differential equation with distribution

Question

I'm trying to solve the differential equation: \begin{equation} \frac{d^{2}y}{dx^{2}}=\delta(x)y \end{equation} My 'guess' was a solution of the following form: \begin{equation} y=A\exp\left[\int_{x_{0}}^{x}dx^{\prime}\int_{x_{0}}^{x^{\prime}}dx^{\prime\prime}\delta(x^{\prime\prime})\right]+B\exp\left[\int_{x_{0}}^{x}dx^{\prime}\delta(x^{\prime})\right] \end{equation} Is there any way to solve such an equation?

EDIT: more interestingly, \begin{equation} \frac{d^{2}y}{dx^{2}}=\delta^{\prime}(x)y \end{equation}

Answer 1

First we use the fact that $f\delta=f(0)\delta.$ We then get a differential equation of the form $y''=c\delta$ with condition $y(0)=c.$

The general solution to $y''=c\delta$ is $$ y(x) = \begin{cases} ax+b, & (x\leq 0) \\ (a+c)x+b, & (x\geq 0) \\ \end{cases} $$

Now we only have to make $y(0)=c$ be satisfied. But this gives $b=c$ so our final general solution is $$ y(x) = \begin{cases} ax+c, & (x\leq 0) \\ (a+c)x+c, & (x\geq 0) \\ \end{cases} $$ which can also be written $$ y(x) = (ax+c) + cx H(x), $$ where $H$ is Heaviside's step function and $a,c$ are constants.

---

A cleaner way to solve $y''=c\delta$:

Taking antiderivatives we get $y'=cH+a,$ where $a$ is some constant. Taking antiderivatives a second time gives $y=cxH+ax+b.$

Then $y(0)=c$ gives $b=c$ as before, so our solution is $$ y(x) = cxH(x)+ax+c. $$

---

Solving $y''=\delta'y$:

First we use that $$ y\delta' = (y\delta)'-y'\delta = y(0)\delta'-y'(0)\delta $$ to write the equation as $$ y'' = y(0)\delta'-y'(0)\delta. $$

Taking antiderivatives first gives $$ y' = y(0)\delta-y'(0)H+a $$ and then $$ y = y(0)H-y'(0)xH+ax+b. $$

Here we get a problem, though: $$ y(0) = y(0)H(0) + b. $$ What value does $H(0)$ have?

The only way I can see to solve this is to have $y(0)=0=b.$ Then we get $$ y = -y'(0)xH+ax. $$ This gives $y' = -y'(0)H+a$ and $y'(0) = -y'(0)H(0)+a$ where we have the same problem. Solving this problem in the same way as the former gives $y'(0)=0=a$ which results in the trivial final solution $$ y(x) \equiv 0. $$

Answer 2

As mentioned by Ninad Munshi in the comment section, this particular differential equation can be solved with the help of the relation $y(x)\delta(x-x_0) \equiv y(x_0)\delta(x)$, because the Dirac delta will map $y(x)$ to either $y(a)$ or zero depending on whether $a$ is contained in the domain of integration of not.

Concretely, we have : $$ y'(x) - y'(a) = \int_a^x \delta(u)y(u) \,\mathrm{d}u = \left\{ \begin{array}{cl} y(0) &\mathrm{if}\; a < 0 < x \\ \frac{1}{2} y(0) &\mathrm{if}\; a=0 \;\mathrm{or}\; x=0 \\ 0 &\mathrm{otherwise} \end{array} \right\} = y(0)(H(x)-H(a)), $$ where $H(x)$ is the Heaviside step function $-$ which is actually to the antiderivative of the Dirac delta. Note that the middle case can use other conventions on an interval $[b,x]$. Integrating once again, we get in the end : $$ y(x) = y(b) + (y'(a) - y(0)H(a))(x-b) + y(0)(\mathrm{ramp}(x)-\mathrm{ramp}(b)), $$ since the ramp function $\mathrm{ramp}(x) = \max(0,x) = xH(x)$ corresponds to the antiderivative of the Heaviside function.