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| − | You can solve for lambda in a way similar to the way Prof. Bell did it at the beginning of class on 10/20/10. From the original problem you get: r^2+8r+(16 | + | You can solve for lambda in a way similar to the way Prof. Bell did it at the beginning of class on 10/20/10. From the original problem you get: r^2+8r+(16+lamda)=0. So r = sqrt(-lambda)-4. As usual, the lambda >0 case gives you non-zero solutions. Since our root is complex of the form -4 plus/minus i*sqrt(lambda), where mu = sqrt(lambda). We can find lambda that allow non-zero solutions like in 7 by using the general form of: |
y = exp(-4*x)*(A*cos(mu*x)+B*sin(mu*x)). Plug in boundary conditions and Ta-Da. | y = exp(-4*x)*(A*cos(mu*x)+B*sin(mu*x)). Plug in boundary conditions and Ta-Da. | ||
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[[2010 MA 527 Bell|Back to the MA 527 start page]] | [[2010 MA 527 Bell|Back to the MA 527 start page]] | ||
Latest revision as of 10:00, 27 October 2010
Homework 9 Collaboration Area
Click on
for some hints about Legendre Polynomials.
Question Page 597, Problem 5:
What do we do with the x in the first term of this problem?
Answer: When you do a Laplace transform wrt t, the x floats along like when you do d/dt(x*t).
Additional question: I get to this point after taking the Laplace Transform:
d/dx*W(x,s) + s/x*W(x,s) = 1/s^2.
Is this correct?
Answer: Yes, that's what I got. Check out lecture 23 around the 28 minute mark to see what to do next.
Message from Bell: When you solve that ODE as a first order linear ODE in the x variable, you should get an integrating factor of x^s. Remember that the arbitrary constant that you get in the solution might depend on s. So I like to write C(s) for the arbitrary constant. You will need to show that C(s) must be zero by showing that
W(0,s)=0.
This follows from one of the boundary conditions given in the problem.
Page 209 Problem 5: I've done this problem 5 times over, (yes, again after reading Prof. Bell's notes) and I keep getting 2/3 out of the integral rather than the necessary 0. Whether it's cos^3 or x^3, I don't see any way to get anything other than Pm*Pn/3*(-1 - 1) = 2/3 * PnPm rather than the 0 we're looking for. Or do I just not understand orthogonality??
New Question: How is everyone else "showing orthogonality?" Are you evaluating all the available (11) polynomials, doing just a few and making a statement, or have you figured out how to integrate (11) itself?
From Bell: My hint about p. 209, problem 5 was to make a change of variables,
x = cos theta.
After that, you'll reduce the orthogonality of P_n(cos theta) with respect to the weight function Sin(theta) on [0,pi] to the orthogonality of the Legendre Polynomials on [-1,1]. The orthogonality of the Legendre Polynomials on the interval [-1,1] is a given fact, so you shouldn't be computing actual integrals to verify the orthogonality for the first few Legendre Polynomials listed on p. 180.
Question Page 209 Problem 7:
Where does the $ \pi $ come from in this solution?
Answer: When you do the positive lambda case, you get A = 0 and let B = 1 => Sin(5*mu) = 0. If mu = m*pi/5, this equation is true. I let B=1 because we cannot have both A and B = 0.
Question: Page 209, Problem 17: For the given equation, shouldn't p=1, q=16, r=1? These values differ from the textbook's values.
Answer: If that were the case, then the equation would be
[py']' + (q+ lambda r) y =
[1 y']' + (16 + lambda) y =
y" + (16 + lambda) y = 0
and it ain't. You need to use problem 6 in the same section to get p,q, and r.
Question: Why isn't q=pg=16*exp(8x)?
Answer: Here is the idea of problem 6. We have the equation
$ y'' + 8 y' + (\lambda + 16)y=0. $
Multiply that equation by p(x). You get
$ py'' + 8p y' + (\lambda p+ 16p)y=0. $
If this were in Sturm-Liouville form, it would look like
$ [py']'+ (q + \lambda r) y = $
$ py'' + p'y' + (q+ \lambda r) y = 0. $
By comparing those two, we see that we need
$ p'=8p $
and q=16p and r=p. Solving the ODE for p yields
$ p(x)=e^{8x}. $
(We can take the arbitrary constant in the solution to be a convenient value because we just want one p(x) that has this property.)
Finally, we get
$ p(x)=e^{8x},\quad q(x)=16e^{8x},\quad\text{and }r(x)=e^{8x}. $
Hmmm. I see what you mean. I think the answer in the back of the book is wrong.--Steve Bell 12:07, 23 October 2010 (UTC)
p. 216, #s 1 and 3:
I am using the hints given, but I'm still not sure I'm doing this correctly. For example, for #1, I've calculated c4 as c4 = [((2*4)+1)/2] * integral from -1 to 1 of (7x^4-6x^2)(P_4(x)) dx, where P_4= (1/8)(35x^4-30x^2+3), and then I would do something similar for C3, C2, C1, and C0. But if I find C4 this way, I'll get an answer where there's an x^9 term, and I don't see how to get an answer in terms for P_4 and P_1, like there is in the back of the book.
Answer: These are definite integrals carried out from -1 to 1, so you should get a numerical value for the solution, which will be the coefficient cm of the mth legendre polynomial. For instance, for m=0, you'll have c_0 =(2*0+1)/2 * integral from -1 to 1 of (7x^4-6x^2)*P_0(x)dx, where P_0(x)=1. The solution of that integral is -3/5, which matches the value for the coefficient of P_0(x) in the back of the book. --Jkusnick 08:47, 25 October 2010 (UTC)
p. 209 #7 & #17:
To verify orthogonality, do you just use the Theorem, or do you have to do the integral?
Answer: The theorem says the eigenfunctions are orthogonal. However, to VERIFY that, you'll have to compute the integrals.
p. 209, #17
Answer:
You can solve for lambda in a way similar to the way Prof. Bell did it at the beginning of class on 10/20/10. From the original problem you get: r^2+8r+(16+lamda)=0. So r = sqrt(-lambda)-4. As usual, the lambda >0 case gives you non-zero solutions. Since our root is complex of the form -4 plus/minus i*sqrt(lambda), where mu = sqrt(lambda). We can find lambda that allow non-zero solutions like in 7 by using the general form of:
y = exp(-4*x)*(A*cos(mu*x)+B*sin(mu*x)). Plug in boundary conditions and Ta-Da.
