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Can DSolve given a general solution to Laplace PDE in spherical coordinates without given boundary conditions? I tried, but not able to get a solution. Do I need some other assumptions? This is what I tried

ClearAll[u,r]
pde= D[(r^2 D[u[r,θ,ϕ],r]),r]+ 1/Sin[θ] D[(Sin[θ]  D[u[r,θ,ϕ],θ]),θ]+ 
        1/Sin[θ]^2 D[u[r,θ,ϕ],{ϕ,2}]==0

sol=DSolve[pde,u[r,θ,ϕ],{r,θ,ϕ},Assumptions->{0<= θ<= Pi}]

With and without assumptions, it returns unevaluated. Used the definition of Laplacian from https://en.wikipedia.org/wiki/Laplace_operator

Mathematica graphics

I did not need the 1/r^2 in the above example I wrote, since RHS is zero.

And just in case, there might be different definitions for this operator, I let Mathematica give the operator

ClearAll[u,r,θ,ϕ]
lap=Laplacian[f[r,θ,ϕ],{r,θ,ϕ},"Spherical"];
DSolve[lap==0,f[r,θ,ϕ],{r,θ,ϕ}]

Adding assumptions such as Assumptions->{r>0,0<= θ<= Pi} did not help.

Maple gives the following solution, without boundary condition specified.

PDE := Diff(r^2*diff(F(r,theta,phi),r),r)
+ 1/sin(theta)*Diff(sin(theta)*diff(F(r,theta,phi),theta),theta)
+ 1/sin(theta)^2*diff(F(r,theta,phi),phi,phi) = 0;

Mathematica graphics

ans := pdsolve(PDE):
pdetest(ans,PDE);

(* 0 *) (*i.e. verified*)

sol:=PDEtools:-build(ans) assuming 0 <= theta, theta <= Pi:
sol:=simplify(sol,size);

Mathematica graphics

Is it possible to obtain the above solution (assuming it is correct) using Mathematica's DSolve? Is there an example anywhere that solves Laplace PDE in spherical coordinates using DSolve I could look at? I googled and did find anything so far.

I am using 11.3 on windows 7.

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  • $\begingroup$ The command of Maple 2018 pdetest(sol,PDE); does not perform 0. In view of it the above Maple result seems to be doubtful. $\endgroup$
    – user64494
    Mar 26, 2018 at 6:09
  • $\begingroup$ I repeat the result of Maple which is required by you in Mathematica seems to be doubtful. The command of Maple pdsolve(PDE) produces $F_1(r)F_2(\phi)F_3(\theta)$ where the functions $F_1,\,F_2,\,F_3$ are solutions of certain ODEs. $\endgroup$
    – user64494
    Mar 26, 2018 at 6:35
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    $\begingroup$ Maple's solution may or may not be correct as one solution, but PDE's with no bc's or ic's have arbitrary functions as part of general solutions. In the above, all the c's look like they are constants, rather than functions, so I don't see how that could be a general solution. Maple may choose a particular form to go for, whereas MMA with the wide open possibilities of a general solution, may choose not to. You could choose forms of solution that derive from separation of variables, but there are way more possibilities than those. $\endgroup$
    – Bill Watts
    Mar 26, 2018 at 7:52
  • $\begingroup$ @Bill Watts: Up to maplesoft.com/support/help/Maple/… , the result of pdsolve(PDE) is not necessarilly a general solution of PDE. $\endgroup$
    – user64494
    Mar 26, 2018 at 12:07

1 Answer 1

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Can DSolve given a general solution to Laplace PDE in spherical coordinates without given boundary conditions?

In Version 12.3 it can

ClearAll[u, r, θ, ϕ, f]
lap = Laplacian[f[r, θ, ϕ], {r, θ, ϕ}, "Spherical"];
DSolve[lap == 0, f[r, θ, ϕ], {r, θ, ϕ}, Assumptions -> {0 <= θ <= Pi}]

Gives

enter image description here

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