Equation resolution
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Hi,
I would like to write this equation in Matlab :
A (dC/dt)+ B (dC/dz)=0
with these conditions :
for t=0 and z>0 : C=C0, for z=0 and t>0 : C=0, for z=L and t>0 : dC/dz=0,
and how to resolve it. thanks for your help in advance
1 个评论
Walter Roberson
2011-4-24
You have not defined the results for t=0 z=0
采纳的回答
Clemens
2011-4-26
I also only see the trivial solution C(z,t) = 0
But let me take some notes and see if a something more general is possible.
Let's write as product: C(z,t) = f(z)g(t) then you got the boundary conditions:
z>0: g(0) = 0;
t>0: f(0)=0;
f'(L) =0;
Then you could choose f any function that solves these conditions.
and solve g...
A f(z) g'(t) + B f'(z) g(t) == 0
simplify to
g'(t) = - B/A f'(z)/f(z) g(t)
So the general solution would be:
g(t) = D * exp(-B/A f'/f t)
but with g(0)=0 it requires D=0
unless one is nitpicking...
- if A=0 and B=0 - you are free to choose g to anything that fullfills the boundaries.
- what if only one is 0: then either g(t)=0 or g'(t) = 0. Both together with your boundary lead to C(z,t) = 0.
- maybe if f=0 or f'=0... no then the outcome for C would C(z,t) = 0
- well actually if you construct f such that it is 0 in the range [0,L] then you could solve g without any more conditions.
- and/or if you make g such that it is 0 for t >=0 you are free to solve f analogous.
but reading the question I think the interesting area would be t>0 and z=[0,L]. So only C(z,t) = 0 in that range.
Or someone with better math skills comes for help :D
2 个评论
mouna
2011-4-26
Clemens
2011-4-29
First things first. When you say "write" there are several ways depending on what you want.
I suppose you are after a numerical solution. You can use bvp4c for that. There is an example in the help.
You can also input it symbolicaly. But I never used that, so no help from me there.
But keep in mind the first post. f is not uniquely defined. So you might get many different solutions.
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