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Messages - Bruce Wu

Pages: 1 2 3 [4]
46
HA3 / Re: HA3-P2
« on: October 08, 2015, 01:11:58 AM »
The yellow diamond is the only place where both initial conditions are satisfied. Will we have to include in our answer that it is where the solution is valid?

47
HA3 / Re: HA3-P6
« on: October 08, 2015, 12:03:01 AM »
Since we have already solved the homogeneous Goursat problem in section 2.3 problem 5, and the contribution from the right hand expression is given as a hint, would it be ok to just add the two known contributions together yielding the final solution without the intermediate steps?

48
Textbook errors / Section 3.1 error in equation 9
« on: October 07, 2015, 07:01:08 PM »
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter3/S3.1.html#mjx-eqn-eq-3.1.9

The $kt$ in the denominator should also be under the square root. Similar error in the line below as a part of remark 2.

49
Quiz 1 / Re: Quiz1-P1
« on: October 02, 2015, 11:19:36 PM »
I think the first one is semilinear because it depends on definition 2 in Chapter 2.1 in textbook. In this case, b=b (x, y) and I don't think similinear is type of linear.

But there is no right hand expression that is a non-zero function of $u$. If there was, say $u_{x}+xu_{y}=u^{2}$ then it would be semilinear.

50
Web Bonus = Sept / Re: Web bonus problem : Week 3 (#2)
« on: September 27, 2015, 05:21:51 PM »
a) By plugging $u=\phi(x-vt)$ into equations (16) and (17) on the problems page, we obtain:
\begin{equation}
v^{2}\phi''-c^{2}\phi''+m^{2}\phi=0
\end{equation}\begin{equation}
v^{2}\phi''-c^{2}\phi''-m^{2}\phi=0
\end{equation}
for (16) and (17), respectively.
Proceeding to solve (1) using methods of linear ODEs yields:
\begin{equation}
\phi''=-\frac{m^{2}}{v^{2}-c^{2}}\phi\Rightarrow \phi(z)=a\sin \left(\frac{m}{\sqrt{v^{2}-c^{2}}}z\right)+b\cos\left(\frac{m}{\sqrt{v^{2}-c^{2}}}z\right)
\end{equation}
Where $a,b$ depend on initial conditions. This is assuming that $v^{2} > c^{2}$, which gives us a periodic solution that is bounded, as desired. Proceeding the same way with (2):
\begin{equation}
\phi''=\frac{m^{2}}{v^{2}-c^{2}}\phi
\end{equation}
If we assume again that $v^{2} > c^{2}$, we will get:
\begin{equation}\large
\phi(z)=ae^{\frac{m}{\sqrt{v^{2}-c^{2}}}z}+be^{-\frac{m}{\sqrt{v^{2}-c^{2}}}z}
\end{equation}
But this solution does not tend to $0$ as $|z|\rightarrow\infty$ unless $a=b=0$, and we do not want the trivial solution. Therefore $v^{2}$ must be less than $c^{2}$. Then:
\begin{equation}
\phi''=-\frac{m^{2}}{c^{2}-v^{2}}\phi\Rightarrow \phi(z)=a\sin \left(\frac{m}{\sqrt{c^{2}-v^{2}}}z\right)+b\cos\left(\frac{m}{\sqrt{c^{2}-v^{2}}}z\right)
\end{equation}
Now $u=\phi(x-vt)$ using $\phi(z)$ in (3) and (6) solve the original problems (16) and (17), respectively.

b) Plugging $u=\phi(x-vt)$ into problem (18):
\begin{equation}
-v\phi'-K\phi'''=0
\end{equation}\begin{equation}
\phi'''+\frac{v}{K}\phi'=0
\end{equation}
If $v$ and $K$ have the same sign, then the characteristic equation will have imaginary roots and give a bounded periodic solution:
\begin{equation}
\phi(z)=a\sin \left(\sqrt{\frac{v}{K}}z\right)+b\cos \left(\sqrt{\frac{v}{K}}z\right)+c
\end{equation}
Same with problem (19):
\begin{equation}
-v\phi'-iK\phi''=0
\end{equation}\begin{equation}
\phi''-\frac{iv}{K}\phi'=0
\end{equation}
In this case, the solution will be suitable regardless of the value of $v$, and the solution will be complex:
\begin{equation}\large
\phi(z)=ae^{\frac{iv}{K}z}+b=a\left(\cos \left(\frac{vz}{K}\right)+i\sin \left(\frac{vz}{K}\right)\right)+b
\end{equation}
Same with problem (20):
\begin{equation}
v^{2}\phi''+K\phi''''=0\Rightarrow\phi''''+\frac{v^{2}}{K}\phi''=0
\end{equation}
Here, since $v$ only appears in terms of its square, we do not need to worry about its value. As long as $K>0$, solving the above ODE gives us a suitable solution:
\begin{equation}
\phi(z)=a\sin \left(\frac{vz}{\sqrt{K}}\right)+b\cos \left(\frac{vz}{\sqrt{K}}\right)+c+d
\end{equation}
Again, $u=\phi(x-vt)$ using $\phi(z)$ in (9), (12), and (14) solve the original problems (18), (19), and (20), respectively.

51
MAT244 Math--Lectures / Re: Integrable/non-integrable systems
« on: December 07, 2014, 12:13:45 AM »
Thanks

52
MAT244 Math--Lectures / Re: Integrable/non-integrable systems
« on: December 06, 2014, 08:52:31 PM »
So please tell me if I am understanding it correctly now,

y = c*x^2 for all real numbers c represents a particular node. However, H(x,y) = y/x^2 = c is not preserved at x = 0, so it is not integrable, even though it has an explicit solution.

53
MAT244 Math--Lectures / Re: Integrable/non-integrable systems
« on: December 06, 2014, 02:09:55 PM »
Then trajectories are level lines of $H$ (or their parts). This precludes nodes and spiral points (and limi cycles—which we have not studied) and allows only saddles and centers (provided at stationary points Hessian of $H$ is non-degenerate).

Also, all diagonal linear 2x2 systems are integrable, but those can be nodes, how is that explained?

54
MAT244 Math--Lectures / Re: Integrable/non-integrable systems
« on: December 06, 2014, 01:41:43 PM »
But from H(x,y) = c, how do we determine the directions of trajectories?

55
MAT244 Math--Lectures / Integrable/non-integrable systems
« on: December 05, 2014, 03:23:08 PM »
What is the difference between integrable and non-integrable systems of first order ODEs?

I remember the professor talking about it in class but I cannot find it in the textbook. I know what it means, but how does this property affect its solutions?

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