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### Messages - Yiyun Liu

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1
##### HA8 / question 1 (a)-(b)
« on: March 19, 2015, 09:10:40 PM »
1.  Find the solutions that depend only on $r$ of the equation
\begin{equation*}
\Delta u:=u_{xx}+u_{yy}=0.
\end{equation*}
2.  Find the solutions that depend only on $\rho$ of the equation
\begin{equation*}
\Delta u:=u_{xx}+u_{yy}+u_{zz}=0.
\end{equation*}
3.  (bonus) In $n$-dimensional case prove that if $u=u(r)$ with  $r=(x_1^2+x_2^2+\ldots+x_n^2)^{\frac{1}{2}}$ then
\begin{equation}
\Delta u = u_{rr}+ \frac{n-1}{r}u_r=0.
\label{equ-H8.1} \end{equation}
4.  (bonus) In $n$-dimensional case prove ($n\ne 2$) that  $u=u(r)$ satisfies Laplace equation as $x\ne 0$ iff  $u=Ar^{2-n}+B$.

$\begin{array}{l} part(a):\\ \Delta u = {u_{rr}} + \frac{1}{r}{u_r} + \frac{1}{{{r^2}}}{u_{\theta \theta }} = 0\\ \Delta u = {u_{rr}} + \frac{1}{r}{u_r} = 0\\ \frac{\partial }{{\partial r}}(r{u_r}) = {u_r} + r{u_{rr}} = r({u_{rr}} + \frac{1}{r}{u_r}) = 0\\ \frac{\partial }{{\partial r}}(r{u_r}) = C\\ u = D\ln (r) + E\\ \\ part(b):\\ \Delta u = {u_{\rho \rho }} + \frac{2}{\rho }{u_\rho } + \frac{1}{{{\rho ^2}}}({u_{\theta \theta }} + \cot (\theta ){u_\theta } + \frac{1}{{{{\sin }^2}(\theta )}}{u_{\theta \theta }}) = 0\\ \Delta u = {u_{\rho \rho }} + \frac{2}{\rho }{u_\rho } = 0\\ \frac{\partial }{{\partial \rho }}({\rho ^2}{u_\rho }) = 2\rho {u_\rho } + {\rho ^2}{u_{\rho \rho }} = {\rho ^2}({u_{\rho \rho }} + \frac{2}{\rho }{u_\rho }) = 0\\ {\rho ^2}{u_\rho } = C,cons\tan ts\\ u = D\frac{1}{\rho } + E \end{array}$

2
##### Test 2 / Re: TT2 problem 5
« on: March 13, 2015, 12:08:13 AM »
$\begin{array}{l} part(1):\\ \\ \hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - 1}^1 {(1 - {x^2})} {e^{ - i\omega x}}dx\\ = \frac{1}{{2\pi }}[\int\limits_{ - 1}^1 {{e^{ - i\omega x}}dx - } \int\limits_{ - 1}^1 {{x^2}} {e^{ - i\omega x}}dx]\\ = \frac{1}{{2\pi }}(\frac{{ - 1}}{{i\omega }}{e^{ - i\omega x}}\mathop |\nolimits_{ - 1}^1 ) - ( - \frac{1}{{i\omega }}{x^2}{e^{ - i\omega x}}\mathop |\nolimits_{ - 1}^1 + \int\limits_{ - 1}^1 {\frac{{2x}}{{i\omega }}} {e^{ - i\omega x}}dx)\\ = - \frac{1}{{2\pi }}\int\limits_{ - 1}^1 {\frac{{2x}}{{i\omega }}} {e^{ - i\omega x}}dx\\ = - (\frac{2}{{\pi {\omega ^2}}}\cos \omega - \frac{2}{{\pi {\omega ^3}}}\sin (\omega ))\\ = \frac{2}{{\pi {\omega ^3}}}\sin (\omega ) - \frac{2}{{\pi {\omega ^2}}}\cos (\omega )\\ \\ part(2):\\ \\ f(x) = \int\limits_{ - 1}^1 {[\frac{2}{{\pi {\omega ^3}}}\sin (\omega ) - \frac{2}{{\pi {\omega ^2}}}\cos (\omega )]} {e^{i\omega x}}d\omega \end{array}$

