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Topics - Shentao YANG

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1
Q-bonus / Q-bonus
« on: December 01, 2016, 08:46:16 PM »
Problem: We need to construct the fastest slide from point $(0,0)$ to $(a,-h)$, If $u(x)$ describes its shape then time is:
$$\begin{equation}
T= \int_0^a  \frac{1}{\sqrt{2gu}} \sqrt{1+u^{\prime\,2}}\,dx.
\label{eq-10.P.3}
\end{equation}$$
Find solution satisfying $u(0)=0$ and $u(a)=-h$

Solution: Let $L = {1 \over {\sqrt u }}\sqrt {1 + {u^{\prime \,2}}} $, since the factor $2g$ does not matter. Note that here $L$ does not depend on $x$, so we have$$u'{{\partial L} \over {\partial u'}} - L = c$$, where $c$ is some constant. Then,
$$\eqalign{
  & {1 \over {\sqrt {1 + {{(u')}^2}} }} = c\sqrt u   \cr
  &  \Rightarrow {1 \over {1 + {{(u')}^2}}} = {c^2}u  \cr
  &  \Rightarrow u' = {{du} \over {dx}} = \sqrt {{1 \over {{c^2}u}} - 1}   \cr
  &  \Rightarrow dx = {{du} \over {\sqrt {{1 \over {{c^2}u}} - 1} }}  \cr
  &  \Rightarrow x = \int {{1 \over {\sqrt {{1 \over {{c^2}u}} - 1} }}du}  \cr} $$
where the solution $u$ should satisfy $u(0)=0$ and $u(a)=-h$

2
Q7 / Q7
« on: November 24, 2016, 08:58:10 PM »
Consider Laplace equation $\Delta u=0$ in the cylinder$\{r\le a,\ 0<z<b,\ 0\le \theta \le 2\pi\}$. Separate variables $u=R(r)Z(z)\Theta(\theta)$.
1. Write down ODE which should satisfy $\Theta$ and solve it (using periodicity).
2. Write down ODE which should satisfy $Z$ and solve it using $Z(0)=Z(b)=0$.
3. Write down ODE which should satisfy $R$.

Ans:
$$\eqalign{
  & \Delta u = {u_{rr}} + {1 \over r}{u_r} + {1 \over {{r^2}}}{u_{\theta \theta }} + {u_{zz}} = 0  \cr
  &  \Rightarrow {{{r^2}R'' + rR'} \over R} + {{\Theta ''} \over \Theta } + {{{r^2}Z''} \over Z} = 0 \cr} $$
(1) Let:
$$\left. \matrix{
  {{\Theta ''} \over \Theta } =  - {m^2} < 0 \hfill \cr
  \Theta (0) = \Theta (2\pi ),\Theta '(0) = \Theta '(2\pi ) \hfill \cr}  \right\}\matrix{
   {{\Theta _{1,m}} = \cos (m\theta )}  \cr
   {{\Theta _{2,m}} = \sin (m\theta )}  \cr

 } $$
(2) Let:
$$\left. \matrix{
  {{{r^2}Z''} \over Z} =  - l \Rightarrow Z'' + {l \over {{r^2}}}Z = 0 \hfill \cr
  Z(0) = Z(b) = 0 \hfill \cr}  \right\}\matrix{
   {{l \over {{r^2}}} = {{{\pi ^2}{n^2}} \over {{b^2}}}} & { \Rightarrow l = {{{r^2}{\pi ^2}{n^2}} \over {{b^2}}}}  \cr
   {z = \sin ({{\pi nz} \over b})} & {n = 1,2,...}  \cr

 } $$
(3) from (1) and (2) we have:
$$\matrix{
   {{r^2}R'' + rR' - \left( {{m^2} + {{{r^2}{\pi ^2}{n^2}} \over {{b^2}}}} \right)R = 0,} & {n = 1,2,...}  \cr

 } $$

3
Chapter 9 / Are these typos?
« on: November 19, 2016, 10:05:43 PM »
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.1.html#mjx-eqn-eq-9.1.15
I think the coefficient for last term should be ${1 \over {2\pi c}}$ instead of ${1 \over {4\pi c}}$.

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter9/S9.1.html#mjx-eqn-eq-9.1.19
I think this equation should be $(\Delta  + {{{\omega ^2}} \over {{c^2}}})v = {1 \over {{c^2}}}f(x)$ instead of  $(\Delta  + {{{\omega ^2}} \over {{c^2}}})v = f(x)$

4
Chapter 8 / HA9, Problem6
« on: November 13, 2016, 01:20:23 PM »

Can any explain in detail the reason behind the hint "solution must be a harmonic polynomial of degree $3$ and it should depend only on $x^2+y^2+z^2$ and $z$"?
Concretely, there are $3$ question need to be addressed:
$(1)$ Why the solution need to be a harmonic polynomial?
$(2)$ Why it should have degree $3$?
$(3)$ Why it should depend only on $x^2+y^2+z^2$ and $z$ but not the other terms?
The solution from the previous year basically made no reasonable explanation and just copied words from the textbook

http://forum.math.toronto.edu/index.php?topic=731.0

5
Chapter 8 / Must harmonic polynomial be homogeneous?
« on: November 12, 2016, 10:05:03 PM »
Why harmonic polynomial of $deg=n$ must also be homogeneous polynomial of $deg=n$?
Say, $\Delta ({x^2} - {y^2} + z) = 2 - 2 + 0 = 0$, but we do not count $ ({x^2} - {y^2} + z)$ as harmonic polynomial of $deg=2$.

