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### Messages - Rouhollah Ramezani

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1
##### Final Exam / Re: problem 5
« on: December 20, 2012, 02:51:13 PM »
By Mean Value property for harmonic functions,
$$u(x)=\frac{1}{\pi r^2}\int_{B_r(x)}u dA$$
where $x \in R^2$ is a point, $B_r(x)$ is the ball of radius $r$ centred at $x$ and the integration is over area.
When $r \to \infty$, integral above becomes the average of $u$ over whole plain. Since this average is a constant, we conclude $u(x)$ should be a constant function.
To state this argument more rigorously;  Let $x_1$ and $x_2$ be distinct in $R^2$ and $|x_1-x_2|=a$:
Then by Mean Value theorem
u(x_2)=\frac{1}{\pi r^2}\int_{B_r(x_2)}u dA

u(x_1)=\frac{1}{\pi (r+a)^2}\int_{B_{r+a}(x_1)}u dA

Where ${B_r(x_2)}$ is the $r$-ball around $x_2$ and ${B_{r+a}(x)}$ is the ball with radius $r+a$ around $x_1$.

Rewriting (2) we get:
$$u(x_1)=\frac{1}{\pi (r+a)^2}\int_{B_{r}(x_2)}u dA + \frac{1}{\pi (r+a)^2}\int_{S}u dA$$
Where $S=B_{r+a}(x_1)$\ $B_{r}(x_2) \in R^2$ is the hashed area in the graph.
We shall prove $u(x_2)-u(x_1)=0$. Since (1) and (2) are true for all $r$, it suffices to show $$lim_{r \to \infty} \frac{1}{\pi r^2}\int_{B_r(x_2)}u dA- \frac{1}{\pi (r+a)^2}\int_{B_{r+a}(x_1)}u dA$$
$$=lim_{r \to \infty} \frac{1}{\pi r^2}\int_{B_r(x_2)}u dA-\frac{1}{\pi (r+a)^2}\int_{B_{r}(x_2)}u dA - \frac{1}{\pi (r+a)^2}\int_{S}u dA=0$$
But first two terms cancel out in limit and remains
$$u(x_2)-u(x_1)=lim_{r \to \infty} \frac{-1}{\pi (r+a)^2}\int_{S}u dA=0$$
Since $u$ is bounded we can write

\begin{aligned}
lim_{r \to \infty} \frac{1}{\pi (r+a)^2}\int_{S}u dA & \leq lim_{r \to \infty} \frac{1}{\pi (r+a)^2} u_{max} S_a \\
& =lim_{r \to \infty} \frac{u_{max}[\pi (r+a)^2-\pi (r)^2]}{\pi (r+a)^2} \\
& =lim_{r \to \infty} \frac{u_{max}(a^2-2ar)}{(r+a)^2}=0 \\
\end{aligned}

Where $u_{max}$ is the upper bound of $u$ and $S_a=\pi (r+a)^2-\pi (r)^2$ is the area of $S$.
Hence $u(x_2)-u(x_1)=0 \qquad \forall x_1,x_2 \in R^2$ and $u$ is constant.

2
##### Home Assignment 2 / Re: Problem 1 -- not done yet!
« on: October 07, 2012, 02:54:17 AM »
Posted by: Rouhollah Ramezani
Â« on: October 01, 2012, 09:00:03 pm Â»

A) is correct

B) definitely contains an error which is easy to fix. Why I know about error? --  solution of RR  is not $0$ as $x=t$

C) Contains a logical error in the domain $\{âˆ’2t<x<2t\}$ (middle sector) which should be found and fixed. Note that the solution of RR there is not in the form $\phi(x+2t)+\psi(x-2t)$

B) is fixed now.
C) is also amended, but I probably failed to spot the "logical error" and it is still there.

PS MathJax is not a complete LaTeX and does not intend to be, so it commands like \bf do not work outside of math snippets (note \bf); MathJax has no idea about \newline as it is for text, not math. For formatting text use either html syntax (in plain html) or forum markup

Yes, I realized that after. And I found out from your other post that we can actually use an html editor+MathJax instead of LaTeX. right?

PPS \bf is deprecated, use \mathbf instead

Wilco.

3
##### Home Assignment 2 / Re: Problem 3
« on: October 01, 2012, 09:50:25 PM »
Should we consider solutions on different intervals (i.e. x>2t and 0<x<2t) since they differ in part (a) and part (d)?

