Messages

31

Home Assignment 5 / Re: Problem 4

« on: October 31, 2012, 09:53:35 PM »

Hopeful solution for 4.b) attached!

32

Home Assignment 5 / Re: Problem 4

« on: October 31, 2012, 09:44:27 PM »

Additional solution for 4.a) (essentially the same as Aida's post here http://forum.math.toronto.edu/index.php?topic=108.msg552#msg552 , but showing more of the sketch, as well as with details on the odd continuation used for sin Fourier series).

33

Home Assignment 5 / Re: Problem 6

« on: October 31, 2012, 09:38:35 PM »

Hopeful solution to 6.e) attached!

34

Home Assignment 5 / Re: Problem 5

« on: October 31, 2012, 09:32:43 PM »

Hopeful solution for 5.a)!

35

Home Assignment 5 / Problem 4

« on: October 31, 2012, 09:32:02 PM »

Hopeful solutions for 4.c)!

edit: Note that sketch is for m=1.

36

Home Assignment 5 / Re: Problem 1

« on: October 31, 2012, 02:27:03 AM »

Jinchao: you can integrate (e^x)sin(x) by parts. Set u = e^x, dv = sinx dx, and go through. You'll have to integrate by parts a second time, but you'll end up with (e^x)sinx integrals on both sides. Hope that helps!

37

Misc Math / Lecture 12 Eqn 9 Question

« on: October 24, 2012, 01:59:26 AM »

Hi all,

Just to clarify - should equation 9 in the notes for lecture 12 (http://www.math.toronto.edu/courses/apm346h1/20129/L12.html#mjx-eqn-eq-9) read

$\lambda_n = - n^2 \pi^2 / l^2$ ?

Cheers,

Ian

38

Term Test 1 / Re: TT1 = Problem 3

« on: October 17, 2012, 01:11:39 AM »

PS. Ian, your posts are virtually useless for a class: too poor handwriting makes it almost impossible to read for anyone who does not know solution. Could you repost?

Sorry!! 

I have tried my best to re-write it nicely (edited original post).

39

Term Test 1 / Re: TT1 = Problem 2

« on: October 16, 2012, 09:53:02 PM »

Ian, while explanation is basically correct I would like to see more convincing arguments. In particular: where solution will be defined uniquely?

With the given conditions, the I think solution is defined for -inf < x < -t, . The given conditions on u and u_t restrict it there, as any wave starting early in time would have to pass through (x,t)=(x,0). 0 < x < -t, however, does not have the required conditions for uniqueness.

40

Term Test 1 / Re: TT1 = Problem 1

« on: October 16, 2012, 09:24:06 PM »

Subqueston (d):

$ \frac{dt}{1} = \frac{dx}{x^2} $
$ t = -x^{-1}+c $
so the general solution is $ u(t,x)=f(t+x^{-1})$.
$u(0,x)=f(x^{-1})=g(x)$
$f(y)=g(y^{-1})$
Since $y=x^{-1}$, so when $x>0$ we have $y>0$ as well.
$u(t,x)=f(t+x^{-1})$
We need $t+x^{-1}>0$
Since $x>0$,
therefore $tx+1>0$
so the domain be defined is $\{(t,x) | tx>-1 \}.

I think Jinchao has the most correct solution.

Qitan, is it possible to only have the one discontinuity in your solution - won't your characteristic curves be "blocked" by the discontinuity at tx=-1, and not able to go any further?

41

Term Test 1 / Re: TT1 = Problem 5

« on: October 16, 2012, 08:25:25 PM »

Solution is attached,

Aida, I think your solution is not correct - the integral from -1 to 0 will only be 0 if x < y in H(x-y), i.e. x < -1.

Also, consider extreme cases - for C(-100), H(-100-y) can never be greater than 0, so a solution with C(x) = 1 cannot be right (unless I've totally misunderstood something).

I believe your solution is not correct; H(x) and H(x-y) has the same value for these two different domain because it is a constant function.

Shouldn't they be different though? For H(x), it's 1 for x>0 and 0 for x<=0, but H(x-y) is 1 for x-y>0, or x>y, and 0 for x<=y.

In any case, how can C(-100) be nonzero? x-y=-100-y>0 is impossible for -1<=y<=1, which is the only area where Q(y) is nonzero.

42

Term Test 1 / Re: TT1 = Problem 3

« on: October 16, 2012, 07:51:52 PM »

Solution is attached,

Aida: I'm not sure your solution is correct: u(0,t)=0 and u_x(0,t)=0 don't necessarily imply that u_xx(0,t) = 0. Consider for example u(x, t) = x^2. There, u_xx(0, t) = 2 despite u(0,t)=0 and u_x(0,t)=0.

Up until crossing out u_xx on the last line, though, I think your solution is still right, and your final answer is definitely right. ;P

43

Term Test 1 / Re: TT1 = Problem 5

« on: October 16, 2012, 07:39:21 PM »

Solution is attached,

Aida, I think your solution is not correct - the integral from -1 to 0 will only be 0 if x < y in H(x-y), i.e. x < -1.

Also, consider extreme cases - for C(-100), H(-100-y) can never be greater than 0, so a solution with C(x) = 1 cannot be right (unless I've totally misunderstood something).

44

Term Test 1 / Re: TT1 = Problem 3

« on: October 16, 2012, 07:32:55 PM »

Solution to Question 3!

Re-wrote solution more nicely at Prof. Ivrii's request. Original at http://i.imgur.com/l4Pw2.jpg

45

Term Test 1 / Re: TT1 = Problem 4

« on: October 16, 2012, 07:03:44 PM »

Solution to 4.a) attached!

Re-written more nicely. Original at http://i.imgur.com/M2yhk.jpg