MaplePrimes - answers and comments on Question, How to plot parameters in system?

Reference

Markiyan Hirnyk — Sun, 09 Dec 2012 18:32:37 Z

Take a look at http://www.mapleprimes.com/questions/137320-Plotting-A-System-Of-Ode#comment137330.

The reading of the answers by Preben Alsholm on ODEs is very useful.

A beginning (and a plot)

Preben Alsholm — Mon, 10 Dec 2012 02:22:54 Z

For each given value of b you are looking for a Lambda s.t. there exists a nontrivial solution g to Eq2 with the homogeneous boundary values as given. dsolve easily solves the boundary value problem for Eq1, so that f will be known once b is given.
It is to be expected that if a Lambda is found there will be infinitely many solutions g.
Thus when turning the problem for g into an initial value problem with one parameter taken to be dg2= g''(0) it may be OK to just ask for a solution having e.g. dg2 = 0.1 (as I have done in the worksheet, see below).

For each b it appears that there will be several eigenvalues Lambda (as is demonstrated for b=0.5 in the worksheet).
This complicates making a decent plot of Lambda versus b as there will be several branches (infinitely many?).

But maybe this can be of some help:
(version 4 now)

NonlinearEigenvalu.mw

no luck

J4James — Sun, 09 Dec 2012 18:58:19 Z

Thanks for your quick response. I tried it but I am getting an empty plot.

Here's my try.

mTp0:=proc(b1) local res,T1,Lambda;
   if not type(b1,numeric) then return 'procname(_passed)' end if;
   res:=dsolve(eval({Eq1,Eq2,bc}, {b=b1}),numeric,output=listprocedure);
   T1:=subs(res,Lambda);
   T1
end proc;
plot(mTp0(b),b=0..0.3);

The out put should look like this. Here in the plot gamma1 (I took it Lambda)

is the lowest possible eignvalue.

Output

Markiyan Hirnyk — Sun, 09 Dec 2012 19:03:15 Z

@J4James I also obtain

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct .
I will think about that.

do you have enough boundary conditions?

PatrickT — Mon, 10 Dec 2012 14:47:29 Z

I've only looked at this very briefly as I couldn't quite understand the question, but for whatever it's worth:

ode1:=diff(f(eta),eta$3)+f(eta)*diff(f(eta),eta$2)-diff(f(eta),eta$1)^2-M^2*diff(f(eta),eta$1)+M^2*b+b^2=0:
ode2:=diff(g(eta),eta$3)+f(eta)*diff(g(eta),eta$2)+(Lambda-2*diff(f(eta),eta$1)-M^2)*diff(g(eta),eta$1)+diff(f(eta),eta$2)*g(eta)=0:
odesys := [ode1,ode2]:
bc := [ f(0) = 0, D(f)(0) = 1, D(f)(N) = b, g(0) = 0, D(g)(0) = 0, D(g)(N) = 0 ];
sys := [op(odesys),op(bc)];
values := [N=10,M=0.1];
params := [b,Lambda];
eval(sys,values);
sysf := eval(eval(sys,values),[b=1,Lambda=1]);
dsolve(sysf,'numeric');

[ seq( dsolve(eval(sys,values),'numeric','output'=listprocedure), params = [b=1,Lambda=1] ) ];
Error, (in dsolve/numeric/bvp/convertsys) too few boundary conditions: expected 8, got 6

Hint

J4James — Mon, 10 Dec 2012 15:19:19 Z

@PatrickT

I do not know but for some reason the last line of your code is not using the params values.

If you try this it will work

b:=1:Lambda:=1:

dsolve(eval(sys,values),'numeric','output'=listprocedure);

Thanks for showing interest