Find an eigenvalue expansion solution for a particular Cauchy problem

Question

I'm having trouble finding an eigenvalue expansion for the solution of:

$$\left\{\begin{array}{c} -u''(x)=f(x) \\ u(0)=0 \\ u'(1)=0 \end{array}\right.$$

because of the $-u$ double prime

Answer 1

You can definitely solve by direct integration: \begin{align} -\int_1^b u''(a)da & = \int_1^b f(a)da \\ -u'(b)+u'(1) & = \int_1^b f(a)da \\ u'(b) & = \int_b^1 f(a)da \\ u(x)-u(0) & = \int_0^x\int_b^1 f(a)da db \\ u(x) & = \int_0^x\int_b^1 f(a)da db. \end{align} However, that does not seem to be what you're being asked to do. It seems that given $L=-\frac{d^2}{dx^2}$, you are asked to find the eigenfunctions of $L$ subject to the endpoint conditions $f(0)=0=f'(0)$. The solutions of $Lf=\lambda f$ subject to to $f(0)=0$, $f'(0)=1$ ($1$ is chosen only as normalization) are $$ f_{\lambda}(x)=\frac{\sin(\sqrt{\lambda} x)}{\sqrt{\lambda}}. $$ This form remains valid for $\lambda=0$ when interpreted as its limiting form: $f_0=\lim_{\lambda\rightarrow 0}f_{\lambda}(x)=x$. Because $f_{\lambda}$ is unique when normalized as stated, there are non-trivial solutions of $-f''=\lambda f$, $f(0)=0$, $f'(1)=0$ only when $f_{\lambda}'(1)=0$: $$ \cos(\sqrt{\lambda}) =0 \iff \sqrt{\lambda}=\pi/2,3\pi/2,5\pi/2,\cdots. $$ So the non-trivial solutions are scalar multiples of $$ \sin(\pi x/2),\sin(3\pi x/2),\sin(5\pi x/2),\cdots,\sin((n+1/2)\pi x),\cdots $$ These functions, when normalized according to $L^2[0,1]$ norm, form an orthonormal basis of $L^2[0,1]$. You can then expand the solution $u$ as $$ u = \sum_{n=1}^{\infty}a_n \sin((n+1/2)\pi x) $$ The endpoint conditions are built into the eigenfunctions. Then $-u''=f$ is solved with $$ f = \sum_{n=1}^{\infty}a_n (n+1/2)^2\pi^2\sin((n+1/2)\pi x) $$ Use eigenfunction orthogonality to solve for $a_m$ from the given functions $f$ by first multiplying by $\sin((m+1/2)\pi x)$ and then integrating both sides: $$ \int_0^1 f(t)\sin((m+1/2)\pi t)dt = a_m(m+1/2)^2\pi^2\int_0^1\sin^2((m+1/2)\pi x)dx. $$ The solution $u$ is $$ u(x) = \sum_{n=1}^{\infty}\frac{\int_0^1f(t)\sin((n+1/2)\pi t)dt}{\int_0^1 \sin^2((n+1/2)\pi t)dt}\frac{\sin((n+1/2)\pi x)}{(n+1/2)^2\pi^2}. $$