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### Messages - Ian Kivlichan

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46
##### 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

47
##### 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).

48
##### 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

49
##### 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

50
##### Term Test 1 / Re: TT1 = Problem 2
« on: October 16, 2012, 06:56:06 PM »
The initial conditions are not sufficient to uniquely define the solution - we need boundary conditions at x = 0.

Consider a counter-example to uniqueness:
1. u = 0 over x > 0 at t = 0
2. a large wave, moving in the -x direction, at t=0 (but still in the region x > 0).

Both 1. and 2. satisfy the given initial conditions, so the solution is not unique. Had we specified Dirichlet, Neumann, etc. conditions, however, one of the two cases would be impossible.

51
##### Term Test 1 / Re: TT1 = Problem 5
« on: October 16, 2012, 06:50:29 PM »
Hopeful solution attached!

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