Toronto Math Forum

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  $u=Ar^{2-n}+B$.

---

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

\[\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} \]

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 T∈t>0, obviously,  M(T) and m(t) satisfies the maximum and minimum principals.

\begin{array}{l}
b)\\
\frac{{dt}}{1} = \frac{{dx}}{{tx}} = \frac{{du}}{0}\quad .o.d.e\\
ln(x) = \frac{{{t^2}}}{2} + C\\
u(x,t) = f(C) = f(lnx - \frac{{{t^2}}}{2}),\quad f\quad arbitary
\end{array}

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 −2x = −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 ∀x ∈ (−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 = −2, 0 ≤ t ≤ 1}, u(−2, t) = 4(t − 1) ≤ 0 , for any 0 ≤ t ≤ 1.  Next consider the right side, where {(x, t)|x = 2, 0 ≤ t ≤ 1}, u(2, t) = −4(t + 1) < 0, for all 0 ≤ t ≤ 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)| − 2 < x < 2, t = 1},
u(x, 1) = −2x − x^2. If the maximum exists, then x ∈ (−2, 2) and u_x (x,1) = 0.
From part(a), u_x (x,1) = −2 − 2x. Taking u_x=0, then x=-1, hence u_max=u(-1,1)=1

part(b):

consider these situations below:
u_t = −2x, u_t(1,1)=2>0
u_xx = −2, u_xx(-1,1)=-2<0
but u(x,t) still satisfies the PDE equation.

(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.

<sol>:The equation is linear,thus,the characteristic equation is below:$$\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { dx }{ 1 } =\frac { dy }{ 3 } =\frac { du }{ xy } \\ \quad \quad \quad \quad \quad Solve\quad theO.D.E:\\ \quad \quad \quad \quad \quad \quad \quad \quad y=3x+C\\ \quad \quad \quad \quad \quad \quad \quad \quad (xy)dx=du\quad →\quad dx(x(3x+C))=du\\ \quad \quad \quad \quad \quad hence,u=x^{ 3 }+\frac { c{ x }^{ 2 } }{ 2 } +K,\quad whereC,\quad K\quad are\quad constants.\\ \quad \quad \quad \quad \quad solve\quad IVP:\\ \quad \quad \quad \quad \quad \quad \quad \quad K=0\\ \quad \quad \quad \quad \quad \quad \quad \quad C=-3x+y\\ \quad \quad \quad \quad \quad \quad \quad \quad u=x^{ 3 }+x^{ 2 }\frac { (-3x+y) }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =-\frac { { x }^{ 3 } }{ 2 } +\frac { y{ x }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad check:\\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ x }=-\frac { 3{ x }^{ 2 } }{ 2 } +xy\\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ y }=\frac { { x }^{ 2 } }{ 2 } \\ \quad \quad \quad \quad \quad \quad \quad \quad u_{ x }+3u_{ y }=xy\\ \\  $$
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{…}. And better to break the source into lines (in the logical places, where \\ are)--V.I.

question 3