Solve the following equation using Green's functions:

$u''-k^2u=f$ with boundary conditions $u(0)-u'(0)=a$ and $u(1)=b$

For this problem, I was going to find the green's function with homogeneous BC's (set both BC's equal to zero), and then I was going to add the solution to the homogeneous equation Lu = 0 (with the BC's given above) to the green's function solution. However, when working out the green's function, I end up with constant that can't be solved. Any help?

  • $\begingroup$ You should post your work so that we can see what, if anything, is wrong. It could be that you made a very simple mistake or something much more complicated and it is difficult for us to tell you what went wrong without seeing your work. $\endgroup$ – Cameron Williams Dec 4 '13 at 17:56
  • $\begingroup$ I would if my work didn't comprise 2+ pages :). Basically, what I did was let Lu = delta(x-t). Then, I solved out everything in the regions x<t and x>t; I get solutions of the form $g(x)=ae^(kx)+be^(-kx)$. After doing so, I apply the BC's to both sides, eliminating 2 of the four constants. Then, I assert continuity at x = t and ensure that the discontinuous derivative jump = 1. I have 2 equations and 2 unknowns, but I can't solve for the constants =/ (it's super messy). $\endgroup$ – Incognito Dec 4 '13 at 18:00

I explained the process for a similar example. Here is a rehash:

  1. Solve the characteristic equation $\lambda^2-k^2=0$
  2. Write down the general solution of homogeneous equation: $y= A\cosh k x+B\sinh kx $.
  3. Find a solution $u$ such that $u(a)=0$ and $u'(a)=1$, where $a\in(0,1)$ is the pole of Green's function. Since the equation is autonomous, solutions can be shifted in time: $$u(x)=k^{-1} \sinh k(x-a) $$
  4. By virtue of 3, the function $u(x)H(x-a)$, where $H$ is the Heaviside function, produces $\delta_a$ when plugged into your equation.
  5. Find a solution of homogeneous equation $y$ such that $y(x)+u(x)H (x-a)$ satisfies the homogeneous boundary condition. This amounts to asking $y(0)=y'(0)$ and $y(1)=-k^{-1} \sinh k(1-a)$. Hence, $$y(x)=-k^{-1} \sinh k(1-a)\cdot \frac{ k \cosh kx + \sinh kx}{k \cosh k + \sinh k }$$
  6. You have Green's function: $$g(x,a) = y(x)+u(x)H (x-a) $$

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