Leplace s Equations. Analyzing the Analyticity of Analytic Analysis DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING. Engineering Math 16.

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Leplace s Analyzing the Analyticity of Analytic Analysis Engineering Math 16.364 1

The Laplace equations are built on the Cauchy- Riemann equations. They are used in many branches of physics such as heat flow, charge and potential field distribution, fluid flow, gravitation and any other field where the dispersal and flow of stuff in a 2D or 3D space, whether the stuff is field strength, chemical concentrations, magnetic field effects, etc. is studied. It is said that the Laplace equations are the most important Partial Differential in physics. Engineering Math 16.364 2

In a nutshell, the Laplace equations state that, for f z analytic in a domain D and f z = u + iv, then u xx + u yy = 0 and v xx + v yy = 0 Engineering Math 16.364 3

This type of PDE is called The Laplacian and is sometimes written: 2 u = u xx + u yy = 0 2 is read nabla squared or (more commonly) del squared. Engineering Math 16.364 4

The Laplace are based on the Cauchy-Riemann equations. Later, it will be proven that the derivative of an analytic function is, itself, analytic. This means that the functions u and v have partial derivatives of all orders and that v xy exists. Engineering Math 16.364 5

We will also use the finding from real-valued calculus that: v xy = v yx and u xy = u yx Engineering Math 16.364 6

With that in mind, consider the following partial derivatives of the C-R equations C-R equations: u x = v y u y = v x Rearrange: u x v y = 0 u y + v x = 0 P. Derivatives with respect to x and y: u xx v yx = 0 u yy + v xy = 0 Add the PDEs: u xx v yx + u yy + v xy = 0 However: v yx = v xy Substitute and simplify to get: u xx + u yy = 0 Engineering Math 16.364 7

Similarly: C-R equations: u x = v y u y = v x Rewrite: v y u x = 0 u y + v x = 0 P. Derivatives with respect to y and x: v yy u xy = 0 u yx + v xx = 0 Add the PDEs: v xx + v yy + u yx u xy = 0 However: u yx = u xy Substitute and simplify to get: v xx + v yy = 0 Engineering Math 16.364 8

Solution of equations that are continuous are called harmonic functions. Harmonic functions u and v that satisfy the C-R equations are said to be harmonic conjugates of each other. Engineering Math 16.364 9

Example: Is this f z analytic? f z = x 2 + y 2 x + ie x sin y u = x 2 + y 2 x and v = e x sin y Are u and v harmonic conjugates of each other? Engineering Math 16.364 10

Example: Consider two functions u = x 2 + y 2 x and v = e x sin y Is u a harmonic function? Is v a harmonic function? Are u and v harmonic conjugates of each other? u x = 2x + 0 1 u y = 0 + 2y 0 u xx = 2 u yy = 2 Since u xx + u yy = 0, u is harmonic. Engineering Math 16.364 11

v = e x sin y v x = e x sin y v y = e x cos y v xx = e x sin y v yy = e x sin y Since v xx + v yy = 0, v is also harmonic. Engineering Math 16.364 12

So, u x, y and v x, y are both harmonic functions. That s fine, but we also want to know if the functions are harmonic conjugates of each other? Rephrasing: Is u + iv an analytic function? To be analytic in a domain, u and v must be harmonic conjugates of each other within that domain, which means the functions must satisfy the Cauchy-Riemann equations within that domain. Engineering Math 16.364 13

Recall from finding the Laplace equations: u x = 2x 1 u y = 2y v y = e x cos y v x = e x sin y Since u x v y, and u y v x, the functions do not satisfy C-R therefore they are not harmonic conjugates. They do not comprise an analytic function. Engineering Math 16.364 14

So, the question arises: If we have a harmonic function such as u = x 2 + y 2 x + y, can we develop an analytic function, f(z), that has this u(x, y) as the real part? That is, can we develop a function v that is a harmonic conjugate of u? Why, yes. We can. By using properties of Laplace and C-R equations, we can generate such a function v. Engineering Math 16.364 15

