Thx for such a cool idea. But how we can adopt this approach for a ODE with numeric solution?

I think that some parts of your code is missing. If you want a response to your

question then you need to upload the complete sheet. Otherwise, it will/is be/- a

waste of time.

Thanks

@Preben Alsholm I have uploaded a reference. please see section 5.

thanks

reference.pdf

@Preben Alsholm I have uploaded a reference. please see section 5.

thanks

reference.pdf

@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

@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

@Preben Alsholm

Well, first of all thanks for such a nice effort. You mentioned correctly that there will be infinite

eigenvalues for a given b but we need to consider the lowest one (including both positive and negative).

The plot for negative eignvalues should look like this

 It will be nice to plot both positive and negative lowest eigenvalues (lambda vs b)  in one figure.

@Preben Alsholm

Well, first of all thanks for such a nice effort. You mentioned correctly that there will be infinite

eigenvalues for a given b but we need to consider the lowest one (including both positive and negative).

The plot for negative eignvalues should look like this

 It will be nice to plot both positive and negative lowest eigenvalues (lambda vs b)  in one figure.

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.  

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.  

@PatrickT 

I am facing the same issue of  "unable to store 'HFloat(1.005717340914193)*y'".

Even I tried to to use approx solution but no luck.

Please have a look. P_new.mw 

Thanks

@PatrickT 

I am facing the same issue of  "unable to store 'HFloat(1.005717340914193)*y'".

Even I tried to to use approx solution but no luck.

Please have a look. P_new.mw 

Thanks

@Preben Alsholm, first of all, M>=0, beta>=0 and alpha>1.

I tried and got the following results. As you mentioned that 

alpha*(D@D)(f)(0)+beta*(D@D)(f)(0)^3 will have multiple solutions but...

PS: The good thing about implicitplot routine is that it gives you the chance to look beyond the

obvious. ic.mw

@Preben Alsholm, first of all, M>=0, beta>=0 and alpha>1.

I tried and got the following results. As you mentioned that 

alpha*(D@D)(f)(0)+beta*(D@D)(f)(0)^3 will have multiple solutions but...

PS: The good thing about implicitplot routine is that it gives you the chance to look beyond the

obvious. ic.mw

You made it very clear but what will be the possible staring point to handle the situation?