Results 1 to 7 of 7

Math Help - IBV

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    IBV

    \text{D.E.}: \ u_t=ku_{xx}, \ 0<x<L, \ t>0
    \displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u_  x(L,t)=0\end{cases}
    \text{I.C.}: \ u(x,0)=f(x) \ \ 0<x<L

    u(x,t)=\varphi(x)T(t)

    \displaystyle\varphi(x)T'(t)=k\varphi''(x)T(t)\Rig  htarrow\frac{\varphi''}{\varphi}=\frac{T'}{Tk}=-\lambda

    \varphi''+\lambda\varphi=0\Rightarrow \varphi=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambd  a})

    \displaystyle T=Ce^{-\lambda kt}

    \varphi_1(0)=1 \ \varphi_2(0)=0
    \varphi_1'(0)=0 \ \varphi_2'(0)=1

    \varphi_1: \ A=1 \ \varphi_2: \ A=0
    \displaystyle\varphi_1': \ B=0 \ \varphi_2': \ B=\frac{1}{\sqrt{\lambda}}

    \displaystyle\varphi=A_2\cos(x\sqrt{\lambda})+B_2\  frac{\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}

    \displaystyle\varphi(0): \ A_2=0

    \displaystyle\varphi': \ B_2\cos(x\sqrt{\lambda})=0

    The eigenvalues must satisfy:

    \displaystyle\cos(x\sqrt{\lambda})=0

    \displaystyle\lambda_n=\pi^2\left(\frac{1+2n}{2L}\  right)^2, \ n\in\mathbb{Z}

    The eigenfunction is:

    \displaystyle\varphi_n(x)=\frac{2L}{\pi(1+2n)}\sin  \left(\frac{\pi x(1+2n)}{2L}\right), \ n\in\mathbb{Z}^+

    Is this much correct?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle u(x,t)=\sum_{n=1}^{\infty}b_n\frac{2L}{\pi(1+2n)}\  exp\left(-\frac{\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\rig  ht)\sin\left(\frac{\pi x(1+2n)}{2L}\right)

    I have shown \displaystyle\int_0^L\varphi_n(x)\varphi_m(x) \ dx=0, \ m\neq n.

    Before I solve for b_n is u(x,t) correct?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,725
    Thanks
    1478
    Quote Originally Posted by dwsmith View Post
    \text{D.E.}: \ u_t=ku_{xx}, \ 0<x<L, \ t>0
    \displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u_  x(L,t)=0\end{cases}
    \text{I.C.}: \ u(x,0)=f(x) \ \ 0<x<L

    u(x,t)=\varphi(x)T(t)

    \displaystyle\varphi(x)T'(t)=k\varphi''(x)T(t)\Rig  htarrow\frac{\varphi''}{\varphi}=\frac{T'}{Tk}=-\lambda

    \varphi''+\lambda\varphi=0\Rightarrow \varphi=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambd  a})

    \displaystyle T=Ce^{-\lambda kt}

    \varphi_1(0)=1 \ \varphi_2(0)=0
    \varphi_1'(0)=0 \ \varphi_2'(0)=1
    Where did \varphi_1 and \varphi_2 suddenly come from?
    Are they e^{\lambda kt}(Acos(x\sqrt{\lambda})) and e^{\lambda kt}(B sin(x\sqrt{\lambda}))?

    If so, you cannot, generally, do that. But here it does not hurt.

    \varphi_1: \ A=1 \ \varphi_2: \ A=0
    \displaystyle\varphi_1': \ B=0 \ \varphi_2': \ B=\frac{1}{\sqrt{\lambda}}

    \displaystyle\varphi=A_2\cos(x\sqrt{\lambda})+B_2\  frac{\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}
    All you really need is that \phi(0, t)= e^{-\lambda kt}(Acos(0)+ Bsin(0))= Ae^{-k\lambda kt}= 0. Since the exponential is never 0, A= 0.

    Then \phi(L, t)= e^{-\lambda kt}B cos(L\sqrt{\lambda})= 0. Again, the exponential is not 0 so we must have either B= 0, which would mean the solution was identically equal to 0, and so not an eigenfunction, or cos(L\sqrt{\lambda})= 0.

