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### Messages - Jessica Chen

Pages: 1 [2]
16
##### Web Bonus Problems / Re: Web Bonus Problem 1
« on: January 24, 2015, 12:42:21 AM »
I interpret this question in the following:

This is a quasilinear equation with coefficient u.
As we solved in HA1, the general solution for $u$ is $u = f(x_0) = f(x-ut)$, now consider the first set of initial data.
Case1. If $x_0 < -a$, then we have $u = -1$ and also $x_0 = x+t < -a$.
Case2. If  $x_0 > a$, then we have $u = 1$ and also $x_0 = x-t > a$.
Case3. If $-a\le x \le a$, then we have $u = \frac{x}{a}$.

x_0 = x - \frac{x}{a}t\rightarrow u = \frac{x}{a+t}

Then the solution looks like
\begin{align}
u(x,t)=&\left\{\begin{aligned}
-1& && x<-a-t,\\
\frac{x}{a+t}& && -a-t\le x \le a+t,\\
1& && x>a+t;
\end{aligned}\right.
\\
\end{align}

17
##### HA1 / Re: HA1 problem 4
« on: January 22, 2015, 11:59:45 PM »
c.
Suppose in general case, $yu x - 4xu y = f(x, y)$
Use polar angles, define

x=r\cos \theta \rightarrow x {\theta} = -r \sin\theta\\
y = 2r\sin \theta \rightarrow y{\theta} = 2r\cos\theta\\
yu x - 4xu y = -2u\theta.

Since trajectories are closed (elliptic shape) $\theta \in (0, 2\pi)$ periodically.
Let $-2u \theta = g(\theta, r)$ then $u(\theta, r) = \int_0^{2\pi}\! g(\theta, r) d\theta$ but g needs to be 0 over the period, i.e $\int_0^{2\pi}\! g(\theta, r) d\theta = 0$.

In part a)
y = -2u\theta\\
r\cos\theta = u
and integral over this period is 0. Thus solution for part a) is valid.

In part b)
x^2 = -2u\theta\\
-\frac{1}{2}r^2 \frac{1}{2}(\cos 2\theta +1) = u\theta\\
u = -\frac{1}{8}r^2\sin 2\theta - \frac{1}{4}r^2\theta = u\\
the integral over period is not zero. Thus the periodic trajectories would not work.

18
##### HA1 / Re: HA1 problem 4
« on: January 22, 2015, 11:23:04 PM »
I got a very nasty equation for part b
b.
First solve a homogeneous: we get $u(x, y) = \phi (4x^2 +y^2)$ and $C = 4x^2 + y^2$
Then to find the general solution:

\frac{d x}{y} = \frac{d x}{x^2} = \frac{d u}{0};\\
\frac{x^2}{y}d x = d u;\\ \text{Since } y = \sqrt{C-4x^2},\\
\frac{x^2}{\sqrt{C-4x^2}}d x=d u\\
\frac{x^2}{\sqrt{C\sqrt{1-\frac{4x^2}{c}}}}dx = du;\\

We substitute

\frac{2x}{\sqrt{c}} = \sin \theta

Then we get:

du=\frac{\frac{\sqrt{c}}{4}\sin^2\theta}{\sqrt{1-\sin^2\theta}}d\sin\theta \frac{\sqrt{2}}{2}\\
du=\frac{\frac{c}{8}\sin^2 \theta }{\cos \theta }d\sin \theta\\
du=\frac{c}{8}\frac{\sin^2 \theta}{\cos \theta} \cos \theta d\theta \\
du=\frac{c}{16}(\theta - \frac{1}{2}\sin 2\theta )\\

Use trig identity we have $u = \frac{c}{16} \arcsin \frac{2x}{\sqrt{c}} - (\frac{1}{2}(2\times\frac{2x}{\sqrt{c}}\sqrt{1-\frac{4x^2}{c}})$
The general solution is $u = \phi(4x^2+y^2)+ \frac{c}{16} \arcsin \frac{2x}{\sqrt{c}} - (\frac{2x}{\sqrt{c}}\sqrt{1-\frac{4x^2}{c}})$

19
##### HA1 / Re: HA1 problem 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.

20
##### HA1 / Re: HA1 problem 1
« on: January 22, 2015, 10:37:03 PM »
i think equation has some typo. should be y instead of 7

Yes you are right! Thanks

21
##### HA1 / Re: HA1 problem 6
« on: January 22, 2015, 10:35:35 PM »
a. This is a quasilinear equation.

\frac{d t}{1} = \frac{d x}{u} = \frac{d u}{0};

Then we get $ut-x= C$ for some arbitrary constant C.
Consider the initial condition $u|_{t=0}=x$, take an initial point $(0, x_0)$ such that

u(x, 0) = x

Therefore we have $u = f(x_0) = f(x-ut)$ along characteristics, so

u(x,y) = f(x-ut) = x-ut\\
u(x, y) = \frac{x}{1+t}

b.
When $t>-1$, the solution is clearly hold.
However when $t<-1$ the solution breaks because the characteristics lines cannot be interpreted if they are intersected or undefined.

22
##### HA1 / Re: HA1 problem 1
« on: January 22, 2015, 10:18:06 PM »
a. This is a first order linear PDE with constant coefficient.

\frac{d x}{4} = \frac{d y}{3} = \frac{d u}{0};

Then we get $u(x, y) = f(3x+4y)$ for some arbitrary $f$.

b.  Use IVP $u|_{x=0}=\sin (y)$, we get

f(4y) = \sin(y)\\f(w) = \sin(\frac{y}{4})

Hence solution is

u(x,y) = f(3x+4y) = \sin(\frac{3}{4}x+y)

c.  With initial condition $u|_{x=0}=y$, $y>0$. Hence the solution is

u(x, y) = \frac{3}{4}x+y.

Since $\forall x>0, y>0\implies \frac{3}{4}x+y >0$.
Thus this solution is defined on the whole domain $\{x>0,y>0\}$.

d.  With initial condition $u|_{x=0}=y$, $y>0$. Hence the solution is

u(x, y) = \frac{3}{4}x+y.

Since $f$ is define when $y>0$, the solution only is defined where $\frac{3}{4}x+y >0$, then when $y> -\frac{3}{4}x$, the solution is defined. However when $y< -\frac{3}{4}x$ we need to impose condition at $y = 0$, we get

f(3x) = x \\
f(w) = \frac{w}{3}\\

Then we get:

u(x,y) = x + \frac {4}{3}y, (y< -\frac{3}{4}x)

Final solution would be:

u(x,y) = \left\{
\begin{array}{l l}