written 7.7 years ago by |
Given imaginary part $v = e^x (xsiny + ycosy)$
$\therefore v_{x} = \frac{\partial V}{\partial x} = e^x (sin y) + (xsin y + ycosy)e^x \\ \therefore v^{2}_{x} = \frac{\partial^2 V}{\partial x^2} = e^x (sin y) + e^x (sin y) + (xsiny + ycosy)e^x \\ v^{2}_{x} = e^x (-xsiny + y(-cosy) - siny - siny) \\ v_{y} = \frac{\partial V}{\partial y} = e^x (-xsiny + y(-csoy) - siny -siny) \\ v_{y} = \frac{\partial V}{\partial y} = e^x (-xsiny - ycosy) - 2siny) \\ v^{2}_{y} = e^x(-xsiny + y(-cosy) - siny - siny) \\ = e^x(-xsiny - ycosy - 2siny)$
$v^{2}_{y} = -2e^x (siny) - e^x (xsiny + ycosy)$
$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0$
V satisfies Laplace equation
Now, let
$\psi_{1}(x,y) = v_{y} = e^x (xcosy + y(-siny) + cosy) \\ \psi_{2}(x,y) = v_{x} = e^x siny + (xsiny + ycosy)e^x$
Now put x = z, y = 0
$\psi_{1}(z, 0) = e^z (z + 1) \\ \psi_{2} (z,0) = 0$
Now, by Milne Thompson method,
$f^1 (z) = \psi_{1}(z,0) + i\psi_{2}(z,0) \\ = e^z (z+1) + i(0) \\ f^1 (z) = e^z(z + 1)$
Integrating above equation,
$F(z) = \int e^z (z + 1)dz \\ = (z+1)e^z + c$
(using Leibnitz rule)
Analytic function,
$F(z) = (z + 1)e^z + c$
Now put z = x + iy
$\therefore u + iv = (x + iy)e^{(x + iy)} + c \\ = (x + iy)e^x * e^{iy} + c \\ = e^x (x + iy) (cosy + isiny) + c$
$U + iV = e^x (xcosy - ysiny) + i(xsiny + ycosy) + c$
The above equation of a is corresponding harmonic conjugate
Analytic function $f(z) = ze^z$
Corresponding harmonic conjugate $w = e^x(xcosy - ysiny)$