To be impeccable, Yes. (but I guess solution where $x>2t$ is too trivial/standard to be of any particular interest here.)

4
##### Home Assignment 2 / Re: Problem4
« on: October 01, 2012, 09:01:58 PM »
a) Both Proofs are straightforward:
$$e_t=u_tu_{tt}+u_xu_{xt}$$
$$=u_tu_{xx}+u_xu_{tx}$$
$$=\frac{\partial u_tu_x}{\partial x}$$
$$=p_x$$
Above we used definition of $p$ and $e$ as well as the original PDE $u_{tt}=u_{xx}$.

$$p_t=u_{tt}u_x+u_tu_{xt}$$
$$=u_{xx}u_x+u_tu_{tx}$$
$$=\frac{\partial}{\partial x}\frac{1}{2}(u_t^2+u_x^2)$$
$$=e_x$$
Again, we used definition of $p$ and $e$ as well as the original PDE $u_{tt}=u_{xx}$.

b) Direct result of differentiating the identities of part (a) is
$$e_{tt}=p_{xt}$$
$$=p_{tx}=e_{xx}$$
$$\rightarrow e \text{ satisfies the PDE.}$$
And
$$p_{xx}=e_{tx}$$
$$=e_{xt}=p_{tt}$$
$$\rightarrow p \text{ satisfies the PDE.}$$

5
##### Home Assignment 2 / Re: Problem 3
« on: October 01, 2012, 09:01:11 PM »
a) Dirichlet boundary condition at $x=0$ means we need an odd continueation of $u|_{t=0}$ and $u_t|_{t=0}$; That is
$$u|_{t=0}=0,\qquad x<0$$
$$u_t|_{t=0}=-1,\qquad x<0$$
By application of d'Alembert's formula, solution to this problem is
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}1\mathrm{d}s+\int_{x-2t}^0-1\mathrm{d}s\Bigr]$$
$$= \frac{1}{2}x$$

b) Neumann boundary condition at $x=0$ means we need an even continueation of $u|_{t=0}$ and $u_t|_{t=0}$; That is
$$u|_{t=0}=0,\qquad x<0$$
$$u_t|_{t=0}=1,\qquad x<0$$
since both functions are even.
By application of d'Alembert's formula, solution to this problem is
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}1\mathrm{d}s+\int_{x-2t}^01\mathrm{d}s\Bigr]$$
$$= t$$

c) Dirichlet boundary condition at $x=0$ means we need an odd continueation of $u|_{t=0}$ and $u_t|_{t=0}$; That is
$$u|_{t=0}=0,\qquad x<0$$
$$u_t|_{t=0}=x,\qquad x<0$$
since both functions are odd.
By application of d'Alembert's formula, solution to this problem is
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}x\mathrm{d}s+\int_{x-2t}^0x\mathrm{d}s\Bigr]$$
$$= \frac{1}{8}\Bigl[(x+2t)^2+(x-2t)^2\Bigr]$$
$$=xt$$

d) Neumann boundary condition at $x=0$ means we need an even continueation of $u|_{t=0}$ and $u_t|_{t=0}$; That is
$$u|_{t=0}=0,\qquad x<0$$
$$u_t|_{t=0}=-x,\qquad x<0$$
By application of d'Alembert's formula, solution to this problem is
$$u(t,x)=\frac{1}{4}\Bigl[\int_0^{x+2t}x\mathrm{d}s+\int_{x-2t}^0-x\mathrm{d}s\Bigr]$$
$$= \frac{1}{8}\Bigl[(x+2t)^2-(x-2t)^2\Bigr]$$
$$=\frac{1}{4}x^2+t^2$$

edit: No need to say all solutions are for $x<2t$.
edit: Fixed integral limits.

