Leplace s Analyzing the Analyticity of Analytic Analysis Engineering Math EECE 3640 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 EECE 3640 2
In a nutshell, the Laplace equations state: 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 EECE 3640 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 EECE 3640 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 EECE 3640 5
We will also use the finding from real-valued calculus that: v xy = v yx and u xy = u yx Engineering Math EECE 3640 6
Derivation of from 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 Substitute: v yx = v xy Cancellation yields: u xx + u yy = 0 Engineering Math EECE 3640 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 Substitute: u yx = u xy Cancellation yields: v xx + v yy = 0 Engineering Math EECE 3640 8
Solution of equations that are continuous are called harmonic functions. Remember: they must be continuous. That means they must be harmonic in a neighborhood of any points that are considered harmonic. Harmonic functions u and v that (together) satisfy the C-R equations are said to be harmonic conjugates of each other. Engineering Math EECE 3640 9
Example: Is this function analytic? f z = x 2 + y 2 x + ie x sin y u = x 2 + y 2 x (real part) and v = e x sin y (imaginary part) Are u and v harmonic conjugates of each other? Engineering Math EECE 3640 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 EECE 3640 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 EECE 3640 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 EECE 3640 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 in a domain. therefore they are not harmonic conjugates. They do not comprise an analytic function. Engineering Math EECE 3640 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 u(x, y) given here 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 EECE 3640 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 = න v y y = න 2x 1 y = 2xy y + h x N.B.: h(x) is a real valued function of x (just x). So, if we find h x then we have found v(x, y) Engineering Math EECE 3640 16
So, we have found that: v = 2xy y + h x dh x Differentiate w.r.t. x : v x = 2y + dx From C-R, we know : v x = u y = 2y 1. By comparing these two values of v x dh x we get = 1. dx Which means Regular anti-derivative h x = න 1 dx = x + c for some arbitrary real constant c. Engineering Math EECE 3640 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 = u + iv = x 2 + y 2 x + y + i 2xy y x + c Engineering Math EECE 3640 18
f z = x 2 + y 2 x + y + i 2xy y x + c Engineering Math EECE 3640 19
Example: Given u x, y = e 4x cos 4y + 2y 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 + 2y u x = 4e 4x cos 4y u y = 4e 4x sin 4y + 2 u xx = 16e 4x cos 4y u yy = 16e 4x cos 4y We see that u xx + u yy = 0, therefore u is harmonic. Engineering Math EECE 3640 20
u = e 4x cos 4y + 2y From C-R, v y = u x = 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. dh x Differentiate w.r.t. x to get: v x = 4e 4x sin 4y + dx Also from C-R, we know v x = u y = 4e 4x sin 4y 2 Engineering Math EECE 3640 21
u = e 4x cos 4y + 2y v x = 4e 4x dh x sin 4y + dx Comparing the result from the first C-R equation with the result from the second, we see that dh x dx = 2 which means h x = 2x + c Engineering Math EECE 3640 22
u = e 4x cos 4y + 2y Substitute h x = 2x + c into equation [1] and get: v = e 4x sin 4y 2x + c Therefore the analytic function comprised of u and v is: f z = e 4x cos 4y + 2y + ie 4x sin 4y 2ix + ic This answers the question that was posed, but let s go further: Engineering Math EECE 3640 23
f z = e 4x cos 4y + 2y + ie 4x sin 4y 2ix f z = e 4x cos 4y + ie 4x sin 4y + 2y 2ix + ic = e 4x cos 4y + i sin 4y 2ix + ( 2i 2 y)) + ic = e 4x e i4y 2i x + +yi + ic = e 4x+i4y 2i x + yi = e 4 x+iy 2i x + iy + ic = e 4z 2iz + ic Engineering Math EECE 3640 24
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 x = a sinh ax cos 5y u xx = a 2 cosh ax cos 5y u = cosh ax cos 5y u y = 5 cosh ax sin 5y u yy = 25cosh ax cos 5y Because u xx = u yy a 2 = 25 a = ±5 Engineering Math EECE 3640 25
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 Engineering Math EECE 3640 26
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 EECE 3640 27
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 >> without trig identities << Engineering Math EECE 3640 28