Регулярные функции с сохранением значения на множестве

[Seergey]

Требуется найти все функции $w(z)=u(x,y)+iv(x,y)$, такие что функция $p(x,y)=\sqrt{u^2(x,y)+v^2(x,y)}$ сохраняла бы постоянное значение на окружности $x^2+y^2=C$

Чтобы существовала регулярная функция необходимо равенство нулю лапласиана $u$ и $v$, а что бы $p(x,y)$ была константой на окружности необходимо чтобы $\frac{\partial p}{\partial u}\frac{\partial u}{\partial t}+ \frac{\partial p}{\partial v}\frac{\partial v}{\partial t}=0, t=x^2+y^2$. Этих условий достаточно, чтобы найти $w(z)$?

[Seergey]

Причем если $u$ и $v$ постоянны на окружности, то модуль конечно же тоже будет постоянным, но ведь есть ещё функции $p$ у которых $u$ и $v$ не постоянны, а сама $ p(u(x,y),v(x,y))$ постоянна. Как найти все?

[Seergey]

$\omega = u(x,y)+i v(x,y)$

$ \begin{cases} P=\sqrt{u^2+v^2} \\ x^2+y^2=\operatorname{const} \\ \end{cases}

$\frac{\partial P}{\partial x}=\frac{\partial P}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial P}{\partial v}\frac{\partial v}{\partial x}$
$\frac{\partial P}{\partial y}=\frac{\partial P}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial P}{\partial v}\frac{\partial v}{\partial y}$

С другой стороны, если $P=\operatorname{const}$ при условии что $x^2+y^2= \operatorname{const}$, то $P$ должна быть функцией $x^2+y^2$, т. е. $P=P(x^2+y^2)=P(t)$, где $t=x^2+y^2$
$\frac{\partial P}{\partial x}=\frac{\partial P}{\partial t}\frac{\partial t}{\partial x}$
$\frac{\partial P}{\partial y}=\frac{\partial P}{\partial t}\frac{\partial t}{\partial y}$

$ + \begin{cases} \frac{\partial^2 P}{\partial x^2}=\frac{\partial^2 P}{\partial t^2}(\frac{\partial t}{\partial x})^2+\frac{\partial P}{\partial t} \frac{\partial^2 t}{\partial x^2}\\ \frac{\partial^2 P}{\partial y^2}=\frac{\partial^2 P}{\partial t^2}(\frac{\partial t}{\partial y})^2+\frac{\partial P}{\partial t} \frac{\partial^2 t}{\partial y^2}\\ \end{cases}
$\frac{\partial^2 P}{\partial x^2}+\frac{\partial^2 P}{\partial y^2}=\frac{\partial^2 P}{\partial t^2} \left((\frac{\partial t}{\partial x})^2+(\frac{\partial t}{\partial y})^2 \right)+\frac{\partial P}{\partial t} \left(\frac{\partial^2 t}{\partial x^2}+\frac{\partial^2 t}{\partial y^2} \right)$ (1)

$+ \begin{cases} \frac{\partial^2 P}{\partial x^2}=\frac{\partial^2 P}{\partial u^2}(\frac{\partial u}{\partial x})^2+\frac{\partial P}{\partial u}\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 P}{\partial v^2}(\frac{\partial v}{\partial x})^2+\frac{\partial P}{\partial v}\frac{\partial^2 v}{\partial x^2} \\ \frac{\partial^2 P}{\partial y^2}=\frac{\partial^2 P}{\partial u^2}(\frac{\partial u}{\partial y})^2+\frac{\partial P}{\partial u}\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 P}{\partial v^2}(\frac{\partial v}{\partial y})^2+\frac{\partial P}{\partial v}\frac{\partial^2 v}{\partial y^2} \\ \end{cases}$

С учетом уравнения Лапласа $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$ и $\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}=0$ получается
$\frac{\partial^2 P}{\partial x^2}+\frac{\partial^2 P}{\partial y^2}=\frac{\partial^2 P}{\partial u^2} \left((\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2 \right)+\frac{\partial^2 P}{\partial v^2} \left((\frac{\partial v}{\partial x})^2+(\frac{\partial v}{\partial y})^2 \right)$

