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May 26, 2005, 11:44 
problem in solving such an equation

#1 
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dy/dt=u(t)S(r)*[1+(dy/dr)^2] where u(t)=U[1+Vsin(w*t)] We are solving for y(r) and U is an constant =37, V >1. S(r)=S*(ra)^(1/3) and S is also constant =3.33. Follwing time depedent boundary conditions is used. dy/dt=0 if u(t)>=S(a) dy/dr=0 if u(t)<=S(a) I do not have problem when I solve this equations when V<1. Also I do not have problem when S(r)=S even if V>1. Please help me thanks 

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May 27, 2005, 00:48 
Re: problem in solving such an equation

#2 
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What kind of numerical scheme are you using to solve the problem ?
Harish 

May 27, 2005, 03:25 
Re: problem in solving such an equation

#3 
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Actually I have used CN, Fully implicit and expilcit with back differencing and central differencing schemes.


May 27, 2005, 05:01 
Re: problem in solving such an equation

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Have you tried to write down the analytic solution  your pde is only first order so you can write down a general expression for the characteristics  from which you should see the source of your problem!
Also isn't S(a)=0 your boundary condition so that dy/dt = 0 on r=a (I assame r=a is the boundary from your notation). 

May 27, 2005, 07:04 
Re: problem in solving such an equation

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I have tried the analytic solutions of such equation. The analytic solution is only possible for the case of V<<1. But the case of V>1 analytical solution is wrong.
Can you give me examples how can analytical solutions helps in such case. Yes you are write about the boundary condition. R 

May 27, 2005, 07:49 
Re: problem in solving such an equation

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I'm not sure what you mean by the analytic solution being wrong  unless you mean you tried to obtain a series solution in V<<1? Otherwise the (correct) analytic solution cannot be wrong since it is the solution to the equation!
What I meant by anaytic solution was that you can reduce your pde, using the method of characteristics, to a set of odes for dt/ds, dr/ds, dy/ds dp/ds and dq/ds where p=dy/dt, q=dy/dr and s is the variable along the characteristic. The analytic solution helps for two reasons: (1) It gives something to test your numerics against and (2) If there is a problem with the existence of the solution (i.e. finite time blowup) then you will be able to see why and in what parameter regions it happens. 

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