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### Messages - Bruce Wu

Pages: 1 2 3 
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 cyclesâ€”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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