Toronto Math Forum

Posted by: Rouhollah Ramezani
« on: October 01, 2012, 09:00:03 pm »

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 $\{−2t<x<2t\}$ (`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.

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
$$p_{xx}=e_{tx}$$
$$=e_{xt}=p_{tt}$$
$$\rightarrow p \text{ satisfies the PDE.}$$

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*}

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.

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.

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 $u(0,0)=f(0)$.