Math Revisited: Harmonic Conjugates And Cauchy-Riemann Equations

Harmonic conjugates and the Cauchy-Riemann equations are among the most important topics in advanced calculus. A useful theorem when looking at the Cauchy –Riemann equations is the following:

 

A function f(z) = u(x,y) + iv(x,y) is analytic in a domain D if and only if v is a harmonic conjugate of u.

 

Consider the function:

f(z) = u(x,y) + iv(x,y) =  (x² – y²) + i2xy

 

So that: u(x,y) =  (x² – y²)     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 -  ¶ u/ ¶ y  =   ¶ v/ ¶ x   and hence  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, to see if it's a harmonic conjugate, switch u(x,y) with  v(x,y) so that:

 

f(z) = u(x,y) + iv(x,y) =   2xy  + i(x² – y²)

 

u(x,y) = 2xy and v(x,y) =  (x² – y²)

 

We first check to see if the u, v functions are analytic

 

Take ¶ u/ ¶ x    =   2y

 

And:  ¶ v/ ¶ y    =   - 2y

 

Since:  ¶ u/ ¶ x  =   ¶ v/ ¶ y    

Thus, it holds only where y = 0,  so then f(x) is differentiable only for points that lie on the x –axis and we conclude the function reversed for conjugates is nowhere analytic. The conclusion is thus that while v is a harmonic conjugate of u throughout the  z-plane, v is not a harmonic conjugate of u.

 

In general, and based on this, one is led to conclude that given a function:

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

 

then f(z) is analytic in some domain D if and only if (-if(z) = v(x,y) –iu(x,y) is also analytic there

Example (2):   Let f(z) = 3x + y + i(3y – x)

 

Show that v is a harmonic conjugate of u and hence the function is analytic in a domain D when u and v are interchanged for f(z). Is the  function also entire?

We have u(x,y) = 3x + y and v(x,y) = (3y – x)

 

Check Cauchy relations:

Take ¶ u/ ¶ x    =   3

 

And:  ¶ v/ ¶ y    =   3

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

 

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

 

¶ v/ ¶ x  =  -1

 

And:  -  ¶ u/ ¶ y   =  - (1) = -1

 

So that -  ¶ u/ ¶ y  =   ¶ v/ ¶ x   and hence  v(x,y) is analytic..

Now, interchange u(x,y) with v(x,y):

 

f(z) = 3x -  y +  i(3y +  x)

 

We have u(x,y) = 3x - y and v(x,y) = (3y +  x)

 

Check the Cauchy relations:

 

Take ¶ u/ ¶ x    =  ¶ v/ ¶ y 

 

And:  ¶ v/ ¶ y    =   3  =  ¶ u/ ¶ x

 

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

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

¶ v/ ¶ x  =  +1

And:  -  ¶ u/ ¶ y   =  - (-1) =  +1

 

So that -  ¶ u/ ¶ y  =   ¶ v/ ¶ x   and hence  v(x,y) is analytic..

 

Since v is a harmonic conjugate of u then the function is analytic in a domain D when u and v are interchanged.

If the function is entire then it also satisfies the LaPlace equation: Ñ ²u = 0

 

Then:

¶ ² u/ ¶ x²  +  ¶ ² u/ ¶ y²       = 0 + 0 = 0

 

And: 

 

¶ ² v/ ¶ x²  +  ¶ ² v/ ¶ y²    =  0 + 0 = 0

 

So the function is also entire on the complex plane.

Practice Problems:

1) Given the function:

f(z) = u(x,y) + iv(x,y) =  cos x cosh y – i(sinx sinh y)

 

a)     Verify the Cauchy-Riemann equations are satisfied

b)     Are they also satisfied for the harmonic conjugate, i.e. when u and v are interchanged?

2)     Let u(x,y) =  (x² – y²) +  2x

 

a)     Show u(x,y) is a harmonic function

b)     Hence or otherwise, find the harmonic conjugate v(x,y) of u.

