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### Messages - Luyu CEN

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1
##### FE / Re: FE6
« on: December 19, 2016, 10:56:31 AM »
Using spherical coodinates, equation (1) can be rewritten as
\begin{align*}
&r u_{tt} - (ru_{rr} + 2u_r) = 0\\
&(r u)_{tt} - (r u)_{rr} = 0
\end{align*}
Let
\begin{equation*}
v = ru
\end{equation*}
And we have
\begin{equation}
v_{tt} - v_{rr} = 0
\end{equation}
We translated the boundary conditions as
\begin{align} &v(r,0)=\left\{
\begin{aligned}
\end{aligned}
\end{align}
In addition, we have a boundary condition from the continuity of u:
\begin{equation}
v|_{r=0} = 0
\end{equation}
Since the boundary conditions are spherical symmetric and only depend on $r$ and $t$, we expect the solution to be a function of $r$ and $t$ also. Thus, we separate variables by $v(r,t) = T(t)T(r)$ and we get
\begin{equation}
\frac{T''(t)}{T(t)} = \frac{R''(r)}{R(r)} = -\lambda
\end{equation}
Note this is a Dirichlet-Dirichlet boundary eigenvalue problem for R(r) with l = 1, we have
\begin{align}
&\lambda_{n} = (\pi n)^2
&R_{n} = \sin(n\pi r)
&& n = 1, 2, ...
\end{align}
Plug $\lambda$ back to (7), and since $v_t(r,0)=0$
\begin{equation}
T(t) = A_n \cos(n\pi t)
\end{equation}
Therefore
\begin{equation}
v = \sum_{n=1}^{\infty}A_n \cos(n\pi t) \sin(n\pi r)
\end{equation}
We solve the coefficients $A_n$ by
\begin{equation*}
A_n = 2\int_{0}^{1}r\sin n\pi r\,dr = -\frac{2\cos(n\pi)}{n\pi}
\end{equation*}
When $r<1$,
\begin{align*}
&v = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{2}{n\pi}\cos(n\pi t)\sin(n\pi r)\\
&u = \frac{1}{r}\sum_{n=1}^{\infty} (-1)^{n+1}\frac{2}{n\pi}\cos(n\pi t)\sin(n\pi r)
\end{align*}
When $r\geq 1$
\begin{equation*}
u = 0
\end{equation*}

2
##### FE / Re: FE3
« on: December 16, 2016, 10:14:12 PM »
By separation of variables,
\begin{equation*}
u(x,t)= X(x) T(t)
\end{equation*}
\begin{align*}
& 4X(x) T''(t)= X''(x)T(t), \\[3pt]
& X(0)T(t)=X(2)T(t)=0,
\end{align*}
\begin{align*}
& 4\frac{ T''(t)}{T(t)}= \frac{X''(x)}{X(x)}, \\[3pt]
& X(0)=X(2)=0.
\end{align*}
\begin{equation}
\end{equation}
This is a Dirichlet-Dirichlet boundary eigenvalue problem for $X(x)$, we have
\begin{align*}
& \lambda_n = \frac{\pi^2n^2}{4}&& n=1,2,\ldots, \\[3pt]
&X_n(x)=\sin (\frac{\pi n x}{2}).
\end{align*}
Then plug $\lambda$ into (5) and solve for $T(t)$,
\begin{equation*}
T_n(t)=A_n \cos(\frac{\pi n}{4}t ) + B_n \sin (\frac{\pi n}{4}t )
\end{equation*}
Therefore,
\begin{equation*}
u_n(x,t)= \bigl( A_n \cos(\frac{\pi n}{4}t )+ B_n \sin (\frac{\pi n}{4}t )\bigr) \cdot
\sin (\frac{\pi n x}{2})
\,\,\,n=1,2,\ldots
\begin{equation}
u(x,t)= \sum_{n=1}^\infty \bigl( A_n \cos(\frac{\pi nt}{4} )+
B_n \sin (\frac{\pi nt}{4} )\bigr) \cdot
\sin (\frac{\pi n x}{2}).
\end{equation}
\begin{align}
&\sum_{n=1}^\infty A_n \sin (\frac{\pi n x}{2})=f(x),\\
&\sum_{n=1}^\infty \frac{\pi n}{4} B_n \sin (\frac{\pi n x}{2} )=0.
\end{align}
Calculate the coefficients $A_n$
\begin{multline*}
A_n = \int_0^2 \sin (\frac{\pi n x}{2}) f(x)\, dx = \int_0^1 x\sin (\frac{\pi n x}{2}) \,dx + \int_1^2 (2-x)\sin (\frac{\pi n x}{2})  \,dx  = \frac{2\sin \frac{\pi n x}{2}}{\bigr(\frac{\pi n}{2}\bigl)^2} = (-1)^{\frac{n-1}{2}} \frac{8}{(\pi n)^2}, \,\,\,\,0 \,\,\,\text{when n is even}
\end{multline*}
I also get $B_n = 0$,
The solution is given by
\begin{equation}
u(x,t)= \sum_{n=1 \,\text{odd} }^\infty (-1)^{\frac{n-1}{2}} \frac{8}{(\pi n)^2} \cos(\frac{\pi nt}{4})\cdot
\sin (\frac{\pi n x}{2}).
\end{equation}

