Find the unique $u:\mathbb{R}\rightarrow\mathbb{R}$ governed by $$-u''(x)+\mu^2u(x)=f(x), x\in\mathbb{R},$$ such that $$\int_\mathbb{R}(|u(x)|^2+|u'(x)|^2+|u''(x)|^2)dx<\infty.$$ Here $\mu$ is a positive constant and $$f(x)=\begin{cases}1&\mbox{if }x\in[-1,1],\\0&\mbox{otherwise}.\end{cases}$$
This question is homework, so I request assistance only, not a full solution. This is for a PDE class I'm taking, but it has been many years since I've worked with ODE's, so I'm struggling with even setting up these types of problems. I recognize that I'll be working on three intervals (namely $(-\infty,-1)$, $[-1,1]$, and $(1,\infty)$), but that's about the extent of my current understanding. Ideally, I'd like some guiding comments that help me to reach a solution that I'll then post to this question.
Suppose, for some $u(x)$, that $$ \int_{\mathbb{R}} ( |u(x)|^2 + |u^\prime(x)|^2 + |u^{\prime\prime}(x)|^2 )\, {\rm d} x < \infty. $$ Then, it follows that $$\int_\mathbb{R}|u(x)|^2dx<\infty.$$ It can be shown that this is a sufficient condition for the existence of the Fourier transform of $u$. Similarly, since $f(x)$ is absolutely integrable, it has a Fourier transform. Now, let $\mathcal{F}(g(x))(s)$ denote the Fourier transform of $g(x)$ with respect to frequency variable $s$. That is, $$\mathcal{F}(g(x))(s)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}g(x)e^{-isx}dx.$$ Then, since $u(x)$ and $f(x)$ both have a Fourier transform, it follows that \begin{align} \tag{$*$} \mathcal{F}(-u''(x)+\mu^2u(x))=\mathcal{F}(f(x)). \end{align} By linearity, the LHS of $(*)$ becomes: \begin{align} \tag{$**$} \mathcal{F}(-u''(x)+\mu^2u(x))=-\mathcal{F}(u''(x))+\mu^2\mathcal{F}(u(x)). \end{align} Now, consider the inverse Fourier transform, given by $$g(x)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\mathcal{F}(g(x))(s)e^{isx}dx.$$ Thus, by the Leibniz integral rule, $$g'(x)=\frac{d}{dt}\!\left(\frac{1}{\sqrt{2\pi}} \int_\mathbb{R}\mathcal{F}(g(x))(s)\,e^{isx}dx \right)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}is \,\mathcal{F}(g(x))(s) \,e^{isx}dx.$$ Therefore, the Fourier transform of $g'(x)$ is $is\mathcal{F}(g(x))(s)$. Hence, $$\mathcal{F}(u''(x))(s)=is\mathcal{F}(u'(x))(s)=(is)^2\mathcal{F}(u(x))(s)=s^2\mathcal{F}(u(x))(s).$$ Thus, $(**)$ reduces to $$(\mu^2+s^2)\mathcal{F}(u(x))(s).$$ Then, on the interval $[-1,1]$, the RHS of $(*)$ is: $$\mathcal{F}(f(x))=\frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{-isx}dx = \left.\frac{1}{\sqrt{2\pi}}\frac{e^{-isx}}{-is}\right|_{x=-1}^{x=1} =\frac{e^{-is}-e^{is}}{\sqrt{2\pi}(-is)}=\frac{e^{is}-e^{-is}}{\sqrt{2\pi}is}$$ $$=\frac{\cos(s)+i\sin(s)-(\cos(s)-i\sin(s))}{\sqrt{2\pi}is}=\frac{2i\sin(s)}{\sqrt{2\pi}is}=\frac{2\sqrt{2}\sin(s)}{2\sqrt{\pi}s}=\sqrt{\frac{2}{\pi}}\frac{\sin(s)}{s}.$$ Therefore, $$\mathcal{F}(u(x))(s)=\sqrt{\frac{2}{\pi}}\frac{\sin(s)}{s(\mu^2+s^2),}$$ so $u$ is given by $$u(x)=\mathcal{F}^{-1}(\mathcal{F}(u(x))(s))=\mathcal{F}^{-1}\left(\sqrt{\frac{2}{\pi}}\frac{\sin(s)}{s(\mu^2+s^2)}\right),$$ where $\mathcal{F}^{-1}$ denotes the inverse Fourier transform. Then, by linearity and the convolution theorem, $$u(x)=\sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{2\pi}}\left(\mathcal{F}^{-1}\left(\frac{\sin(s)}{s}\right)*\mathcal{F}^{-1}\left(\frac{1}{\mu^2+s^2}\right)\right).$$ Thus, since $\mu$ is a constant, the inverse Fourier transform of $\frac{1}{\mu^2+s^2}$ is the same as the inverse Fourier transform of $\frac{\mu}{\mu^2+s^2}$, which is given by the Laplace distribution, $\sqrt{\frac{\pi}{2}}e^{-\mu|x|}$. Then, since $\frac{\sin(s)}{s}$ is the Fourier transform of $f(x)$, it follows that $$\mathcal{F}^{-1}\left(\frac{\sin(s)}{s}\right)=f(x).$$ Therefore, $$u(x)=\sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{2\pi}}\sqrt{\frac{\pi}{2}}\left(f(x)*e^{-\mu|x|}\right)=\frac{1}{\sqrt{2\pi}}(f(x)*e^{-\mu|x|}).$$