| Author | Message | ||
     Fernando Buezas (fernandob) Member Username: fernandob Post Number: 4 Registered: 02-2007 |
Dear Nelson: please doest exist anyway to solve eigenvalues's problem like L(u) + f(lambda)u = 0 ? For example f(lambda)=lambda^2 thankyou | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 788 Registered: 06-2003 |
FlexPDE solves the system A*u + Lambda*B*u = 0 But you can subsequently compute Lambda1 = sqrt(Lambda) to solve the effective system A*U + Lambda1^2*B*u. | ||
| abbasali Member Username: abbasali Post Number: 7 Registered: 04-2010 |
Hi dear Dr. Nelson I am a mathematics student and I have derived some abstract mathematics result about an eigenvalue problem. A simple form of this problem is as follows TITLE 'nonlinear eigenvalue problem' select modes=1 ERRLIM =.001 VARIABLES u DEFINITIONS v r1=3 r=10 e1=3 e2=0 k1=1* USTEP(r1-(x-e1)^2-(y-e1)^2 ) k2=1*USTEP(10-(x-e2)^2-(y-e2)^2 ) EQUATIONS Div(grad(u))-v*u=(-k1*lambda*u)-((k2-k1)*(lambda^2)*u) BOUNDARIES REGION 1 'blob1' { the outter blob } v=.7 START(r+e1,0) VALUE(u)=0 ARC(CENTER=e1,e1) ANGLE=360 REGION 2 'blob2' { the embedded blob } v=.1 START(r1+e2,0) ARC(CENTER=e2,e2) ANGLE=360 PLOTS surface(u) painted SUMMARY REPORT(INTEGRAL(u^2,'blob2')) REPORT(INTEGRAL((k2-k1)*u^2,'blob1')) report(lambda) END I would like to know is this type of equations which depends linearly on the eigenparameter in a subset of the domain and quadratically elsewhere , after some simplifications, models any physical phenomenon? regards. | ||
| rgnelson Moderator Username: rgnelson Post Number: 1416 Registered: 06-2003 |
FlexPDE does not know how to do nonlinear eigenvalue problems. It solves the system A*u - Lambda*B*u = 0 In your case, the Lambda^2 will be treated as if it were Lambda. In Buezas' system, there was only a single reference to lambda^2, so it could be replaced by a redefined lambda1. Since your system includes both first and second powers, this kind of redefinition will not reduce the system to a single power of lambda. There may be ways to redefine the system to do what you want, but I don't know what they are right now .... |