# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment 1 => Topic started by: James McVittie on September 20, 2012, 06:05:25 PM

Title: Problem 4
Post by: James McVittie on September 20, 2012, 06:05:25 PM
For Part (c) of Problem 4, does it refer to the solution of (5) or (6) or both?
Title: Re: Problem 4 [corrected]
Post by: Victor Ivrii on September 20, 2012, 07:42:53 PM
To both: for one of them solution does not exist.

There was an error in the left-hand expression, now it has been corrected
Title: Re: Problem 4
Post by: Peishan Wang on September 21, 2012, 08:38:37 PM
From the auxiliary equations dx/a=dy/b=du/f, we can either express du in terms of dx and integrate over x, or express du in terms of dy and integrate over y. But sometimes these two approaches give different results. Then can we say that the general solution does not exist?
Title: Re: Problem 4
Post by: Victor Ivrii on September 22, 2012, 02:15:54 AM
In this settings you need to select a solution having certain properties

It may happen that

• Is what one expects
• Unusually narrow

Title: Re: Problem 4
Post by: Peishan Wang on September 23, 2012, 04:04:44 AM
I followed the normal steps and found the general solutions to both equations. I cannot figure out why one of them do not exist. Is it because some function is not defined? Get lost in part (c)...
Title: Re: Problem 4
Post by: Victor Ivrii on September 23, 2012, 04:08:42 AM
I followed the normal steps and found the general solutions to both equations. I cannot figure out why one of them do not exist. Is it because some function is not defined? Get lost in part (c)...

You need to ask yourself: does a solution you found satisfy all conditions of the problem.
Title: Re: Problem 4
Post by: Qitan Cui on September 23, 2012, 04:21:38 AM
But the only condition I have is the original PDE and my general solution just well satisfies that equation. What other conditions should I have? Thanks!
Title: Re: Problem 4
Post by: Victor Ivrii on September 23, 2012, 05:39:08 AM
But the only condition I have is the original PDE and my general solution just well satisfies that equation. What other conditions should I have? Thanks!

It was a general remark. In Problem 5 there is no other explicit condition. However one needs to take a look on the general solution.
Title: Re: Problem 4
Post by: Aida Razi on September 24, 2012, 09:02:35 PM
Full solution is attached!
Title: Re: Problem 4
Post by: Rouhollah Ramezani on September 25, 2012, 08:48:36 PM
a) This is a first order linear inhomogeneous PDE. We begin by examining characteristic lines:

\begin{equation*}
\frac{\,dx}{y}=\frac{\,dy}{-x}=\frac{\,du}{xy}
\end{equation*}
First equation implies $x^2+y^2=C$, for some constant $C$. Therefore $u(x,y)=\phi(x^2+y^2)$, for some arbitrary $\phi$ is solution to homogeneous equation. From second equation we get $\,dx=\frac{\,du}{x}$ which implies $u=\frac{x^2}{2}$ is a particular solution to the inhomogeneous equation.
The general solution would be:
$$u(x,y)=\phi(x^2+y^2)+\frac{x^2}{2}$$
where $\phi$ is arbitrary.

b) General solution to the homogeneous equation is identical to (a). Remaining task is to solve for $u(x,y)$ in $\frac{\,dx}{y}=\frac{\,du}{x^2+y^2}$ and find a particular solution. Substituting $y^2$ with $C-x^2$ we get the following ODE:
$$C\frac{\,dx}{\sqrt{C-x^2}}=\,du$$
Integerating both sides we get
\begin{equation*}
u=\int C\frac{\,dx}{\sqrt{C-x^2}}  \end{equation*}
$$=C\int \frac{\,dx}{\sqrt{C}\sqrt{1-(\frac{x}{\sqrt{C}})^2}} \\ =C\arcsin{\frac{x}{\sqrt{C}}} \\ = (x^2+y^2)\arcsin{\frac{x}{\sqrt{x^2+y^2}}}$$
General solution is
$$u(x,y)=\phi(x^2+y^2)+(x^2+y^2)\arcsin{\frac{x}{\sqrt{x^2+y^2}}}$$
where $\phi: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function.

c) In the latter case,  if $(x,y)=(0,0)$ is in domain of $u$, general solution does not exist as $\arcsin{\frac{x}{\sqrt{x^2+y^2}}}$ is not well-defined.
Title: Re: Problem 4
Post by: Victor Ivrii on September 26, 2012, 02:11:08 AM

c) In the latter case,  if $(x,y)=(0,0)$ is in domain of $u$, general solution does not exist as $\arcsin{\frac{x}{\sqrt{x^2+y^2}}}$ is not well-defined.

Not persuasive: may be we just were not smart enough? You need to demonstrate that solution really does not exist and explain what is an obstacle.

Hint Use polar coordinates
Title: Re: Problem 4
Post by: Peishan Wang on September 26, 2012, 03:53:33 AM
Using polar coordinates would make things much easier (see attachments). However we need to be careful here since arccos gives us the angle from 0 to pi. When y<0 (i.e. theta > pi), it will become 2pi - arccos.

