Obtain solution of boundary problem as linear operator. I'm kinda stuck with a problem right now. I have the boundary problem
$$\left\{
\begin{array}{l}
-u''(x)+\mu u(x)= f(x), \quad x\in (0,T) \\
u'(0)=u'(T)=0
\end{array}
\right.$$
and I have to obtain the values of $\mu\in\mathbb{R}$ such that the equation has solution for every $f\in L^2(0,T)$, and write that solution as $u=Kf$, where $K$ is an operator of the form
$$[Ku](x)=\int_0^T k(x,y)f(y)dy.$$
The problem is that my formation on differential equations is quite limited, and the only example the proffesor gave us is for the case $\mu=0$, which is quite trivial. Can I get help from you guys? 
Thanks a lot! 
 A: Consider the Neumann eigenfunctions of $-\frac{d^2}{dx^2}$ on $(0,T)$, i.e. the functions $u_n$ such that $-u_n'' = \lambda_n u_n$ and $u_n'(0) = u_n'(T) = 0$. It is well known that 
$$\lambda_n = \left(\dfrac{n \pi}{T}\right)^2, u_n = \cos\left(\dfrac{n\pi x}{T}\right),$$
and further, that these functions form a basis for $L^2(0,T)$ (i.e. you can expand any function in $L^2$ in a Fourier series using this basis).
So, if we assume the solution $u$ can be written $u \sim \sum a_n u_n$ (with the $a_n$ yet to be determined), and expand $f$ in this basis: $f \sim \sum b_n u_n$ (the $b_n$ can be determined using the usual F.S. formulae), we check
$$
\sum a_n (\mu + \lambda_n)u_n = - u'' + \mu u = f = \sum b_n u_n
$$
and so $a_n = \dfrac{b_n}{\mu + \lambda_n}$. Can you see from this which values of $\mu$ might give you problems?
To write a solution operator, the main idea is to use the definition of the Fourier transform coefficients:
$$
 u = \sum a_n u_n(x) = \sum_n \dfrac{b_n}{\mu + \lambda_n} u_n(x) = \sum_n \dfrac{\int_0^T f(y) \cos(n\pi y /T)\,dy }{\mu + \lambda_n}\cos(n\pi x/T)
$$
and if you formally exchange the limit and the integral:
$$
\int_0^T \left( \sum_n \dfrac{1}{\mu + \lambda_n} \cos(n\pi y /T)\cos(n\pi x/T)\right)f(y)\,dy
$$
In reality, one has to proceed more carefully with integrals and infinte sums, but I'm not sure what is expected of you here. You might want to read about the Dirichlet kernel.
A: For any second order system like this, it is worth knowing that you can choose any linearly independent solutions $\phi$, $\psi$ of the homogenous equation $-u''+\mu u=0$, and obtain a particular solution $u$ of $-u''+\mu u=f$ as
$$
      u=\frac{1}{w(\phi,\psi)}\left[\phi\int \psi f\,dx - \psi\int \phi f\,dx\right]     
$$
where $w(\phi,\psi)=\phi\psi'-\phi'\psi$ is the Wronskian. The Wronskian is constant because
$$
              w(\phi,\psi)' = (\phi\psi'-\phi'\psi)'=(\phi\psi''-\phi''\psi)=0.
$$
To see why this works, remember that the Wronskian is constant, and look at
$$
\begin{align}
   u' & = \frac{1}{w(\phi,\psi)}\left[\phi'\int\psi f\,dx-\psi'\int\phi f\,dx\right] \\
   u'' & =\frac{1}{w(\phi,\psi)}\left[\phi''\int\psi f\,dx-\psi''\int\phi f\,dx+(\phi'\psi-\phi\psi')f\right]
\end{align}
$$
Therefore, $-u''+\mu u=f\;$ because $-\phi''+\mu \phi=0$ and $-\psi''+\mu\psi=0$. The general solution of $-u''+\mu u=f$ involves two constants $A$, $B$:
$$
     u=\frac{1}{w(\phi,\psi)}\left[\phi\int \psi f\,dx - \psi\int \phi f\,dx\right]+A\phi+B\psi.
$$
So you can always solve the equation. The only issue is the endpoint conditions. There is a standard trick that works nicely for the endpoint conditions. You choose $\phi$ to satisfy the right endpoint condition, and $\psi$ to satisfy the left, and then you form
$$
         u = \frac{1}{w(\phi,\psi)}\left[\phi(x)\int_{0}^{x}f(t)\psi(t)\,dt+\psi(x)\int_{x}^{T}f(t)\,\phi(t)\,dt\right]
$$
That takes care of the constants. This is because there are constants $C$ and $D$ such that
$$
        u(0)=C\psi(0),\;\;u'(0)=C\psi'(0)\\ u(T)=D\phi(0),\;\;u'(T)=D\phi'(0).
$$
For your case that means $\psi'(0)=0$ and $\phi'(T)=0$ are required. Two such solutions are
$$
       \phi(x) = \cosh(\sqrt{\mu}(x-T)),\;\;\; \psi(x)=\cosh(\sqrt{\mu}x).
$$
You don't care about normalizing $\phi$, $\psi$ in some way because this happens automatically when dividing by the Wronskian. The Wronskian is constant and, therefore,
may be evaluated at the endpoint $T$ to find its constant value:
$$
\begin{align}
     W(\phi,\psi)  = & \phi\psi'-\phi'\psi \\
        = & +\sqrt{\mu}\cosh(\sqrt{\mu}(x-T))\sinh(\sqrt{\mu}x) \\
          & -\sqrt{\mu}\sinh(\sqrt{\mu}(x-T))\cosh(\sqrt{\mu}x) \\
        = & W(\phi,\psi)|_{x=T}=\sqrt{\mu}\sinh(\sqrt{\mu}T).
\end{align}
$$
Now you see when you can and cannot solve this problem in a general way. There is no general solution for all $f$ whenever $\sinh(\sqrt{\mu}T)=0$, which happens for
$$
              \mu = -n^{2}\pi^{2}/T^{2},\;\;\; n=0,1,2,3,\cdots.
$$
The final answer that you seek is
$$
\begin{align}
       u = \frac{1}{\sqrt{\mu}\sinh(\sqrt{\mu}T)}
       & \left[\cosh(\sqrt{\mu}(x-T))\int_{0}^{x}\cosh(\sqrt{\mu}t)f(t)\,dt\right. \\
       & \left. +\cosh(\sqrt{\mu}x)\int_{x}^{T}\cosh(\sqrt{\mu}(t-T))f(t)\,dt \right]
\end{align}
$$
You can easily express this answer as $u = \int_{0}^{T}K(x,t)f(t)\,dt$.
