Author Topic: HA1 problem 4  (Read 562 times)

Victor Ivrii

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HA1 problem 4
« on: January 20, 2015, 06:51:24 AM »
Solutions to be posted as a "Reply" only after January 22, 21:00

a.  Find the general solution of
\begin{equation}
yu_x-4xu_y=y;
\label{eq-HA1.5}
\end{equation}
b.  Find the general solution of
\begin{equation}
yu_x-4xu_y=x^2;
\label{eq-HA1.6}
\end{equation}
c.  In one instanse solution does not exist. Explain why.

Biao Zhang

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Re: HA1 problem 4
« Reply #1 on: January 22, 2015, 10:19:41 PM »
HA1-4

Jessica Chen

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Re: HA1 problem 4
« Reply #2 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:
\begin{equation}
\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;\\
\end{equation}

We substitute
\begin{equation}
\frac{2x}{\sqrt{c}} = \sin \theta

\end{equation}
Then we get:
\begin{equation}
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 )\\
\end{equation}
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 $u = \phi(4x^2+y^2)+ \frac{c}{16} \arcsin \frac{2x}{\sqrt{c}} - (\frac{2x}{\sqrt{c}}\sqrt{1-\frac{4x^2}{c}})$

« Last Edit: January 23, 2015, 11:42:12 PM by Jessica Chen »

Jessica Chen

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Re: HA1 problem 4
« Reply #3 on: January 22, 2015, 11:59:45 PM »
c.
Suppose in general case, $yu x - 4xu y = f(x, y)$
Use polar angles, define
\begin{equation}
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.
\end{equation}
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) \begin{equation}
y = -2u\theta\\
r\cos\theta = u
\end{equation} and integral over this period is 0. Thus solution for part a) is valid.

In part b) \begin{equation}
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\\
\end{equation} the integral over period is not zero. Thus the periodic trajectories would not work.
« Last Edit: January 23, 2015, 11:48:07 PM by Jessica Chen »

Victor Ivrii

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Re: HA1 problem 4
« Reply #4 on: January 23, 2015, 11:13:13 AM »
Jessica, fix your post (remove some underscores and add some backslashes)

Do not use * for multiplication as it reserved for convolution which we will study later. Use a \cdot b or a\times b producing $a\cdot b$ or $a\times b$