Messages

76

HA3 / Re: HA3-P3

« on: October 03, 2015, 05:52:31 PM »

Just wondering, is there a typo in the original problem? The equation given is: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{tt} \end{equation}
But I think the last term should be with respect to x: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{xx}
\end{equation}

77

Quiz 1 / Re: Quiz1-P1

« on: October 02, 2015, 07:46:33 PM »

I think that the first one should be considered linear homogeneous. We see in section 1.3 Classification of Equations in the textbook [http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter1/S1.3.html] that a linear equation may have variable coefficients. In this case, the x is a variable coefficient, but the equation is still linear.

At least, this is what I think, but I might be wrong. Does anyone else have any idea?

You are right. V.I.

78

Quiz 1 / Quiz 1 - P3

« on: October 02, 2015, 12:35:31 AM »

This problem was: \begin{equation} \begin{cases}
u_x + 3u_y = xy \\
u|_{x=0} = 0
\end{cases}
\end{equation}
And my solution is: \begin{equation}
\frac{dx}{1} = \frac{dy}{3} = \frac{du}{xy}
\end{equation}

\begin{equation}
3dx = dy
\end{equation}

\begin{equation}
3x - y = C
\end{equation}

\begin{equation}
du=xydx
\end{equation}

\begin{equation}
du = x(3x-C)dx
\end{equation}

\begin{equation}
du = (3x^2-Cx)dx
\end{equation}

\begin{equation}
u = x^3 - \frac{C}{2}x^2 + \phi(3x-y)
\end{equation}

\begin{equation}
u = x^3 - \frac{3x-y}{2}x^2 + \phi(3x-y)
\end{equation}

With initial condition, we have:
\begin{equation}
u|_{x=0}=\phi(-y)=0
\end{equation}

So: \begin{equation} \phi = 0 \end{equation}

So the final solution will be: \begin{equation}
u = x^3 - \frac{3x-y}{2}x^2
\end{equation}

79

HA2 / Re: HA2-P1

« on: October 01, 2015, 01:53:36 PM »

a)
i. We have characteristics: \begin{equation}
\frac{dt}{2} = \frac{dx}{3}
\end{equation}
So:
\begin{equation}
3t = 2x + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(3t - 2x)
 \end{equation}


ii. We have characteristics: \begin{equation}
\frac{dt}{1} = \frac{dx}{t}
\end{equation}
So:
\begin{equation}
\frac{t^2}{2} = x + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(\frac{t^2}{2} - x)
 \end{equation}


iii. We have characteristics: \begin{equation}
\frac{dt}{1} = \frac{dx}{x}
\end{equation}
So:
\begin{equation}
t = \ln(x) + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(t - \ln(x))
 \end{equation}


iv. We have characteristics: \begin{equation}
\frac{dt}{1} = \frac{dx}{x^2}
\end{equation}
So:
\begin{equation}
t = \frac{-1}{x} + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(t + \frac{1}{x})
 \end{equation}


v. We have characteristics: \begin{equation}
\frac{dt}{1} = \frac{dx}{x^3}
\end{equation}
So:
\begin{equation}
t = \frac{-1}{2x^2} + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(t + \frac{1}{2x^2})
 \end{equation}

80

Textbook errors / Error in HA 2 Problem 1?

« on: October 01, 2015, 01:22:09 PM »

Hi Professor,

HA2 Problem 1 a) has as the fifth problem, \begin{equation}
u_x + x^3u_x = 0
\end{equation}

Did you perhaps mean \begin{equation}
u_t + x^3u_x = 0
\end{equation}

to make an equation involving partial derivatives with respect to t AND x, like the other problems?

81

Web Bonus = Sept / Re: Web bonus problem : Week 3 (#3)

« on: September 29, 2015, 09:49:35 PM »

I tried to solve this problem but I could not get all the way through. I will post what I was able to get.

We start with the equation \begin{equation}\label{eq:problem}
u_{tt} - c^2u_{xx} + \sin(u) = 0
\end{equation}
The equation can be solved by the use of the characteristic coordinates, as discussed in class. As a reminder, we define: \begin{equation} \begin{cases}
\zeta = x + ct \\
\eta = x - ct
\end{cases} \end{equation}
Following the logic used in lecture/the online textbook, we can see that \begin{equation} x = \frac{1}{2}(\zeta + \eta) \end{equation} and \begin{equation} t = \frac{1}{2c}(\zeta - \eta) \end{equation}
By the chain rule (omitting several steps, as they are given in the textbook), we can show that:
\begin{equation}
-4c^2u_{\zeta\eta} = -\frac{1}{4}(c\partial_x + \partial_t)(c\partial_c - \partial_t) = u_{tt} - c^2u_{xx} = -\sin(u)
\end{equation}
We can see that in the characteristic coordinates, the original equation \eqref{eq:problem} becomes: \begin{equation}
u_{\zeta\eta} = \frac{1}{4c^2}\sin(u)
\end{equation}

Now here I get a little stuck. I have tried to progress with the solution after some thought but I am not sure if it will make sense.

Rewriting the equation, we have: \begin{equation}
4c^2\frac{\partial u}{\sin(u)} = \partial_\zeta\partial_\eta
\end{equation}
So integrating both sides, we get: \begin{equation}
-4c^2\ln(\cot(u) + \csc(u)) = \phi(\zeta)\partial_\eta + C
\end{equation}

Now here is where I really get stuck. I guess I have to integrate again, but this will lead to a nasty solution. For now I will leave the post in case someone is able to add on to the solution or provide a hint. I will return to the problem after thinking about it for a while.