3
##### FE / Re: FE2
« on: December 15, 2016, 06:36:09 PM »
This is 1D heat equation on half-line problem with Dirichlet boundary condition.
\begin{equation}
U(x,t)=\int
_{-\infty}^\infty G(x,y,t) F(y)\,dy.
\end{equation}
where $F(y)$ is the function by taking the odd continuation of $f(y)$ and
\begin{equation}
G(x,y,t)=
\frac{1}{2\sqrt{k\pi t}}e^{-\frac{(x-y)^2}{4kt}}
\end{equation}
Note that $f(y)$ is actually an odd function itself and substitute k with $\frac{1}{4}$, we have
\begin{equation}
U(x,t)=\int
_{-\infty}^\infty G(x,y,t) f(y)\,dy = \int
_{-\infty}^\infty \frac{1}{\sqrt{\pi t}}e^{-\frac{(x-y)^2}{t}} ye^{-y^2}\,dy
\end{equation}
\begin{multline*}
\int
_{-\infty}^\infty \frac{1}{\sqrt{\pi t}}e^{-\frac{(x-y)^2}{t}} ye^{-y^2}\,dy
= \frac{1}{\sqrt{\pi t}}e^{-\frac{x^2}{1+t}}\int
_{-\infty}^\infty ye^{-\frac{(1+t)(y-\frac{x}{1+t})^2}{t}} \,dy
\\=\frac{1}{\sqrt{\pi t}}e^{-\frac{x^2}{1+t}}\int
_{-\infty}^\infty \bigl(\frac{x}{1+t}e^{-\frac{(1+t)(y-\frac{x}{1+t})^2}{t}} + (y-\frac{x}{1+t})e^{-\frac{(1+t)(y-\frac{x}{1+t})^2}{t}}\bigr)\,dy
\\= \frac{1}{\sqrt{\pi t}}\sqrt{\frac{t}{1+t}}\frac{\sqrt{\pi}}{2}\frac{x}{1+t}e^{-\frac{x^2}{1+t}}(erf(\infty)-erf(-\infty)) + \frac{1}{\sqrt{\pi t}}e^{-\frac{x^2}{1+t}}\sqrt{\frac{t}{1+t}}e^{-\frac{(1+t)(y-\frac{x}{1+t})^2}{t}}|_{-\infty}^{\infty} \\= \frac{xe^\frac{-x^2}{1+t}}{\sqrt{1+t}(1+t)}
\end{multline*}
Therefore, the solution is given by
\begin{equation}
u(x,t) = \frac{xe^\frac{-x^2}{1+t}}{\sqrt{1+t}(1+t)}
\end{equation}

