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Thread: PDE - Fourier Method

  1. #1
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    PDE - Fourier Method

    Obtain all solutions of the equation partial ^2 u/partial x^2 - partial u/partial y = u of the form u(x,y)=(A cos alpha x + B sin alphax)f(y) where A, B and alpha are constants. Find a solution of the equation for which u=0 when x=0; u=0 when x = pi, u=x when y=1.

    The solution is u = -2 summuation from n=1 to infinity (((-1)^n)/n)e^((1+n)(1-y)) sin nx.

    I believe the next step is to use u(x,Y) = X(x)Y(y) so the equation then becomes (1/x) partial ^2 X/partial x^2 - (1/y)partial Y/partial y = u. Then I get lost, can I get some help on how I would solve this problem?
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  2. #2
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    Quote Originally Posted by walter9459 View Post
    Obtain all solutions of the equation partial ^2 u/partial x^2 - partial u/partial y = u of the form u(x,y)=(A cos alpha x + B sin alphax)f(y) where A, B and alpha are constants. Find a solution of the equation for which u=0 when x=0; u=0 when x = pi, u=x when y=1.

    The solution is u = -2 summuation from n=1 to infinity (((-1)^n)/n)e^((1+n)(1-y)) sin nx.

    I believe the next step is to use u(x,Y) = X(x)Y(y) so the equation then becomes (1/x) partial ^2 X/partial x^2 - (1/y)partial Y/partial y = u. Then I get lost, can I get some help on how I would solve this problem?
    You have $\displaystyle \frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial y} = u$. By the method of Fourier let $\displaystyle u(x,y) = X(x)Y(y)$ and so $\displaystyle \frac{X''}{X} - \frac{Y'}{Y} = 1$. Thus, $\displaystyle \frac{X''}{X}= 1 + \frac{Y''}{Y}$. The LHS is a function of $\displaystyle x$ and RHS is a function of $\displaystyle y$ which means that $\displaystyle \frac{X''}{X} = 1 + \frac{Y'}{Y} = k $ where $\displaystyle k$ is some real number. This leads us to the equations, $\displaystyle X '' - k X = 0$ and $\displaystyle Y' = (k-1)Y$.
    ---
    In the second equation, $\displaystyle Y' = (k-1)Y$ the solution is $\displaystyle Y = ce^{(k-1)y}$ where $\displaystyle c$ is any real number. In the first equation, $\displaystyle X '' - k X = 0$ things are more serious. There are three possible cases: $\displaystyle k>0,k<0,k=0$. Two of these cases would lead to only trivial solutions ($\displaystyle k>0,k=0$), I leave that up to you to try to do. Only in the case $\displaystyle k<0$ do we get interesting solutions. If $\displaystyle k<0$ then $\displaystyle X = a\sin(\sqrt{-k}x) + b\cos(\sqrt{-k}x)$. Thus, the product solutions are $\displaystyle (a\sin (\sqrt{-k}x)+b\cos (\sqrt{-k}x))e^{(k-1)y}$ (we absorbed the constant $\displaystyle c$). We are told that this should be zero when $\displaystyle x=0$, substiting $\displaystyle x=0$ we get $\displaystyle be^{(k-1)y} = 0 \implies b=0$. Thus, the solutions must have the form $\displaystyle a\sin(\sqrt{-k}x) e^{(k-1)y}$. Then we are told at $\displaystyle x=\pi$ we get zero, so $\displaystyle a\sin (\sqrt{-k}\pi) e^{(k-1)y} = 0 \implies \sqrt{-k}\pi = \pi n, n\in \mathbb{Z}$. Therefore, $\displaystyle k=-n^2,n>0$. Therefore, the functions $\displaystyle a_n \sin (nx) e^{-(n^2+1)y}$ satisfy the PDE and the boundary value problem.
    ---

    The idea is that $\displaystyle \sum_{n=1}^{\infty} a_n \sin(nx) e^{-(n^2+1)y}$ solves this PDE and we try to pick $\displaystyle a_n$ so that at $\displaystyle y=1$ we get $\displaystyle x$ i.e. $\displaystyle \sum_{n=1}^{\infty} a_n e^{-(n^2+1)}\sin (nx) = x$. Can you solve for the coefficients $\displaystyle a_n$ to make this statement true?
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  3. #3
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    Sorry, but I got a little confused on calculating a_n.
    a_n = 2/pi integral o to pi f(x) sin rx dx
    a_n = 2/pi inegral o to pi sin (nx)e^((1+n^2))sin(rx)dx


    I know from what our teacher showed us in class but did not understand how he did this that e^((1=n^2)y) goes to other side of the equal side so that you have
    a_n e^((1+n^2)y) = 2/pi integral o to pi sin (nx) sin (rx) dx
    a_n = 2/pi (e^(1+n^2))/e^((1+n^2)y) integral o to pi sin nx sin rx dx
    a_n = 2/pi e^((1+n^2) (1-y)) integral o to pi sin nx sin rx dx

    Please let me know if I am on the right track!

