Green's function for the Helmholtz equation

[DrClaude]

Homework Statement

Show that
$$
G(x,x') = \left\{ \begin{array}{ll} \frac{1}{2ik} e^{i k (x-x')} & x > x' \\ \frac{1}{2ik} e^{-i k (x-x')} & x < x' \end{array} \right.
$$
is a Green's function for the 1D Helmholtz equation, i.e.,
$$
\left( \frac{\partial^2}{\partial x^2} + k^2 \right) G(x,x') = \delta(x-x')
$$

Homework Equations

See above.

The Attempt at a Solution

I am having problems making a Dirac delta appear. I get that the first derivative is discontinuous, but the second derivative is continuous. I don't see any singularity appearing when putting the Green's function into the Helmholtz equation.

Any help appreciated.

 

[Dick]

Homework Statement

Show that
$$
G(x,x') = \left\{ \begin{array}{ll} \frac{1}{2ik} e^{i k (x-x')} & x > x' \\ \frac{1}{2ik} e^{-i k (x-x')} & x < x' \end{array} \right.
$$
is a Green's function for the 1D Helmholtz equation, i.e.,
$$
\left( \frac{\partial^2}{\partial x^2} + k^2 \right) G(x,x') = \delta(x-x')
$$

Homework Equations

See above.

The Attempt at a Solution

I am having problems making a Dirac delta appear. I get that the first derivative is discontinuous, but the second derivative is continuous. I don't see any singularity appearing when putting the Green's function into the Helmholtz equation.

Any help appreciated.

You should show some of your work. But take the example of $f(x)=|x|$. The first derivative is discontinuous at 0. Away from 0 the second derivative is zero. If you want to integrate the second derivative to get the first derivative you need to put a delta function at 0. A discontinuity in the first derivative means "delta function" in the second.

 

[DrClaude]

You should show some of your work.

You are right, I should've have set a better example.

But take the example of $f(x)=|x|$. The first derivative is discontinuous at 0. Away from 0 the second derivative is zero. If you want to integrate the second derivative to get the first derivative you need to put a delta function at 0. A discontinuity in the first derivative means "delta function" in the second.

I see now that I have
$$
G(x,x') = \frac{1}{2ik} e^{i k |x-x'|}
$$
so
$$
\frac{\partial^2}{\partial x^2} e^{i k |x-x'|} = 2 i k \delta(x-x') - k^2 e^{i k |x-x'|}
$$
and $G(x,x')$ is indeed a Green's function.