# Math Help - PDE - Fourier Method

1. ## 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?

2. Originally Posted by walter9459
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 $\frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial y} = u$. By the method of Fourier let $u(x,y) = X(x)Y(y)$ and so $\frac{X''}{X} - \frac{Y'}{Y} = 1$. Thus, $\frac{X''}{X}= 1 + \frac{Y''}{Y}$. The LHS is a function of $x$ and RHS is a function of $y$ which means that $\frac{X''}{X} = 1 + \frac{Y'}{Y} = k$ where $k$ is some real number. This leads us to the equations, $X '' - k X = 0$ and $Y' = (k-1)Y$.
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In the second equation, $Y' = (k-1)Y$ the solution is $Y = ce^{(k-1)y}$ where $c$ is any real number. In the first equation, $X '' - k X = 0$ things are more serious. There are three possible cases: $k>0,k<0,k=0$. Two of these cases would lead to only trivial solutions ( $k>0,k=0$), I leave that up to you to try to do. Only in the case $k<0$ do we get interesting solutions. If $k<0$ then $X = a\sin(\sqrt{-k}x) + b\cos(\sqrt{-k}x)$. Thus, the product solutions are $(a\sin (\sqrt{-k}x)+b\cos (\sqrt{-k}x))e^{(k-1)y}$ (we absorbed the constant $c$). We are told that this should be zero when $x=0$, substiting $x=0$ we get $be^{(k-1)y} = 0 \implies b=0$. Thus, the solutions must have the form $a\sin(\sqrt{-k}x) e^{(k-1)y}$. Then we are told at $x=\pi$ we get zero, so $a\sin (\sqrt{-k}\pi) e^{(k-1)y} = 0 \implies \sqrt{-k}\pi = \pi n, n\in \mathbb{Z}$. Therefore, $k=-n^2,n>0$. Therefore, the functions $a_n \sin (nx) e^{-(n^2+1)y}$ satisfy the PDE and the boundary value problem.
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The idea is that $\sum_{n=1}^{\infty} a_n \sin(nx) e^{-(n^2+1)y}$ solves this PDE and we try to pick $a_n$ so that at $y=1$ we get $x$ i.e. $\sum_{n=1}^{\infty} a_n e^{-(n^2+1)}\sin (nx) = x$. Can you solve for the coefficients $a_n$ to make this statement true?

3. 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!

4. Originally Posted by walter9459
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, $\sum_{n=1}^{\infty}a_n e^{-(n^2+1)}\sin (nx) = x$ for $0.

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

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

This simplifies to, $u(x,y) = 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}e^{(1+n^2)(1-y)}\sin (nx)$.

5. 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!

6. 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!

7. Originally Posted by walter9459
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 $x^2\frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2}\cdot \frac{\partial^2 u}{\partial t^2}$ with $u(a,t)=u(2a,t)=0$ is not sufficient infromation.
Are you told any initial conditions like $u(x,0)$ or $u_t(x,0)$?

8. No, I provided all the info the problem has.