Finding Green's Function for BVP

Question

Find Green's function for the boundary value problem

$$\begin{cases} y''+ \lambda y =0 \quad\text{ in } [0,1]\\ y(0)=0, y(1)=0 \end{cases}$$ is given for $\lambda > 0$, $\lambda \neq n^{2} \pi^{2}$ where $(n=1,2,3,...).$

Answer 1

As there is no forcing (RHS $= 0$), the Green's function is just the zero function. I presume that the correct RHS should be $\delta(x-\xi)$, the point forcing at $x=\xi$.

First you need to a test for the existence of Green's function: check the completely homogeneous problem and see if it only admits the trivial zero solution.

The homogeneous problem has solution of the form

\begin{equation} y\left(x\right)=c_{1}\cos\left(\sqrt{\lambda}x\right)+c_{2}\sin\left(\sqrt{\lambda}x\right) \end{equation}

Using the boundary data,

\begin{equation} 0=y\left(0\right)\implies c_{1}=0 \\ 0=y\left(1\right)\implies c_{2}\sin\left(\sqrt{\lambda}\right)=0\implies c_{2}=0 \end{equation} because $\lambda\neq n^{2}\pi^{2}$. Hence the solution to the homogeneous problem is indeed the trivial solution. There exists a Green's function.

Consider the Green's function \begin{equation} G\left(x,\xi\right)=\begin{cases} G_{-}\left(x,\xi\right), & 0\leq x<\xi\leq1\\ G_{+}\left(x,\xi\right) & 0\leq\xi<x\leq1 \end{cases} \end{equation} where $G_{-}\left(x,\xi\right)$ and $G_{+}\left(x,\xi\right)$ solve the homogeneous ODE and also satisfy the following four conditions,

\begin{equation} G_{-}\left(0\right)=0 \\ G_{+}\left(1\right)=0 \\ \text{Continuity of } G \text{ at some point } x=\xi \\ \text{Jump discontinuity of } \frac{dG}{dx} \text{ at some point } x=\xi \text{ such that } \left[\frac{dG}{dx}\right]_{\xi^{-}}^{\xi^{+}} = 1 \\ \end{equation}

We already have a general form of the solution to the homogeneous problem, i.e. \begin{equation} G_{-}\left(x\right)=A\cos\left(\sqrt{\lambda}x\right)+B\sin\left(\sqrt{\lambda}x\right) \\ G_{+}\left(x\right)=A\cos\left(\sqrt{\lambda}x\right)+D\sin\left(\sqrt{\lambda}x\right) \end{equation} where $A,B,C,D$ are constants to be determined by the above four conditions.

The boundary conditions force $A=C=0$. Continuity and jump condition yield two equations with two unknowns, \begin{equation} B\sin\sqrt{\lambda}\xi=D\sin\sqrt{\lambda}\left(\xi-1\right)\\ \sqrt{\lambda}D\cos\left(\sqrt{\lambda}\left(\xi-1\right)\right)-\sqrt{\lambda}B\cos\left(\sqrt{\lambda}\xi\right)=1 \end{equation}

After using some trigonometric identities in solving this system(the angle difference formula for $\sin\left(X-Y\right)=\sin X\cos Y-\cos X\sin Y$), we have

\begin{equation} B=\frac{\sin\left(\sqrt{\lambda}\left(\xi-1\right)\right)}{\sqrt{\lambda}\sin\sqrt{\lambda}}\\ D=\frac{\sin\left(\sqrt{\lambda}\xi\right)}{\sqrt{\lambda}\sin\sqrt{\lambda}} \end{equation}

Plugging these two constants back into the definition of $G$, we finally have constructed the Green's function, \begin{equation} G\left(x,\xi\right)=\begin{cases} \frac{\sin\left(\sqrt{\lambda}\left(\xi-1\right)\right)\sin\left(\sqrt{\lambda}x\right)}{\sqrt{\lambda}\sin\sqrt{\lambda}}, & 0\leq x<\xi\leq1\\ \frac{\sin\left(\sqrt{\lambda}\left(x-1\right)\right)\sin\left(\sqrt{\lambda}\xi\right)}{\sqrt{\lambda}\sin\sqrt{\lambda}} & 0\leq\xi<x\leq1 \end{cases} \end{equation}