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Find eigenvalues and eigenfunctions of

$$ \begin{cases} y'' + \lambda y = 0 \\ y(0) = y'( \pi ) = 0 \end{cases} $$

Also, express $f(x) = x$ as a series of eigenfunctions of the above problem.

Attempt

I write down characteristic equation $\mu^2 + \lambda = 0$, thus $\mu = \sqrt{ - \lambda} $.

If $\lambda > 0$, then

$$y(x) = C_1 \cos ( \sqrt{ \lambda } x ) + C_2 \sin ( \sqrt{ \lambda } x ) $$

Since $y(0) = 0$, then $C_1 = 0$, thus $y(x) = C_2 \sin( \sqrt{ \lambda } x) $. Next, $y'(x) = \sqrt{ \lambda } C_2 \cos ( \sqrt{ \lambda } x ) $ and

$$ y'( \pi ) = \sqrt{ \lambda } C_2 \cos ( \sqrt{ \lambda } \pi ) = 0 \implies \sqrt{ \lambda } \pi = \frac{ (2n + 1) \pi }{2} \implies \boxed{ \lambda_n = \frac{ (2n + 1)^2 }{4} } \; \; \text{eigenvalues} $$

Thus, eigenfunctions are $$ \boxed{ y_n(x) = C_2 \sin \left( \frac{ (2n - 1)^2 }{4} x \right) } $$

If $\lambda < 0$, then

Write $\mu = \Lambda \in \mathbb{R} $, thus

$$ y(x) = C_1 e^{\Lambda x} + C_2 e^{- \Lambda x} \implies y(0) = C_1 +C_2 = 0$$

and

$$ y'(pi) = 0 \implies \Lambda C_1 e^{ \Lambda \pi } - \Lambda C_2 e^{ - \Lambda \pi } = 0 $$

Thus, we have the system

$$ \left( \begin{matrix} 1 & 1 \\ \Lambda e^{\Lambda \pi } & - \Lambda e^{- \Lambda pi } \end{matrix} \right) \left( \begin{matrix} C_1 \\ C_2 \end{matrix} \right) = 0 $$

but the above matrix is nonsingular, thus there are no solutions to the system, thus there are no eigenvalue and no eigenfunctions in this situation.

Question: How can we express $f(x) = x$ as a series of eigenfunctions? Also, is my above procedure correct? Any feedback would be extremely appreciated.

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Your procedure is correct, regarding the eigenfunctions of the BVP. As for the last part, it is well known that the eigenfunctions form a basis of the function space on which the operator is defined. So, in our problem $\displaystyle x=\sum_{n=1}^{\infty}c_ny_n(x)$ and in order to find $c_n,~n\geq 1$ you should make use of the orthogonality property: $$\int\limits_{0}^{\pi}y_n(x)y_m(x)dx=\frac{\pi}{2},~n\neq m.$$ Multiply every part of the equation by $y_m(x)$ and integrate on $[0,\pi]$ (all the conditions needed to pass the integral through the sum are fullfilled), so you get:

$$c_m\int\limits_{0}^{\pi}y_m^2(x)dx=\int\limits_{0}^{\pi}xy_m(x)dx,$$

and all you have to do is to evaluate the integrals. In fact, $\displaystyle \int\limits_{0}^{\pi}y_m^2(x)dx=\pi/2$ and for the next one you will need integration by parts.

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