Help with a Helmholtz-like equation

[DrJekyll]

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

<br /> (\partial_x^2-K^2)^2 G(\vec{x},\vec{x}&#039;)=\delta(\vec{x}&#039;-\vec{x})<br />

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

<br /> A \left[ e^{-K(\vec{x}-\vec{x}&#039;)}H(\vec{x}-\vec{x}&#039;) + e^{K(\vec{x}-\vec{x}&#039;)}H(\vec{x}&#039;-\vec{x}) \right]<br /> + B \left[ (\vec{x}-\vec{x}&#039;) e^{-K(\vec{x}-\vec{x}&#039;)}H(\vec{x}-\vec{x}&#039;) + (\vec{x}&#039;-\vec{x}) e^{K(\vec{x}-\vec{x}&#039;)}H(\vec{x}&#039;-\vec{x}) \right]<br />
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.

 

[klondike]

<br /> (\partial_x^2-K^2)^2 G(\vec{x},\vec{x}&#039;)=\delta(\vec{x}&#039;-\vec{x})<br />

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

This gets a bit tricky since once I start taking derivatives I get derivatives of the delta function.

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^{&#039;}(t)f(t)dt=-f^{&#039;}(t)

 

[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).

 

[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
<br /> (\partial_x^2 - K^2)G(x,x&#039;) = F(x)<br />
is
<br /> -\int dx&#039; F(x&#039;) \frac{e^{-K |x-x&#039;|}}{2 K}<br />

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.

 

[klondike]

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

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

 

[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.

 

[klondike]

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.

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