Mumbai University > Information Technology, Computer Engineering > Sem 3 > Applied Mathematics 3

Marks: 5M

Year: Dec 2013

We have seen above the orthogonal trajectories of u = c1 are given by v = c2 where v is the harmonic conjugate of u.

$ :. u= e^{-x} cos y + xy$

Differentiating w.r.t x&y

$u_x = - e^{-x} cos y + y \hspace{0.2cm} and \hspace{0.2cm} u_y = - e^x sin y +x$

Also $f^1(z) = u_x + iv_x = u_x – iu_y$ [By C-R equations]

$= (-e^{-x} cos y + y) – i (-e^{-x} sin y + x)$

By Milne – Thompson’s method , we replace x by z and y by zero.

$:. f^1(z) = -e^{-z} – iz$

By integration $f(z) = e^{-z} – i \frac{z^2}{2} +c$

$:. f (z) = e^{-(x+iy)} – i\frac{(x+iy)^2}{2} +c\\ = e^{-x} e^{-iy} \frac{-i}{2} (x^2 +2 i xy – y^2 )+c\\ = e^{-x} (cos y – i sin y ) \frac{-i}{2} (x^2 – y^2) +xy$

Now only considering imaginary part

$v= e^{-x} sin y \frac{-1}{2} (x^2 – y^2)$

:. The required orthogonal trajectories are

$e^{-x} sin y +\frac{1}{2} (x^2 – y^2) = c.$