Solvability Condition

Question

How do we find the solvability condition for the given equation,

$$-u''-u=f,\hspace{0.5cm}-\pi<x<\pi,\hspace{0.5cm}u(\pi)-u(-\pi)=0,\hspace{0.5cm}u'(\pi)-u'(-\pi)=0$$

Answer 1

You are asking that $u$ should have a continuously differentiable $2\pi$-periodic extension. In that case the Fourier series of $$ u(x)=\sum_{k\in\Bbb Z}c_ke^{ikx} $$ has to satisfy $$ \sum_{k\in\Bbb Z}c_k(k^2-1)e^{ikx}=f(x) $$ and by extracting the coefficients $$ 2\pi(k^2-1)c_k=\int_{-\pi}^\pi e^{-ikx}f(x)dx $$ This directly tells you that this is only possible if $$ 0=\int_{-\pi}^\pi e^{\pm ix}f(x)dx $$

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One could also get this directly from the solution formula $$ u(x)=-u(-\pi)\cos(x)-u'(-\pi)\sin(x)-\int_{-\pi}^x\sin(x-t)f(t)\,dt $$ with \begin{align} u'(x)&= u(-\pi)\sin(x)-u'(-\pi)\cos(x)-\int_{-\pi}^x\cos(x-t)f(t)\,dt \\ u''(x)&= u(-\pi)\cos(x)+u'(-\pi)\sin(x)-f(x)+\int_{-\pi}^x\sin(x-t)f(t)\,dt \end{align} so that indeed $u''(x)+u(x)=-f(x)$ and the boundary conditions read \begin{align} u(-\pi)=u(\pi)&=u(-\pi)-\int_{-\pi}^\pi\sin(\pi-t)f(t)\,dt \\ u'(-\pi)=u'(\pi)&=u'(-\pi)-\int_{-\pi}^\pi\cos(\pi-t)f(t)\,dt \end{align}