    Thanks!
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  4. #4
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    Quote Originally Posted by walter9459 View Post
    Sorry, but I got a little confused on calculating a_n.
    a_n = 2/pi integral o to pi f(x) sin rx dx
    a_n = 2/pi inegral o to pi sin (nx)e^((1+n^2))sin(rx)dx
    You want to get, $\displaystyle \sum_{n=1}^{\infty}a_n e^{-(n^2+1)}\sin (nx) = x$ for $\displaystyle 0<x<\pi$.

    The Fourier sine series for $\displaystyle f(x) = x$ on $\displaystyle (0,\pi)$ is $\displaystyle \sum_{n=1}^{\infty}b_n \sin (nx)$ where $\displaystyle b_n = \frac{2}{\pi}\int_0^{\pi} x\sin(nx)dx$.
    Therefore, $\displaystyle b_n = (-1)^{n+1}\cdot\tfrac{2}{n}$ which means $\displaystyle a_n e^{-(n^2+1)} = (-1)^{n+1}\cdot \tfrac{2}{n} \implies a_n = (-1)^{n+1} \cdot e^{(n^2+1)}\cdot \tfrac{2}{n}$.

    The solution is therefore, $\displaystyle u(x,y) = 2\sum_{n=1}^{\infty} (-1)^{n+1} \cdot e^{(n^2+1)}\cdot \tfrac{1}{n}e^{-(n^2+1)y}$

    This simplifies to, $\displaystyle u(x,y) = 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}e^{(1+n^2)(1-y)}\sin (nx)$.
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  5. #5
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    Thank you! I kept at it after I submitted my reply and I finally figured it out. But I was hoping you would reply so I could verify if the way I did it was correct and it was so it at least tells me I am on the right track!

    Thank you! All your help is greatly appreciated!
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  6. #6
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    I hope I am not breaking the rules by asking you questions regarding my other problems. But I do not want you to providing me the answer but just clear up some questions I have. The problem is x^2partial ^2u/partial^2x = 1/c^2partial ^2u/partial^2t c=constant, u=0 for x=a and x=2a. w^2=c^2(lamda^2 + 1/4), lamda= (n*pi)/loge 2, n being a positive integer. U(x,t)=sin[lamda loge (x/a)](x/a)^1/2(A cos wt + B sin wt).

    Finally I attempted the first step and I came up with x^2 X" = (1/c^2)X = k^2 and T" = T = k^2. Which then become x^2X" - (1/c^2)(k^2)X = 0 and T" - (k^2)T = 0.

    Second step is
    k=0 -------> X(x) = A(x) + B
    u(a,t) = (A(a) + B)y = 0
    --------------> A(a)y + By = 0 y<>0 --------> A= -B.

    u(2a,t)= (A(2a) + B)y = 0
    --------------> A(2a)y + By = 0
    A(2a)y - A(2a)y = 0 -------> A = 0

    Am I correct so far?

    Thank you again for all your help!
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  7. #7
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    Quote Originally Posted by walter9459 View Post
    I hope I am not breaking the rules by asking you questions regarding my other problems. But I do not want you to providing me the answer but just clear up some questions I have. The problem is x^2partial ^2u/partial^2x = 1/c^2partial ^2u/partial^2t c=constant, u=0 for x=a and x=2a. w^2=c^2(lamda^2 + 1/4), lamda= (n*pi)/loge 2, n being a positive integer. U(x,t)=sin[lamda loge (x/a)](x/a)^1/2(A cos wt + B sin wt).
    Solving $\displaystyle x^2\frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2}\cdot \frac{\partial^2 u}{\partial t^2}$ with $\displaystyle u(a,t)=u(2a,t)=0$ is not sufficient infromation.
    Are you told any initial conditions like $\displaystyle u(x,0)$ or $\displaystyle u_t(x,0)$?
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  8. #8
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    No, I provided all the info the problem has.
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