4
##### FE / Re: FE4
« on: December 14, 2016, 04:56:26 PM »
Yes thanks. I modify my solution to solve for the sector instead of the half disk
\begin{equation}
\Delta =\partial_r^2 + \frac{1}{r}\partial_r +
\frac{1}{r^2}\partial_\theta^2
\end{equation}
Let's plug in $u = R(r)P(\theta)$ and I get
\begin{equation*}
R''(r)P(\theta) + \frac{1}{r}R' (r)P(\theta) + \frac{1}{r^2} R(r) P''(\theta)=0
\end{equation*}
which could be rewritten as
\begin{equation*}
\frac{r^2 R''(\rho) + r R' (r)}{R(r)}+
\frac{P''(\theta)}{P(\theta)}=0
\end{equation*}
The two terms must be constant
\begin{align}
&r^2 R'' +r R' = \lambda R\\[3pt]
&P(\theta)=-\lambda P(\theta).
\end{align}
The initial conditions can be translated as
\begin{align}
& u =1\qquad &&\text{for  }  r = 4,\\
& u=0 &&\text{for  }  \theta = -\frac{5\pi}{6}, \frac{5\pi}{6} \end{align}
Therefore the boundary condition for P is $P(-\frac{5\pi}{6}) = P(\frac{5\pi}{6}) = 0$ and this is a Dirichlet-Dirichlet boundary problem with $l = \frac{5\pi}{3}$.
Thus we have
\begin{align}
&\lambda_{n} = (\frac{3n}{5})^2
&P_{n} = \sin(\frac{3n}{5}(\theta + \frac{5\pi}{6}))
&& n = 1, 2, ...
\end{align}
\begin{equation}
r^2 R'' +r R' - (\frac{3n}{5})^2 R = 0
\end{equation}
The Euler equation for this is $k(k-1) + k - (\frac{3n}{5})^2 = 0$ and we have $k = \frac{3n}{5}, -\frac{3n}{5}$
Since we are looking for continuous solutions,
\begin{equation}
R_n = A_n r^{\frac{3n}{5}}\\
u = \sum_{n = 1}^{\infty}A_n r^\frac{3n}{5} \sin(\frac{3n}{5}(\theta + \frac{5\pi}{6}))
\end{equation}
Apply the initial condition
\begin{equation}
\sum_{n = 1}^{\infty}A_n 4^\frac{3n}{5} \sin(n(\theta + \frac{5\pi}{6})) = 1
\end{equation}
Then we calculate the coefficients by
\begin{equation}
A_n=\frac{2}{4^\frac{3n}{5}\frac{5\pi}{3}}\int_{-\frac{5\pi}{6}}^{\frac{5\pi}{6}}\sin(\frac{3n}{5}(\theta+\frac{5\pi}{6}))\,d\theta=\frac{2}{4^\frac{3n}{5} \frac{5\pi}{3}}\int_0^{\frac{5\pi}{3}}\sin(\frac{3n}{5}x) \,dx=\frac{2}{4^\frac{3n}{5}\frac{5\pi}{3}\frac{3n}{5}}(\cos(\frac{3n}{5}x)|_{\frac{5\pi}{3}}^0)=\frac{1}{4^{\frac{3n}{5}-1}\frac{3\pi}{5} \frac{3n}{5}}\,\,\,\text{n odd, 0 otherwise}
\end{equation}
Therefore
\begin{equation}
u = \sum_{n = 1\,\,n\,\text odd}^{\infty} \frac{1}{4^{\frac{3n}{5}-1}\frac{3\pi}{5} \frac{3n}{5}}r^{\frac{3n}{5}} \sin(n(\theta + \frac{5\pi}{6})) = \sum_{n = 1\,\,n\,\text odd}^{\infty} \frac{100}{9n\pi4^{\frac{3n}{5}}} r^{\frac{3n}{5}} \sin(n(\theta + \frac{5\pi}{6}))
\end{equation}

