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 R May 26, 2005 11:44

problem in solving such an equation

Hi, I have problem in solving such an equation

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*(r-a)^(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.

 harish May 27, 2005 00:48

Re: problem in solving such an equation

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

-Harish

 R May 27, 2005 03:25

Re: problem in solving such an equation

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

 Tom May 27, 2005 05:01

Re: problem in solving such an equation

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).

 R May 27, 2005 07:04

Re: problem in solving such an equation

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

 Tom May 27, 2005 07:49

Re: problem in solving such an equation

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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