Elliptic pDE with two closed boundaries
- To: mathgroup at smc.vnet.net
- Subject: [mg43413] Elliptic pDE with two closed boundaries
- From: CAP F <Ferdinand.Cap at eunet.at>
- Date: Fri, 29 Aug 2003 07:16:23 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Selwyn Hollis,
Thank you again for your interest and critics [Messages
43294,43380,43381] concerning my notebook c37.I have to apologize for
the missing of a more extensive description concerning the notebook.I
have not been aware of the fact, that mathgroup users not yet having
read the pages 216 -219 of my book (F. Cap, Mathematical Methods in
Physics and Engineering with Mathematica, CRC Press, 2003,
ISBN01584884029) would have difficulties to understand, that the
notebook has the intention to demonstrate that variational methods are
able to solve elliptic partial differential equations with inhomogeneous
boundary values on two different closed boundary curves. Since you are a
specialist in the field (sorry, I did not read your book on the same
matters) I am glad to have discussion with you. I agreee that the very
low order (3rd) solution by Wallnoefer, Ref [4.17], is very rough.
To apologize for the missing explanations in the notebook, I send an
improved version.I hope this answers all open questions?
(* c37: Two boundaries for the Laplace equation see Fig.4.14 in the
book \
Mathematical Methods in Physics and Engineering with Mathematica, CRC
Press,
ISBN 1584884029.There are two different closed boundary curves,
a square of length a=
4 and a circle of radius 1. On both boundaries the boundary
condition u[
boundary]=
x^2+y2 is valid.
The solution method is the RITZ variational method.It gives
the \
very approximative solution f - only 3rd approximation*)
Clear[f,P1,Q,CG,QG,QQG,A,U];(* f is formula (4.7.22)*)f[x_,y_]=
x^2+y^2+(x^2+
y^2-1)*(x^2-4)*(y^2-4)*(0.0969+0.0521*(x+y)-0.0673*(x^2+y^2)-0.0474*
x*y+0.0131*(x^3+y^3)+0.0186*(x^2*y+x*y^2))
P1=Plot3D[f[x,y],{x,-2.,2.},{y,-2.,2.},Shading\[Rule]False,
ClipFill\[Rule]None, PlotRange\[Rule]{0.,8.},
PlotPoints\[Rule]100](*this gives Fig 4.14*)
In[7]:=
f[2.,2]
In[10]:=
Sqrt[0.5]
In[11]:=
f[0.707107,0.707107]
In[4]:=
g[x_,y_]=Simplify[D[f[x,y],{x,2}]+D[f[x,y],{y,2}]]
In[6]:=
Plot3D[Evaluate[g[x,y],{x,-2.,2.},{y,-2.,2.},
Shading\[Rule]False, (*ClipFill\[Rule]None,*)
PlotRange\[Rule]{0.,8.},
PlotPoints\[Rule]100]]
CG=Graphics[Circle[{0.,0.},1.],AspectRatio\[Rule]Automatic]
Show[CG]
Q=Line[{{-2.,-2.},{2.,-2.},{2.,2.},{-2.,2.},{-2.,-2.}}];
QG=Graphics[Q,AspectRatio->Automatic]
Show[QG]
QQG=Show[QG,CG]
(* The RITZ trial function G[x,
y], (4.7.22) and (4.7.23) does not satisfy the Laplace equation*)
G[x_,y_]=x^2+
y^2+(x^2+y^2-1)*(x^2-4)*(y^2-4)*(co+c1*(x+y)+c2*(x^2+y^2)+c3*x*y+
c4*(x^3+y^3)+c5*(x^2*y+x*y^2))
FullSimplify[D[G[x,y],{x,2}]+D[G[x,y],{y,2}]]