1) f(z) = u(x,y) + iv(x,y) =
cos x cosh y – i(sinx sinh y)
We have to verify the Cauchy- Riemann equations. We note
that:
u(x,y) = cos x cosh
y and v(x,y) = - sinx sinh y
Then:
¶ u/ ¶ x = - sinx cosh y = ¶ v/ ¶ y
And:
¶ v/ ¶ x = - cos x sinh y = - ¶ u/ ¶ y
So the Cauchy_Riemann equations are satisfied
b2) Let
u(x,y)
= (x2 – y2) + 2x
Show
u(x,y) is a harmonic function
Solution:
If it’s harmonic then we must have:
¶ 2 u/ ¶ x2 + ¶ 2
u/ ¶ y2 = 0
Take the 1st, 2nd partials:
¶ u/ ¶ x = 2x + 2
and ¶ u/ ¶ y = -2y
¶ 2
u/ ¶ x2 = 2
and ¶ 2
u/ ¶ y2 = = -2
Therefore:
¶ 2 u/ ¶ x2 + ¶ 2
u/ ¶ y2 = (2) +
(-2) = 0
SO that
u(x,y) is a harmonic function.
b) Hence or otherwise find the harmonic conjugate v(x,y)
We’ve already shown:
¶ u/ ¶ x = 2x + 2
and ¶ u/ ¶ y = -2y
By the Cauchy –Riemann equations:
¶ v/ ¶ x =
- ¶ u/ ¶ y = 2y
And:
¶ u/ ¶ x = ¶ v/ ¶ y = 2x + 2
Take the differential using the chain rule:
dv = (¶ v/ ¶ x ) dx +
(¶ v/ ¶ y ) dy
Substitute from Cauchy-Riemann equations:
dv = 2y (dx) + (2x + 2) dy
Integrating:
v = 2 xy + 2xy + 2y
Or:
v = 4xy + 2y = 2y (2x +
1) + C
No comments:
Post a Comment