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?
Show Posts
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
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.
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.
\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.