    \displaystyle\varphi(0): \ A_2=0

    \displaystyle\varphi': \ B_2\cos(x\sqrt{\lambda})=0

    The eigenvalues must satisfy:

    \displaystyle\cos(x\sqrt{\lambda})=0
    You mean cos(L\sqrt{\lambda})= 0

    \displaystyle\lambda_n=\pi^2\left(\frac{1+2n}{2L}\  right)^2, \ n\in\mathbb{Z}
    The eigenfunction is:

    \displaystyle\varphi_n(x)=\frac{2L}{\pi(1+2n)}\sin  \left(\frac{\pi x(1+2n)}{2L}\right), \ n\in\mathbb{Z}^+

    Is this much correct?

    Thanks.
    Yes, that is correct and your second post is correct.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by HallsofIvy View Post
    Where did \varphi_1 and \varphi_2 suddenly come from?
    Are they e^{\lambda kt}(Acos(x\sqrt{\lambda})) and e^{\lambda kt}(B sin(x\sqrt{\ambda}))?

    If so, you cannot, generally, do that. But here it does not hurt.
    My book solves all PDEs in the same manner. I don't know why it is done that way but it works in all cases.

    Here is what the book says about \varphi 1 and 2.

    The theorem in the theory of ODE which explicitly states the proposition above is: if the coefficients c_0(x), \ c_1(x), \ c_2(x) are continuous on the interval [a,b] and c_0(x) is different from zero throughout the interval, then for any fixed x_0 in the interval [a,b] and for any constants \alpha, \ \beta. The DE has one and only one solution \varphi(x)=\varphi(x,\lambda) satisfying at x_0 the initial conditions \varphi(x_0)=\alpha, \ \varphi'(x_0)=\beta.
    Moreover, \varphi(x,\lambda) and \varphi'(x,\lambda) are continuous and continuously differentiable functions of both x for a\leq x\leq b and all lambda.
    It is convenient to single out the two special solutions \varphi_1=\varphi_1(x,\lambda), \ \varphi_2=\varphi_2(x,\lambda) of which satisfy the DE
    \varphi_1(0)=1, \ \varphi_2(0)=0
    \varphi'_1 (0)=0, \ \varphi'_2(0)=1.

    We call 1 and 2 the basic solutions at x nought.
    Last edited by dwsmith; March 19th 2011 at 03:48 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle \int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=\int_0^Lb_m\sin^2\left(\frac{\pi x(1+2m)}{L}\right) \ dx

    \displaystyle\int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=b_m\frac{L}{2}\Rightarrow b_m=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx
    Last edited by dwsmith; March 20th 2011 at 11:29 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    For the case f(x)=1

    \displaystyle b_m=\frac{2}{L}\int_0^L\sin\left(\frac{\pi x(1+2m)}{L}\right) \ dx=\frac{2}{L}\left[\frac{-2L}{\pi(1+2m)}\cos\left(\frac{\pi x(1+2m)}{2L}\right)\right]_0^L

    \displaystyle b_m=\frac{4}{\pi(1+2m)}, \ \ m=n

    \displaystyle u(x,t)=\frac{8}{\pi^2}\sum_{n=0}^{\infty}\frac{L}{  (1+2n)^2}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)

    If everything above is correct, can I do this?

    \displaystyle\sum_0^{\infty}\frac{1}{(1+2m)^2}=\fr  ac{\pi^2}{8}

    \displaystyle  u(x,t)=\frac{8}{\pi^2}\sum_{n=0}^{\infty}\frac{1}{  (1+2n)^2}\cdot L\sum_{n=0}^{\infty}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi  x(1+2n)}{2L}\right)

    \displaystyle\Rightarrow u(x,t)=L\sum_{n=0}^{\infty}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi  x(1+2n)}{2L}\right)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Ok, I pretty sure the above answer is wrong.

    I now have:

    \displaystyle u(x,t)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{1+  2n}\exp\left[\frac{-\pi^2}{4}\left(\frac{1+2n}{L}\right)^2kt\right]\sin\left(\frac{\pi x(1+2n)}{2L}\right)

    How do I show this is true for all x with the IC:

    \displaystyle u(x,0)=\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{1}{1+  2n}\sin\left(\frac{\pi x(1+2n)}{2L}\right)=1
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum