Solving $-u''+u=\delta'(x-1)$ using the Fourier transform 
Using Fourier transforms, solve the following boundary value problem $$-u''+u=\delta'(x-1)$$ where $\delta$ stands for the Dirac delta function, with $u(x) \to 0$ as $\lvert x \rvert \to \infty$.


I applied Fourier transforms to both sides of the equation and I arrived to the conclusion that the Fourier transform of the solution must be
$$u_F(k)=\frac{ike^{-ik}}{\sqrt{2\pi}(k^2+1)}$$
Then, using a Foruier transforms table, I get the following solution $u(x)=e^{x-1}/2$ if ($x<1$) and $u(x)=-e^{1-x}/2$ if ($x\geq 1$), which is the derivative of $e^{-\lvert x-1\rvert}/2$. As this function has Fourier transform $\frac{e^{-ik}}{\sqrt{2\pi}(k^2+1)}$, its derivative has Fourier transform $ik\frac{e^{-ik}}{\sqrt{2\pi}(k^2+1)}$, and hence should be the solution to the boundary problem. The problem is this function is not even continuous at $x=1$. What exactly is it that I'm doing wrong? Or, if I've done everything right so far, how can I fix this discontinuity problem?
 A: Your answer is correct. And mathematicians have developed a nice theory that can give meaning to derivatives of discontinuous functions: the theory of distributions. In fact, the solution you obtained satisfies the equation in the sense of distributions; see https://en.wikipedia.org/wiki/Distribution_(mathematics).
According to the theory of distributions, a function $u$ is a distributional solution to the equation $-u''+u=\delta'(x-1)$ if it satisfies
$$
\int_{-\infty}^{\infty}u(x)(-\phi''(x)+\phi(x))\, dx+\phi'(1)=0
$$
for any "nice" function $\phi$ (smooth in particular). $\phi$ is called a "test function". You can check that your solution satisfies the integral equation above by using integration by parts:
\begin{align}
\int_{-\infty}^{\infty}u(x)\phi''(x)\, dx
& =\int_{-\infty}^{1}u(x)\phi''(x)\, dx+\int_{1}^{\infty}u(x)\phi''(x)\, dx \\
& =-\int_{-\infty}^{1}u'(x)\phi'(x)\, dx+u(1-0)\phi'(1)-\int_{1}^{\infty}u'(x)\phi'(x)\, dx-u(1+0)\phi'(1) \\
& =\int_{-\infty}^{1}u''(x)\phi(x)\, dx+u(1-0)\phi'(1)-u'(1-0)\phi(1) \\
& \quad +\int_{1}^{\infty}u''(x)\phi(x)\, dx-u(1+0)\phi'(1)+u'(1-0)\phi(1).
\end{align}
Here, $u(1\pm 0)$ are the right and the left limit of $u$ at $x=1$. The rest of the calculations are left for you to work on your own.

[Another approach] Let $v(x)=u(x)+H(x-1)$, where $H(x-1)$ is the Heaviside function
$$
H(x-1)=
\begin{cases}
1 & (x>1), \\
0 & (x<1).
\end{cases}
$$
Then you see that $v(x)$ is not only continuous but also differentiable; in fact, we have $v'(x)=e^{|x-1|}/2$. Moreover, the second derivative exists except for $x=1$ and is equal to $u(x)$:
$$
v''(x)=u(x)=
\begin{cases}
\displaystyle \frac{e^{x-1}}{2} & (x>1), \\
\displaystyle -\frac{e^{1-x}}{2} & (x<1).
\end{cases}
$$
Finally, since $d/dx[H(x-1)]=\delta(x-1)$, we conclude that
$$
u(x)=v''(x)=u''(x)+\delta'(x-1).
$$
A: *

*Let us follow Paul Garrett's advice and instead consider the shifted problem $$-u''+u=\delta' ,\qquad  u(\pm\infty)~=~0.\tag{1}$$
to reduce clutter. (It is straightforward to shift back the solution in the end.) Since the RHS of the ODE (1) is odd, the unique solution $u$ is odd.


*A moment of thought suggests that the odd solution with the given boundary conditions is
$$u(x)~=~-\frac{1}{2} {\rm sgn}(x) e^{-|x|}
~=~\frac{1}{2}\frac{de^{-|x|}}{dx}. \tag{2}$$
In particular it is discontinuous at the origin.


*Heuristic differentiation twice confirms the solution (2):
$$u^{\prime}(x)~\stackrel{(2)}{=}~(\frac{1}{2}{\rm sgn}(x)^2-\delta(x)) e^{-|x|}
~=~\frac{1}{2} e^{-|x|}-\delta(x), \tag{3}$$
$$u^{\prime\prime}(x)~\stackrel{(3)}{=}~-\frac{1}{2}{\rm sgn}(x) e^{-|x|}-\delta^{\prime}(x) 
~\stackrel{(2)}{=}~u(x)-\delta^{\prime}(x). \tag{4}$$
The heuristic calculations (3) & (4) are not kosher within the framework of standard distribution theory since that theory only define how to multiply distributions with $C^{\infty}$-functions.


*Alternatively, the Fourier transform
$$ \hat{u}(k)~=~\int_{\mathbb{R}}\!\mathrm{d}x~ e^{-ikx}u(x)\tag{5}$$
heuristically leads to
$$(k^2+1)\hat{u}(k)~\stackrel{(1)+(5)}{=}~\int_{\mathbb{R}}\!\mathrm{d}x~ e^{-ikx}\delta^{\prime} (x)~=~ik. \tag{6}$$
The problem here is that $x\mapsto e^{-ikx}$ plays the role of a test function in the evaluation of the Dirac delta distribution, and test functions are supposed to go to $0$ for $|x|\to\infty$.


*The inverse Fourier transform yields the solution (2)
$$\begin{align} u(x)~=~&\int_{\mathbb{R}}\!\frac{\mathrm{d}k}{2\pi} e^{ikx}\hat{u}(k)\cr
 ~\stackrel{(6)}{=}~&\int_{\mathbb{R}}\!\frac{\mathrm{d}k}{2\pi} e^{ikx}\frac{ik}{k^2+1}\cr
~=~&\frac{d}{dx} \int_{\mathbb{R}}\!\frac{\mathrm{d}k}{2\pi} e^{ikx}\frac{1}{k^2+1}\cr
~=~&\frac{1}{2}\frac{d}{dx} \int_{\mathbb{R}}\!\frac{\mathrm{d}k}{2\pi i} e^{ikx}\left(\frac{1}{k-i}-\frac{1}{k+i}\right)\cr
~=~&\frac{1}{2}\frac{de^{-|x|}}{dx},
 \end{align}\tag{7}$$
where we in the last equality closed the $k$-contour in the upper (lower) $k$-plane for $x>0$ ($x<0$), respectively, and used the residue theorem.


*Finally, a rigourous approach is to rewrite the ODE (1) as an integral equation
$$ -u(x)+\int_{-\infty}^x\!\mathrm{d}y\int_{-\infty}^y\!\mathrm{d}z~u(z)~\stackrel{(1)}{=}~H(x). \tag{8}$$
It is straightforward to check that eq. (2) is a solution to eq. (8).
