Results 1 to 5 of 5

Thread: Non-Homogenous Heat Equation

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    10

    Non-Homogenous Heat Equation

    Consider the following non-homogenous heat equation on $\displaystyle 0 \leq x \leq \pi$

    $\displaystyle u_t = k u_{xx} - 1$ with $\displaystyle u(x,0) = 0, u(0,t) = 0, u(\pi, t) = 0$

    Find a solution of the form

    $\displaystyle \displaystyle \sum_1^{\infty} b_n(t) \phi_n (x)$

    where $\displaystyle \phi_n(x)$ are the eigenfunctions of an appropriate homogenous problem, and find explicit expressions for $\displaystyle b_n(t)$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Aug 2010
    Posts
    10
    So I think

    $\displaystyle \phi_n(x) = \sin \frac{n \pi x}{L}$

    so I find solutions in the form

    $\displaystyle \displaystyle u(x,t) = \sum_1^{\infty} b_n (t) \ \sin \frac{n \pi x}{L}$

    Am I on the right track? Is the eigenfunction correct?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    You are on the right track. You'll need a Fourier sine series for -1 and then sub in your expansion and equate coefficients.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2010
    Posts
    10
    Quote Originally Posted by JoernE View Post
    So I think

    $\displaystyle \phi_n(x) = \sin \frac{n \pi x}{L}$

    so I find solutions in the form

    $\displaystyle \displaystyle u(x,t) = \sum_1^{\infty} b_n (t) \ \sin \frac{n \pi x}{L}$

    Am I on the right track? Is the eigenfunction correct?
    Okay, ignore this last post. I have a different method....

    Let $\displaystyle u = w(x,t) + v(x)$

    Then

    $\displaystyle \displaystyle \frac{\partial u}{\partial t} = \frac{\partial w}{\partial t}$ and $\displaystyle \displaystyle \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 v}{\partial x^2}$

    Sub these back into the PDE to obtain

    $\displaystyle \displaystyle \frac{\partial w}{\partial t} = k \frac{\partial^2 w}{\partial x^2} + k \frac{\partial^2 v}{\partial x^2} - 1 $

    Choose $\displaystyle \displaystyle k \frac{\partial^2 v}{\partial x^2} - 1 = 0 $

    So

    $\displaystyle \displaystyle V'' = \frac{1}{k}, \ V(0) = V(\pi) = 0$

    $\displaystyle \displaystyle V' = \frac{1}{k} + A$

    $\displaystyle \displaystyle V = \frac{1}{k} + Ax + B$


    $\displaystyle \displaystyle V(0) = \frac{1}{k} + B = 0 \ \Rightarrow \ B = - \frac{1}{k} $

    $\displaystyle \displaystyle V(\pi) = \frac{1}{k} + A \pi + B \ \Rightarrow \ A = -\frac{1}{k \pi} - \frac{B}{\pi}$

    $\displaystyle \displaystyle V(x) = 0$??

    Also since $\displaystyle \displaystyle k \frac{\partial^2 v}{\partial x^2} - 1 = 0 $ we have

    $\displaystyle \displaystyle \frac{\partial w}{\partial t} = k \frac{\partial^2 w}{\partial x^2}$ which is homogeneous.

    $\displaystyle \displaystyle u(0, t) = v(0) + w(0, t) = 0$

    $\displaystyle \displaystyle \Rightarrow \ w(0,t) = 0$

    $\displaystyle \displaystyle u(x, 0) = v(x) + w(x,0) = 0$

    $\displaystyle \displaystyle \Rightarrow \ w(x,0) = 0$

    $\displaystyle \displaystyle u(\pi, t) = v(\pi) + w(\pi, t) = 0$

    $\displaystyle \displaystyle \Rightarrow \ w(\pi, t) = 0$

    Then we have

    $\displaystyle \displaystyle w_t = k w_{xx}$ with $\displaystyle \displaystyle w(x,o) = 0, w(0,t) = 0, w(\pi, t) = 0$.

    Since this is homogenous, we can solve by letting

    $\displaystyle \displaystyle w = XT$

    $\displaystyle \displaystyle \frac{T'}{kT} = \frac{X''}{X} = - \lambda$

    $\displaystyle \displaystyle X = A \cos(\sqrt{\lambda}t) + B \sin(\sqrt{\lambda}t)$

    $\displaystyle \displaystyle X(0) = A = 0$

    but now I get a little lost.....I think I know where to go, just not exactly how to get there. Already have calculated that $\displaystyle v(x) = 0$, which will lead to what I need to prove (I think).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2010
    Posts
    10
    Quote Originally Posted by JoernE View Post

    $\displaystyle \displaystyle X = A \cos(\sqrt{\lambda}t) + B \sin(\sqrt{\lambda}t)$
    This should be $\displaystyle \displaystyle X = A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x)$

    so

    $\displaystyle \displaystyle X(\pi) = B \sin (\sqrt{\lambda} \pi) = 0$

    $\displaystyle \displaystyle \ \sqrt{\lambda} \pi = \frac{n \pi }{2}$

    $\displaystyle \displaystyle \ \sqrt{\lambda} = \frac{n}{2}$

    $\displaystyle \displaystyle \ \lambda_n = \big( \frac{n}{2} \big)^2$ for $\displaystyle n = 1,2,...$

    So

    $\displaystyle \displaystyle \phi_n(x) = \sin \big(\frac{n \pi x}{2} \big) $

    and so

    $\displaystyle \displaystyle u(x,t) = v(x) + \sum_1^{\infty} b_n \sin \big(\frac{n \pi}{2} \big) $

    $\displaystyle \displaystyle \ = \displaystyle \sum_1^{\infty} b_n(t) \phi_n (x)
    $

    and I think

    $\displaystyle \displaystyle \ b_n = \frac{1}{\pi} \int_0^{\pi} \sin(\sqrt{\lambda} x) dx$

    Could someone please tell me if this is now correct?

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. homogenous equation
    Posted in the Differential Equations Forum
    Replies: 12
    Last Post: Feb 22nd 2011, 02:06 AM
  2. Tricky homogenous equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Sep 16th 2010, 07:49 PM
  3. Heat problem with a heat equation
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Nov 25th 2009, 09:40 AM
  4. Homogenous Equation Help
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Sep 27th 2009, 11:39 AM
  5. Two dimensional, non-homogenous heat equation
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: Jun 29th 2009, 07:45 AM

Search Tags


/mathhelpforum @mathhelpforum