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Math Help - PDE double checking, unsure of answer

  1. #1
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    PDE double checking, unsure of answer

    Hi,

    I've done this question but I'm unsure if I used the boundary conditions right. The question is:

    Find u(x,t) given u_xx=0 0<x<1, t>0 with boundary conditions u(0,t) = sin(t), u(1,t) = 0.5 t>0

    To solve this question I found u = cx + f(t) where f(t) is any differentiable equation of t by integrating twice. I then used the boundary conditions to find c in terms of t. Is this correct?

    Thank you!
    Last edited by angels_symphony; January 20th 2008 at 09:22 PM.
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  2. #2
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    What equation are you solving?
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    What equation are you solving?
    oh gosh sorry! the equation is u_xx=0
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    If u_xx = 0 then u_x = k + f(y) for any function f(y). That means u = kx + f(t) for any function f(t).

    The boundary condition say u(0,t)= sin t and u(1,t) = .5, thus, k*0 + f(t) = sin t and so f(t) = sin t. Thus, u(x,t) = kx + sin t using the second boundary condition we have u(1,t) = k + sin t = .5 for t>0, which seems impossible no constant k will make this possible because t varies.
    Yea, thats pretty much what I found, I ended up with a constant in terms of t, which is why I was unsure if I was using the boundary conditions correct. (First time solving this type of equation and I don't have any examples to follow from) But thank you anyways!
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  5. #5
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    I thought that maybe I was sepposed to use the boundary conditions to find a value of t...meaning that in f(t) t had to be something certain...or am I just going way off track?
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    If u_xx = 0 then u_x = k + f(y) for any function f(y). That means u = kx + f(t) for any function f(t).

    The boundary condition say u(0,t)= sin t and u(1,t) = .5, thus, k*0 + f(t) = sin t and so f(t) = sin t. Thus, u(x,t) = kx + sin t using the second boundary condition we have u(1,t) = k + sin t = .5 for t>0, which seems impossible no constant k will make this possible because t varies.
    u_{xx} = 0.

    Therefore u_x = f(t).

    Therefore u(x, t) = x f(t) + g(t).


    B.C. u(0,t)= \sin t: \, \sin t = g(t).

    Therefore u(x, t) = x f(t) + \sin t.


    B.C. u(1,t) = 0.5: \,  0.5 = f(t) + \sin t \therefore f(t) = 0.5 - \sin t.

    Therefore u(x, t) = x (0.5 - \sin t) + \sin t = \frac{x}{2} - x \sin t + \sin t.


    The checking (satisfies the P.D.E., satisfies the given B.C.'s) I leave to angels sympthony.

    Mind you, this is only how a *ahem* physicist would do it ......
    Last edited by mr fantastic; January 21st 2008 at 01:31 AM. Reason: Content unchanged. Changed formatting.
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