) e^(-w(0)) = F1 (w)
   Find equation for u(H,w) = A(w) e^(w(H)) + B(w) e^(-w(H)) = F2 (w)
   Solve for coefficients A(w) and B(w)  in terms of F1(w) and F2(w) 
   Substitute back into u(y, w) = A(w) e^(wy) + B(w) e^(-wy).
   Reorganize to isolate terms with F1(w) and F2(w)
   Simplify using definition of sinh (x).   [sinh x = [e^(x) - e^(-x)]/2]

It is a very long and drawn out process, so I feel like there is some trick I haven't used in the past to deal with this more elegantly.  >.>

richyw · Answer

thanks. although I probably would have been better off just memorizing the solution! haha

richyw · Answer

well better off in terms of cramming for my final next week. exams suck all the joy out of math!

accessdenied · Answer

Heh, sure!  It's always nice to get ahead of the exams and be able to play with the maths afterwards rather than worry.  :P