6
##### Home Assignment 2 / Re: Problem 1
« on: October 01, 2012, 09:00:03 PM »
A) The problem as is has a unique solution. No extra consitions are necessary. Since $x>3t$, we are confident that $x-2t$ is always positive. I.e. initial value functions are defined everywhere in domain of $u(t,x)$. Using d'Alembert's formula we write:
\begin{equation*}
u(t,x)=\frac{1}{2}\Bigl[e^{-(x+2t)}+e^{-(x-2t)}\Bigr]+\frac{1}{4}\int_{x-2t}^{x+2t}e^{-s}\mathrm{d}s
\end{equation*}
$$= \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
Stated in this way, $u(t,x)$ is determined uniquely in its domain. OK

B) In this case, we need an extra boundary condition at $u_{|x=t}=0$ to find the unique solution:
For $x>2t$ general formula for $u$ is as part (A):
$$u(t,x) = \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
For $t<x<2t$, story is different:
$$u(t,x)= \phi(x+2t)+\psi(x-2t)$$
where $\phi(x+2t)= \frac{1}{4}\bigl[e^{-(x+2t)}+1\bigr]$ is determined by initial conditions at $t=0$. To find $\psi(x-2t)$, we impose $u_{|x=t}=0$ to solution:
$$u_{|x=t}=\phi(3t)+\psi(-t)=0$$
$$\Rightarrow \psi(s)=-\phi(-3s)$$
$$=\frac{-1}{4}\bigl[e^{3s}+1\bigr]$$
Hence the general solution for $t<x<2t$ is:
$$u(t,x)= \frac{1}{4}\bigl[e^{-(x+2t)}+1\bigr]-\frac{1}{4}\Bigl[e^{3x-6t}+1\Bigr]$$
Note that we would not be able to determine $\psi$ if we did not have the extra condition  $u|_{x=t}$. Also note that $u_x|_{x=t}=\frac{1}{2}(e^{-3t}) \neq 0$. This means the problem would have been overdetermined, without any solution, if we considered boundary condition $u|_{x=t}=u_x|_{x=t}=0$.

C) In this case we need to impose the strongest boundary condition to get the unique solution:
Case $x>2t$ is identical to part (A) and (B):
$$u(t,x) = \frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}$$
We find the solution in region $-3t<x<-2t$ by imposing boundary conditions $u|_{x=-3t}=u_x|_{x=-3t}=0$ to $u(t,x)= \phi(x+2t)+\psi(x-2t)$. This gives
$$\phi(-t)+\psi(-5t)=0$$
$$\phi'(-t)+\psi'(-5t)=0$$
Differentiating first equation and adding to the second we get $-4\psi'(-5t)=0$. Therefore $\psi(s)=C$, $\phi(s)=-C$ and $u(t,x)$ is identically zero.Note that we would not be able to find $\phi$ and $\psi$ uniquely, if we did not have both boundary conditions at $x=-3t$.

From continuety of $u$ in $t>0$, $x>-3t$, we conclude $u|_{x=-2t}=0$. This helps us to find solution for $-2t<x<2t$. Analogous to part (B) we write:
$$u(t,x)= \phi(x+2t)+\psi(x-2t)$$
Impose $u|_{x=-2t}=0$ to $u$ to get $\psi(-4t)=-\phi(0)=C$. Therefore solution here is:
$$u(t,x)=\frac{1}{4}e^{-(x+2t)}+C$$
By continuity at $x=2t$, we get $C=\frac{3}{4}$. General solution for part (C) can be explicitly formulated as
\begin{equation*}
u(x,y)=
\left\{\begin{aligned}[h]
&\frac{1}{4}e^{-(x+2t)}+\frac{3}{4}e^{-(x-2t)}, & x>2t\\
&\frac{1}{4}e^{-(x+2t)}+\frac{3}{4}, & -2t<x<2t\\
&0, & -3t<x<-2t\\
\end{aligned}
\right.
\end{equation*}

7
##### Home Assignment 1 / Re: Problem 4
« 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?

8
##### Home Assignment 1 / Re: Problem 4
« 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:

u(r,\theta)=r^2\theta+\phi(r)

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:

u_{rr}=2\theta+\phi''(r) \\

\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.

9
##### Home Assignment 1 / Re: Problem 6
« on: September 25, 2012, 09:58:02 PM »
a) Directly implied by definition of $\rho(x,t)$ is $N(t,a,b)=\int_{a}^{b}\rho(t,x) dx$.

b) by definition of $q$ and conservation of cars we have:
$$\frac{\partial N}{\partial t}(t,a,b)=\lim_{h \rightarrow 0} \frac{N(t+h,a,b)-N(t,a,b)}{h} \\ =\lim_{h \rightarrow 0} \frac{h(q(t,a)-q(t,b))}{h} \\ =q(t,a)-q(t,b)$$

c) Differentiating integral form of $N(t,a,b)$ with respect to $t$
$$\frac{\partial N}{\partial t}=\int_{a}^{b}\rho_t(t,x) dx$$
making it equal to result of part (b) we get the integral form of "conservation of cars":
$$\int_{a}^{b}\rho_t(t,x) dx=q(t,a)-q(t,b)$$

d) RHS of above equation can be expressed as $\int_{a}^{b}-q_x(t,x) dx$. Therefore
\int_{a}^{b}\rho_t(t,x) dx=\int_{a}^{b}-q_x(t,x) dx