3
##### HA6 / Re: HA6 problem 3
« on: March 05, 2015, 09:07:14 PM »
$\begin{gathered} part(a): \hfill \\ \hfill \\ f(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}} \hfill \\ \hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - \infty }^\infty {{e^{\frac{{ - \alpha {x^2}}}{2}}}} {e^{ - i\omega x}}dx = \frac{1}{{\sqrt {2\pi \alpha } }}{e^{\frac{{ - {\omega ^2}}}{{2\alpha }}}} \hfill \\ g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x) = \frac{1}{{2i}}f(x)({e^{i\beta x}} - {e^{ - i\beta x}}) \hfill \\ thus,\hat g(\omega ) = \frac{1}{{2i}}\left[ {(\hat f(\omega - \beta ) - (\hat f(\omega + \beta )} \right] \hfill \\ = \frac{1}{{2i\sqrt {2\pi \alpha } }}\left( {{e^{\frac{{ - {{(\omega - \beta )}^2}}}{{2\alpha }}}} - {e^{\frac{{ - {{(\omega + \beta )}^2}}}{{2\alpha }}}}} \right) \hfill \\ similar, \hfill \\ g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x) = \frac{1}{2}f(x)({e^{i\beta x}} + {e^{ - i\beta x}}) \hfill \\ \hat g(x) = \frac{1}{2}\left[ {(\hat f(\omega - \beta ) + (\hat f(\omega + \beta )} \right] = \frac{1}{{2\sqrt {2\pi \alpha } }}\left( {{e^{\frac{{ - {{(\omega - \beta )}^2}}}{{2\alpha }}}} + {e^{\frac{{ - {{(\omega + \beta )}^2}}}{{2\alpha }}}}} \right) \hfill \\ \hfill \\ part(b); \hfill \\ \hfill \\ f(x) = x{e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x) \hfill \\ let,f(x) = xg(x),g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\cos (\beta x),hence, \hfill \\ \hat f(\omega ) = i\frac{{d\hat g(x)}}{{d\omega }} = \frac{i}{{2\sqrt {2\pi \alpha } }}(\frac{{ - (\omega - \beta )}}{\alpha }{e^{^{\frac{{ - {{(\omega - \beta )}^2}}}{{2\alpha }}}}} - \frac{{(\omega + \beta )}}{\alpha }{e^{\frac{{ - {{(\omega + \beta )}^2}}}{{2\alpha }}}}) \hfill \\ likewise, \hfill \\ f(x) = x{e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x) \hfill \\ f(x) = xg(x),where,g(x) = {e^{\frac{{ - \alpha {x^2}}}{2}}}\sin (\beta x),hence, \hfill \\ \hat f(\omega ) = i\frac{{d\hat g(x)}}{{d\omega }} = \frac{1}{{2\sqrt {2\pi \alpha } }}(\frac{{ - (\omega - \beta )}}{\alpha }{e^{^{\frac{{ - {{(\omega - \beta )}^2}}}{{2\alpha }}}}} + \frac{{(\omega + \beta )}}{\alpha }{e^{\frac{{ - {{(\omega + \beta )}^2}}}{{2\alpha }}}}) \hfill \\ \hfill \\ \end{gathered}$

4
##### HA6 / Re: HA6 problem 4
« on: March 05, 2015, 09:00:43 PM »
\begin{gathered}
part(a): \hfill \\
\hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - \infty }^\infty  {f(x){e^{ - i\omega x}}dx}  \hfill \\
= \frac{1}{{2\pi }}\int\limits_{ - a}^a {{e^{ - i\omega x}}dx}  \hfill \\
= \frac{1}{{2\pi }}(\frac{{{e^{i\omega a}} - {e^{ - i\omega a}}}}{{i\omega }}) \hfill \\
= \frac{{\sin (\omega a)}}{{\pi \omega }} \hfill \\
\hfill \\
part(b): \hfill \\
\hfill \\
take  f(x) = xg(x),where,  g(x) = 1 \hfill \\
hence,    \hat f(\omega ) = i\frac{{d\hat g(\omega )}}{{d\omega }} = i\frac{{d(\frac{{\sin (\omega a)}}{{\pi \omega }})}}{{d\omega }} = i(\frac{{a\cos (a\omega )}}{{\pi \omega }} - \frac{{\sin (a\omega )}}{{\pi {\omega ^2}}}) \hfill \\
\hfill \\
part(c): \hfill \\
\hfill \\
IFT: \hfill \\
f(x) = \int\limits_{ - \infty }^\infty  { \hat f(\omega ){e^{i\omega x}}d\omega  = } \int\limits_{ - \infty }^\infty  {\frac{{\sin (\omega a)}}{{\pi \omega }}} {e^{i\omega x}}d\omega  \hfill \\
thus,\int\limits_{ - \infty }^\infty  {\frac{{\sin (xa)}}{x}{e^{i\omega x}}} dx = \pi ,      |\omega | \leqslant a,if  a = 1,and     \omega  = 0. \hfill \\
\end{gathered} \]

