Author Topic: TT1 Problem 3 (noon)  (Read 760 times)

Victor Ivrii

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TT1 Problem 3 (noon)
« on: October 19, 2018, 04:05:37 AM »
(a) Show that $v(x,y)= xe^x \sin (y) +y e^x\cos(y) $ is a harmonic function.

(b) Find the harmonic conjugate function $u(x,y)$.

(c) Consider $u(x,y)+iv(x,y)$ and write it as a function $f(z)$ of $z=x+iy$.

Ye Jin

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Re: TT1 Problem 3 (noon)
« Reply #1 on: October 19, 2018, 09:41:04 AM »
(a) WTS $v_{xx} + v_{yy}=0$
     $v_x = e^xsiny+xe^xsiny+ye^xcosy$, $v_{xx}=e^xsiny+e^xsiny+xe^xsiny+ye^xcosy$
     $v_y =xe^xcosy+e^xcosy-ye^xsiny$, $v_{yy}=-xe^xsiny-e^xsiny-e^xsiny-ye^xcosy$
     so, $v_{xx} + v_{yy}=0$

(b) Since v is harmonic, then it is analytic.
     $v_x=u_y, -v_y=u_x$
     $u=\int v_x dy=\int e^xsiny+xe^xsiny+ye^xcosy dy$
       $= -e^xcosy-xe^xcosy+e^x(ysiny+cosy)+h(x)$
     $u_x=-e^xcosy-e^xcosy-xe^xcosy+e^x(ysiny+cosy)+h^{'}(x)$
           $=-e^xcosy-xe^xcosy+e^xysiny+h^{'}x$
     so,$-xe^xcosy-e^xcosy+ye^xsiny=-e^xcosy-xe^xcosy+e^xysiny+h^{'}(x)$
     so, $h^{'}(x)=0$
     h(x)=c
    $u=-xe^xcosy+e^xysiny+c$
   
(c) $f(z)=u+iv= -xe^xcosy+e^xysiny+c+ixe^xsiny+iye^xcosy$
                       $=-e^xcosy(x-iy)+e^xsiny(y+ix)+c$
                       $=-e^xcos(y)\bar{z}+e^xsin(y)i\bar{z}+c$
                       $=\bar{z}e^x(cosy+isiny)+c$
                       $=\bar{z}e^{Rez}e^{Imz}+c$
                       $=\bar{z}e^z+c$


     
« Last Edit: October 19, 2018, 09:54:44 AM by Ye Jin »

Meng Wu

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Re: TT1 Problem 3 (noon)
« Reply #2 on: October 19, 2018, 09:59:30 AM »
For part$(b)$:
CR-equation is:
$$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

Chae Young Oh

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Re: TT1 Problem 3 (noon)
« Reply #3 on: October 19, 2018, 09:43:23 PM »
I think Ye Jin's CR equations are correct, because we want to find the harmonic conjugate of v (so vx=uy,−vy=ux) , not the harmonic conjugate of u (in that case ux=vy, uy=-vx).

Meng Wu

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Re: TT1 Problem 3 (noon)
« Reply #4 on: October 19, 2018, 11:26:20 PM »
.
« Last Edit: October 20, 2018, 10:09:34 PM by Meng Wu »

Victor Ivrii

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Re: TT1 Problem 3 (noon)
« Reply #5 on: October 20, 2018, 03:41:34 PM »
Since (c) says that $u+iv$ must be analytic,
$$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$

Also $\bar{z}e^{z}$ is neither analytic, nor anti-analytic (satisfying "adjoint" CR)
« Last Edit: October 21, 2018, 01:46:23 AM by Victor Ivrii »

Meng Wu

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Re: TT1 Problem 3 (noon)
« Reply #6 on: October 20, 2018, 10:45:12 PM »
Since $(c)$ gives $u(x,y)+iv(x,y)$, the CR equation is definitely $\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}; \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$.
$$\\$$
$(b).$
$$-\frac{\partial v}{\partial x}=-(e^x\sin(y)+xe^x\sin(y)+ye^x\cos(y))=\frac{\partial u}{\partial y}$$
$$\begin{align}\Rightarrow u(x,y)&=\int (-e^x\sin(y)-xe^x\sin(y)-ye^x\cos(y))dy \\&=
xe^x\cos(y)-ye^x\sin(y)+h(x)\end{align}$$
Hence,
$$\begin{align}\frac{\partial u}{\partial x}&=e^x\cos(y)+xe^x\cos(y)-ye^x\sin(y)+h'(x)\\&=\frac{\partial v}{\partial y}=xe^x\cos(y)+e^x\cos(y)-ye^x\sin(y)\end{align}$$
$$\Rightarrow h'(x)=0\\\Rightarrow h(x)=C$$
where $C$ is an arbitrary real constant.$\\$
Therefore, $$u(x,y)=xe^x\cos(y)-ye^x\sin(y)+C$$
$$\\$$
$(c).$
$$\\$$
$$\begin{align}u(x,y)+iv(x,y)&=xe^x\cos(y)-ye^x\sin(y)+C+i(xe^x \sin (y) +y e^x\cos(y))\\&=xe^x\cos(y)+ixe^x\sin(y)+iye^x\cos(y)-ye^x\sin(y)+C\\&=xe^{x+iy}+iye^{x+iy}+C\\&=e^{x+iy}(x+iy)+C\end{align}$$
Therefore, $$\begin{align}f(z)&=ze^z+C\end{align}$$