Trouble proving an identity

[lavster]

Homework Statement

required to prove:

$$(\nabla^2+k^2)\frac{e^{ikr}}{r}=-4\pi\delta (r)$$

Homework Equations

im assuming we are working in spherical coordinates (not sure - could be cylindrical/2D polar)
laplacian for spherical (considering its only a function of r) is $$\frac{1}{r^2}\frac{d}{dr}r^2\frac{d}{dr}$$
quoitent rule

The Attempt at a Solution

i get LHS = $$-k^2\frac{exp{ikr}}{r^2}+k^2\frac{exp{ikr}}{r}$$ for all r using the quotient rule :(


Also, i do notice that for r=0 this equation will blow up so maybe the normal approach using the quotient rule etc doesn't work at this point. i then thought that maybe use an identity which has the expression as an integrand as $$4\pi$$ quite often comes out due to the angular part of the integral, eg da = $$r^2sin \theta d\theta d\phi$$but i can't think of any.
any hints?

thanks

 

[mighty2000]

for the post and for giving me the opportunity to prove this interesting equation.

Firstly, it is important to clarify that we are indeed working in spherical coordinates. This is because the Laplacian operator, \nabla^2, is defined differently in different coordinate systems. In cylindrical coordinates, it is given by \frac{1}{r}\frac{d}{dr}r\frac{d}{dr}+\frac{1}{r^2}\frac{d^2}{d\phi^2}+\frac{d^2}{dz^2}, while in spherical coordinates it is given by \frac{1}{r^2}\frac{d}{dr}r^2\frac{d}{dr}+\frac{1}{r^2}\frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{d}{d\theta}+\frac{1}{r^2\sin^2\theta}\frac{d^2}{d\phi^2}.

Now, let's take a closer look at the left-hand side of the equation we are trying to prove:

(\nabla^2+k^2)\frac{e^{ikr}}{r} = \frac{1}{r^2}\frac{d}{dr}r^2\frac{d}{dr}\frac{e^{ikr}}{r}+k^2\frac{e^{ikr}}{r}

Using the product rule and the chain rule, we can simplify this expression to:

=\frac{1}{r^2}\left(2r\frac{d}{dr}\frac{e^{ikr}}{r}+r^2\frac{d^2}{dr^2}\frac{e^{ikr}}{r}\right)+k^2\frac{e^{ikr}}{r}

Now, let's focus on the first term within the parentheses. Using the quotient rule and the product rule, we can further simplify this to:

2\frac{d}{dr}\frac{e^{ikr}}{r}+r\frac{d^2}{dr^2}\frac{e^{ikr}}{r} = \frac{2}{r}\frac{d}{dr}e^{ikr}+\frac{d}{dr}\left(\frac{1}{r