Math Revisited: The Cauchy-Riemann Equations And Harmonic Functions

Time to revisit one of the more fundamental aspects of advanced calculus. 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 the  function.

  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.

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?