Given: u = x 2 + y 2 x + y Find v that is a harmonic conjugate of u: From C-R, v y = u x = 2x 1 and v x = u y = 2y 1 v is the anti-partial derivative of v y : v = v y y = 2x 1 y = 2xy y + h x N.B.: h(x) is a real valued function of x. So, if we find h x then we have found v Engineering Math 16.364 16

dh x Differentiate: v x = 2y + dx Also from C-R, v x = u y = 2y 1. By comparing these two values of v x dh x we get = 1. dx Which means h x = 1 dx = x + c for some arbitrary real constant c. Engineering Math 16.364 17

Finally, we have found that v = 2xy y x + c is the harmonic conjugate of u = x 2 + y 2 x + y Together, they form the analytic function f z = x 2 + y 2 x + y + i 2xy y x + c Engineering Math 16.364 18

Example: Given u x, y = e 4x cos 4y First, determine that u x, y a harmonic function. If so, find v that is a harmonic conjugate of u: u = e 4x cos 4y u x = 4e 4x cos 4y u xx = 16e 4x cos 4y u y = 4e 4x sin 4y u yy = 16e 4x cos 4y We see that u xx + u yy = 0, therefore u is harmonic. Engineering Math 16.364 19

u = e 4x cos 4y From C-R, u x = v y = 4e 4x cos 4y Anti-partial: v = v y y = e 4x sin 4y + h x [1] This is v x, y but we have to figure out what h x is. Differentiate w.r.t. x to get: v x = 4e 4x sin 4y + Also from C-R, we know v x = u y = 4e 4x sin 4y dh x dx Engineering Math 16.364 20

u = e 4x cos 4y v x = 4e 4x sin 4y + dh x dx Comparing the result from the first C-R equation with the result from the second, we see that dh x = 0 which means h x = c dx for some real constant c. Engineering Math 16.364 21

u = e 4x cos 4y Substitute h x = c into equation [1] and get: v = e 4x sin 4y + c This is the harmonic conjugate of u(x, y). Therefore the analytic function comprised of u and v is: f z = e 4x cos 4y + ie 4x sin 4y + ic This answers the question that was posed, but let s go further: Engineering Math 16.364 22

f z = e 4x cos 4y + ie 4x sin 4y f z = e 4x cos 4y + ie 4x sin 4y = e 4x cos 4y + i sin 4y = e 4x e i4y = e 4x+i4y = e 4 x+iy = e 4z (pretty neat, huh? ) Engineering Math 16.364 23

Example Given: u = cosh ax cos 5y Find a so that u(x, y) is harmonic. Then, find a harmonic conjugate v x, y. u = cosh ax cos 5y u x = a sinh ax cos 5y u y = 5 cosh ax sin 5y u xx = a 2 cosh ax cos 5y u yy = 25cosh ax cos 5y Because u xx = u yy a 2 = 25 a = ±5 Engineering Math 16.364 24

u = cosh ax cos 5y We are asked to find a harmonic conjugate, so we don t have to test all cases. We will use a = +5. u x = v y = 5 sinh 5x cos 5y v = v y y = 5 sinh 5x cos 5y y v = 5 sinh 5x 1 5 sin 5y + h x v = sinh 5x sin 5y + h x We almost have v. We still have to figure out h x. Engineering Math 16.364 25

u = cosh ax cos5 y v = sinh 5x sin 5y + h x v x = 5 cosh 5x sin 5y + dh x dx Since, for, a = +5, we know v x = u y = 5 cosh 5x sin 5y we see that dh x dx = 0 h x = c Engineering Math 16.364 26

u = cosh ax cos 5y If we take c = 0, we will get a harmonic conjugate. So, University of Massachusetts, Lowell f z = u + iv = cosh 5x cos 5y + i sinh 5x sin 5y. f z is an analytic function. For extra fun, show that f z = cosh 5z Engineering Math 16.364 27