# Laplace's Equation,Boundary Conditions Etc.

1. Sep 22, 2011

### Anamitra

Let us consider Laplace's equation in the spherical coordinates [with azimuthal symmetry]. It's solution is given by:

psi[r,theta]=summation over n [from zero to infinity] [An r^n+ Bn r^{-
(n+1)}]Pn[Cos(theta)] ----------- (1)
An and Bn are arbitrary constants and Pn stands for the Legendre Polynomials.
It does not seem to cover all types of boundary conditions even on the
surface of a sphere.
Let us think of a function f(theta). One can perform a Fourier -Legendre transformation/expansion on f(theta)[http://mathworld.wolfram.com/Fourier-LegendreSeries.html ] keeping in mind that the Legendre polynomials form a complete ,orthogonal set.

We may think of a function for a spherical boundary: f(theta)=1000 theta^5+200 theta
+50sin^(2/5)[theta]+7Cos^(6/7)[theta]
Expansion in terms of Legendre's polynomials might be difficult[rather
impossible] in terms of Pn[Cos(theta)] though we may carry out an expansion in terms of Pn(theta)

[If expansion is possible [in terms of Pn[Cos(theta)]we may try out a proper fit with equation (1) for the boundary condition. Otherwise we may go in for numerical methods for intractable functions on the boundary]

We may use numerical methods for boundary conditions that do not match
against the "known solutions".
Interestingly the numerical methods will consider the PDE directly and the boundary conditions without any regard for known solutions! One may try out this exercise for any type of PDE that has"known solutions"

An Analytical Consideration:
We may consider a trial soln of the form:
psi=[An f(psi) r^n +Bn r^{-(n+1}}]Pn [Cos[theta]]----(2)
An ,Bn are constants. Pn stands for the Legendre Polynomials.Substituting the above "psi" into the original equation[Laplace's Equation with azimuthal symmetry] we get a differential equation connecting f',f,psi,theta and r.During the formation of the equation
we consider psi as a function of theta and r.Accordingly we perform the partial differentiation
But,while solving the resulting equation to get f, we may take psi,theta and r independent of each other.[ie, theta and r are taken as constants]. Thus, we will solve an ordinary [second order]differential equation wrt the variable psi[to get f]
The function f ,obtained, will of course be a function of psi ,theta and r. If it is used in (2) and differentiated, considering psi dependent on r and theta,we will get the same differential
equation[ODE] we solved taking theta and r as constants!Thus we may extend the solution set.
[Summation has to be applied on the new trial solutions]

Last edited: Sep 22, 2011
2. Oct 9, 2011

### Anamitra

Lets try to expand the function $${f}{(}{x}{)}{=}{x^4-3x^2+x}$$ in terms of Pn[Cos x] and see if it is possible:
We write:
$${x^4-3x^2+x}{=}{C}_{0}{P}_{0}{(}{Cos}{x}{)}{+}{C}_{1}{P}_{1}{(}{Cos} {x}{)}{+}{ C}_{2}{P}_{2}{(}{Cos}{(}{x}{)}{)}{+} {C}_{3}{P}_{3}{(}{Cos x}{)}$$
$${+}{ C}_{4}{P}_{4}{(}{Cos x}{)}{+}{ C}_{5}{P}_{5}{(}{Cos x}{)}{+}{…………………..}$$ ----------------------- (1)
Now,
Multiplying both sides of (1) by Pm(Cos(x)) Sin(x) and integrating from 0 to piwe have,
$$\int{(}{x^4-3x^2+x}{)}{P}_{m}{(}{Cos x}{)}{Sin}{(}{x}{)}{dx}{=}\Sigma{C}_{i}\int{P}_{m}{(}{Cos}{x}{)}{P}_{i}{(}{Cos}{x}{)}{Sin}{(}{x}{)}{dx}$$ --------------------- (2)
[Integration extends from 0 to pi=3.142. Summation includes values of i from 0 to infinity]
We substitute on the RHS of (2):
t=Cos x
$${RHS}{=}\Sigma{C}_{i}\int{P}_{m}{(}{t}{)}{P}_{i}{(}{t}{)}{dt}$$
Integration runs from -1 to+1
$${RHS}{=}{C_{m}}\frac{2}{2m+1}$$
Therefore,
$$C_{m}{=}{(}{m+.5}{)}\int{(}x^4-3x^2+x{)}{P_{m}}{(}{Cos x}{)}{Sin}{(}{x}{)}{dx}$$
Integration is from 0 to pi
$${C_{0}}{=}{6.253306}$$
$${C_{1}}{=}{-15.499888}$$
$${C_{2}}{=}{14.470519}$$
$${C_{3}}{=}{-8.332501}$$
$${C_{4}}{=}{5.054528}$$
$${C}_{5}{=}{-3.485501}$$
We may substitute the values of the constants in the RHS of equation (1)
Using x=0.5 we get -0.1875 on the LHS of (1)--- while on the RHS we get 307.401595 [using the first 6 terms of the expansion.
[Using Pn(x) instead of Pn(Cos x) we get :
$${x^4-3x^2+x}{=}{-}\frac{4}{5}{P}_{0}{(}{x}{)}{+}{P}_{1}{(}{x}{)}{-}\frac{10}{7}{P}_{2}{(}{x}{)}{+}\frac{8}{35}{P}_{4}$$
For x=0.5 we have -0.1875 on both sides.]

Last edited: Oct 9, 2011