Fundamental Solution to 2nd Order ODE

Question

I'm currently doing a problem with the fundamental solution for $$-u''+k^2u=f(x) \quad , \quad -\infty < x < \infty$$ I'm wondering if fundamental solutions are supposed to satisfy the homogeneous equation or the above equation? Thanks.

Answer 1

Typically a "fundamental solution" or "Green's function" is a function that solves the equation $-u'' + k^2u = \delta(x).$

Answer 2

The fundamental solution, as mentioned, satisfies $-u''+k^2 u = \delta_{y}(x)$. To the left or to the right of $y$, the fundamental solution satisfies $-u''+k^2u=0$. The fundamental solution needs to be continuous across $y$, and, in order to have the $\delta$ function behavior, there is a discontinuity in the first derivative of the solution with a jump of $-1$ so that $-u''$ pop's out the delta function at $y$.

The only requirements at $\pm\infty$ is that the fundamental solution should be in $L^2(\mathbb{R})$. Assume $k > 0$. In this case, that means $$ u(x) = \left\{\begin{array}{cc}C_1e^{kx}, & -\infty < x < y \\ C_2e^{-kx}, & y < x < \infty.\end{array}\right. $$ The constants $C_1$, $C_2$ are chosen so that $u$ is continous at $y$ and so that the derivative of $u$ has a jump of $+1$ as $x$ crosses $x$ from below to above $y$. That is, $$ C_2 e^{-ky} - C_1 e^{ky} = 0 \\ -kC_2 e^{-ky}-kC_1 e^{ky} = 1. $$ Multiplying the first equation by $k$ and adding gives $$ -2kC_1e^{ky} = 1 \implies C_1 = -\frac{1}{2k}e^{-ky}. $$ Multiplying the first by $k$ and subtracting instead gives $$ 2kC_2e^{-ky} = -1 \implies C_2 = -\frac{1}{2k}e^{ky} $$ Therefore, $u$ can be written as $$ u_y(x) = -\frac{1}{2k}e^{-k|x-y|} $$