f(z) = f(x + iy) = u(x,y) + iv(x,y)

which is taken to be differentiable at the point z

_{0}= x_{0 }+ iy_{0}- then the partial derivatives of u and v exist at the point (x_{0 }, y_{0}) and satisfy the equations:
a) ¶ u/ ¶ x = ¶ v/ ¶ y and b) ¶ v/ ¶ x = - ¶ u/ ¶ y

Which are the

*Cauchy- Riemann equations*. If such condition is met then the function is said to be analytic in the region, Â.
An additional condition likely to be examined is whether the function is

*harmonic*. If it is, then u and v also have continuous 2^{nd}partial derivatives. Then, we can differentiate both sides of (a) with respect to x, and (b) with respect to y to obtain:
1) ¶

^{2}u/ ¶ x^{2}= ¶^{2}v/ ¶ x ¶ y and
2) - ¶

^{2}u/ ¶ y^{2}= ¶^{2}v/ ¶ y ¶ x
From which:

¶

^{2}u/ ¶ x^{2}= - ¶^{2}u/ ¶ y^{2}or ¶^{2}u/ ¶ x^{2}+ ¶^{2}u/ ¶ y^{2}= 0
Which is known as

**.**Laplace ’s equation
Similarly, we can perform an analogous process for v (differentiating both sides of (a) with respect to y and (b) with respect to x) to arrive at:

¶

^{2}v/ ¶ x^{2}+ ¶^{2}v/ ¶ y^{2}= 0
If then this equation is satisfied, v is harmonic.

Let’s look at an example using the function.

We had: f(z) = (x

^{2 }– y^{2}) + i2xy
So that: u(x,y) = x

^{2}- y^{2}
And v(x,y) = 2x y

Now, we first check to see if the eqns. are analytic

Take ¶ u/ ¶ x = 2x

And: ¶ v/ ¶ y = 2x

Since: ¶ u/ ¶ x = ¶ v/ ¶ y then u(x.y) is analytic

Now check the other function, v(x.y)

¶ v/ ¶ x = 2y

And: - ¶ u/ ¶ y = - (-2y) = 2y

So that v(x,y) is analytic..

Note: If f(z) is analytic everywhere in the complex plane it is said to be

*an entire function*.
Now, check to see whether the functions are

*harmonic*.
For u(x,y) this means we need: ¶

^{2}u/ ¶ x^{2}+ ¶^{2}u/ ¶ y^{2}= 0
Since: ¶ u/ ¶ x = 2x, then ¶

^{2}u/ ¶ x^{2}= 2
Since: ¶ u/ ¶ y = -2y then ¶

^{2}u/ ¶ y^{2}= -2
Then:

¶

^{2}u/ ¶ x^{2}+ ¶^{2}u/ ¶ y^{2}= 2 + (-2) = 0
For v(x.y): we need ¶

^{2}v/ ¶ x^{2}+ ¶^{2}v/ ¶ y^{2}= 0
Since: ¶ v/ ¶ x = 2y then ¶

^{2}v/ ¶ x^{2}= 2
Since: ¶ v/ ¶ y = 2x then ¶

^{2}v/ ¶ y^{2}= 2
Then: ¶

^{2}v/ ¶ x^{2}+ ¶^{2}v/ ¶ y^{2}= 2 + 2 = 4
So, v(x,y) is evidently not harmonic since the sum of the 2

^{nd}partials are not equal zero.__:__

*Problems for the Math Maven*
1) Given the function: u(x,y) = x

^{3}– 3xy^{2}
Show the function is harmonic on the entire complex plane.

2) Given the function: u(x.y) = exp(-x) [x sin y – y cos y]

a) Show u(x,y) is harmonic

b) Find v(x,y) such that f(z) = u + iv is analytic

_{}

^{}

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