Register to reply

Du/dt = A*d2 u/dy2

by timsea81
Tags: du or dt, u or dy2
Share this thread:
timsea81
#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/
Phys.Org News Partner Science news on Phys.org
Scientists discover RNA modifications in some unexpected places
Scientists discover tropical tree microbiome in Panama
'Squid skin' metamaterials project yields vivid color display
mfb
#2
Jan29-13, 01:09 PM
Mentor
P: 12,053
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
#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!


Register to reply