# ODE with Dirac Delta and conditions at infinity

1. Mar 9, 2015

### Gallo

I'm trying to solve the following equation (even if I'm not sure if it's well posed)

$$\partial_{x} \, y(x) + a(x)\, y(x) = \delta(x)$$

with $\quad \lim_{x \rightarrow \pm \infty}y(x) = 0$

It would be a classical first order ODE If it were not for the boundary conditions and the Dirac Delta. How should I take account of them?

2. Mar 9, 2015

### RUber

One idea might be to integrate over x.

3. Mar 9, 2015

### Gallo

I'm not sure, what you meant....I would get some kind of integral equation ..

4. Mar 9, 2015

### RUber

Right - often solutions will involve an integral of some sort.
Can you solve the homogeneous problem $y_x + ay = 0?$
Then, for the dirac delta, you enforce continuity in the function at x=0, and a jump condition in the first derivative at x=0.

5. Mar 9, 2015

### Dick

It is a classical ODE except for what must happen at x=0. y(x) must have a jump discontinuity at x=0 assuming a(x) is regular function. Just solve y'(x)+a(x)*y(x)=0 in the domains x>0 and x<0 and patch in the discontinuity.

6. Mar 9, 2015

### RUber

Right...thanks Dick. Sorry for the error in my last post. First order ODE has jump in the 0th derivative, i.e. the function itself. I have been spending too much time in 2nd order problems.

7. Mar 9, 2015

### Gallo

Thanks, following your hint this is I would go:

if $y_{\pm}$ solve $y'_{\pm}(x)+a(x)*y_{\pm}(x)=0$ in $\pm x> 0$ then $y_{\pm} = C_{\pm} \, e^{-A_{\pm}(x)}$ and a solution is given by

$$y(x) = y_{-}(x) + H(x) (u_{+}(x) - u_{-}(x))$$

where

$$A_{+}(x) = \int_{0^{+}}^{x} a(z) dz$$

$$A_{-}(x) = \int_{x}^{0^{-}} a(z) dz$$

and $H(x)$ is the Heaviside step function.

Now I need a jump of one in 0, so I impose $C_{+} - C_{-} = 1$

How do I fix the remaining constant? Moreover I guess that at infinity $A_{\pm}(x)$ has to blow up

8. Mar 10, 2015

### Dick

I don't think you want to reverse the limits in your definition of $A_{-}(x)$. But otherwise, I generally agree. I don't think you have enough information to fix the constants any better than that.

Last edited: Mar 10, 2015
9. Mar 10, 2015

### Ray Vickson

The Dirac delta right-hand-side is exactly what you would use when calculating a Green's Function. Since $\delta(x) = 0$ when $x \neq 0$, you need $y'(x) + a(x) y(x) = 0$ for $x < 0$ and for $x > 0$. If you integrate the de from $-\epsilon$ to $+\epsilon$, you get
$$\int_{-\epsilon}^{+\epsilon} y'(x) \, dx + \int_{-\epsilon}^{+\epsilon} a(x) y(x) \, dx = 1.$$
If $y(x)$ is bounded around $x = 0$, the second term above vanishes in the limit $\epsilon \to 0$, while the first term is $y(+\epsilon) - y(-\epsilon)$, which approaches $y(+0) - y(-0)$ as $\epsilon \to 0$. In other words, $y(x)$ has a jump discontinuity of magnitude 1 as $x$ passes from left to right through 0.

10. Mar 10, 2015

### Gallo

Thanks, that's what I thought!