1
Test 1 / Re: TT1 Problem 3
« on: February 12, 2015, 10:29:04 PM »
Solution to Question 3
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.
Pages: [1]
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
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 x−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)
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.
$y=3x−C$
Then again from the Integral Lines:
dx(xy)=du
dx(x(3x−C))=du
u=x^3−C/2x^2+C1
Then by the initial condition:
u(x=0)=0
C1=0
Therefore,C=3x−y
Therefore,
u(x,y)=x^3−1/2x^2C
$u(x,y)=x^3−\frac{1}{2}x^2(3x−y)$
$u(x,y)=−\frac{1}{2}x^3+\frac{1}{2}x^2y$
$y=3x−C$
Then again from the Integral Lines:
dx(xy)=du
dx(x(3x−C))=du
u=x^3−C/2x^2+C1
Then by the initial condition:
u(x=0)=0
C1=0
Therefore,C=3x−y
Therefore,
u(x,y)=x^3−1/2x^2C
$u(x,y)=x^3−\frac{1}{2}x^2(3x−y)$
$u(x,y)=−\frac{1}{2}x^3+\frac{1}{2}x^2y$
6
HA1 / Re: HA1 problem 6
« on: January 23, 2015, 10:24:52 AM »
b） when t < 0 t not equal 0
when t > 0 all t are fine
when t > 0 all t are fine
7
HA1 / Re: HA1 problem 2
« on: January 23, 2015, 10:20:06 AM »
c） 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: [1]