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you forget $c$ when you are doing the integration in the second part
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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 $u(a)=-h$
$$\begin{equation}
T= \int_0^a \frac{1}{\sqrt{2gu}} \sqrt{1+u^{\prime\,2}}\,dx.
\label{eq-10.P.3}
\end{equation}$$
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 $u(a)=-h$
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FE / Re: Final exam coverage
« on: November 25, 2016, 08:03:59 PM »The website says HA 1-10 is a good preparation. Does that mean the final exam doesn't cover chapter 10? (i.e variational methods)Same question also for Chapter 11 that we are going to cover next week (there is even no HA for Chapter 11)...
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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:
$$\matrix{
{{r^2}R'' + rR' - \left( {{m^2} + {{{r^2}{\pi ^2}{n^2}} \over {{b^2}}}} \right)R = 0,} & {n = 1,2,...} \cr
} $$
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:
$$\matrix{
{{r^2}R'' + rR' - \left( {{m^2} + {{{r^2}{\pi ^2}{n^2}} \over {{b^2}}}} \right)R = 0,} & {n = 1,2,...} \cr
} $$
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Chapter 8 / Re: HA 10, problem 3c
« on: November 23, 2016, 08:22:53 PM »
I think you can just plug in $x=\rho \sin(\phi)\cos(\theta)$, etc and the resulting $u$ will automatically satisfy the laplacian in spherical coordinate.
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Chapter 8 / Re: HA 10, problem 3a
« on: November 22, 2016, 09:34:57 AM »
I think you need to plug in the equation to solve for $C_0$, for general $C_0$ function $u$ may not be harmonic.
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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 $(\Delta + {{{\omega ^2}} \over {{c^2}}})v = f(x)$
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 $(\Delta + {{{\omega ^2}} \over {{c^2}}})v = f(x)$
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TT2 / Re: TT2-P5
« on: November 17, 2016, 08:25:50 PM »So there is no problem in not defining $\theta(0)$.Well...I think the main problem is that Fourier transformation is defined on the whole real line. So, if $\theta$ is not defined at $0$, then it may be questionable to do all the later calculation.
Also...the domain of Heaviside function does contain $0$...
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TT2 / Re: TT2-P5
« on: November 17, 2016, 09:39:15 AM »
First of all, I think the description of the problem is a little bit problematic since it does not define $\theta(0)$. I would use the convention that $\theta(0)=0$
Here is a reference: https://en.wikipedia.org/wiki/Heaviside_step_function
For
$${f_ + }(x) = {e^{ - \varepsilon |x|}}\theta ( + x) = \left\{ {\matrix{
{{e^{ - \varepsilon x}}} & {x \ge 0} \cr
0 & {x < 0} \cr
} } \right.$$
$${{\hat f}_ + }(k) = {1 \over {2\pi }}\int_0^\infty {{e^{ - (\varepsilon + ik)x}}dx = } \left. { - {{{e^{ - (\varepsilon + ik)x}}} \over {2\pi (\varepsilon + ik)}}} \right|_0^\infty = {1 \over {2\pi (\varepsilon + ik)}}$$
Similarly,
$${{\hat f}_ - }(k) = {1 \over {2\pi }}\int_{ - \infty }^0 {{e^{(\varepsilon - ik)x}}dx = } \left. {{{{e^{(\varepsilon - ik)x}}} \over {2\pi (\varepsilon - ik)}}} \right|_{ - \infty }^0 = {1 \over {2\pi (\varepsilon - ik)}}$$
Therefore,
$${f_ + }(x) = \int_{ - \infty }^\infty {{1 \over {2\pi (\varepsilon + ik)}}} {e^{ikx}}dk$$
$${f_ - }(x) = \int_{ - \infty }^\infty {{1 \over {2\pi (\varepsilon - ik)}}} {e^{ikx}}dk$$
And
$$\hat g(x) = {{\hat f}_ + }(k) + {{\hat f}_ - }(k) = {1 \over {2\pi (\varepsilon + ik)}} + {1 \over {2\pi (\varepsilon - ik)}} = {{2\varepsilon } \over {2\pi ({\varepsilon ^2} + {k^2})}}$$
$$\hat h(x) = {{\hat f}_ + }(k) - {{\hat f}_ - }(k) = {1 \over {2\pi (\varepsilon + ik)}} - {1 \over {2\pi (\varepsilon - ik)}} = {{ - 2ik} \over {2\pi ({\varepsilon ^2} + {k^2})}}$$
Here is a reference: https://en.wikipedia.org/wiki/Heaviside_step_function
For
$${f_ + }(x) = {e^{ - \varepsilon |x|}}\theta ( + x) = \left\{ {\matrix{
{{e^{ - \varepsilon x}}} & {x \ge 0} \cr
0 & {x < 0} \cr
} } \right.$$
$${{\hat f}_ + }(k) = {1 \over {2\pi }}\int_0^\infty {{e^{ - (\varepsilon + ik)x}}dx = } \left. { - {{{e^{ - (\varepsilon + ik)x}}} \over {2\pi (\varepsilon + ik)}}} \right|_0^\infty = {1 \over {2\pi (\varepsilon + ik)}}$$
Similarly,
$${{\hat f}_ - }(k) = {1 \over {2\pi }}\int_{ - \infty }^0 {{e^{(\varepsilon - ik)x}}dx = } \left. {{{{e^{(\varepsilon - ik)x}}} \over {2\pi (\varepsilon - ik)}}} \right|_{ - \infty }^0 = {1 \over {2\pi (\varepsilon - ik)}}$$
Therefore,
$${f_ + }(x) = \int_{ - \infty }^\infty {{1 \over {2\pi (\varepsilon + ik)}}} {e^{ikx}}dk$$
$${f_ - }(x) = \int_{ - \infty }^\infty {{1 \over {2\pi (\varepsilon - ik)}}} {e^{ikx}}dk$$
And
$$\hat g(x) = {{\hat f}_ + }(k) + {{\hat f}_ - }(k) = {1 \over {2\pi (\varepsilon + ik)}} + {1 \over {2\pi (\varepsilon - ik)}} = {{2\varepsilon } \over {2\pi ({\varepsilon ^2} + {k^2})}}$$
$$\hat h(x) = {{\hat f}_ + }(k) - {{\hat f}_ - }(k) = {1 \over {2\pi (\varepsilon + ik)}} - {1 \over {2\pi (\varepsilon - ik)}} = {{ - 2ik} \over {2\pi ({\varepsilon ^2} + {k^2})}}$$
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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
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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 $deg=2$.
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 $deg=2$.
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Chapter 6 / Re: Domain of the theta function in section 6.4
« on: November 05, 2016, 11:14:41 PM »
Just to make sure, in the "membrane" case you describe, we basically do not need the periodic assumption of $\Theta$ (Except the boundary condition), right?
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Chapter 6 / Re: Question about the deviation of laplacian
« on: November 05, 2016, 11:10:46 PM »
Misprint at where? I guess the link is pointing to a wrong equation, equation $(6)$ is correct, but $(6)'$, I guess, leave out a square.
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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
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
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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 ${\rho ^2}{\sin}(\varphi )$
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 ${\rho ^2}{\sin}(\varphi )$
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