6
Chapter 6 / Domain of the theta function in section 6.4
« on: November 05, 2016, 07:14:10 PM »
Can any explain why we need the $\Theta $ function in section 6.4 (and onward) defined on $[0,2\pi ]$ instead of on $[0,2\pi )$ so that we can remove the periodic assumption of the $\Theta $ function and the boundary conditions related to $\theta $.

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.4.html

From Wikipedia, a standard convention for defining polar coordinate system to achieve Uniqueness of polar coordinates is restrict the domain to $[0, 2\pi)$ or $(−\pi, \pi]$.

https://en.wikipedia.org/wiki/Polar_coordinate_system

7
Chapter 6 / Question about the deviation of laplacian
« on: November 05, 2016, 06:51:32 PM »
Can any one explain in detail to me how we get these two formula in section 6.3:
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.3.html

$$\int\!\!\!\int \Delta  u \cdot v\,dxdy =  - \int\!\!\!\int \nabla  u \cdot \nabla v\,dxdy$$

$$\int\!\!\!\int\!\!\!\int
 \Delta  u \cdot v{\rho ^2}\sin (\phi )\,d\rho d\phi d\theta  =  - \int\!\!\!\int\!\!\!\int
 ( {u_\rho }{v_\rho } + {1 \over {{\rho ^2}}}{u_\phi }{v_\phi } + {1 \over {{\rho ^2}\sin (\phi )}}{u_\theta }{v_\theta }){\rho ^2}\sin (\phi )\,d\rho d\phi d\theta  = \int\!\!\!\int\!\!\!\int
 ( {({\rho ^2}\sin (\phi ){u_\rho })_\rho } + {(\sin (\phi ){u_\phi })_\phi } + {({1 \over {\sin (\phi )}}{u_\theta })_\theta })v\,d\rho d\phi d\theta .$$

By the way, I think the equation $(6)'$ in
 http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.3.html#mjx-eqn-eq-6.3.6
is wrong, I guess the denominator of the last term should be ${\rho ^2}{\sin ^2}(\varphi )$ instead of ${\rho ^2}{\sin}(\varphi )$

8
Chapter 3 / A tentative solution to Assignment 5, Problem3(g)
« on: October 16, 2016, 12:12:08 PM »
The problem is here: http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter3/S3.2.P.html
Here is my tentative solution, hope everyone for correction:
consider:
$$w(x,t) = \left\{ {\matrix{
   {u(x,t) - 1} & {0 < x < \infty } & {t < 1}  \cr
   {u(x,t)} & {0 < x < \infty } & {t > 1}  \cr

 } } \right.$$
Then we have:
$${w_t} = {u_t}, {w_{xx}} = {u_{xx}}$$
Therefore:

$$\matrix{
   {{w_t} = k{w_{xx}}} & {t > 0} & {x > 0}  \cr
   {w\left| {_{t = 0} = u(x,0) - 1 =  - 1} \right.} & {} & {x > 0}  \cr
   {w\left| {_{x = 0} = \matrix{
   {\left\{ {\matrix{
   {u(0,t) - 1 = 1 - 1 = 0} & {t < 1}  \cr
   {u(0,t) = 0} & {t > 1}  \cr

 } } \right\}} & { = 0}  \cr

 } } \right.} & {t > 0} & {}  \cr

 } $$
which is the Dirichlet boundary condition.
Solving this problem wrt $w(x,t)$, I get (The calculation may be wrong):
$$w = 1 - erf({x \over {\sqrt {4kt} }})$$
Therefore:
$$u = \left\{ {\matrix{
   {2 - erf({x \over {\sqrt {4kt} }})} & {x > 0} & {0 < t < 1}  \cr
   {1 - erf({x \over {\sqrt {4kt} }})} & {x > 0} & {t > 1}  \cr

 } } \right.$$
However, in this case $u(x,t)$ is not continuous at $t = 1$.
Hope everyone for correction / verification.

9
Chapter 3 / Are there typos?
« on: October 09, 2016, 01:29:01 PM »
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter3/S3.2.html#mjx-eqn-eq-3.2.15
I guess the second term should be ${G_N}(...)$ instead of ${G_N}_y(...)$. As far as I can understand, we have already cancelled out the ${G_N}_y(...)$ term under the context of Neumann Boundary condition.
By the way, I cannot understand why there is a $k$ in this equation, can any one explain it to me?

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter3/S3.2.html#mjx-eqn-eq-3.2.24
I guess this one should be
${(4\pi kt)^{{{ - n} \over 2}}}$ instead of ${(2\sqrt {\pi kt} )^{{{ - n} \over 2}}}$

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