What is the Green’s function of the boundary value problem $ \frac{\mathrm d^2 y}{\mathrm d x^2}-\frac{1}{x}\frac{\mathrm dy}{\mathrm dx}=1,\quad y(0)=y(1)=0, $
this boundary problem is not self adjoint, so please help me how to solve it.
What is the Green’s function of the boundary value problem $ \frac{\mathrm d^2 y}{\mathrm d x^2}-\frac{1}{x}\frac{\mathrm dy}{\mathrm dx}=1,\quad y(0)=y(1)=0, $
this boundary problem is not self adjoint, so please help me how to solve it.
First note that the solution to the homogeneous problem is $y(x) = a + b x^2.$
We wish to solve $\begin{equation*} \frac{d^2}{dx^2}G(x,t) - \frac{1}{x} \frac{d}{dx} G(x,t) = \delta(x-t),\tag{1} \end{equation*}$ where $G$ satisfies the boundary conditions $G(0,t) = G(1,t) = 0$. Therefore, $G(x,t) = \begin{cases} a(t)x^2, & x
Integrating (1) from $t-\epsilon$ to $t+\epsilon$ we find the jump condition
$\left.\frac{d}{dx}G(x,t)\right|_{x=t+\epsilon} - \left.\frac{d}{dx}G(x,t)\right|_{x=t-\epsilon} = 1.$ Integrating once more we find the continuity condition
$\left.G(x,t)\right|_{x=t+\epsilon} - \left.G(x,t)\right|_{x=t-\epsilon} = 0.$ These two conditions determine $a(t)$ and $b(t)$. After a little work one finds $G(x,t) = \begin{cases} \displaystyle\frac{(t^2-1)x^2}{2t}, & x
Although it is often used, I don't remember ever seeing the above method given a proper name. I call it the jump method. There are at least two other common approaches to getting Green's functions. One involves an eigenfunction expansion and the other the fundamental solution.
Denoting $y'(x) = v(x)$, we have that $v'(x) - \dfrac{v(x)}{x} = 1$ Let $v(x) = x g(x)$. Then we get that $xg'(x) + g(x) - g(x) = 1\implies g'(x) = \dfrac1x$ Hence, we have $g(x) = \log(x) + c_1 \implies v(x) = x \log(x) + c_1 x$ Hence, we now need to solve for $y'(x) = x \log(x) + c_1 x$ $y(x) = \int x \log (x) + c_2 x^2 + c_3 = \dfrac{x^2}4 \left(2 \log(x) - 1 \right) + k_1 x^2 + k_2$ $y(0) = 0 \implies k_2 = 0$ $y(1) = -\dfrac14 + k_1 = 0 \implies k_1 = \dfrac14$ Hence, $y(x) = \dfrac{x^2 \log(x)}2$