а также применяя условия Коши-Римана $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ и $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$
$\frac{\partial^2 P}{\partial x^2}+\frac{\partial^2 P}{\partial y^2}= \left((\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2 \right) \left(\frac{\partial^2 P}{\partial u^2}+\frac{\partial^2 P}{\partial v^2} \right)$ (2)

т. е., учитывая (1) и (2)
$\left((\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2 \right) \left(\frac{\partial^2 P}{\partial u^2}+\frac{\partial^2 P}{\partial v^2} \right)=\frac{\partial^2 P}{\partial t^2} \left((\frac{\partial t}{\partial x})^2+(\frac{\partial t}{\partial y})^2 \right)+\frac{\partial P}{\partial t} \left(\frac{\partial^2 t}{\partial x^2}+\frac{\partial^2 t}{\partial y^2} \right)$

Для исключения $\frac{\partial u}{\partial x}$ и $\frac{\partial u}{\partial y}$ необходимо решить линейную систему
$ \begin{cases} \frac{\partial P}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial P}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial P}{\partial t}\frac{\partial t}{\partial x} \\ \frac{\partial P}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial P}{\partial v}\frac{\partial v}{\partial y}=\frac{\partial P}{\partial t}\frac{\partial t}{\partial y} \\ \end{cases}
$ \begin{cases} \frac{\partial u}{\partial x}=\frac{\frac{\partial P}{\partial t}\frac{\partial t}{\partial x}-\frac{\partial P}{\partial v}\frac{\partial v}{\partial x}}{\frac{\partial P}{\partial u}} \\ \frac{\partial u}{\partial y}=\frac{\frac{\partial P}{\partial t}\frac{\partial t}{\partial y}-\frac{\partial P}{\partial v}\frac{\partial v}{\partial y}}{\frac{\partial P}{\partial u}} \\ \end{cases}

Применяя условия Коши-Римана получается
$ \begin{cases} \frac{\partial u}{\partial x}=\frac{\frac{\partial P}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial P}{\partial v}\frac{\partial u}{\partial y}}{\frac{\partial P}{\partial u}} \\ \frac{\partial u}{\partial y}=\frac{\frac{\partial P}{\partial t}\frac{\partial t}{\partial y}-\frac{\partial P}{\partial v}\frac{\partial u}{\partial x}}{\frac{\partial P}{\partial u}} \\ \end{cases}
$ \begin{cases} \frac{\partial u}{\partial x}=\frac{\frac{\partial P}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial P}{\partial v}\frac{\frac{\partial P}{\partial t}\frac{\partial t}{\partial y}-\frac{\partial P}{\partial v}\frac{\partial u}{\partial x}}{\frac{\partial P}{\partial u}}}{\frac{\partial P}{\partial u}} \\ \frac{\partial u}{\partial y}=\frac{\frac{\partial P}{\partial t}\frac{\partial t}{\partial y}-\frac{\partial P}{\partial v}\frac{\frac{\partial P}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial P}{\partial v}\frac{\partial u}{\partial y}}{\frac{\partial P}{\partial u}}}{\frac{\partial P}{\partial u}} \\ \end{cases}
$ \begin{cases} \frac{\partial u}{\partial x}=\frac{\frac{\partial P}{\partial t}(\frac{\partial P}{\partial u}\frac{\partial t}{\partial x}+\frac{\partial P}{\partial v}\frac{\partial t}{\partial y})}{(\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2}=\frac{\frac{\partial P}{\partial x}\frac{\partial P}{\partial u}+\frac{\partial P}{\partial y}\frac{\partial P}{\partial v}}{(\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2} \\ \frac{\partial u}{\partial y}=-\frac{\frac{\partial P}{\partial t}(\frac{\partial P}{\partial v}\frac{\partial t}{\partial x}-\frac{\partial P}{\partial u}\frac{\partial t}{\partial y})}{(\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2}=\frac{-\frac{\partial P}{\partial x}\frac{\partial P}{\partial v}+\frac{\partial P}{\partial y}\frac{\partial P}{\partial u}}{(\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2} \\ \end{cases}