5
##### Test 1 / Re: TT1 Problem 4
« on: February 12, 2015, 11:28:26 PM »
c)
as the principle of maximum and minimum stated that the max and min must occur either on the bottom t=0 or the sides edges at x=0,L including the corners (0,T),(L,T) for some Tâˆˆt>0, obviously,  M(T) and m(t) satisfies the maximum and minimum principals.

6
##### Test 1 / Re: TT1 Problem 1
« on: February 12, 2015, 11:22:11 PM »
\begin{array}{l}
c)\\
u(x,0) = {e^{ - {x^2}}}\\
f(lnx) = {e^{ - {x^2}}},\\
f(x) = {e^{ - {e^{2x}}}}\\
u(x,t) = {e^{ - {e^{2lnx - {t^2}}}}} = {e^{ - {x^2}{e^{ - {t^2}}}}}
\end{array}
Since the arbitrary function f is uniquely determined by the initial condition, the solution is fully determined and the problem is well-posed.

7
##### Test 1 / Re: TT1 Problem 1
« on: February 12, 2015, 10:49:23 PM »
\begin{array}{l}
b)\\
\frac{{dt}}{1} = \frac{{dx}}{{tx}} = \frac{{du}}{0}\quad .o.d.e\\
ln(x) = \frac{{{t^2}}}{2} + C\\
\end{array}

8
##### HA3 / Re: HA3 problem 1
« on: February 05, 2015, 09:02:10 PM »
Part(a)
Dirichlet boundary
\begin{array}{l}
{u_t} = k{u_{xx}},  x > 0,t > 0\\
u{|_{t = 0}} = g(x)\\
u{|_{x = 0}} = 0\\
u(x,t) = \int\limits_0^\infty  {G(x - y,t)g(y)dy - \int\limits_0^\infty  {G(x + y,t)g(y)dy} } \\
{\rm{           = }}\frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {{e^{\frac{{ - {{(x - y)}^2}}}{{4kt}}}}} g(y)dy - \frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {{e^{\frac{{ - {{(x + y)}^2}}}{{4kt}}}}} g(y)dy\\
{\rm{          = }}\frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {({e^{\frac{{ - {{(x - y)}^2}}}{{4kt}}}} - {e^{\frac{{ - {{(x + y)}^2}}}{{4kt}}}}) }g(y)dy\\

\end{array}

part(b):
Neumann boundary
\begin{array}{l}
{u_t} = k{u_{xx}},  x > 0,t > 0\\
u{|_{t = 0}} = g(x)\\
{u_x}{|_{x = 0}} = 0\\
u(x,t) = \int\limits_0^\infty  {G(x - y,t)g(y)dy + \int\limits_0^\infty  {G(x + y,t)g(y)dy} } \\
{\rm{           = }}\frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {{e^{\frac{{ - {{(x - y)}^2}}}{{4kt}}}}} g(y)dy + \frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {{e^{\frac{{ - {{(x + y)}^2}}}{{4kt}}}}} g(y)dy\\
{\rm{          = }}\frac{1}{{\sqrt {4k\pi t} }}\int\limits_0^\infty  {({e^{\frac{{ - {{(x - y)}^2}}}{{4kt}}}} + {e^{\frac{{ - {{(x + y)}^2}}}{{4kt}}}}} )g(y)dy\\
\\

