# Help with a Helmholtz-like equation

1. Sep 14, 2010

### DrJekyll

I am trying to find the solution to a problem defined as follows:

$$(\partial_x^2-K^2)^2 G(\vec{x},\vec{x}')=\delta(\vec{x}'-\vec{x})$$

where K is simply a constant and x is three dimensional.

$$A \left[ e^{-K(\vec{x}-\vec{x}')}H(\vec{x}-\vec{x}') + e^{K(\vec{x}-\vec{x}')}H(\vec{x}'-\vec{x}) \right] + B \left[ (\vec{x}-\vec{x}') e^{-K(\vec{x}-\vec{x}')}H(\vec{x}-\vec{x}') + (\vec{x}'-\vec{x}) e^{K(\vec{x}-\vec{x}')}H(\vec{x}'-\vec{x}) \right]$$
Here, H is the step function.

and I am trying to find A and B such that the first equation is satisfied. Does anyone have any advice on how to proceed?

My initial guess was to apply $$(\partial_x^2-K^2)^2$$ to my proposed solution and then solve for A and B. This gets a bit tricky since once I start taking derivatives I get derivatives of the delta function.

2. Sep 17, 2010

### klondike

Hmm, I have never had to deal with problem like that. It seems to me a nonlinear operator.

But I frequently encountered derivatives of the delta function. I would integrate both sides such that the derivatives of the Dirac Delta is part of the integrand. and then

$$\int^{+\infty}_{-\infty}\delta^{'}(t)f(t)dt=-f^{'}(t)$$

3. Sep 18, 2010

### MathematicalPhysicist

Why do you think it's nonlinear?

It's actually linear, I mean we have a fourth order PDE:

$$(\partial_x^2-K^2)^2= \partial_x^4 -2\partial_x^2 K^2 +K^4$$
remember that K is a scalar operator and thus it commutes with any other operator.

As for the question itself, I guess you know the recipe for finding Green function,
Do I need to repeat it? (cause I myself kinda keep forgetting).

4. Sep 18, 2010

### DrJekyll

I'm no good with Green's functions, hence the post. However I did come up with a solution:

First, I know the solution to
$$(\partial_x^2 - K^2)G(x,x') = F(x)$$
is
$$-\int dx' F(x') \frac{e^{-K |x-x'|}}{2 K}$$

So we just need to look for the function that produces -1/2K times the exponentials and step functions when operated on by the above.

Turns out that this produces $$A = \frac{1}{4K^3}$$ and $$B=\frac{1}{4K^2}$$

Maple spat this answer out earlier, but it is hard to trust maple unless you already know the answer.

5. Sep 18, 2010

### klondike

Do you mean $$(\frac{\partial^{2}}{\partial x^{2}})^{2}=\frac{\partial^{4}}{\partial x^{4}}$$?

6. Sep 20, 2010

### MathematicalPhysicist

Yes.
$$(\frac{\partial^{2}}{\partial x^{2}})^2=(\frac{\partial^2}{\partial x^2})(\frac{\partial^2}{\partial x^2})$$.

Think of it as multiplication of operators.

7. Sep 20, 2010

### klondike

Ok, Thanks. I thought it was the power of the second partial derivative.

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