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Like Mickey said, there is brief explanation about the steady state solution at page 3. | Like Mickey said, there is brief explanation about the steady state solution at page 3. | ||
Revision as of 18:07, 1 December 2013
Homework 12 collaboration area
I am not sure how to start on problem number 10 on page 567. Any hint? Thanks!
From djkees:
I think you start at (7) on 560 and change the BC's, and that the initial temperature of the bar is 100.
From Mickey Rhoades Mrhoade
Look at the lecture notes from 11-20 about 1/4 of the way down. Dr. Bell shows you how to solve this problem. You set U(x,t) = u(x,t) - uinfinity U(x,t) is the homogeneous solution from problem #7 and the steady state will be a line from 100 to zero with 50 at the midpoint. Your uinfinity will be 50 at L/2 = 5 and then you just plug in for U(x,t) for x = 5 and the corressponding t values.
From Eun Young:
Like Mickey said, there is brief explanation about the steady state solution at page 3.
$ u_{\infty} = u(0,t) + \frac x L (u(L,t) - u(0,t)) $.
Let $ U(x,t) = u(x,t) - u_{\infty} $.
Then, you can see that U satisfies the heat equation with homogeneous boundary conditions.
I.e. $ \frac{\partial U}{\partial t} = c^2 \frac{\partial ^2 U}{\partial x^2} $
$ U(0,t) = U(L,t) = 0 $ .
You can compute an initial condition U(x,0) by letting t=0.
Then, now you can get U(x,t) using (9) and (10).
Since you know U(x,t) and $ u_{\infty} $ now, you can get u(x,t).
From Farhan: This might be a silly question: In the last step of finding a solution to a wave or heat equation, why do we take a SERIES of the eigen functions, and then incorporate the initial condition to get the solution of the entire problem. I know that, sum of the solutions (eigen functions) is also a solution to the PDE, but in the last step, what if we work with ONLY ONE eigen function and impose the initial condition? Will that be wrong?
Farhan, I think the series of the eigenfunctions is needed to satisfy both the boundary conditions and the initial conditions (as stated on p 548, a single solution will generally not satisfy the initial conditions). I think it would be hard to come up with a single function that satisfied both (other than the zero function). Please correct me if my thinking is wrong here! -Mjustiso
For number 11, page 566 (12.5, 12.6) There is a note in the that says "An is given by (2) in section 11.3" When I go to section 11.3 on page 492, I see (2) and it says y"+0.05y'+25y=r(t)
I am not sure how this applies to the problem. Maybe the reference is incorrect. Can anyone help on this one?
Al
From Eun Young:
I think it's a typo. See (5*) and (6*) in section 11.2 instead of (2) in section 11.3.
Thanks Eun,
Can you please give me a hint on how to apply this to the problem? I was thinking (5*) and (6*) would be A sub n (An) = something, so we could use "An" in the problem.
Al
From Mickey Rhoades Mrhoade
Al - Dr. Bell solves this exact problem at the end of the lecture notes from 11-18. -Mick
From Eun Young:
I think there is one more typo. $ u(x,t)=f(x) \Rightarrow u(x,0) = f(x) . $.
Show that u(x,t) is equal to the given form using separation of variables: u(x,t) = F(x)G(t).
Then, use u(x,0)=f(x) to see An is the coefficient of cosine fourier series of f(x) given by (5*) and (6*).
In regard to problem #4 on page 574 (12.7). How can we integrate (e^-v) Cos (pv)dv? I first looked to see if the functions were even or odd but e^-v is neither so no simplification there. Integration by parts yields another integral that needs integration by parts and it looks never ending. So I am thinking there must be an identity or another method to do this. Any ideas?
Al
From Jake Eppehimer:
If you do int by parts twice, you see that you get almost the same thing as the starting integral. So then you just move it to the other side of the equation and solve for it, as if the integral was a variable. A(w) should be 2/[pi(w^2+1)] and B(w) is 0.
