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### Messages - James McVittie

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1
##### Home Assignment 4 / Re: Problem 1
« on: October 24, 2012, 09:30:43 PM »
Solution to Problem 1(d)

To show that eigenfunctions corresponding to different eigenvalues are orthogonal, we evaluate the following:

$$(\lambda_{n}-\lambda_{m})\intop_{0}^{l}X_{n}(x)X_{m}(x)dx$$

Notice that we can make a simple substitution, apply the Fundamental Theorem of Calculus using the boundary conditions. Then,

$$(\lambda_{n}-\lambda_{m})(X_{n}(x)X_{m}(x))=X_{n}"(x)X_{m}-X_{n}(x)X"_{m}(x)=(X_{n}'(x)X_{m}(x)-X_{n}(x)X'_{m}(x))'$$

Plugging into the original integral, we obtain:

$$\intop_{0}^{l}(X_{n}'(x)X_{m}(x)-X_{n}(x)X'_{m}(x))'dx=X_{n}'(l)X_{m}(l)-X_{n}(l)X'_{m}(l)-X_{n}'(0)X_{m}(0)+X_{n}(0)X'_{m}(0)=0$$

Therefore, the eigenfunctions corresponding to different eigenvalues are orthogonal.

2
##### Home Assignment 4 / Re: Problem 3
« on: October 22, 2012, 11:58:32 AM »
Are we allowed to assume that solutions are real or must we always assume in greatest generality that the solution could be complex?
Thanks

3
##### Home Assignment 4 / Problem 3
« on: October 20, 2012, 10:01:33 PM »
What is implied by the word "weird", is it just something unexpected that comes up or something that hasn't been discussed in the course? Thanks!

4
##### Home Assignment 4 / Re: Problem 1
« on: October 20, 2012, 09:07:18 AM »
What does OX stand for?

5
##### Misc Math / Lecture Notes 7 - Code problem
« on: October 13, 2012, 05:56:12 PM »
In Lecture Notes 7 - 1D Wave Equation: IBVP, there was code below equation (10) and (15) that didn't get turned into text. It was $Oct$ and (\ref{eq-4}), what do these stand for? Thank you!

6
##### APM346 Misc / Homework Assignment 4
« on: October 10, 2012, 01:55:34 PM »
Homework 4 was taken down off the website. Is it still due next week or was it only meant as a set of review questions.
Thanks!

7
##### APM346 Misc / HW Solutions Posting on Forum
« on: October 02, 2012, 02:08:48 PM »
I was just wondering what the policy was about posting solutions to problem sets on the forum. On the general announcement, it indicates that posting solutions is to be done no earlier than 1 day after the deadline (Tuesday night), yet all the solutions are already posted.

8
##### Home Assignment 2 / Re: problem 1 typo?
« on: September 29, 2012, 08:13:56 AM »
Can we assume that for part (C) of Problem 1 that the Cauchy conditions are evenly reflected for x < 0?

9
##### APM346 Misc / TA office hours
« on: September 28, 2012, 02:20:00 PM »
What are the TA office hours? On the course outline they are still listed as TBD....

10
##### Misc Math / Wave Equation Solution over R
« on: September 26, 2012, 02:02:55 PM »
(With all appropriate boundary conditions defined) When the wave equation is defined for x>0, we can find a solution to the wave equation for x < ct and 0 < x < ct. If the equation is defined for all x, how do the solutions change?

11
##### Misc Math / Re: Heat Wave Equation Solution Integral
« on: September 26, 2012, 12:41:27 PM »
Sorry I meant the Heat Equation. What does the integral of the solution to that equation represent?

12
##### Misc Math / Heat Wave Equation Solution Integral
« on: September 25, 2012, 08:04:29 PM »
After we determined the solution to the heat wave equation, we took the integral from negative infinity to some x. The solution was in the form of a Normal Distribution function where in statistics the integral indicates the probability that we randomly choose a value within that interval. I was wondering what are the physical implications of this same integral in the wave equation? Does the area under the solution indicate the energy in the system up to that point or does it have some other meaning?

13
##### Home Assignment 1 / Re: Problem 3
« on: September 25, 2012, 07:29:46 AM »
Find the solution:

$u_{x}+3u_{y}=xy$

$u_{x=0}=0$

By examining Integral Lines:

$\frac{1}{dx}=\frac{3}{dy}=\frac{xy}{du}$ <very ugly form. Never use it $\color{blue} {\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}}$

Then from the first equality:

$3x=y+C$ where C is some constant.

$y=3x-C$

Then again from the Integral Lines:

$dx(xy)=du$

$dx(x(3x-C))=du$

$u=x^{3}-\frac{C}{2}x^{2}+C_{1}$

Then by the initial condition:

$u(x=0)=0$

$C_{1}=0$

Therefore,$C=3x-y$

Therefore,

$u(x,y)=x^{3}-\frac{1}{2}x^{2}C$

$u(x,y)=x^{3}-\frac{1}{2}x^{2}(3x-y)$

$u(x,y)=-\frac{1}{2}x^{3}+\frac{1}{2}x^{2}y$

Check:

$u_{x}(x,y)=-\frac{3}{2}x^{2}+xy$

$u_{y}(x,y)=\frac{1}{2}x^{2}$

$u_{x}+3u_{y}=xy$

14
##### Home Assignment 1 / Re: Problem 5
« on: September 23, 2012, 07:42:54 PM »
I was trying to figure out the "typo'ed" equation using material from the other 1D wave equation lectures but to no avail. Would you be able to explain how to approach the old equation or possibly post an outline of the solution?

15
##### Home Assignment 1 / Re: Problem 5
« on: September 23, 2012, 04:37:05 PM »
Professor Ivrii, could you please confirm this?

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