# Solution of differential equation with Dirac Delta

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

• The equation you've written reduces to $y''(x)=-y'(0)\delta(x)$, since $y(x)\delta(x)$ is supported, as a distribution, at the origin. – Jonas Dahlbæk Jul 8 '14 at 13:12
• The new equation reduces to $y''+y=-y'(0)\delta$, which is probably a better candidate for periodic boundary conditions. – Jonas Dahlbæk Jul 8 '14 at 13:18
• True, but strictly speaking, I don't expect $y^\prime(0)$ to be defined, and I can't treat it like an arbitrary constant. I only expect $y(x)$ to be continuous at $0$. Normally in such scenarios you integrate in an $\epsilon$ neighborhood of the interface to obtain relations between the derivatives at both sides, but that isn't going to work here.. So how do I proceed? – SarthakC Jul 8 '14 at 13:18

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.

• Um.. I guess you're right. But could you tell me how to approach this for the case of periodic boundary conditions? I've edited the question accordingly. I realize this can be solved for some cases, but can it be solved for periodic boundary conditions? – SarthakC Jul 8 '14 at 13:05

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.

• Yes, but the derivative would NOT be continuous at $x=0$, right? If it were, then $y^{\prime\prime}$ would have at most a jump discontinuity, and hence the LHS would have a jump discontinuity while the RHS is a delta function.. – SarthakC Jul 8 '14 at 13:23
• The idea is that I expect there to be a change in the derivative around $x=0$. What i don't get is how to quantify it. As you've done, the change in derivative turns out to depend on $y^\prime(0)$, which I don't know/isn't defined.. – SarthakC Jul 8 '14 at 13:25
• Hmm. Could you clarify on your last step? Why is the RHS zero? Ideally wherever you have written $f_\epsilon$ there should be a limit of $\epsilon \rightarrow 0$, right? And in the last step you swap the limits of $x\rightarrow 0$ and $\epsilon \rightarrow 0$. I don't think that is justified, because your final result is saying that $ey$ is differentiable at $0$, which would be equivalent to saying that $y$ is differentiable at $0$ (since $e$ is just $e^{ix}$) function (am I right in saying that?). But we know $y$ is not. I think swapping the limits over there might be unjustified.. – SarthakC Jul 8 '14 at 14:10
• Well. $y^{\prime\prime}$ was integrable, right? Just discontinuous. I believe the problem earlier was the swapping of the limits. In your new edit, I don't quite see how to proceed from where you've left to getting the final solution.. Could you just point a few more steps? What do I do with the $y^\prime(0)$ term? If i integrate your last equation from $-1$ to some arbitrary $s$, and apply continuity at $s=0$ on $y$, then that will give me $y^\prime(0)=0$, and $y$ continuous and differentiable everywhere. Is that even correct? – SarthakC Jul 8 '14 at 14:29
• Umm.. Yes, I already got that $y(x)$ has a ramp function type singularity(by which you mean something like $|x|$, right?). But thanks for showing this more rigorously. However, I'm still no where closer in getting the solution, right? Being able to get rid of the $y^\prime(0)$ term has already been my headache for a long time... – SarthakC Jul 8 '14 at 15:01