Solving $-u''+u=\delta'(x-1)$ using the Fourier transform

Question

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 $\lvert x \rvert \to \infty$.

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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?

Answer 1

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.

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[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 $$ u(x)=v''(x)=u''(x)+\delta'(x-1). $$

Answer 2

1. 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.

2. 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.

3. 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.

4. 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 $|x|\to\infty$.

5. 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.

6. 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).