MAT334-2018F > Term Test 1

TT1 Problem 3 (noon)

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Victor Ivrii:
(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:
(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.
     $v_x=u_y, -v_y=u_x$
     $u=\int v_x dy=\int e^xsiny+xe^xsiny+ye^xcosy dy$
       $= -e^xcosy-xe^xcosy+e^x(ysiny+cosy)+h(x)$
     so, $h^{'}(x)=0$
(c) $f(z)=u+iv= -xe^xcosy+e^xysiny+c+ixe^xsiny+iye^xcosy$


Meng Wu:
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:
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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