MAT389 Fall 2016, Problem Set 4

Transcription

1 MAT389 Fall 2016, Problem Set 4 Harmonic conjugates 4.1 Check that each of the functions u(x, y) below is harmonic at every (x, y) R 2, and find the unique harmonic conjugate, v(x, y), satisfying v(0, 0) = v 0. Express the resulting holomorphic (entire, in fact) functions, f(z) = u(x, y) + iv(x, y), in terms of z. (i) u(x, y) = ax + by + c (where a, b, c R) and v 0 = 0, (ii) u(x, y) = x 2 y 2 2x and v 0 = 1, (iii) u(x, y) = y 3 3x 2 y and v 0 = 0, (iv) u(x, y) = x 4 6x 2 y 2 + y 4 and v 0 = 0, (v) u(x, y) = e 2y cos 2x and v 0 = 1. Hint: for expressing the entire function in (v) in terms of the complex variable z, you will need the definition of the exponential function e z = e x+iy = e x e iy = e x (cos y + i sin y) (i) Since u(x, y) is linear in both x and y, it is clear that u xx = u yy = 0, proving the harmonicity of u(x, y) over the whole of R 2. Using the Cauchy Riemann equations we can find the partial derivatives of a putative harmonic conjugate v(x, y): v x = u y = b, v y = u x = a. Integrating the first of these equations gives v(x, y) = bx + φ(y) (since we are integrating with respect to x, the constant of integration can depend on y). Differentiating with respect to y, we obtain v y = φ (y), which, put together with the second of the Cauchy Riemann equations above, yields φ (y) = a, or φ(y) = ay + d for some real constant d. The initial condition then implies 0 = v(0, 0) = ( bx + ay + d) = d. (x,y)=(0,0) All in all, we have f(z) = u(x, y) + iv(x, y) = ax + by + c + i( bx + ay) = (a ib)z + c. (ii) We have u xx = 1 and u yy = 1, so u(x, y) satisfies the Laplace equation. From the Cauchy Riemann equations, we get Following the same steps as in (i), we have v x = u y = 2y, v y = u x = 2x 2. v = 2xy + φ(y) v y = 2x + φ (y) φ (y) = 2 φ(y) = 2y + C (C R)

2 From the initial condition v(0, 0) = 1 we calculate C = 1. Finally, f(z) = u(x, y) + iv(x, y) = x 2 y 2 2x + i(2xy 2y + 1) = z 2 2z + i. (iii) Once again, the procedure is exactly the same as in (i): u xx + u yy = 6y + 6y = 0. v x = u y = 3x 2 3y 2, v y = u x = 6xy. v = x 3 3xy 2 + φ(y) = v y = 6xy + φ (y) = φ (y) = 0 = φ(y) = C (C R) v(0, 0) = 0 = C = 0. f(z) = u(x, y) + iv(x, y) = y 3 3x 2 y + i(x 3 3xy 2 ) = i(x + iy) 3 = iz 3. (iv) Yeah, this is getting boring... u xx + u yy = 12x 2 + ( 12x 2 ) = 0. v x = u y = 12x 2 y 4y 3, v y = u x = 4x 3 12xy 2. v = 4x 3 y 4xy 3 + φ(y) = v y = 4x 3 12xy 2 + φ (y) = φ (y) = 0 = φ(y) = C (C R) v(0, 0) = 0 = C = 0. f(z) = u(x, y) + iv(x, y) = x 4 6x 2 y 2 + y 4 + i(4x 3 y 4xy 3 ) = (x + iy) 4 = z 4. (v) Okay, at least this one is a little different! u xx + u yy = 4e 2y cos 2x + 4e 2y cos 2x = 0. v x = u y = 2e 2y cos 2x, v y = u x = 2e 2y sin 2x. v = e 2y sin 2x + φ(y) = v y = 2e 2y sin 2x + φ (y) = φ (y) = 0 = φ(y) = C (C R) v(0, 0) = 1 = C = 1. f(z) = u(x, y) + iv(x, y) = e 2y cos 2x + i(1 e 2y sin 2x) = e 2y e 2xi + i = e 2iz + i