В итоге получается диффур для нахождения $P(t(x,y))$
$\left[ \left(\frac{\frac{\partial P}{\partial x}\frac{\partial P}{\partial u}+\frac{\partial P}{\partial y}\frac{\partial P}{\partial v}}{(\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2} \right)^2+ \left(\frac{-\frac{\partial P}{\partial x}\frac{\partial P}{\partial v}+\frac{\partial P}{\partial y}\frac{\partial P}{\partial u}}{(\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2} \right)^2 \right] \left(\frac{\partial^2 P}{\partial u^2}+\frac{\partial^2 P}{\partial v^2} \right)=\frac{\partial^2 P}{\partial t^2} \left((\frac{\partial t}{\partial x})^2+(\frac{\partial t}{\partial y})^2 \right)+\frac{\partial P}{\partial t} \left(\frac{\partial^2 t}{\partial x^2}+\frac{\partial^2 t}{\partial y^2} \right)$

$\left(\frac{\partial P}{\partial t} \right)^2 \left(\frac{\partial^2 P}{\partial u^2}+\frac{\partial^2 P}{\partial v^2} \right) \left((\frac{\partial t}{\partial x})^2+(\frac{\partial t}{\partial y})^2 \right)=\left[\frac{\partial^2 P}{\partial t^2} \left((\frac{\partial t}{\partial x})^2+(\frac{\partial t}{\partial y})^2 \right)+\frac{\partial P}{\partial t} \left(\frac{\partial^2 t}{\partial x^2}+\frac{\partial^2 t}{\partial y^2} \right) \right] \left( (\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2 \right)$

По условию сохранения модуля
$ \begin{cases} P=\sqrt{u^2+v^2} \\ x^2+y^2=t \\ \end{cases}

$ \begin{tabular}{lll} $\frac{\partial P}{\partial u}=\frac{u}{\sqrt{u^2+v^2}}$ & $\frac{\partial t}{\partial x}=2x$ & $(\frac{\partial t}{\partial x})^2=4x^2$ \\ $\frac{\partial^2 P}{\partial u^2}=\frac{v^2}{(u^2+v^2)^{\frac{3}{2}}}$ & $\frac{\partial^2 t}{\partial x^2}=2$ & $(\frac{\partial t}{\partial y})^2=4y^2$ \\ $\frac{\partial P}{\partial v}=\frac{v}{\sqrt{u^2+v^2}}$ & $\frac{\partial t}{\partial y}=2y$ & $(\frac{\partial P}{\partial u})^2=\frac{u^2}{u^2+v^2}$ \\ $\frac{\partial^2 P}{\partial v^2}=\frac{u^2}{(u^2+v^2)^{\frac{3}{2}}}$ &$\frac{\partial^2 t}{\partial y^2}=2$ & $(\frac{\partial P}{\partial v})^2=\frac{v^2}{u^2+v^2}$ \\ \end{tabular}$

В частном случае, при сохранении модуля, диффур становится однородным, второго порядка
$-4\frac{\partial P}{\partial t}-4\frac{\partial^2 P}{\partial t^2}(x^2+y^2)+\frac{4(\frac{\partial P}{\partial t})^2(x^2+y^2)}{\sqrt{u^2+v^2}}=0$
$\frac{\partial^2 P}{\partial t^2}t-(\frac{\partial P}{\partial t})^2\frac{t}{P}+\frac{\partial P}{\partial t}=0$
Замена: $\frac{\partial P}{\partial t}=P(t) z(t) \qquad \frac{\partial^2 P}{\partial t^2}=\frac{\partial P}{\partial t} z(t)+P(t)\frac{\partial z}{\partial t}=P(t) z^2(t)+P(t)\frac{\partial z}{\partial t}$

$\frac{\partial z}{\partial t} tP-z^2 \frac{t}{P}P^2+Ptz^2+P z=0$
$z=\frac{c_1}{t}$
$\frac{\partial P}{\partial t}=P\frac{c_1}{t}$

$P=c_2t^{c_1}=c_2(x^2+y^2)^{c_1}$

Для нахождения $u$ и $v$ лучше перейти в полярную систему координат, поскольку интегралы будут легче в пять раз
$u=u(\rho,\varphi) \qquad v=v(\rho,\varphi)$

