| Line 4: | Line 4: | ||
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)? | 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 C's: u(0,t) = 0, u(L,t) = 0 | ||
| + | |||
| + | Initial C's: 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 An. | ||
| + | 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. Happy Thanksgiving. | ||
[[2010 MA 527 Bell|Back to the MA 527 start page]] | [[2010 MA 527 Bell|Back to the MA 527 start page]] | ||
Revision as of 20:14, 25 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 C's: u(0,t) = 0, u(L,t) = 0
Initial C's: 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 An. 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. Happy Thanksgiving.