3 4.2 Let u, v : Ω R 2 R be harmonic functions. Show that v is a harmonic conjugate to u if and only if u is a harmonic conjugate to v. Hint: Cauchy Riemann. Nuff said. Saying that v is a harmonic conjugate to u is, by definition, saying that f(z) = u + iv is holomorphic: the partial derivative u x, u y, v x and v y exist and are continuous on Ω, and they satisfy the Cauchy Riemann equations u x = v y and u y = v x. Define g(z) = U + iv = v iu. We claim that g(z) is holomorphic in other words, that u is a harmonic conjugate to v. Indeed, the partial derivatives U x, U y, V x and V y exist and are continuous on Ω: after all, they coincide with the partial derivatives of u and v up to a sign. Moreover, the Cauchy Riemann equations for g are satisfied on Ω: U x = v x = u y = V y, U y = v y = u x = V y. There is an even more straightforward way of going about this problem. You need only observe that g(z) = if(z) (and f(z) = ig(z)), so g is holomorphic if and only if f is. There are three reasons I directed you towards using the Cauchy Riemann equations instead of the second, most direct, approach above. The first approach shows you an interesting connection between the sign in the Cauchy Riemann equations and that in the statement of this problem; the second links that sign to the fact that i 2 = 1. It also allows me to insist on the fact that the Cauchy Riemann equations are only necessary conditions for differentiability: adding the continuity of the first partial derivatives gives a sufficient condition. The next problem. 4.3 Let u, v : Ω R 2 R be harmonic functions, and suppose v is a harmonic conjugate to u. Under which conditions is u (not u) also a harmonic conjugate to v? If v is a harmonic conjugate to u (i.e., if u + iv is holomorphic), then u x = v y and u y = v x. If, moreover, u is a harmonic conjugate to v (i.e., v + iu is holomorphic), then v x = u y and v y = u x. Out of these four equations, we can get the following ones: u x = u x, u y = u y, v x = v x, v y = v y Taking the right-hand sides of these equations to their respective left-hand sides, and dividing by 2, we get u x = 0, u y = 0, v x = 0, v y = 0. In other words, both u and v must be (locally) constant.

4 Steady temperatures The variation over time of temperature in a given (homogeneous, isotropic) material is controlled by the so-called heat equation, T t = α T, where the temperature T is a function of space and time, is the Laplacian, and α is a positive constant known as the thermal diffusivity of the material in question. Time-independent temperature distributions are then solutions to the Laplace equation T = 0. Complex Analysis can be used to find solutions of the above in two-dimensional contexts e.g., a thin layer of the material that is insulated in the vertical direction). 4.4 Let Ω R 2 be the infinite strip {(u, v) R 2 0 < v < 1}, and notice that its boundary Ω consists of two pieces: the lines v = 0 and v = 1. Consider the boundary-value problem given by T = 0 on Ω, T (u, 0) = T 0, T (u, 1) = T 1 (1) Check that a solution (which is, in fact, unique) to (1) is given by the real part of the function (which is holomorphic on Ω). F (w) = T 0 i(t 1 T 0 )w Comment: there is, of course, no need to appeal to Complex Analysis to find the solution to (1). By symmetry, it cannot depend on the coordinate u along the strip; only on the coordinate v across the strip. Then T = 0 is reduced to T vv = 0, which says that T must be a linear function of v. Making it satisfy the boundary conditions fixes which linear function it is. Since F (w) is holomorphic on Ω, its real part, T (u, v) = Re F (w) = T 0 + (T 1 T 0 )v, is harmonic on Ω. Moreover, T (u, 0) = T 0 + (T 1 T 0 )0 = T 0 and T (u, 1) = T 0 + (T 1 T 0 )1 = T 1. T = T 1 v T = T 0 + (T 1 T 0 )v T = T 0 u

