### Author Topic: HA1 problem 2  (Read 1720 times)

#### Victor Ivrii

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##### HA1 problem 2
« on: January 20, 2015, 06:47:46 AM »
Solutions to be posted as a "Reply" only after January 22, 21:00

a. Find the general solution of

xu_x+4 yu_y=0
\label{eq-HA1.2}

in $\{(x,y)\ne (0,0)\}$; when this solution is continuous at $(0,0)$?

b. Find the general solution of

xu_x-4yu_y=0
\label{eq-HA1.3}

in $\{(x,y)\ne (0,0)\}$; when this solution is continuous at $(0,0)$?

c. Explain the difference between (\ref{eq-HA1.2}) and (\ref{eq-HA1.3}).

#### Biao Zhang

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##### Re: HA1 problem 2
« Reply #1 on: January 22, 2015, 10:18:30 PM »
not sure this one is right

#### Jessica Chen

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##### Re: HA1 problem 2
« Reply #2 on: January 22, 2015, 10:51:12 PM »
Not sure if I interpreted it right either.

c.
The difference between a and b is
part a) the characteristic lines look like a parabola, all trajectories have (0, 0) as the limit point;
part b) the characteristic lines look like a delta function, only (x=0, y=0) has (0,0) as the limit point.
« Last Edit: January 23, 2015, 12:00:33 AM by Jessica Chen »

#### Ping Wei

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##### Re: HA1 problem 2
« Reply #3 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

#### Victor Ivrii

I just concretize c: it is node vs saddle (remember ODE class?). So solution in (a) is $f(y/x^4)$ and it is continuous iff it is constant; solution in (b) is $f(yx^4)$ and it is  continuous iff $f$ is continuous.