$ \begin{cases} x= \rho \cos \varphi$ \\ y= \rho \sin \varphi$ \\ \end{cases}$

$P(\rho,\varphi)=P(\rho)=c_2 \rho^{2c_1}$

Условия Коши-Римана в полярной системе координат
$\rho \frac{\partial u}{\partial \rho}=\frac{\partial v}{\partial \varphi}$
$\frac{\partial u}{\partial \varphi}=-\frac{1}{\rho }\frac{\partial v}{\partial \rho}$

Функции $\frac{\partial u}{\partial \rho}(\frac{\partial P}{\partial u},\frac{\partial P}{\partial v},\frac{\partial P}{\partial \rho},\frac{\partial P}{\partial \varphi})$ и $\frac{\partial u}{\partial \varphi}(\frac{\partial P}{\partial u},\frac{\partial P}{\partial v},\frac{\partial P}{\partial \rho},\frac{\partial P}{\partial \varphi})$ также незначительно изменятся
$ \begin{cases} \frac{\partial u}{\partial \rho}=\frac{\rho\frac{\partial P}{\partial \rho}\frac{\partial P}{\partial u}+\frac{\partial P}{\partial \varphi}\frac{\partial P}{\partial v}}{\rho \left((\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2 \right)} \\ \frac{\partial u}{\partial \varphi}=\frac{-\rho \frac{\partial P}{\partial \rho}\frac{\partial P}{\partial v}+\frac{\partial P}{\partial \varphi}\frac{\partial P}{\partial u}}{(\frac{\partial P}{\partial u})^2+(\frac{\partial P}{\partial v})^2} \\ \end{cases}$

$ \begin{tabular}{ll} $P=c_2 \rho^{2c_1}$ \\ $\frac{\partial P}{\partial \rho}=2c_1c_2 \rho^{2c_1-1}$ & $\frac{\partial P}{\partial \varphi}=0$ \\ $\frac{\partial P}{\partial u}=\frac{u}{\sqrt{u^2+v^2}}$ & $\frac{\partial P}{\partial v}=\frac{v}{\sqrt{u^2+v^2}}$ \end{tabular}$

$ \begin{cases} \frac{\partial u}{\partial \rho}=\frac{\rho \frac{u}{\sqrt{u^2+v^2}}2c_1c_2 \rho^{2c_1-1}}{\rho}=2c_1 \frac{u}{\rho} \\ \frac{\partial u}{\partial \varphi}=-\rho \frac{v}{\sqrt{u^2+v^2}}2c_1c_2 \rho^{2c_1-1}=-2c_1\sqrt{{c_2}^2\rho^{4c_1}-u^2} \\ \end{cases}$$

С точностью до знаков, пусть $2c_1=c_1; {c_2}^2=c_2$
$ \begin{cases} \frac{\partial u}{\partial \rho}=c_1 \frac{u}{\rho} \\ \frac{\partial u}{\partial \varphi}=-c_1\sqrt{c_2\rho^{2c_1}-u^2} \\ \end{cases}$$

Чтобы найти $u(\rho,\varphi)$ необходимо решить ДУ в частных производных
$(\frac{\partial u}{\partial \varphi})^2={c_1}^2(c_2\rho^{2c_1}-u^2)$
$\partial \varphi^2=\frac{\partial u^2}{{c_1}^2(c_2\rho^{2c_1}-u^2)}$
$\partial \varphi=\pm \frac{\partial u}{c_1\sqrt{c_2\rho^{2c_1}-u^2}}$

Ответ от выбора знака не зависит
$\varphi=c_1^{-1}\arctg (\frac{u}{\sqrt{c_2\rho^{2c_1}-u^2}})+\widetilde c_3(\rho)$
$u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3(\rho))$

$\frac{\partial u}{\partial \rho}=c_1\sqrt{c_2}\rho^{c_1-1}\sin(c_1 \varphi+c_3(\rho))+\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3(\rho))\frac{\partial c_3(\rho)}{\partial \rho}=c_1 \frac{u}{\rho}=c_1 \frac{\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3(\rho))}{\rho}$
$\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3(\rho))\frac{\partial c_3(\rho)}{\partial \rho}=0$
$\frac{\partial c_3(\rho)}{\partial \rho}=0$
$c_3(\rho)=c_3$