\end{array}

9
##### HA3 / Re: HA3 problem 5
« on: February 05, 2015, 09:01:29 PM »
Part(a)
\begin{array}{l}
\mathop u\nolimits_t  =  - 2x,\\
\mathop u\nolimits_x  =  - 2t - 2x\\
\mathop u\nolimits_{xx}  =  - 2
\end{array}
plugging in the equation, so âˆ’2x = âˆ’2x. u(x,t) is the solution which satisfies the equation

part(a):
By principle of maximum theorem. If a maximum is at an interior point, then ux=ut=0,then x=t=0, which lies on the bottom boundary, for u = 0 âˆ€x âˆˆ (âˆ’2, 2).  That is there is no maximum in the interior. Now consider case of the two side edges, firstly on the left side, where {(x, t)|x = âˆ’2, 0 â‰¤ t â‰¤ 1}, u(âˆ’2, t) = 4(t âˆ’ 1) â‰¤ 0 , for any 0 â‰¤ t â‰¤ 1.  Next consider the right side, where {(x, t)|x = 2, 0 â‰¤ t â‰¤ 1}, u(2, t) = âˆ’4(t + 1) < 0, for all 0 â‰¤ t â‰¤ 1.Therefore we can conclude that the maximum value of u on the bottom and sides is 0. Consider the case of the top, where {(x, t)| âˆ’ 2 < x < 2, t = 1},
u(x, 1) = âˆ’2x âˆ’ x^2. If the maximum exists, then x âˆˆ (âˆ’2, 2) and u_x (x,1) = 0.
From part(a), u_x (x,1) = âˆ’2 âˆ’ 2x. Taking u_x=0, then x=-1, hence u_max=u(-1,1)=1

part(b):

consider these situations below:
u_t = âˆ’2x, u_t(1,1)=2>0
u_xx = âˆ’2, u_xx(-1,1)=-2<0
but u(x,t) still satisfies the PDE equation.

10
##### HA2 / Re: HA2 problem 4
« on: January 29, 2015, 09:01:43 PM »
$\begin{array}{l} part(a):\\ \\ \frac{{\partial e}}{{\partial t}} = {u_t}{u_{tt}} + {c^2}{u_x}{u_{xt}}\\ \frac{{\partial \rho }}{{\partial x}} = c{u_{tx}}ux + c{u_t}{u_{xx}},since\quad {u_{tt}} = {c^2}{u_{xx}}\\ \Rightarrow \frac{{\partial e}}{{\partial t}} = {c^2}{u_t}{u_{xx}} + {c^2}{u_x}{u_{xt}}\quad (1)\\ c\frac{{\partial \rho }}{{\partial x}} = {c^2}{u_{tx}}ux + {c^2}{u_t}{u_{xx}} = \frac{{\partial e}}{{\partial t}} = (1)\\ \frac{{\partial \rho }}{{\partial t}} = c{u_{tt}}{u_x} + c{u_t}{u_{xt}}\quad (2)\\ \frac{{\partial e}}{{\partial x}} = {u_t}{u_{xt}} + {c^2}{u_x}{u_{xx}}\quad \\ \Rightarrow \frac{{\partial e}}{{\partial x}} = {u_t}{u_{xt}} + {u_x}{u_{tt}}\\ c\frac{{\partial e}}{{\partial x}} = c{u_t}{u_{xt}} + c{u_x}{u_{tt}} = \frac{{\partial \rho }}{{\partial t}} = (2)\\ \\ part(b):\\ \\ from\quad (a)\quad known\quad that\quad c\frac{{\partial \rho }}{{\partial x}} = \frac{{\partial e}}{{\partial t}},\quad c\frac{{\partial e}}{{\partial x}} = \frac{{\partial \rho }}{{\partial t}}\\ thus,\quad c\frac{{{\partial ^2}\rho }}{{\partial xt}} = \frac{{{\partial ^2}e}}{{\partial {t^2}}}\quad (1),\quad \frac{{{\partial ^2}\rho }}{{\partial {t^2}}} = c\frac{{{\partial ^2}e}}{{\partial xt}}\quad (2)\\ likewise,\quad \frac{{{\partial ^2}e}}{{\partial xt}} = c\frac{{{\partial ^2}\rho }}{{\partial {x^2}}}\quad (3),\quad \frac{{{\partial ^2}\rho }}{{\partial xt}} = c\frac{{{\partial ^2}e}}{{\partial {x^2}}}(4)\\ from\quad (1),(4):\quad {e_{tt}} = {c^2}{e_{xx}}\\ from\quad (2)(3):\quad {\rho _{tt}} = {c^2}{\rho _{xx}}\\ As\quad shown,\quad both\quad e(x,t)\quad and\quad \rho (x,t)\quad from\quad the\quad same\quad wave\quad equation. \end{array}$

