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Math Help - PDE boundary conditions

  1. #1
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    PDE boundary conditions

    Hi, this is the question



    Consider
    u_t=u_xx 0<x<l, t>0 for u(x,t)
    with boundary conditions u(0,t) = f(t) and u(l,t) = g(t) for t>0
    and initial condition u(x,0) = h(x) for 0<x<l.

    Show that this problem can be transformed into a new equations problem with a new dependent varaible v(x,t) with a 2nd order partial differential equation, vanishing boundary conditions (v(0,t) = v(l,t) = 0, t>0) and an initial condition.

    Assume differentiability when needed, f and g are defined for all t needed.




    The main problem I have with this question is I'm not sure how to make the boundary conditions vanish...

    Thanks!!
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  2. #2
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    Quote Originally Posted by angels_symphony View Post
    Hi, this is the question



    Consider
    u_t=u_xx 0<x<l, t>0 for u(x,t)
    with boundary conditions u(0,t) = f(t) and u(l,t) = g(t) for t>0
    and initial condition u(x,0) = h(x) for 0<x<l.

    Show that this problem can be transformed into a new equations problem with a new dependent varaible v(x,t) with a 2nd order partial differential equation, vanishing boundary conditions (v(0,t) = v(l,t) = 0, t>0) and an initial condition.

    Assume differentiability when needed, f and g are defined for all t needed.
    Suppose that,
    \frac{\partial u}{\partial t} = \frac{\partial ^2 u}{\partial x^2}
    With boundary value problem u(0,t) = f(t) \mbox{ and }u(L,t) = g(t) \mbox{ for }t\geq 0.
    And initial value problem u(x,0) = h(x) \mbox{ for }0\leq x\leq L.

    Write the function u(x,t) as v(x,t)+w(x,t) such that v(x,t) satisfies the boundary value problem. For example, v(x,t) = f(t) + \frac{x}{L}[g(t) - f(t)]. And w(x,t) is the modified function that is needed to solve the equation.

    Substitute that into the equation,
    \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 w}{\partial x^2} = \frac{\partial v}{\partial t} + \frac{\partial w}{\partial t}.

    Thus,
    \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)] .
    With, w(0,t) = w(L,t) = 0 \mbox{ for }t\geq 0.
    And, w(x,0) = h(x) - v(x,0) = h(x) - f(0) + \frac{x}{L}[ g(0)-f(0)].

    Therefore, we have transformed an equation with a non-homogenous boundary value problem into an equation with a homogenous boundary value problem but it has an 'uglier' form. Meaning we have to solve the inhomogenous heat equation.

    Now there is a method for solving (with homogenous boundary),
    \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = \mu (x,y).
    For a given differenciable function \mu (x,y).
    But that is a totally different question.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Suppose that,
    \frac{\partial u}{\partial t} = \frac{\partial ^2 u}{\partial x^2}
    With boundary value problem u(0,t) = f(t) \mbox{ and }u(L,t) = g(t) \mbox{ for }t\geq 0.
    And initial value problem u(x,0) = h(x) \mbox{ for }0\leq x\leq L.

    Write the function u(x,t) as v(x,t)+w(x,t) such that v(x,t) satisfies the boundary value problem. For example, v(x,t) = f(t) + \frac{x}{L}[g(t) - f(t)]. And w(x,t) is the modified function that is needed to solve the equation.

    Substitute that into the equation,
    \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 w}{\partial x^2} = \frac{\partial v}{\partial t} + \frac{\partial w}{\partial t}.

    Thus,
    \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)] .
    With, w(0,t) = w(L,t) = 0 \mbox{ for }t\geq 0.
    And, w(x,0) = h(x) - v(x,0) = h(x) - f(0) + \frac{x}{L}[ g(0)-f(0)].

    Therefore, we have transformed an equation with a non-homogenous boundary value problem into an equation with a homogenous boundary value problem but it has an 'uglier' form. Meaning we have to solve the inhomogenous heat equation.

    Now there is a method for solving (with homogenous boundary),
    \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = \mu (x,y).
    For a given differenciable function \mu (x,y).
    But that is a totally different question.

    Hi,

    I'm not quite understanding this part:

    Thus,
    \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)] .
    With, w(0,t) = w(L,t) = 0 \mbox{ for }t\geq 0.
    And, w(x,0) = h(x) - v(x,0) = h(x) - f(0) + \frac{x}{L}[ g(0)-f(0)].

    Why do the boundaries becomes zero? Sorry I'm pretty new to this and I'm not really understanding my professors notes!
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  4. #4
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    Quote Originally Posted by angels_symphony View Post
    \frac{\partial^2 w}{\partial x^2} - \frac{\partial w}{\partial t} = f'(t) + \frac{x}{L} [ g'(t) - f'(t)] .
    This comes from v(x,t) = f(t) + \frac{x}{L}[g(t) - f(t)] and thus \frac{\partial^2 v}{\partial x^2} = 0 and \frac{\partial v}{\partial t} = f'(t) + \frac{x}{L}[g'(t) - f'(t)].

    With, w(0,t) = w(L,t) = 0 \mbox{ for }t\geq 0.
    Because v(x,t)+w(x,t) = u(x,t) thus v(x,t)+w(x,t) solve this equation. Note v(0,t) = f(t) which means v(0,t)+w(0,t) = u(0,t) \implies f(t) + w(0,t) = f(t) \implies w(0,t) = 0 for t\geq 0. Similarly w(L,t) = 0 for t\geq 0.

    And, w(x,0) = h(x) - v(x,0) = h(x) - f(0) + \frac{x}{L}[ g(0)-f(0)].
    Again u(x,t) = v(x,t) + w(x,t) thus u(x,0) = v(x,0) + w(x,0) \implies h(x) - v(x,0) = w(x,0). Note, v(x,0) = f(0) + \frac{x}{L}[g(0)-f(0)].
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