Difference between revisions of "HWK12MA527Fall2013" - Rhea

Latest revision as of 16:36, 3 December 2013

Homework 12 collaboration area

From djkees:

I am assuming on Q3 of page 574 that it is referencing (13) in 11.7?

I assumed the same, A(w) is of that form. -Mboersma

I am not sure how to start on problem number 10 on page 567. Any hint? Thanks!

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) - u_infinity  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 u_infinity 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 djkees:

I have a solution using the method presented here/in class, but when I run actual values into it, the midpoint temperature eventually reduces to 0. Shouldn't it be instead reducing to 50?

From czehrung:

djkees, did you recombine your solution of U(x,t) into the original "trick" equation given in the 12-2 lecture?

You should come up with something like u(x,t)= U(x,t)+u"infinity"(x). That solution should converge to 50 degrees at 5cm after a length of time.

From djkees:

Thank you for your recommendation, I see my error now. Thank you very much.

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.

From Collier

We are given $ f(x)=exp(-|x|) $ which is an even function. It is with cos(pv) for A(p), therefore we can simplify the integral to be twice the positive integral. At this point, the absolute value is not needed, because it is only being evaluated for positive v's. $ \int_0^\infty \! exp(-v)*cos(pv) \, \mathrm{d}v $. This integral is a pretty common definite integral and can be found in most integral tables, or you can use the indefinite form that is the last entry in the integral table on the inside front cover of our textbook.

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