Let us consider Laplace's equation in the spherical coordinates [with azimuthal symmetry]. It's solution is given by:(adsbygoogle = window.adsbygoogle || []).push({});

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]

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# Laplace's Equation,Boundary Conditions Etc.

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