Author Topic: HA2-P5  (Read 3361 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2599
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii

Chi Ma

  • Full Member
  • ***
  • Posts: 16
  • Karma: 0
    • View Profile
Re: HA2-P5
« Reply #1 on: September 29, 2015, 12:58:57 AM »
Problem 5a
$$yu_x-xu_y=x\\
\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x}\\
x^2+y^2 = C\\
u = -y + \phi(x^2+y^2)$$

Problem 5b
$$yu_x-xu_y=x^2\\
\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x^2}\\
x^2+y^2 = C\\
\left\{\begin{array}{2}
x = r\cos\theta\\
y = r\sin\theta\\
\end{array}\right.\\
du = -xdy = -r^2\cos^2\theta d\theta\\
u = - \frac{r^2}{2} \left(\theta + \frac{1}{2}\sin2\theta\right) + \psi(r)$$

Problem 5c. I think there is a typo in equation (14). The coefficient on $u_x$ should be $y$. Yes indeed. V.I.
$$yu_x+xu_y=x\\
\frac{dx}{y} = \frac{dy}{x} = \frac{du}{x}\\
x^2-y^2 = C\\
u = y + \zeta(x^2-y^2)$$

Problem 5d
$$yu_x+xu_y=x^2\\
\frac{dx}{y} = \frac{dy}{x} = \frac{du}{x^2}\\
x^2-y^2 = C,\qquad -\infty < C < \infty$$

For $0 < C < \infty$, let $C = r^2$.
$$\left\{\begin{array}{2}x = r\sec\theta\\
y = r\tan\theta\\
\end{array}\right.\\
du = xdy = r^2\sec^3\theta d\theta\\
u =  \frac{r^2}{2} \left(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|\right) + \psi(r)$$

For $-\infty < C < 0$, let $C = -r^2$.
$$\left\{\begin{array}{2}
x = r\tan\theta\\
y = r\sec\theta\\
\end{array}\right.\\
du = xdy = r^2\tan^2\theta\sec\theta d\theta = r^2\left(\sec^3\theta - \sec\theta\right)d\theta\\
u =  \frac{r^2}{2} \left(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|\right) + \phi(r)$$

At ${x = \pm y}$, $u$ is undefined and discontinuous.

« Last Edit: September 30, 2015, 05:18:04 PM by Victor Ivrii »

Rong Wei

  • Sr. Member
  • ****
  • Posts: 43
  • Karma: 0
    • View Profile
Re: HA2-P5
« Reply #2 on: September 30, 2015, 12:09:10 AM »
For 5b, I know derivative of arcsin(x) is 1/(1-x^2)^(1/2), can I use this to calculate the integral ? sorry about my typo, I haven't download matlab yet, I will download it~

Chi Ma

  • Full Member
  • ***
  • Posts: 16
  • Karma: 0
    • View Profile
Re: HA2-P5
« Reply #3 on: September 30, 2015, 08:11:18 AM »
I think so. You probably will have 2 equations for the positive and negative square roots.

Rong Wei

  • Sr. Member
  • ****
  • Posts: 43
  • Karma: 0
    • View Profile
Re: HA2-P5
« Reply #4 on: October 01, 2015, 01:43:07 PM »
you are so right! see you in class!

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2599
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: HA2-P5
« Reply #5 on: October 01, 2015, 01:53:38 PM »
Introduce a natural parameter on the circle: $x=r\cos(\theta)$, $x=r\sin(\theta)$--it will make integral much easier.

5b
The trouble with here is that you get expression through $\theta$ and this is bad! Polar angle is not defined uniquely as long as you can merrily go around an origin and we are looking for a single-valued continuous solutions. This phenomena may occur as characteristics are closed curves so we need to integrate the r.h.e. along it.

If integral is $0$ as it happens with r.h.e. $x$ or $y$ or $xy$ etc, increment of $u$ is $0$ and it is a single-valued function but for r.h.e. like $1$ or $x^2$ integral is not $0$ and the required solution does not exist at all (not that we cannot find it—it simply does not exist!)

5d
It simple to use often not $\sec$ and $\tan$ but hyperbolic functions $\cosh$ and $\sinh$
« Last Edit: October 03, 2015, 06:01:17 AM by Victor Ivrii »