Maple Packages for Section 6.4

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restart;

>	with(plots):

Warning, the name changecoords has been redefined

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Laplace's Equation on a Ring

We find a function u which satisfies

                                u = 0,  for 1 r 2,  - .

with

                                 u(1, ) = u(2, ) = sin( ) for  - .

We have seen that the two differential equations associated with this problem are

       '' +  = 0,  (- ) =  ( ), and  '(- ) =  '( ).

and

     r( rR ') ' -  R = 0.

The solutions for the first equation, with eigenvalues and eigenfunctions are

       = 0,      = 1.

       = ,     ( ) = cos(n )  or  sin(n ).

The second equation is now    R '' + r R ' - R = 0.

This has led to two familiar ordinary differential equations which have solutions

               R(r) = 1 and ln(r),     = 1, in case n = 0,

and

              R(r) =  and ,  = sin(n ) and cos(n ) in case .

Combine these to get

      u(r, ) =  +    +

Use the boundary conditions to get that all a 's are zero, and all b 's are zero, except

             and    .

We solve these equations.

>	solve({b[1]+b[-1]=1,2*b[1]+1/2*b[-1]=1},{b[1],b[-1]});

We can now make u.

>	u:=(r,theta)->(r+2/r)/3*sin(theta);

We check the boundary conditions and that u  solves Laplace's Equation.

>	u(1,theta); u(2,theta);

>	diff(r*diff(u(r,theta),r),r)/r+diff(u(r,theta),theta,theta)/r^2;

>	simplify(%);

It is interesting to view the graph of this solution for Laplace's Equation on a ring, keeping in mind the maximum principal.

>	cylinderplot([r,theta,u(r,theta)],r=1..2,theta=0..2*Pi,           orientation=[-5,70],axes=NORMAL, shading=zhue, lightmodel=light1, style=patchnogrid);

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Laplace's Equation on a Half Disk

We suppose that u satisfies  u = 0 for 0 < r < 1, with u( r, 0) = 0  = u(r, ),  u( 1,  ) = .  

We know what is the PDE:  

                      =  0,

and that this leads to the ordinary differential equations with boundary conditions

       '' +  = 0,    (0) = 0,   ( ) = 0

    r( rR ') ' -  R = 0.

It follows that     = n,   ( ) = sin(n ), and R(r) =  .

     General solution is

               u(r,  ) =  .

The requirement that 1 = u( 1,  ) = .

We computer the a 's.

>	f:=theta->theta^4*(Pi-theta);

We compute 10 coefficients, but printing them does not seem to serve a useful purpose.

>	for n from 1 to 10 do      a[n]:=int(f(theta)*sin(n*theta),theta=0..Pi)/                 int(sin(n*theta)^2,theta=0..Pi): end do: n:='n':

Here is the definition of u.

>	u:=(r,theta)->sum(a[n]*r^n*sin(n*theta),n=1..10);

We draw the graph of u.

>	cylinderplot([r,theta,u(r,theta)],r=0..1,theta=0..Pi,           orientation=[-18,75],axes=NORMAL, shading=zhue, numpoints=1000);