More Difficult Contour Integrals

Time for some more contour integrals. We have seen earlier examples of contour line integrals and now we look at a more detailed example:

We want to integrate around the closed contour  for which f(z) = 3x + 2iy:

We check first to see if the function is analytic using the Cauchy –Riemann relations:

Recall that we require:  (see e.g.   http://brane-space.blogspot.com/2013/10/looking-at-harmonic-conjugates-and.html   )

i) ¶ u/ ¶x = ¶ v/ ¶y  and ii) ¶ u/ ¶y   =  - ¶ v/ ¶x

For f(z) = 3x + 2iy  =  u(x,y) + iv(x,y)

We have:  ¶ u/ ¶x =  3   and ¶ v/ ¶y    = 2,    so  ¶ u/ ¶x ¹ ¶ v/ ¶y 

And:

¶ u/ ¶y   = 0 =     - ¶ v/ ¶x

Since condition (i) is not fulfilled the function is not analytic, hence we cannot evaluate using Cauchy’ theorem , i.e. 

 ò  _C  f(z) dz = 0

So we must integrate line segment by line segment, viz.

I  =   ò _C  f(z) dz  =   ò _C  (u + iv) (dx + idy)  =     

 ò  _C  (3x + 2iy) (dx + idy)

=    ò  _C  (3x dx  -   2y dy)  +  i ò  _C  ( 2y dx + 3x dy) 

I =     å³ _{n = 1}     [ò  _C  (3x dx  -   2y dy)   + iò  _C  ( 2y dx + 3x dy) 

On C1:  0 < x  < 1, y = 3, dy = 0

ò  _C1  (3x dx  -   2y dy)  +  i ò  _C1  ( 2y dx + 3x dy)   

=  ∫ ₀ ¹    3x dx  +   i∫ ₀ ¹    6 dy  =    [3/2 x²] ₀ ¹       +   i[6y] ₀ ¹    

=  3/2 + 6i

On C2: 3 < y  < 5, x = 1, and dx = 0

ò  _C2  (3x dx  -   2y dy)  +  i ò  _C2  ( 2y dx + 3x dy)   

=   ò₃ ⁵   ( -2y ) dy  +   i ò₃ ⁵     3 dy  =      [- y²] ₃ ⁵      +   i[3y] ₃ ⁵

= - 16 + 6i

On C3: 1 < t  < 0, x = t, and dx = dt, y = 2t +3, dy = 2dt 

Then:

ò  _C3  (3x dx  -   2y dy)  +  i ò  _C3  ( 2y dx + 3x dy)   

=   ∫ ₁ ⁰   [ 3t  dt  - 2(2t + 3) 2 dt] + i  ∫ ₁ ⁰   [ 2(2t + 3) dt + 3t (2dt)]

= - ∫ ₁ ⁰    (5t  + 12)  dt   + i∫ ₁ ⁰    (10t  + 6)  dt  

=   -  [5t² / 2 + 12t] ₁ ⁰  + i[5t²  + 6t] ₁ ⁰     =  29/2 – 11i

Then the contour integral value is:

I =  ( 3/2 + 6i)  + (-16 + 6i) + (29/2 – 11i) =  0 + i

Problems  for Math Mavens:

1) For the closed path shown in the diagram below, evaluate the contour integral. I.

Let f(z) = z ²

   2) For the example shown in the blog post (Fig. 1) , what would the value of I be if:

f(z) = 3x + i3y?

The Cauchy –Riemann Equations: Gateway to Advanced Complex Analysis

Back in March we examined a few examples of complex functions. Basically, it entailed writing them in assorted different forms, including complex polar. We operated more or less at the cusp between beginner complex analysis and more meaty stuff – peculiar to intermediate analysis. Now, in this post, I want to look at the interface between basic and more advanced complex analysis as embodied in the Cauchy – Riemann equations.

 

Basically, full details are beyond the scope of one or even a few blog posts, but interested readers are invited to get any number of texts which treat the topic, including ‘Complex Analysis for Mathematics and Engineering’, ‘Complex Variables and Applications’, and ‘Applied Complex Variables’ (Harper Collins College outline).

 

Basically, given a function defined: 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 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/ ¶ x²  =    ¶ ² v/ ¶ x  ¶ y    and

 

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

 

From which:

 ¶ 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.

 

Let’s look at an example, using a function from the March 19 blog.

 

We had: f(z) = (x² – y²) + i2xy

 

So that: u(x,y) =  x² -  y²

 

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/ ¶ 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

 

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) = 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.