| Line 6: | Line 6: | ||
Answer: | Answer: | ||
| − | Boundary | + | Boundary Conditions: u(0,t) = 0, u(L,t) = 0. In this problem, L=1. |
| − | Initial | + | Initial Conditions: u(x,0)= f(x), which can be seen from the diagram as |
| − | + | ||
| − | You'll get an ugly integral evaluation but most terms cancel and it leaves you with 3 sine terms that then go into the fourier series | + | f(x) = x - 1/4 for 1/4 < x < 1/2 and f(x) = -x + 3/4 for 1/2 < x < 3/4 |
| + | |||
| + | so you'll have to split up the integral when calculating A_n. | ||
| + | And yes, the last Initial Condition is | ||
| + | |||
| + | d(u)/dt(x,0) = g(x) = 0. | ||
| + | |||
| + | You'll get an ugly integral evaluation but most terms cancel and it leaves you with 3 sine terms that then go into the fourier series. | ||
| Line 18: | Line 24: | ||
How do we show that the constant = B^4 without any boundary conditions to work with? | How do we show that the constant = B^4 without any boundary conditions to work with? | ||
| + | Answer. The beta to the fourth power is just a way to name the positive | ||
| + | constant lambda to make the solutions easier to write. There will also | ||
| + | be the cases lambda=0 and lambda negative (minus beta to the fourth) to | ||
| + | deal with. You won't use boundary conditions to eliminate solutions | ||
| + | until problem 15. | ||
Question, Page 552, Problem 5: | Question, Page 552, Problem 5: | ||
Revision as of 06:46, 28 November 2010
Homework 13 collaboration area
Question, Page 546, Problem 9:
Are the Boundary Conditions u(1/4,t) = 0 and u(3/4,t)=0 or u(0,t)=1/4 and u(L,t)=3/4? Also, can anyone help explain what f(x) is in this problem? I think g(x) = 0, but I'm not sure about f(x)?
Answer: Boundary Conditions: u(0,t) = 0, u(L,t) = 0. In this problem, L=1.
Initial Conditions: u(x,0)= f(x), which can be seen from the diagram as
f(x) = x - 1/4 for 1/4 < x < 1/2 and f(x) = -x + 3/4 for 1/2 < x < 3/4
so you'll have to split up the integral when calculating A_n. And yes, the last Initial Condition is
d(u)/dt(x,0) = g(x) = 0.
You'll get an ugly integral evaluation but most terms cancel and it leaves you with 3 sine terms that then go into the fourier series.
Question, Page 547, Problem 15:
How do we show that the constant = B^4 without any boundary conditions to work with?
Answer. The beta to the fourth power is just a way to name the positive constant lambda to make the solutions easier to write. There will also be the cases lambda=0 and lambda negative (minus beta to the fourth) to deal with. You won't use boundary conditions to eliminate solutions until problem 15.
Question, Page 552, Problem 5:
How do we show Pn? I think I understand that this is part of calculating lambdas using the Sturm-Louiville, but I haven't been able to figure it out.
Question, Page 548, Problem 16:
What about f(x) for this problem? I am really having a hard time identifying the f(x) for these problems. (Actually, this entire section in general) Does anybody know of a good reference for example problems?
