Author Topic: Problem 4  (Read 8453 times)

James McVittie

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Problem 4
« 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?

Victor Ivrii

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Re: Problem 4 [corrected]
« Reply #1 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

Peishan Wang

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Re: Problem 4
« Reply #2 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?

Victor Ivrii

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Re: Problem 4
« Reply #3 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 broad
  • Unusually narrow

Your task is to determine what happens here.

Peishan Wang

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Re: Problem 4
« Reply #4 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)...

Victor Ivrii

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Re: Problem 4
« Reply #5 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.

Qitan Cui

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Re: Problem 4
« Reply #6 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!

Victor Ivrii

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Re: Problem 4
« Reply #7 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.

Aida Razi

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Re: Problem 4
« Reply #8 on: September 24, 2012, 09:02:35 PM »
Full solution is attached!

Rouhollah Ramezani

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Re: Problem 4
« Reply #9 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.

Victor Ivrii

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Re: Problem 4
« Reply #10 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
« Last Edit: September 26, 2012, 03:35:48 AM by Victor Ivrii »

Peishan Wang

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Re: Problem 4
« Reply #11 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.
« Last Edit: September 26, 2012, 03:55:16 AM by Peishan Wang »

Victor Ivrii

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Re: Problem 4
« Reply #12 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)

Rouhollah Ramezani

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Re: Problem 4
« Reply #13 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.

Victor Ivrii

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Re: Problem 4
« Reply #14 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)
« Last Edit: September 26, 2012, 04:42:53 PM by Victor Ivrii »