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Re: PDE, laplace, exact, should be simple...

Hello Steve
thanks for your detailed answer to my query. [ Incidentally I should
apologise for the inexplicable profusion of equal signs in my
postings, I'm almost afraid to type another one. ]
I noticed that another maths package was able to arrive, with a little
prompting, at the correct solution and wondered if I was missing
something obvious using Mathematica. I'll study your comments closely,
many thanks

On 9 June 2010 12:19, schochet123 <schochet123 at> wrote:
> Depending on the generality you are trying to achieve, this problem is
> very far from simple.
> If all one wanted was to obtain the solution Sin[Pi x] Sinh[Pi y]/
> Sinh[Pi] for the specific problem
> {D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0 , u[0, y] == 0, u[x=
, 0]
> == 0, u[1, y] == 0, u[x, 1] =Sin[x]}
> then one could define
> myDSolve[D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0 , u[0, y] ==
= 0,
> u[x, 0] == 0, u[1, y] == 0, u[x, 1] ==Sin[Pi x]},u,{x,y}]={=
> -
>>Function[{x,y}, Sin[Pi x]Sinh[Pi y]/Sinh[Pi] ]}}
> Why can't the built-in DSolve find that answer? I am not from Wolfram,
> but it seems to me that DSolve doesn't attempt to find check whether
> specific functions are solutions, because there are infinitely many
> equations that have explicit solutions and looking for all of them
> would take too long.  As a simple example, consider a homogeneous
> linear variable-coefficient single ODE in the variable x. Whenever the
> sum of all the coefficients equals zero then E^x is a solution. You
> can easily add an appropriate set of boundary conditions that make E^x
> be the unique solution. However, if the ODE is complicated enough then
> Mathematica will not find that solution. If it were to look for such
> solutions, then why not look for the solutions E^(2 x) or E^(k x) for
> arbitrary k or arbitrary polynomial solutions, or ...
> The upshot is that DSolve uses a set of algorithms that solve entire
> classes of problems.
>  What class of problems does the above problem belong to?
> 1) If you want to solve the 2-D Laplace equation on any rectangle with
> Dirichlet boundary conditions (u= something) on all sides, with three
> conditions of the form u==0 and the fourth of the form u==f, wher=
e =
> f
> is c Sin[k( x-x0)]] or c Sin[k (y-y0)]  and vanishes at the endpoints
> of the boundary interval, then you need to check that the boundary
> conditions are given for two values of each variable, that  three of
> the four conditions say that u equals zero, and that the fourth is of
> the above form. You can then write a function myDSolve that will give
> the solution u[x_,y_]=f[x] Sinh[k (y-y0)]/Sinh[k (y1-y0)] where the
> boundary value f is taken on at y==y1, and the value zero at y==y=
> except that you may need to switch the roles of x and y.
> 2) If you want to solve the Laplace equation in arbitrary dimensions
> then there are analogous but more complicated formulas.
> 3) If you have nonzero boundary values on all sides then in dimension
> d the solution will be a sum of 2^d terms of the above form.
> So it should be possible to write a Mathematica program that will find
> solve problems of generality 1-3. However:
> 4) If you want to allow the boundary data f to be an arbitrary smooth
> function that vanishes at the endpoints of the boundary interval then
> you need to calculate its Fourier Sine coefficients and form an
> infinite series of solutions of the above form. In general Mathematica
> will not be able to calculate Integrate[f[x] Sin[k x],{x,0,Pi}] to
> obtain those coefficients.
> Moreover, even when Mathematica does calculate the above integral,
> substituting specific values for k may yield 0/0 and hence give the
> answer Indeterminate. For example try calculating the general formula
> for the Fourier Sine coefficients on the interval [0,Pi] of the
> function f[x_]= x Sin[3 x]. For this particular function it is easy to
> see that this problem occurs only for k==3, but in general it is
> probably not possible to determine what the bad values of k are.
> 5) If you want to allow more general boundary conditions and more
> general PDEs you will find that in general you cannot calculate
> explicitly the appropriate eigenfunctions to use in the series
> expansion, at which point you are stuck.
> So (Disclaimer once again: I am not from Wolfram so this is just a
> guess)  the reason DSolve does not find your  solution is apparently
> that generality levels 1-3 seem too specific to bother implementing,
> while levels 4-5 are too difficult.
> Steve
> On Jun 2, 9:05 am, peter lindsay <plinds... at> wrote:
>> forgive the simplicity of this:
>> D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0
>> BCs={u[0, y] == 0, u[x, 0] == 0, u[1, y] == 0, u[x, 1] ==
> ==
>>  Sin[=F0 x]}
>> DSolve etc, etc, etc...
>> A solution is Sin[Pi x] Sinh[Pi y]
>> How can I get mathematica to come up with this gem ?
>> thanks, and sorry again for any stupidity on my part
>> Peter Lindsay

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