Solving Cauchy-Riemann Partial Differential Equations

Cauchy-Riemann Partial Differential Equations are at the core of many problems in mathematical physics. What are the Cauchy- Riemann equations?  

Given an equation of form:

f(z) = f(x + iy)   = u(x,y) + iv(x,y)

 

which is taken to be differentiable at the point z ₀ = x₀ + iy₀  - then the partial derivatives of u and v exist at the point (x₀ ,  y₀ ) 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 concerns 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/ ¶ x²  =    ¶ ² v/ ¶ x  ¶ y    and

2)  - ¶ ² u/ ¶ y²  =    ¶ ² v/ ¶ y  ¶ x  

 

From which a special example is obtained:

 ¶ ² u/ ¶ x²  =   - ¶ ² u/ ¶ y²       or    ¶ ² u/ ¶ x²  +  ¶ ² u/ ¶ y²       = 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/ ¶ x²  +  ¶ ² v/ ¶ y²       = 0

If then this equation is satisfied, v is harmonic.

We next look at the function.

  f(z) = (x² – y²) + i2xy

So that: u(x,y) =  x² -  y²

 

And v(x,y) =  2x y

 

Next, we 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 we see:  -  ¶ u/ ¶ y   =  - (-2y) = 2y

 

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

  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) we need: ¶ ² u/ ¶ x²  +  ¶ ² u/ ¶ y²       = 0

Since: ¶ u/ ¶ x    =   2x, then   ¶ ² u/ ¶ x²  =   2

 

Since: ¶ u/ ¶ y   =   -2y then  ¶ ² u/ ¶ y²    =   -2 

Then: ¶ ² u/ ¶ x²  +  ¶ ² u/ ¶ y²       = 2 + (-2) = 0

 For v(x.y): we need ¶ ² v/ ¶ x²  +  ¶ ² v/ ¶ y²       = 0

 Since: ¶ v/ ¶ x  =  2y then  ¶ ² v/ ¶ x²  =   2

Since: ¶ v/ ¶ y    =   2x   then  ¶ ² v/ ¶ y²      = 2

Then: ¶ ² v/ ¶ x²  +  ¶ ² v/ ¶ y²         = 2 + 2 = 4

 

 Evidently, v(x,y) isn't harmonic since the sum of the 2^nd partials doesn't equal zero.

 

Suggested Problems:

1)     Given the function: u(x,y) = x³ – 3xy²

 

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

 
3) Let f(z) = exp(x) cos(y) + i(exp(x)sin(y) = u(x,y) + iv(x,y)

a) Determine if the function is analytic for both u and v.

b) Determine if the function is harmonic for both u and v.