Solution of differential equation with Dirac Delta

Question

Is it possible to solve a differential equation of the following form?

$\partial_x^2y + \delta(x) \partial_x y = 0$

where $\delta(x)$ is the dirac delta function. I need the solution for periodic boundary conditions from -1 to 1, but I'll be fine if you direct me for any sort of boundary conditions..

I've realised that I can solve this for some types of boundary conditions. What i'd be really interested in is how to do this for periodic boundary conditions...

Technically, if I approach the problem by splitting the regions $x<0$ and $x>0$ and solve in each part separately, I can solve it and get linear equations in both regions. This will give me $4$ variables. Periodicity, and periodicity of the derivative will give me 2 equations. Continuity at $x=0$ will give me one more. How do i relate the derivative around the $x=0$ interface?

A little background: If there was no delta function, but rather say some gaussian approximation, I would be expect to be able to solve it, but I don't see why I can't get the information of the derivative around $x=0$ when i put in a dirac delta function. My actual problem is reasonably more complicated but this is the quickest simple example I could reduce my problem to. If I try to integrate in an epsilon region around $0$, then I end up with an expression in $y^\prime(0)$, which isn't defined.

If it helps, i'm actually interested in a physical problem, so this delta function is just an approximation, but i'm unable to take any function in it's place whose limit I can make tend to a delta function..

Any help or direction would be greatly appreciated!

EDIT: I guess I should make my actual problem a bit clearer as well. In this simplistic case i can simplify the differential equation by means of the substitution of $z=y^\prime$, however I can't really do that in my original equation. I'm basically interested in some technique by which I can get the information for the change in derivative of the function around the delta function. A better example might be $\partial_x^2y + \delta(x) \partial_x y +y= 0$

Answer 1

Set $z=\partial_x y$. Then $z=z(x)$ satisfies the equation $$ \partial_x z+\delta(x)\,z=0, $$ with solution $$ z(x)=z(0)\,\mathrm{e}^{-\int_0^x\delta(\xi)\,d\xi}. $$ Hence, in order to obtain $y=y(x)$ you should integrate $z$ once.

Answer 2

We can reduce to a first order equation: $$ e^{ix}(i\partial_x)e^{-2ix}(-i\partial_x)e^{ix} y=y'(0)\delta. $$ Now, since the RHS is supported at the origin, we obtain $$ \partial_x [e^{-2ix}(-i\partial_x)e^{ix} y]=-iy'(0)\delta, $$ or $$ \partial_x [e^{-2ix}(-i\partial_x)e^{ix} y+iy'(0) H]=0, $$ where $H$ is the heaviside step function. It follows that $$ e^{-2ix}(-i\partial_x)e^{ix} y+iy'(0) H=c $$ for some constant $c\in\mathbb C$. Thus, $$ \partial_x e^{ix}y(x)=ie^{2ix}(c-iy'(0)H(x)). $$

The problem, then, is that this equation implies that $y'$ has a step-function type singularity at the origin, where it jumps an amount $y'(0)$. But $y'(0)$ is not well-defined then, and thus neither is our starting equation! This stems from the term $y'\delta$. If you expect $y'$ to not be continuous at $0$, it doesn't make sense to multiply the delta-distribution with $y'$, since $$ (y'\delta)(g)=\delta(g y')=y'(0)g(0). $$ Instead, let us formulate a different equation, $$ y''+y=-C_y\delta $$ for some $C_y\in\mathbb C$ independent from $y'(0)$, say for instance $C_y=(y'(0+)+y'(0-))/2$. For a continuous function $y$, $C_y=y'(0)$, but in your case $C_y$ is well-defined even though $y$ is not assumed continuous. Other choices would be possible candidates as well.