Consider the equation $L u=f, \lambda_{1}(u)=0, \lambda_{2}(u)=0,$ where $\mathrm{L}$ is a second order linear differential operator. A Green's function $g(x, y)$ for $\mathrm{L}$ must satisfy $\lambda_{1}(g(x, y))=0, \lambda_{2}(g(x, y))=0,$ where $y$ is fixed, $\lambda_{1}$ and $\lambda_{2}$ are liner functionals, and $g$ is considered as a function of $x$.
- List the other properties $g(x, y)$ must satisfy.
- Consider the equation $u^{\prime \prime}(x)=f(x), u(0)=0, \int_{0}^{1} u(t) d t=0 .$ Find the Green's function for this equation. (Hint: the Green's function has the form $u_{1}(\cdot) u_{2}(\cdot)$ where $u_{1}$ is a solution to $u^{\prime \prime}(x)=0, u(0)=0$ while $u_{2}$ is a solution to $u^{\prime \prime}(x)=0$. $\int_{0}^{1} u(t) d t=0 .$)
- Write down a solution to $u^{\prime \prime}(x)=f(x), u(0)=0, \int_{0}^{1} u(t) d t=0$
Based on my knowledge about green's function, I think
$g(x,y)$ must satisfy
- $Lg(x,y)=0, x\not=y$.
- $\frac{\partial g(x,y)}{\partial x}\big|_{x=y^+}-\frac{\partial g(x,y)}{\partial x}\big|_{x=y^-}=1$
- $g(x,y)$ is continuous at $x=y$
- $g(0,y)=0$
$\int_0^1 g(x,y)dx=0$
- By the usual way I use to find green function, I let
$$ g(x,y)=\begin{cases} c_1x+c_2,\text{if } x<y\\ d_1x+d_2, \text{if } y<x \end{cases} $$ combined the other properties of $g(x,y)$ mentioned in 1, I get $$ g(x,y)=\begin{cases} \frac{1}{2}y^2x,\text{if } x<y\\ (\frac{y^2}{2}+1)x-y, \text{if } y<x \end{cases} $$
Question:a) Is the above answer right? b) if I use the hint in Q2, how can I find the green's function? c) the answer to question Q3 is just $u=\int_0^1 g(x,y)f(y) dy$?