@dharr But this doesn't work for multiple subs, such as

plot([Th-273,subs(Tc=400,n*alpha*dT/2),subs(Tc=500,n*alpha*dT/2),Th=300..700]);
 

But still i didnt get it. For the two arbitrary constants, I am using bcs:=phi2(1)=0,phi2(S0)=-N4*Phi;

My question is, Is using laplace option (sol := dsolve({ode, bcs}, phi2(x),method=laplace):) in maple for such ode with such boundary conditions ok or not?

But still i didnt get it. For the two arbitrary constants, I am using bcs:=phi2(1)=0,phi2(S0)=-N4*Phi;

My question is, Is using laplace option (sol := dsolve({ode, bcs}, phi2(x),method=laplace):) in maple for such ode with such boundary conditions ok or not?

In which circumstances, we can treat the real part of the complex number as a real solution ?

In which circumstances, we can treat the real part of the complex number as a real solution ?

In figure below, sigma has multiple solutions. How we will check that the positive

roots for sigma (labeled as 1) is a valid one?

In the next figure, I have used higher digits compare to the above figure, which helped to

aviod the numerical noise but it shows another branch of solution for sigma (labeled as 3). How to check

whether,the solution we got are correct or not?

for equations and its graphical out please see the below attached files.

20110818_section4-5.mws

20110818_section4-5.txt

@pagan 

Thanks man. I will try my best. I accept my "so-called" mistake. The problem is I am using the same

paramters maybe thats why it seems to you as a repetition. By the way if some one already give

me the answer then why I will ask it for "allegedly" more than 5 times.

Any way if you would have spend this time on helping me with the problem, that would have given

me alot of peace and relief.

I will be very very very careful in future.

Thanks. 

("|:|")

kh2n

Furthermore, for R3[-7.45,-0.678] Eq1 has multiple solutions for S0.   

@pagan 

I am still facing some issues with it. So I tried again to make it more clear to get the desired results.

Well, If am bothering too much, so I am sorry for that. But I really need some guidance. 

The Key difference is that we have to use the value of S0 from Eq1 into Eq2 .

@Axel Vogt 

thanks for your help.

@Axel Vogt 

thanks for your help.

@Axel Vogt 

I got your point quite clear. If we do like this

S0:=0.9:

temp6c1:= subs(parvals,subs(params,temp6b)):

plots:-implicitplot(temp6c1,R3=-8..0,Sigma=-40..40,gridrefine=2);

In what way this plot can be related to locate that zero (It is located at the peak 

in ur log plot )

@Axel Vogt 

I got your point quite clear. If we do like this

S0:=0.9:

temp6c1:= subs(parvals,subs(params,temp6b)):

plots:-implicitplot(temp6c1,R3=-8..0,Sigma=-40..40,gridrefine=2);

In what way this plot can be related to locate that zero (It is located at the peak 

in ur log plot )

@Axel Vogt 

The question is  

temp6c1:= subs(parvals,subs(params,temp6b)):

I want to plot temp6c1 for Sigma(R30,S0).

@Axel Vogt 

The question is  

temp6c1:= subs(parvals,subs(params,temp6b)):

I want to plot temp6c1 for Sigma(R30,S0).