du/dt = A*d2 u/dy2


by timsea81
Tags: du or dt, u or dy2
timsea81
timsea81 is offline
#1
Jan29-13, 12:40 PM
P: 90
du/dt = A*d^2u/dx^2
Where u=u(x,t) and A is a constant.
The boundary conditions are:

u=f(x) when t=0
u=0 when x=0, independent of t
u=V when x=L, independent of t, where V and L are both non-zero constants.

Is there a general solution to this, or do I need to solve it numerically?

I tried solving it by separation of variables similar to what was done here:
http://en.wikipedia.org/wiki/Heat_eq...Fourier_series

But in the link the last boundary condition is u=0 when x=L, which allows us to conclude something about lambda that I can't with a non-zero boundary condition.

If it is separable, we have u(x,t)=X(x)T(t)=V when x=L independent of t. Following the method used in the link, to the point where we apply the different boundary conditions and my problem becomes different than what is in the link:

u(x,t) = [B sin(rootlambda x)] * [A e^-(lambda alpha t)]

Since u(L,t) is a non-zero constant independent of t, I think that I need to conclude A=0 leading to a trivial solution. I got this far with it before and concluded that the solution was not separable and therefore can only be evaluated numerically. Do you agree, or am I missing something?

Here is a link to another forum where I am looking for the same answer:
http://www.reddit.com/r/cheatatmathh.../dudt_ad2udy2/
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mfb
mfb is offline
#2
Jan29-13, 01:09 PM
Mentor
P: 10,766
The transformation ##u \to u-\frac{V}{L}x## will not influence your differential equation, but give u(L)=0. Solve the equation for the modified u, and add that linear part afterwards?
timsea81
timsea81 is offline
#3
Jan29-13, 01:35 PM
P: 90
Quote Quote by mfb View Post
The transformation ##u \to u-\frac{V}{L}x## will not influence your differential equation, but give u(L)=0. Solve the equation for the modified u, and add that linear part afterwards?
Interesting. Let's develop that a little:

u=u(x,t)
du/dt = A*(d^2u/dx^2)
u(x,0) = f(x)
u(0,t) = 0
u(L,t) = V

Let w(x,t) = u(x,t) - (V/L)*x

Now,
w(x,0) = u(x,t) - (V/L)*x = f(x)-[(V/L)*x]
We can just call this g(x) so that's fine

w(0,t) = u(0,t) - (V/L)*0 = 0-0 = 0
w(L,t) = u(L,t) - (V/L)*L = V-V = 0

dw/dt = du/dt - 0 = du/dt
dw/dx = du/dx - (V/L)
d^2w/dt^2 = d^2u/dt^2 - 0 = d^2u/dt^2

du/dt = A*(d^2u/dx^2) is given in the problem, and from above, this means that

dw/dt = A*(d^2w/dx^2)

So I solve dw/dt = A*(d^2w/dx^2) the same way they do for the heat equation on wikipedia, then use the relation

u(x,t) = w(x/t) + (V/L)*x

To find u(x,t)

I think that will work! Thanks!


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