Please let me know if there's anything wrong with my posted solution.
Title: Re: Problem 4
Post by: Victor Ivrii on September 26, 2012, 05:19:38 AM
Peishan, the question now boils down to: why solution with r.h.e. $x^2+y^2$ does not exist and one really needs to use polar coordinates (or to use equivalent geometric motivation)
Title: Re: Problem 4
Post by: Rouhollah Ramezani on September 26, 2012, 04:09:03 PM

You need to demonstrate that solution really does not exist and explain what is an obstacle.

Hint Use polar coordinates

Writing solution in polar coordinates we get:
\begin{equation}
u(r,\theta)=r^2\theta+\phi(r)
\end{equation}
The key point here is the fact that at $r=0$, $\theta$ can be anything and $(r,\theta)$ still expresses the same point. Let's suppose a general solution exists and aim for contradiction. Taking second derivative of $u(r,\theta)$ vis.a.vis $r$ at certain points we get:
\begin{equation}
u_{rr}=2\theta+\phi''(r) \\
\end{equation}
\Rightarrow \left\{ \begin{aligned} u_{rr}(r_0,0)-u_{rr}(0,0)=\phi''(r_0)\\ u_{rr}(r_0,\theta)-u_{rr}(0,\theta)=\phi''(r_0) \end{aligned} \right.
But $u_{rr}(0,0)=u_{rr}(0,\theta)$ as they both represent the same point. Therefore $u_{rr}(r,\theta)=\phi''(r)$ and is independent of $\theta$ which contradicts $(1)$. Alternatively, we can continue and conclude since $u_{rr}$ is a function of $r$ only so should be $u$, in contradiction to general solution attained before.

In other words, $u$ can not be general solution because $u_{rr}$ is not well-defined at the origin.
Title: Re: Problem 4
Post by: Victor Ivrii on September 26, 2012, 04:34:04 PM
Here we are talking about first order equation. There is no wave equation. Stick to the problem! How our original equation
\begin{equation}
yu_x-xu_y=f(x,y)
\label{eq-1}
\end{equation}
looks in the polar coordinates?

Hint: the l.h.e. is $-\partial_\theta u$. Prove it using chain rule  $u_\theta = u_x x_\theta + u_y y_\theta$ and calculate $x_\theta$, $y_\theta$.
Note that $\theta$ is defined modulo $2\pi \mathbb{Z}$ and all functions must be $2\pi$-periodic with respect to $\theta$ (assuming that we consider domains where one can travel around origin)
Title: Re: Problem 4
Post by: Rouhollah Ramezani on September 26, 2012, 05:30:52 PM

Hint: the l.h.e. is $-\partial_\theta u$. Prove it using chain rule  $u_\theta = u_x x_\theta + u_y y_\theta$ and calculate $x_\theta$, $y_\theta$.

$$x=r\cos{\theta} \rightarrow x_{\theta}=-r\sin{\theta}\\ y=r\sin{\theta} \rightarrow y_{\theta}=r\cos{\theta}\\ \Rightarrow yu_x-xu_y=r\sin{\theta}u_x-r\cos{\theta}u_y \\ = -u_x x_\theta - u_y y_\theta \\ =-u_\theta$$

Note that $\theta$ is defined modulo $2\pi \mathbb{Z}$ and all functions must be $2\pi$-periodic with respect to $\theta$ (assuming that we consider domains where one can travel around origin)

Do you mean:
$$-u_\theta=r^2 \\ \Rightarrow u=-r^2\theta$$ and since it's not periodic we are done?
Title: Re: Problem 4
Post by: Victor Ivrii on September 26, 2012, 09:40:44 PM

Do you mean:
$$-u_\theta=r^2 \\ \Rightarrow u=-r^2\theta$$ and since it's not periodic we are done?

Basically yes, except certain things should be clarified: since our trajectories are closed each is parametrized by some parameter ($\theta$) running from $0$ to $T$ (in our case $T=2\pi$ but it may depend on trajectory) and equation looks like $\partial_\theta u= g(\theta,r)$. So we are looking for periodic solution
$u(\theta,r)=\int g(\theta,r)\,d\theta$.

Note that primitive of periodic function $g(\theta)$  is periodic if and only if average of $g$ over period is $0$:
$\int_0^T g(\theta)\,d\theta=0$. Otherwise this primitive is the sum of a periodic function and a linear function.

Finally, in (a) $g(\theta)=r^2 \sin(\theta)\cos(\theta)$ and integral over period is $0$; in (b) $g(\theta)=r^2$ and  integral over period is not $0$.

So, the source of trouble is: periodic trajectories.
Title: Re: Problem 4
Post by: Di Wang on October 14, 2012, 10:25:34 PM
is there any clear solution for part c of question four. thank you? :)
Title: Re: Problem 4
Post by: Victor Ivrii on October 15, 2012, 02:57:14 AM
is there any clear solution for part c of question four. thank you? :)

See above