### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Ping Wei

Pages: 
1
##### Test 1 / Re: TT1 Problem 3
« on: February 12, 2015, 10:29:04 PM »
Solution to Question 3

2
##### Test 1 / Re: TT1 Problem 4
« on: February 12, 2015, 10:02:14 PM »
Ut= 6x , Ux= 2x^2+6t, Uxx= 6x  so Ut - Uxx = 6x -6x =0

3
##### Test 1 / Re: TT1 Problem 4
« on: February 12, 2015, 10:00:21 PM »
a), u(L,T) is maximum value
b), u(0,0) is minimum value

4
##### HA2 / Re: HA2 problem 1
« on: January 29, 2015, 09:42:54 PM »
A) The problem always has  unique solution. No extra consitions are necessary. Since x>4t, we are confident that xâˆ’3t is always positive. I.e. initial value functions are defined everywhere in domain of u(t,x). Using d'Alembert's formula we write:
u(t,x)=1/3cos(x+3t)+2/3cos(x-3t)

5
##### HA1 / Re: HA1 problem 3
« on: January 23, 2015, 10:33:14 AM »
By examining Integral Lines: $\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}$; then we got $3x=y+C$ where $C$ is some constant.
$y=3xâˆ’C$

Then again from the Integral Lines:

dx(xy)=du

dx(x(3xâˆ’C))=du

u=x^3âˆ’C/2x^2+C1

Then by the initial condition:

u(x=0)=0

C1=0

Therefore,C=3xâˆ’y

Therefore,

u(x,y)=x^3âˆ’1/2x^2C

$u(x,y)=x^3âˆ’\frac{1}{2}x^2(3xâˆ’y)$

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

6
##### HA1 / Re: HA1 problem 6
« on: January 23, 2015, 10:24:52 AM »
bï¼‰ when t < 0 t not equal 0
when t > 0 all t are fine

7
##### HA1 / Re: HA1 problem 2
« on: January 23, 2015, 10:20:06 AM »
cï¼‰ The difference between two cases is that in one of them all trajectories have (0,0) as the limit points and in another only those with x=0 or y=0

Pages: