## 5 Reputation

4 years, 168 days

## Revised question...

@tomleslie Thank you for your advice. I have revised my problem and now have its formulation as follows:

restart;
V:=5*10^(-5):
A:=2*10^(-4):
L:=0.02:
C0:=0.01:
d:=0.25:
eq1:=CL(t)*V+CR(t)*V+A*int(CS(x,t),x=0..L)=C0*V:
eq2:=CL(t)=eval(CS(x,t),x=0):
eq3:=CR(t)=eval(CS(x,t),x=L):
eq4:=CL(0)=C0:
eq5:=CL(infinity)=C0/2:
eq6:=CL(infinity)=CR(infinity):
eq7:=V*(C0-CL(t))=A*d*int(eval(diff(CS(x,tau),x),x=0),tau=0..t);
eq8:=V*CR(t)=A*d*int(eval(diff(CS(x,tau),x),x=L),tau=0..t);
eq9:=diff(CS(x,t),t)=d*diff(CS(x,t),x,x);
eq10:=CS(x,0)=C0;

I did't realize about boundary condition for CS(x,0). Sorry for wasting your time. In problem formulation I changed eq5 and eq6 because it originally should be for infinite time, but for this particular problem 10000 sec is an infinite time. eq 10 was wrong, sorry for that. Now it is correct.

## sol...

@tomleslie I added 3rd boundary condition. Is it correct? How to proceed futher?

 > restart; V:=5*10^(-5):
 > A:=2*10^(-4):
 > L:=0.02:
 > C0:=0.01:
 > d:=0.25:
 > eq1:=CL(t)*V+CR(t)*V+A*int(CS(x,t),x=0..L)=C0*V:
 > eq2:=CL(t)=eval(CS(x,t),x=0):
 > eq3:=CR(t)=eval(CS(x,t),x=L):
 > eq4:=CL(0)=C0:
 > eq5:=CL(10000)=C0/2:
 > eq6:=CL(10000)=CR(10000):
 > eq7:=V*(C0-CL(t))=A*d*int(eval(diff(CS(x,tau),x),x=0),tau=0..t);
 > eq8:=V*CR(t)=A*d*int(eval(diff(CS(x,tau),x),x=L),tau=0..t);
 > eq9:=diff(CS(x,t),t)=d*diff(CS(x,t),x,x);
 > eq10:=CS(x,0)=0;
 (1)
 > eq7a:=subs(eq2, eq7); eq8a:=subs(eq3, eq8); eq9a:=subs(eq9,eq10);
 (2)
 > # # Notice in the above that the lhs of eq(7a) and eq(8a) # appear to be functions of 't', but the corresponding # right hand sides are not! This could only be true if # CS(0,t) is a constant and CS(0.02, t) is (another) # constant: I'm not convincd that this is what the OP # really means. # # Continue (carefully) for a little #   sols:=pdsolve(eq9); # # which shows that CS(x, t) is separable, ie can be # written as _F1(x)*_F2(t). Furthermore the subsidiary # conditions can be solved to give a "general" solution #   genSol:= CS(x,t)=rhs(dsolve(op([2,1,1],sols))*dsolve(op([2,1,2],sols)));
 (3)
 > # # Suubstitute this general solution in the "boundary # conditions" of eq7a and eq8a #   eq7b:=simplify(subs([genSol,eval(genSol, x=0)], eq7a));   eq8b:=simplify(subs([genSol,eval(genSol, x=0.02)], eq8a)); eq9b:=simplify(subs([genSol,eval(genSol, t=0)], eq9a));
 (4)
 > # # Note the lhs of eq7b *seems* to be a function of 't', but the # rhs is not. Not possible. The only(?) way this can be avoided # is if _C1^2=-_C1*_C2, or _C2=-_C1, so make this substitution # in eq7b to get eq7c. Make the same substitution in eq8b, and # also use eq7c, to eliminate _C1 from eq8b, arriving at eq8c #   eq7c:=isolate(simplify(subs( _C2=-_C1, eq7b)), _C1^2);   eq8c:= subs( eq7c, simplify(subs(_C2=-_C1, eq8b)));
 > # # Note that the lhs of eq8c is a function of 't' whereas the # rhs is not. The only way(?) this could be resolved is if # _c1=0, but this collapses the general solution (genSol above) # to C(x,t)=0. SO this would seem to be about as far as it is # worth going #
 > pdsolve([eq7b,eq8b,eq9b]);
 >

## 3rd condition...

@tomleslie I see. So that is  CS(x,0)=0

## Drawing...

@tomleslie Her eis the drawing of the process

@tomleslie  I m kind of confused. eq9 is second order PDE. Why it requires 3 boundary conditions???Moreover, why eq2 and  eq 3 with eq-ns 1,4,5,6 cannot be considered as complex boundary conditions?

## confirmation...

@tomleslie  Yes, that is