5
##### FE / Re: FE4
« on: December 14, 2016, 12:58:27 PM »
\begin{equation}
\Delta =\partial_r^2 + \frac{1}{r}\partial_r +
\frac{1}{r^2}\partial_\theta^2
\end{equation}
Let's plug in $u = R(r)P(\theta)$ and I get
\begin{equation*}
R''(r)P(\theta) + \frac{1}{r}R' (r)P(\theta) + \frac{1}{r^2} R(r) P''(\theta)=0
\end{equation*}
which could be rewritten as
\begin{equation*}
\frac{r^2 R''(\rho) + r R' (r)}{R(r)}+
\frac{P''(\theta)}{P(\theta)}=0
\end{equation*}
The two terms must be constant
\begin{align}
&r^2 R'' +r R' = \lambda R\\[3pt]
&P(\theta)=-\lambda P(\theta).
\end{align}
The initial conditions can be translated as
\begin{align}
& u =1\qquad &&\text{for  }  r = 4,\\
& u=0 &&\text{for  }  \theta = -\pi/6, 5\pi/6 \end{align}
Therefore the boundary condition for P is $P(-\frac{\pi}{6}) = P(\frac{5\pi}{6}) = 0$ and this is a Dirichlet-Dirichlet boundary problem with $l = \pi$.
Thus we have
\begin{align}
&\lambda_{n} = n^2
&P_{n} = sin(n(\theta + \frac{\pi}{6}))
&& n = 1, 2, ...
\end{align}
\begin{equation}
r^2 R'' +r R' - n^2 R = 0
\end{equation}
The Euler equation for this is $k(k-1) + k - n^2 = 0$ and we have $k = n, -n$
Since we are looking for continuous solutions,
\begin{equation}
R_n = A_n r^n
u = \sum_{n = 1}^{\infty}A_n r^n sin(n(\theta + \frac{\pi}{6}))
\end{equation}
Apply the initial condition
\begin{equation}
\sum_{n = 1}^{\infty}A_n 4^n sin(n(\theta + \frac{\pi}{6})) = 1
\end{equation}
Then we calculate the coefficients by
\begin{equation}
A_n=\frac{2}{4^n \pi}\int_{-\frac{\pi}{6}}^{\frac{5\pi}{6}}\sin(n(\theta+\frac{\pi}{6}))\,d\theta=\frac{2}{4^n \pi}\int_0^{\pi}\sin(nx) dx=\frac{2}{n4^n \pi}(\cos(nx)|_{\pi}^0)=\frac{1}{n4^{n-1} \pi}\,\,\,\text{n odd, 0 otherwise}
\end{equation}
Therefore
\begin{equation}
u = \sum_{n = 1\,\,n\,\text odd}^{\infty} \frac{1}{n4^{n-1} \pi}r^n sin(n(\theta + \frac{\pi}{6}))
\end{equation}

