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?
Best Answer
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 $$ u(x)=v''(x)=u''(x)+\delta'(x-1). $$