# Solution to 2D Laplacian in polar coordinates yields complex solution

I was studying a textbook: Physical Foundations of Technical Acoustics (referenced by a research paper) and I tried to solve the Laplacian of a wave equation by following the steps given. It was equation (2.35) from page 29:

$$\nabla^2 \Phi = {\frac 1r^2} {\frac\partial {\partial r}} \left(r^2 {\frac{\partial \Phi} {\partial r}}\right)+ {\frac 1{r^2\sin \theta}} {\frac\partial {\partial \theta}} \left(\sin \theta {\frac{\partial \Phi} {\partial \theta}}\right)$$

I did follow the separation of variables method, but I hit the roadblock when I was putting the solutions back together, following equation (2.42) from page 32. The book uses the spherical wave of zero order and substitutes $j_0 ,p_0,$ and $h_0^{(2)} (kr)$.

Using equation (2.42a), $$h_m^{(2)} (kr) = j_m(kr)+i n_m(kr),$$

I obtained: $${\frac {A_0}{kr}(\sin(kr)-i\cos (kr))e^{i\omega t}}$$

Which, in complex notation: $${\frac {A_0}{kr}(ie^{-ikr})e^{i\omega t}}$$

(Why is $i$ attached to the exponent $e$? According to the results below, it shouldn't be there.)

The results disagrees with the solution given by equation (2.43) on page 34: $${\frac {a\Phi_\textrm{max}}{r}e^{i(\omega t-kr-\phi)}}$$

Now the question: Why is the solution different and my answer has an imaginary unit, which makes the solution imaginary? Did I make a fatal mistake?

Side Question: How did $\frac {A_0}{kr}$ become $\frac {a\Phi_\textrm{max}}{r}$ (where did the $k$ go?) and how does the $\phi$ show up in the solution?

It seems that if I put the value of the Bessel function $j_0$ into the Neumann function in equation (2.42a) and vice versa, I will get the solution similar to the book's solution except with the constants. This also goes for another solution in the book with first order waves.

Glossary: $j_m$ Bessel function, $n_m$ Neumann function, $h_m^{(2)}$ Hankel function.

Note: I have given a books.google link for the book. The pages I referenced are accessible.

• Are you sure it's not a typo where they have -kr instead of -ikr? The key difference is, ikr would oscillate in space like a wave should, whereas -kr just drops off with distance and all oscillation is in phase with time. That sounds like what you get when a wave is evanescent, such as in a reflecting boundary layer where the phase at the boundary is picked out. So aside from the mathematics, you should be able to tell from the physical situation. – Ken G Oct 9 '16 at 16:20
• I obtained: A0kr(sin(kr)−icos(kr))eiωt is it the possible functions $\Phi$ obtained by using variable seperable? – NIkhil Reddy Ramolla Oct 9 '16 at 16:26
• What you have written is not the wave equation, but an expression for the laplacian. Can you clarify your question? – anon01 Oct 9 '16 at 19:29
• @KenG Are you referring to equation 2.43? I missed the parenthesis on the power of e, which I have fixed. – user3680043 Oct 10 '16 at 8:01
• @ConfusinglyCuriousTheThird It is only the Laplacian part of the wave equation that is being discussed in that part of the book. I'm going to edit the question to clarify. – user3680043 Oct 10 '16 at 8:04