6
##### FE / Re: FE1
« on: December 14, 2016, 12:11:48 PM »
The general solution for u is
\begin{equation*}u = \phi(x+3t) + \psi(x-3t)\end{equation*}
Now impose the initial conditions (2) and (3)
\begin{align}
&\phi(x) + \psi(x) = f(x)\\
&\phi'(x) + \psi'(x) = g(x)\end{align}
Solve the above to give
\begin{align}
&\phi(x) = \frac{1}{2}f(x) + \frac{1}{6}\int_{0}^{x} g(x') dx'  \\
&\psi(x)  = \frac{1}{2}f(x) - \frac{1}{6}\int_{0}^{x} g(x') dx'\end{align}
Therefore, when x > 3t,
\begin{equation}u(x) = \phi(x+3t) + \psi(x-3t) =  \frac{1}{2}[f(x+3t) + f(x-3t)] + \frac{1}{6}\int_{x-3t}^{x+3t} g(x') dx'\end{equation}
To find what u is when x<3t, impose initial condition (4)
\begin{equation*}\phi'(3t) + \psi'(-3t) = h(t)\end{equation*}
Which implies
\begin{equation*}\phi(3t) - \psi(-3t) = 3\int_{0}^{t} h(t') dt' + C\end{equation*}
Let $x =-3t, x<0, t = -\frac{x}{3}$
\begin{equation*}\phi(-x) - \psi(x) = 3\int_{0}^{-x/3} h(t')dt' + C\\
\psi(x) = \phi(-x) - 3\int_{0}^{-x/3} h(t')dt' + C\end{equation*}
Then for u to be continuous, $\psi(x)$ must be continuous at 0,
\begin{equation} \psi(0_{+}) = \frac{1}{2}f(0) = \phi(0) + C =\frac{1}{2}f(0)+C=\psi(0_{-})\end{equation}
We get C = 0
Therefore when x< 3t,
\begin{equation} u = \frac{1}{2}[f(x+3t)+f(3t-x)] + \frac{1}{6}[\int_{0}^{x+3t} g(x') dx' + \int_{0}^{3t-x} g(x') dx' ]- 3\int_{0}^{t-x/3} h(t')dt'\end{equation}
Now plug in $f, g, h$, we get
\begin{align}
&u = \cos (x+3t) + 3\cos (x-3t), x>3t\\[2pt]
&u = \cos (x+3t) + 2\cos (3t-x) + 1, 0<x<3t
\end{align}

7
##### Chapter 8 / Typo
« on: November 23, 2016, 04:26:33 PM »
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.9

If we follow from equation(7) http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.7
(9) should be
\begin{equation}
\sin^2(\phi)
\Phi'' +\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi,
\end{equation}
\begin{equation}
\sin^2(\phi)
\Phi'' +2\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi,
\end{equation}
There will be a coefficient 2 if we substitute $\Phi(\phi)=L(\cos(\phi))$ into the equation and use the chain rule to get its derivatives but I think not now.

8
##### Chapter 9 / Typos?
« on: November 23, 2016, 01:36:19 PM »
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.1.html#mjx-eqn-eq-9.1.8
The coefficient should be $\frac{1}{4\pi c^2 t}$ instead of $\frac{1}{4\pi c t}$

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.1.html#mjx-eqn-eq-9.1.18
$v^\pm_\omega$ should be$-\frac{1}{4\pi}\iiint |x-y|^{-1}e^{\mp i\omega c^{-1}|x-y|}f(y)\, dy$ instead of
$v^\pm_\omega= -\frac{1}{4\pi}\iiint |x-y|^{-1}e^{\mp i\omega c^{-1}|x-y|}\, dy.$

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.2.html#mjx-eqn-eq-9.2.3
This one I am not sure but I think the coefficient $\frac{1}{2}$ is missing if we follow equation (2)
Otherwise we could not use the quadratic form in (4) which leads to theorem 1 in this section.
Therefore, it should be
\begin{equation}
\iint_{\Sigma} \Bigl(
\frac{1}{2}\bigl(u_{t}^2+|\nabla u|^2\bigr)\nu_t -c^2 u_t \nabla u \cdot \nu_x\Bigr) \,d\sigma=0
\end{equation}
\begin{equation}
\iint_{\Sigma} \Bigl(
\bigl(u_{t}^2+|\nabla u|^2\bigr)\nu_t -c^2 u_t \nabla u \cdot \nu_x\Bigr) \,d\sigma=0
\end{equation}
And the equation after (5) http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.2.html#mjx-eqn-eq-9.2.5 also needs to be changed.

9
##### Chapter 6 / Re: Change of Variable
« on: November 08, 2016, 08:56:21 PM »
Fixed!