11
##### HA2 / Re: HA2 problem 2
« on: January 29, 2015, 09:01:08 PM »
(a): $v = ur$,  then  $u = \frac{v}{r}$ and
\begin{align}
&u_t = \frac{ v_t }{r},\notag \\
&u_{tt}  = \frac{1}{r}{v_{tt}}\label{A}\\
&u_r  =  - \frac{v}{{{r^2}}} + \frac{{{v_r}}}{r},\label{B}\\
&u_{rr}  = \frac{{2v}}{{{r^3}}} - \frac{{2{v_r}}}{{{r^2}}} + \frac{{{v_{rr}}}}{r}.\label{C}
\end{align}
Plugging (\ref{A})--(\ref{C})  into  the equation we have
$$\frac{{{v_{tt}}}}{r} = {c^2}(\frac{{2v}}{{{r^3}}} - \frac{{2{v_r}}}{{{r^2}}} + \frac{{{v_{rr}}}}{r} + \frac{{2{v_r}}}{{{r^2}}} - \frac{{2v}}{{{r^3}}}) \frac{{{v_{tt}}}}{r} = {c^2}(\frac{{{v_{rr}}}}{r})$$
Hence,      ${v_{tt}} = {c^2}{v_{rr}}$.

(b) From (a),  known  that ${v_{tt}} = {c^2}{v_{rr}}$:
$$v(x,t) = f(x + ct) + g(x - ct), u =r^{-1} \left[ f(r + ct) + g(r - ct) \right]$$
where $f$ and $g$ are  arbitrary functions.

(c) $\phi (r) = v(r,0) = ru(r,0) = r\Phi ( r )$, $\varphi (r) = {v_t}(r,0) = r{u_t}(r,0) = r\Psi (r )$,
$$u(r,t) = \frac{1}{2}\left[ {(r + ct)\Phi (r + ct) + (r - ct)\Psi (r - ct)} \right] + \frac{1}{{2cr}}\int\limits_{r - ct}^{r + ct} {s\Phi (s)ds} .$$

(d)  If  this solution  is  continuous  at $r=0$, $\lim_{r \to 0} u(r,t)$ should  exist; so, $\lim_{r \to 0} \left[ {f(r + ct) + g(r - ct)} \right] = 0$. If $\lim_{r \to 0} \left[ {f(r + ct) + g(r - ct)} \right] \ne 0$, then  $\lim_{r \to 0} u(r,t)=\infty$.

Therefore, $f(ct) + g( - ct) = 0$, that is, $f(ct) = - g( - ct)$. So $g(r - ct) = - f(ct - r)$,
$$u(r,t) = r^{-1}\left[ f(r + ct) - f(ct - r)\right].$$

I fixed LaTeX. Please do not use the crapware you used to produce that code! It was produced by some aliens and is not for human use! V.I.

12
##### HA2 / Re: HA2 problem 3
« on: January 29, 2015, 09:00:33 PM »
(a) Let $0 < 3t < x$. Then
\begin{equation*}
u(x,t) = \frac{1}{6}\int_{x - 3t}^{x + 3t} {\cos x'dx'}
= \frac{1}{6}\left[ {\sin \left( {x + 3t} \right) - \sin (x - 3t)} \right]
= \frac{1}{3}\cos x\sin 3t.
\end{equation*}
Let $0 < x < 3t$. Then
\begin{equation*}
u(x,t) = \frac{1}{6}\int_{3t - x}^{x + 3t} {\cos x'dx'}  = \frac{1}{6}\left[ {\sin (x + 3t) - \sin (3t - x)} \right]
= \frac{1}{3}\sin x\cos 3t.
\end{equation*}

