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**Home Assignment 5 / Re: Problem 4**

« **on:**October 31, 2012, 09:53:35 PM »

Hopeful solution for 4.b) attached!

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Hopeful solution for 4.b) attached!

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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).

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Hopeful solution to 6.e) attached!

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Hopeful solution for 5.a)!

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Hopeful solutions for 4.c)!

edit: Note that sketch is for m=1.

edit: Note that sketch is for m=1.

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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!

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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

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

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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).

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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.

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Subqueston (d):I think Jinchao has the most correct solution.

$ \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 \}.

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?

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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.

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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

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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).

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Solution to Question 3!

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

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