# Finding Green’s function

I tried to solve my homework, but unfortunately got stuck at some point. Could you please help me to get through it?

For a given boundary value problem: $$Ly =y'''+y''\\ y(0)+y'(0) = y''(0)=y(1) = 0$$ Establish whether there is a Green's function and if so, construct it.

I start with $$G'''(x,s) + G''(x,s) = \delta(x-s)\\\text{If } x \neq s:G(x,s) = c_1e^{-x}+c_2x+c_3$$

$$For \ x < s : \begin{cases} a_2+a_3 = 0\\ a_1 = 0 \end{cases} (\text{the boundary condition at } x=0)\\ For \ x > s : \frac{b_1}{e}+b_2+b_3 = 0\ (\text{the boundary condition at } x=1)$$ $$G(x,s) = \begin{cases} G_1(x,s)=a_2x+a_3 \ \ \ \text{for } x<s\\ G_2(x,s)=b_1e^{-x}+b_2x+b_3\ \ \ \text{for } x>s \end{cases}$$

Added:

Continuity of the Green’s function permits $$G_1(s,s) =G_2(s,s)\\ a_2s+a_3 = b_1e^{-s}+b_2s+b_3\\ s(a_2-b_2)+(a_3-b_3) =b_1e^{-s}$$ While the jump discontinuity condition gives $$\begin{cases} \lim_{\epsilon \rightarrow 0}\left.\frac{dG(x,s)}{dx}\right|_{s-\epsilon}^{s+\epsilon}= 1 \\ \lim_{\epsilon \rightarrow 0}\left.\frac{d^2G(x,s)}{dx^2}\right|_{s-\epsilon}^{s+\epsilon}= 1 \end{cases} \\ \begin{cases} \frac{dG_2(s,s)}{dx}-\frac{dG_1(s,s)}{dx}=1 \\ \frac{d^2G_2(s,s)}{dx^2}-\frac{d^2G_1(s,s)}{dx^2}=1 \end{cases}\\ \begin{cases} b_2-b_1e^{-s}-a_2=1 \\ b_1e^{-s}=1 \end{cases} \\ \begin{cases} b_1 = e^{s} \\ b_2-a_2 = 2 \end{cases}$$ Now: $$\begin{cases} a_1 =0 \\ a_2 = -a_3 \\ b_1 = e^{s} \\ b_2-a_2 = 2 \\ s(a_2-b_2)+(a_3-b_3) = 1 \\ e^{s-1} +b_2 +b_3 =0 \end{cases} \\ \left\{\begin{matrix} a_1 = 0\\ a_2 = -\frac{2es+e+e^s}{2es}\\ a_3 =\frac{2es+e+e^s}{2es}\\ b_1 = e^s\\ b_2 = \frac{2es-e-e^s}{2es}\\ b_3 = -\frac{(e^s+e)(2s-1)}{2es} \end{matrix}\right. \\ G(x,s) = \begin{cases} \frac{2es+e+e^s}{2es}(1-x)\ \ \ \text{for } x<s\\ e^{s-x}+\frac{2es-e-e^s}{2es}x-\frac{(e^s+e)(2s-1)}{2es}\ \ \ \text{for } x>s \end{cases}$$

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## 1 Answer

While it is correct that the Green's function must satisfy what you have written, do not forget the extra conditions that are necessary:

1) $g(x,s)$ must be continuous in $x$ and $s$.

2) $LG(x,s)=0$ for $x\neq s$

3) $DG(x,s)=0$ for $s\neq 0$, where $D$ satisfies $Dy=0$, the boundary condition operator, which in your case is the pair of boundary conditions.

4) Derivative Jump Condition: $\lim_{a\rightarrow 0^+}G'(a,s)-\lim_{a\rightarrow 0^-} G'(a,s)=1/p(s)$ where $p$ comes from the Sturm-Liouville operator form of $L=\frac{d}{dx}\left[p(x)\frac{d}{dx}\right]+q(x)$. Be sure to correctly take the limits from the right and left respectively!

5) $G(x,s)=G(s,x)$ for every $x,s$.

In particular, conditions 3 and 4 will give three equations to solve for your constants

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