Author Topic: Problem 2  (Read 4703 times)

Peishan Wang

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Problem 2
« on: September 21, 2012, 01:36:09 PM »
Could you give some hints to Q2? There's no way that I can make the general solution continuous at (0,0)..... Thanks a lot!

Victor Ivrii

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Re: Problem 2
« Reply #1 on: September 21, 2012, 01:57:34 PM »
First, I need to apologize: this problem contained a misprint which I just corrected. The source of errors is simple: everything was done in the extreme rush and nobody checked it.

Now hint: it may happen in some problems that the solution (having certain properties) does not exist or must have a very special form. Then the your presented solution should state this.

I am not claiming that this is the case with this problem but your statement is wrong anyway: there is at least one continuous solution $u=0$ identically.

« Last Edit: September 21, 2012, 07:59:02 PM by Victor Ivrii »

Peishan Wang

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Re: Problem 2
« Reply #2 on: September 21, 2012, 07:19:46 PM »
Thank you for your hint but I still didn't get the point..

For example if the general solution has the form f(x/y), how can I make them continuous at (0,0)? Thanks!!
« Last Edit: September 21, 2012, 07:51:12 PM by Peishan Wang »

Victor Ivrii

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Re: Problem 2
« Reply #3 on: September 21, 2012, 07:30:41 PM »
Thank you for your hint but I still didn't get the point..

For example if the general solution has the form $f(x/y)$, how can I make it continuous at (0,0)? Thanks!!

You are almost done (but check the general solutions!) Think - and don't post solutions!!!
« Last Edit: September 22, 2012, 12:37:35 PM by Victor Ivrii »

Fatima Yousuf

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Re: Problem 2
« Reply #4 on: September 22, 2012, 09:34:52 AM »
Hi, for number 2, say we get $u = f(g(x,y))$ for our general solution when $(x,y) \ne (0,0)$. Then are we just finding a $u(0,0)$ that is equal to the limit of $u = f(g(x,y))$ as $(x,y)$ approaching $(0,0)$ in order to make u continuous at $(0,0)$?
« Last Edit: September 22, 2012, 12:38:43 PM by Victor Ivrii »

Victor Ivrii

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Re: Problem 2
« Reply #5 on: September 22, 2012, 12:40:30 PM »
Hi, for number 2, say we get $u = f(g(x,y))$ for our general solution when $(x,y) \ne (0,0)$. Then are we just finding a $u(0,0)$ that is equal to the limit of $u = f(g(x,y))$ as $(x,y)$ approaching $(0,0)$ in order to make u continuous at $(0,0)$?

Well, you need to be sure that this limit exists, right?

Kun Guo

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Re: Problem 2
« Reply #6 on: September 22, 2012, 07:11:15 PM »
I am not sure what does it mean by 'explain the difference'. Is it the difference between the solutions or the condition under which they are continuous at (0,0)?

Victor Ivrii

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Re: Problem 2
« Reply #7 on: September 22, 2012, 10:31:22 PM »
I am not sure what does it mean by 'explain the difference'. Is it the difference between the solutions or the condition under which they are continuous at (0,0)?

The solutions of (a) and (b) are drastically different. Why?

Aida Razi

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Re: Problem 2
« Reply #8 on: September 24, 2012, 09:01:39 PM »
Full solution is attached!

Rouhollah Ramezani

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Re: Problem 2
« Reply #9 on: September 25, 2012, 04:38:04 PM »
a) This is a first order linear PDE. We begin by writing equation of characteristic lines:
\begin{equation*}
\frac{dy}{4y}=\frac{dx}{x}
\rightarrow \ln{y}=4\ln{x}+C
\end{equation*}
\begin{equation*}
\Rightarrow \frac{y}{x^4}=C
\end{equation*}

This concludes:
\begin{equation*}
u(x,y)=f(\frac{y}{x^4})
\end{equation*}
where $f: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function.
For $u(x,y)$ to be continuous at $(x,y)=(0,0)$, we should have $$u(0,0)=\lim_{x,y\rightarrow0}{u(x,y)}=0$$For this limit to exist, $\lim_{x,y\rightarrow 0}f(\frac{y}{x^4})$ should exist, meaning $f$ should tend to the limit-value regardless of the path on $x$-$y$ plane in which $x$ and $y$ tend to zero. In particular when we choose $y=Cx^4$ for some C, we get $\lim_{x,y\rightarrow 0}{f(\frac{y}{x^4}})=f(C)$. This being true $\forall C\in \mathbb{R}$ concludes $f$ is a constant function. The only continuous function satisfying PDE is the identically constant function.

b) Analogous to part (a), we start with writing equation of characteristic lines:
\begin{equation*}
\frac{dy}{-4y}=\frac{dx}{x}
\rightarrow \ln{y}=-4\ln{x}+C
\end{equation*}
\begin{equation*}\Rightarrow yx^4=C
\end{equation*}
This concludes:
  \begin{equation*}
u(x,y)=f(yx^4)\end{equation*}
where $f: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function.
This time, $$\lim_{x,y\rightarrow 0}f(yx^4)=\lim_{c \rightarrow 0}f(c)$$
For $u(x,y)$ to be continuous we just need to define $u(0,0)=\lim_{c \rightarrow 0}f(c)$. The easiest way to do is of course by getting $f$ continuous at zero and letting $u(0,0)=f(0)$.

« Last Edit: September 25, 2012, 06:45:53 PM by Victor Ivrii »

Victor Ivrii

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Re: Problem 2
« Reply #10 on: September 25, 2012, 06:53:47 PM »
The difference between two cases is that in one of them all trajectories have $(0,0)$ as the limit points and in another only those with $x=0$ or $y=0$ (node vs saddle).

Actually since in the saddle case $x^4y=C$ for $C\ne 0$ consists of two disjoint parts (with $x>0$ and with $x<0$), the values of $u$ on these parts are not necessarily equal and $u=f(x^4y)+x|x|^{-1}g(x^4y)$ with $g(0)=0$.