So,
\begin{equation*}
u(x,t)=\left\{ \begin{aligned}
&\frac{1}{6}\cos x\sin 3t && 0<3t<x,\\ & \frac{1}{3}\sin x\cos 3t. &&0<x<3t. \end{aligned}\right. \end{equation*} (b) Let0 < 3t < x$, then$u(x,t) = \frac{1}{3}\sin 3t\cos x$. Let$0 < x < 3t$. Then \begin{equation*} u(x,t) = \frac{1}{6}\int_0^{3t + x} {\cos x'dx' + \frac{1}{6}} \int\limits_0^{3t - x} {\cos x'dx'} = \frac{1}{6}\sin (x + 3t) + \frac{1}{6}\sin (3t - x) = \frac{1}{3}\sin 3t\cos x \end{equation*} (c)$0 < 3t < x$. Then \begin{equation*} u(x,t) = \frac{1}{6}\int\limits_{x - 3t}^{x + 3t} {\sin x'dx'} = - \frac{1}{6}\cos (x + 3t) + \frac{1}{6}\cos (x - 3t) = \frac{1}{3}\cos x\cos 3t. \end{equation*} Let$0 < x < 3t$. Then \begin{equation*} u(x,t) = \frac{1}{6}\int_{3t - x}^{x + 3t} \sin x'dx' = \frac{1}{6}\left[ {\cos (3t - x) - \cos (x + 3t)} \right] = \frac{1}{3}\cos x\cos 3t \end{equation*} (d)$0 < 3t < x$, then$u(x,t) = \frac{1}{3}\cos x\cos 3t$. Let$0 < x < 3t\$. Then
\begin{equation*}
u(x,t) = \frac{1}{6}\int\limits_0^{x + 3t} {\sin x'dx' + \frac{1}{6}} \int\limits_0^{3t - x} {\sin x'dx'}
= \frac{1}{3} - \frac{1}{6}\left[ {\cos (x + 3t) + \cos (3t - x)} \right]
= \frac{1}{3}(1 - \cos x\cos 3t)
\end{equation*}

I fixed LaTeX: too short lines and too small vertical spacing made what you wrote difficult to read. --V.I.

13
##### HA1 / Re: HA1 problem 3
« on: January 28, 2015, 06:26:39 PM »
<sol>:The equation is linear,thus,the characteristic equation is below:$$\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { dx }{ 1 } =\frac { dy }{ 3 } =\frac { du }{ xy } \\ \quad \quad \quad \quad \quad Solve\quad theO.D.E:\\ \quad \quad \quad \quad \quad \quad \quad \quad y=3x+C\\ \quad \quad \quad \quad \quad \quad \quad \quad (xy)dx=du\quad â†’\quad dx(x(3x+C))=du\\ \quad \quad \quad \quad \quad hence,u=x^{ 3 }+\frac { c{ x }^{ 2 } }{ 2 } +K,\quad whereC,\quad K\quad are\quad constants.\\ \quad \quad \quad \quad \quad solve\quad IVP:\\ \quad \quad \quad \quad \quad \quad \quad \quad K=0\\ \quad \quad \quad \quad \quad \quad \quad \quad C=-3x+y\\ \quad \quad \quad \quad \quad \quad \quad \quad u=x^{ 3 }+x^{ 2 }\frac { (-3x+y) }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =-\frac { { x }^{ 3 } }{ 2 } +\frac { y{ x }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad check:\\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ x }=-\frac { 3{ x }^{ 2 } }{ 2 } +xy\\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ y }=\frac { { x }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ x }+3u_{ y }=xy\\ \\$$
I just try whether what I typed could be shown

Yes--but you need to surround it by double dollars (I did it for you) . Still this is extremely bad code with all these \quad and many {} are completely unnecessary; also long pieces of text (like the whole line) should not be in formulas and shorter ones should be tagged as \text{â€¦}. And better to break the source into lines (in the logical places, where \\ are)--V.I.

14
##### Web Bonus Problems / Re: Web Bonus Problem 2
« on: January 27, 2015, 09:39:39 PM »
proï¼ŒI think there is a type error of the hint you provide, should x=(xi+eta)/2 instead of xi=(xi+eta)/2 ?
thx

Thanks, corrected. V.I.

15
##### HA1 / Re: HA1 problem 3
« on: January 22, 2015, 09:06:03 PM »
question 3

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