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?
You have . By the method of Fourier let and so . Thus, . The LHS is a function of and RHS is a function of which means that where is some real number. This leads us to the equations, and .
Originally Posted by walter9459
In the second equation, the solution is where is any real number. In the first equation, things are more serious. There are three possible cases: . Two of these cases would lead to only trivial solutions ( ), I leave that up to you to try to do. Only in the case do we get interesting solutions. If then . Thus, the product solutions are (we absorbed the constant ). We are told that this should be zero when , substiting we get . Thus, the solutions must have the form . Then we are told at we get zero, so . Therefore, . Therefore, the functions satisfy the PDE and the boundary value problem.
The idea is that solves this PDE and we try to pick so that at we get i.e. . Can you solve for the coefficients to make this statement true?
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!
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!
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!
No, I provided all the info the problem has.