10
##### Chapter 6 / Change of Variable
« on: November 04, 2016, 01:20:01 AM »
\begin{equation*}x=r\cos (\theta)\end{equation*}, \begin{equation*}y=r\sin(\theta)\end{equation*}
\begin{equation*}
\left\{
\begin{aligned}
&r=\sqrt{x^2+y^2},\\
&\theta = \arctan \bigl(\frac{y}{x}\bigr);
\end{aligned}\right.
\end{equation*}

\begin{equation}
\left\{\begin{aligned}
&r_x=\frac{x}{\sqrt{x^2+y^2}}=\frac{r\cos(\theta)}{r}=\cos(\theta),\\
&r_y=\frac{y}{\sqrt{x^2+y^2}}=\frac{r\sin(\theta)}{r}=\sin(\theta),\\
&\theta_x = \frac{1}{(\frac{y}{x})^2+1} \cdot -\frac{y}{x^2}=-\frac{y}{y^2+x^2}=-\frac{r\sin(\theta)}{r^2}=-r^{-1}\sin(\theta),\\
&\theta_y = \frac{1}{(\frac{y}{x})^2+1} \cdot \frac{1}{x}=\frac{x}{y^2+x^2}=\frac{r\cos(\theta)}{r^2}=r^{-1}\cos(\theta).
\end{aligned}\right.
\end{equation}
By the chain rule
\begin{equation}
\left\{
\begin{aligned}
&\partial_x = r_x\partial_r  + \theta_x\partial_\theta  =\cos(\theta)\partial_r - r^{-1}\sin(\theta)\partial_\theta,\\
&\partial_y = r_y\partial_r  + \theta_y\partial_\theta  = \sin(\theta)\partial_r + r^{-1}\cos(\theta)\partial_\theta
\end{aligned}\right.
\end{equation}
Plug the above into the laplacian and expand the squares
\begin{multline*}
\Delta = \partial_x^2+ \partial_y^2=
\bigl(\cos(\theta)\partial_r - r^{-1}\sin(\theta)\partial_\theta\bigr)^2 +
\bigl(\sin(\theta)\partial_r +
r^{-1}\cos(\theta)\partial_\theta\bigr)^2 =\\
\bigl(\cos(\theta)\partial_r\cos(\theta)\partial_r -\cos(\theta)\partial_r\cdot r^{-1}\sin(\theta)\partial_\theta -r^{-1}\sin(\theta)\partial_\theta \cos(\theta)\partial_r+r^{-1}\sin(\theta)\partial_\theta r^{-1}\sin(\theta)\partial_\theta\bigr) + \\
\bigl(\sin(\theta)\partial_r\sin(\theta)\partial_r+ \sin(\theta)\partial_r r^{-1}\cos(\theta)\partial_\theta + r^{-1} \cos(\theta) \partial_\theta \sin(\theta)\partial_r+r^{-1}\cos(\theta)\partial_\theta r^{-1}\cos(\theta)\partial_\theta\bigr).
\end{multline*}
By the product rule, we have
\begin{multline*}
\Delta=
\bigl(\cos^2(\theta)\partial_r^2 + \frac{1}{r^2}\cos(\theta)\sin(\theta)\partial_\theta +r^{-1}\sin^2(\theta)\partial_r-2r^{-1}\sin{\theta}\cos{\theta}\partial_r\partial_\theta+ r^{-2}\cos^2(\theta)\partial_\theta^2 - r^{-2}\cos(\theta)\sin(\theta)\partial_\theta\bigr) +\\
\bigl(\sin^2(\theta)\partial_r^2 - \frac{1}{r^2}\sin(\theta)cos(\theta)\partial_\theta + r^{-1}\cos^2(\theta)\partial_r + 2r^{-1}\sin{\theta}\cos{\theta}\partial_r\partial_\theta +
r^{-2}\sin^2(\theta)\partial_\theta^2+r^{-2}\sin(\theta)\cos(\theta)\partial_\theta\bigr)=\\
\partial_r^2 +\frac{1}{r}\partial_r +\frac{1}{r^2}\partial_\theta^2.
\end{multline*}

11
##### Chapter 2 / The definition of semilinear equation in 2.1 of textbook
« on: September 27, 2016, 05:22:03 PM »