$u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3)$

Осталось найти $v(\rho,\varphi)$
$\frac{\partial u}{\partial \rho}=c_1\sqrt{c_2}\rho^{c_1-1}\sin(c_1 \varphi+c_3)=\frac{1}{\rho}\frac{\partial v}{\partial \varphi}$

$v(\rho,\varphi)=\int{c_1\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3)d\varphi}=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4(\rho)$

$\frac{\partial v}{\partial \rho}=-c_1\sqrt{c_2}\rho^{c_1-1}\cos(c_1 \varphi+c_3)+\frac{\partial c_4(\rho)}{\partial \rho}=-\frac{1}{\rho}\frac{\partial u}{\partial \varphi}=-c_1\frac{\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)}{\rho}$
$\frac{\partial c_4(\rho)}{\partial \rho}=0$
$c_4(\rho)=c_4$

С учетом начального условия $P(\rho^2)=\operatorname{const}$ из множества аналитических функций $v$ необходимо выбрать такие, которые в паре с функцией $u$ постоянны вдоль любой линии семейства окружностей
$\begin{cases} u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \\ v(\rho,\varphi)=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4 \\ \sqrt{u^2+v^2}=\operatorname{const} \\ \rho^2=\operatorname{const} \\ \end{cases}$
$\begin{cases} u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \\ v(\rho,\varphi)=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4 \\ u^2+v^2=\operatorname{const} \\ \rho^2=\operatorname{const} \\ \end{cases}$
$\begin{cases} u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \\ v(\rho,\varphi)=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4 \\ d \left( \left(u(\rho=\operatorname{const},\varphi) \right)^2+ \left(v(\rho=\operatorname{const},\varphi) \right)^2 \right)=0 \\ \end{cases}$
$\begin{cases} u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \\ v(\rho,\varphi)=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4 \\ \frac{\partial \left(u(\rho=\operatorname{const},\varphi) \right)^2}{\partial \varphi}+\frac{\partial \left(v(\rho=\operatorname{const},\varphi)^2 \right)}{\partial \varphi}=0 \\ \end{cases}$
$\begin{cases} u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \\ v(\rho,\varphi)=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4 \\ \frac{\partial \left(\left(\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \right)^2 \right)}{\partial \varphi}+\frac{\partial \left( \left(\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4 \right)^2 \right)}{\partial \varphi}=0 \\ \end{cases}$
$\begin{cases} u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \\ v(\rho,\varphi)=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4 \\ -2c_1\sqrt{c_2}c_4\rho^{c_1}\sin(c_1 \varphi+c_3) =0 \\ \end{cases}$
$\begin{cases} u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \\ v(\rho,\varphi)=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)+c_4 \\ c4=0 \\ \end{cases}$

$\begin{cases} u(\rho,\varphi)=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3) \\ v(\rho,\varphi)=-\sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3) \\ \end{cases}$

$\omega=\sqrt{c_2}\rho^{c_1}\sin(c_1 \varphi+c_3)-i \sqrt{c_2}\rho^{c_1}\cos(c_1 \varphi+c_3)=$
$=-\sqrt{c_2}\rho^{c_1} i(\cos(c_1 \varphi+c_3)+i \sin(c_1 \varphi+c_3))=$
$=-\sqrt{c_2} i e^{\ln \rho^{c_1}}e^{i(c_1 \varphi+c_3)}=-\sqrt{c_2} i e^{\ln \rho^{c_1}}e^{i c_1 \varphi}e^{c_3}=$
$=-\sqrt{c_2} e^{c_3} i (e^{\ln \rho}e^{i \varphi})^{c_1}=(-\sqrt{c_2}i\cos (c_3)+\sqrt{c_2}\sin(c_3))e^{c_1 \ln z}$

$\omega=(c_1+c_2i)e^{c_3 \ln z}$

Волковыский 1.182, 1.183, 1.184, 1.185, 1.186