Help with E/M Potential using Laplace Solution

[Fjolvar]

I'm struggling to learn how to find the potential using laplace solution. I know that X(x) can be rewritten in terms of C1e^kx + D1e-kx OR C1 cosh kx + D1 sinh kx OR cos kx + sin kx... but when do you know how to use which form. I understand it partially but not fully. And then how do you consider Y(y)? Given its a 2-dim problem in cartesian. Any information or references would be immensely appreciated!

 

[Fjolvar]

C1e^kx + D1e-kx can be rewritten as C1 cosh kx + D1 sinh kx, correct? Depending on the geometry of the problem, but cos kx + sin kx cannot be rewritten right? You need one sinusoidal and one hyperbolic function in the 2dim case, is that right?

 

[Born2bwire]

Cosine and sine can be related to cosh and sinh by scaling the argument by the imaginary number i. So there really is no difference between either one of the three formulations because in the end, all it will change is the phase of your solved modes k. But as long as you properly solve for the coefficients and k then you should come up with equivalent solutions for either of the three cases because all three form equivalent bases for the solutions to Laplace's equation in terms of the modes k. Although, some cases can be easier than others if you know more about the conditions of the problem. For example, if you know that the boundary conditions at infinity are Dirichlet with values of zero then using the exponential form is desirable because we can quickly see that the coefficient on the exp(kx) term must be zero otherwise the solution would blow up. This is not so readily seen if we took the sinusoidal or even hyperbolic sinusoidal cases.

 

[Nabeshin]

If it's truly a 2-dimensional problem you just suppress the 3rd component (you can equally well do the formalism of separation of variables in only 2D, it turns out to be the same just with only two functions).

As far as choosing sin/cos or sinh/cosh, this kind of thing is revealed only by the boundary conditions. Let's say you know that X(x) is sins and cosines. Then the other component (assuming 2D) must be sinh/cosh (or the specific case of exponentials). This follows directly from the laplace equation, where you know that not all of the coefficients can be real/imaginary, but there must be some mixing.

Laplace equation is just all about the boundary conditions.

 

[PJC01]

I can understand your struggle in learning how to find the potential using Laplace solution. Finding the potential using Laplace solution involves solving a partial differential equation, which can be a complex process. However, there are some steps you can follow to make the process easier.

Firstly, it is important to understand the boundary conditions of the problem. This will help you determine the appropriate form for X(x) and Y(y). For example, if the boundary conditions are such that X(x) should approach zero at infinity, then you can use the form C1e^kx + D1e^-kx.

Secondly, the form of X(x) and Y(y) should satisfy the Laplace equation, which is a second-order partial differential equation. This equation can be solved using separation of variables, where you assume that the solution can be written as a product of functions of x and y.

Thirdly, once you have determined the appropriate forms for X(x) and Y(y), you can use the boundary conditions to solve for the constants C1, D1, C2, and D2. This will give you the complete solution for the potential.

As for considering Y(y), it depends on the geometry of the problem. If the problem is two-dimensional and in Cartesian coordinates, then Y(y) can be considered as a function of y only.

I understand that this may still be a complex process, and it is always helpful to refer to textbooks or online resources for further clarification. Some good references for learning about Laplace solution and potential include "Introduction to Electrodynamics" by David J. Griffiths and "Classical Electrodynamics" by John David Jackson.

I hope this information helps you in your learning process. Keep practicing and seeking help when needed, and you will surely master the concept of Laplace solution and potential.