Perturbation expansion Q. who get the best answer?

Question

know $y''+(1+f(x)\epsilon)y=0$ with $y(0)=1$ and $y(\pi)=0$ find perturbation expansion. How can I find the expansion after I get the y(x)? Many thanks

Answer 1

The problem does not have a solution. This assertion is inspired by @Dylan's sharp observation below.

Suppose the contrary. The equation has to hold at $\epsilon=0$. We then have $$y''+y =0,\quad y(0)=1,\ y(\pi)=0.$$ Then $y(x)=a\cos(x)+b\sin(x)$ for some constant $(a,b)$. But $y(0)=a=1$ and $y(\pi)=-a=0$, a contradiction.

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Here is the general solution and condition for the existence of a solution. This is

Suppose the boundary is at $(x_1,x_2)$ and $y(x_1)=\alpha_1$, $y(x_2)=\alpha_2$.

At $\epsilon=0$, we should have $$y''+y =0,\quad y(x_1)=\alpha_1,\ y(x_2)=\alpha_2.$$ $$y(x)=a\cos(x)+b\sin(x)$$ for some constant $(a,b)$. We should have $$\begin{bmatrix} \cos(x_1) & \sin(x_1) \\ \cos(x_2) & \sin(x_2) \end{bmatrix} \begin{bmatrix} a \\ b\end{bmatrix} = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} $$ For the above linear algebra system to have solution for arbitrary $(\alpha_1, \alpha_2)$, $\tan(x_1)\neq\tan(x_2)$.

(to be continued)

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The following is the general solution for the non-degenerate conditions prescribed above.

Let $y = \sum_{n=0}^\infty\epsilon^n y_n$, where $y_0(x_1)=\alpha_1,\,y_0(x_2)=\alpha_2$ and $y_n(x_1)=y_n(x_2)=0,\ \forall n\in\mathbf N$. Substitute it into the original equation. We have $$\sum_{n=0}^\infty \epsilon^{n+1} (y_{n+1}''+ y_{n+1}+ f(x)y_n)=0.$$ Setting the coefficients of $\epsilon^{n+1}$ to zero, or \begin{align} y_0''+y_0&=0,\quad y_0(x_1)=\alpha_1,\ y_0(x_2)=\alpha_2,\\ y_n''+y_n+ f(x)y_{n-1}&=0,\quad y_n(x_1)=y_n(x_2)=0,\quad \forall n\in\mathbf N. \end{align} You have the recursive equations to solve for $y$. The first boundary value problem is solved by a $\cos$ function. The second boundary value problem can be solved by many techniques, such as variational method, Fourier series expansion, e.t.c.