Since $a$ and $b$ are arbitrary, $(1)$ implies $\rho_t=-q_x$. The PDE $\rho_t+q_x=0$ is  conservation of cars equation.

e) Letting $q=c\rho$, we will get a first order linear PDE with constant coeffiecient:
$$\rho_t+c\rho_x=0 \\ \rightarrow \rho(t,x)=\phi(x-ct)$$
The last equation means density of cars is moving at some constant pace $c$ in time, i.e. all cars are driving at speed $c$.
Intuitively we know that driver's speed has negative correlation with traffic density. A more realistic choice for $c$ is to let it be a monotone decreasing function of $\rho$. In this case however, conservation of cars equation is not linear anymore.  This is discussed in detail by Prof. Ivrii in last year's forum.

10
##### Home Assignment 1 / Re: Problem 4
« 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.

11
##### Home Assignment 1 / Re: Problem 5
« on: September 25, 2012, 05:27:11 PM »
a) This is a 1-D wave PDE general solution of which is discussed in the class.

\begin{equation*}
u(t,x)=\phi(x+3t) + \phi(x-3t)
\end{equation*}
b) Using D'Alemblert formula we get:
\begin{equation*}
u(t,x)=\frac{1}{2} \bigl[ (x+3t)^2-(x-3t)^2 \bigr] + \frac{1}{6}\int_{x-3t}^{x+3t} s\,ds \\
=\frac{1}{2}\bigl[ (x+3t)^2-(x-3t)^2 \bigr]+\frac{1}{12}(x+3t)^2-\frac{1}{12}(x-3t)^2 \\
=\frac{7}{6}(x+3t)^2+\frac{5}{6}(x-3t)^2
\end{equation*}

c) We impose Goursat problem boundary conditions to general solution and get:
$$\phi(6t)+\psi(0)=t \\ \phi(0)+\psi(-6t)=2t$$
Letting $t=0$ in first equation and subtracting it from the second we get $\phi(0)=\psi(0)=0$. Therefore $$\phi(t)=\frac{t}{6} \\ \psi(t)=\frac{-t}{3}$$
Final solution for Goursat problem is $$u(t,x)=\frac{-1}{6}x+\frac{3}{2}t$$

12
##### Home Assignment 1 / Re: Problem 2
« 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 $u(0,0)=f(0)$.

13
##### Home Assignment 1 / Re: Problem 1
« on: September 25, 2012, 04:28:16 PM »
a,b) This is a first order linear PDE with constant coefficient. We begin by writing  equation of characteristic lines:
\begin{equation*}
\frac{dy}{-3}=dx
\Rightarrow y=-3x+C
\end{equation*}
This implies $u(x,y)=\phi(3x+y)$, for some arbitrary $\phi$. Letting $u|_{x=0}=e^{-y^2}$ we get:
$$\phi(y)=e^{-y^2}$$
Hence solution to IVP is$$u(x,y)=e^{-(3x+y)^2}$$
c) With initial condition $u|_{x=0}=y$ we have $\phi(y)=y$, $y>0$. Hence the solution is $$u(x,y)=3x+y$$
Since $3x+y>0$ $\forall x,y>0$, this solution is valid everywhere in the domain of $u(x,y)$.

d) In this case, the solution is defined only where $3x+y>0$ i.e. where $y>|3x|$. To find solution on the whole domain, we need to impose another condition e.g. at $y=0$. Letting $u|_{y=0}=x$ we get:
$$\phi(3x)=x, (x<0) \rightarrow \phi(x)=\frac{x}{3}, (x<0)$$ $$\Rightarrow u(x,y)=x+ \frac{y}{3}$$
Above equation is constrained to $x+ \frac{y}{3}<0$.
Final solution would be:
\begin{equation*}
u(x,y)= \left\{
\begin{aligned}
&3x+y,&y>|3x|\\
&x+ \frac{y}{3}, &y<|3x|
\end{aligned}
\right.
\end{equation*}

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