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Du/dt = A*d2 u/dy2 
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#1
Jan2913, 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 nonzero 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 nonzero 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 nonzero 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/ 


#2
Jan2913, 01:09 PM

Mentor
P: 11,631

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?



#3
Jan2913, 01:35 PM

P: 90

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 = 00 = 0 w(L,t) = u(L,t)  (V/L)*L = VV = 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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