5 4.5 Show that the transformation w = f(z) = 1 π ( ) log z + i Arg z defines a holomorphic function on the upper half-plane H of the z-plane (), and that it maps said upper half-plane to the infinite strip Ω in the w-plane. Where are the two pieces of Ω coming from? Hint: remember Problem 3.10(ii). We have already shown in Problem 4.2(ii) that f(z) is holomorphic on the upper-half plane (in fact, on a strictly larger open set, but who cares): u(r, θ) = log r and v(r, θ) = θ, as well as their partial derivatives of the first order, are continuous over the whole of the upper half-plane H, and satisfy the Cauchy Riemann equations in polar form, u r = 1 r v θ, 1 r u θ = v r. In polar coordinates, the upper half-plane can be described by the inequalities r > 0 and 0 < θ < π. Thus (1/π) log r sweeps the whole real line, while θ/π moves from 0 to 1. In particular, the rays θ = θ 0 (with 0 < θ 0 < π constant) in the z-plane get sent to horizontal lines v = θ 0 /π in the w-plane. By examining the limits θ 0 0 and θ 0 π, it is clear that v = 0 and v = 1 come from the positive and the negative real axes, respectively. 4.6 Show that the real part of F f provides a solution to the boundary-value problem T = 0 on H, T (x, 0) = { T0 if x > 0 T 1 if x < 0 (2) Comment: (2) is not as easy as (1), though you could still make a symmetry argument to conclude that the solution should be independent of the radial coordinate and depend only on the angular one. Notice that the radial (resp., angular) coordinate in (2) corresponds to the u (resp. v) coordinate in (1) through the transformation w = f(z). We have T (x, y) = Re(F f)(z) = T 0 + (T 1 T 0 ) Arg z π = T 0 + (T 1 T 0 ) θ π Since F f is a holomorphic function on H, its real part satisfies the Laplace equation. Along the positive real axis, we have θ = 0, and so T (x, 0) = T 0 if x > 0. The negative real axis is given by θ = π, which yields T (x, 0) = T 1 for x < 0.

6 T = T 0 + (T 1 T 0 ) Arg z π y T = T 1 w = 1 π ( ) log z + i Arg z T = T 0 x v T = T 1 T = T 0 + (T 1 T 0 )v T = T 0 u Electrostatic potentials Electric and magnetic fields (in three-dimensional vacuum) obey Maxwell s equations: E = ρ E = B ɛ 0 t ( ) E B = 0 B = µ 0 J + ɛ 0 t where E and B denote the electric and magnetic fields, respectively; ρ and J are the electric charge density (charge per unit volume) and electric current density (current per unit area), respectively; and ɛ 0 and µ 0 are universal constants known as the permittivity and permeability of free space. In static field configurations with zero electric charge and density, the electric and magnetic fields decouple: the electric field only enters into the first two equations; the magnetic field, only into the last two. For E, we have E = 0, E = 0

7 In this situation, we can show that E is (minus) the gradient of some function V, which we call the electrostatic potential 1. It satisfies the Laplace equation, for V = ( V ) = E = 0 In problems with translational symmetry along the vertical axis, V only depends on the x and y coordinates. Since it is a harmonic function, we might be able to use Complex Analysis! 4.7 If you have taken a course on electromagnetism, you have probably seen (even if you don t remember right now) that the electrostatic potential created by an infinite wire is given by V = λ 2πɛ 0 log r 0 r, where λ is the linear charge density of the wire, r is the distance to it, and r 0 is an arbitrary constant. In terms of the three-dimensional picture, we are placing the wire along the z-axis, and r is the radial coordinate in a cylindrical coordinate system. Now think of that same wire inside a cylinder of unit radius and parallel to it. If we keep the surface of the cylinder at a constant value of the potential (physically, this means that the electric field is perpendicular to it), the potential function above also provides a solution to the Laplace equation V = 0 on the inside of the cylinder (except at the center of it where the wire is located), and satisfies a Dirichlet boundary condition: V r=1 = V 0. (i) Calculate the value of V 0 in terms of r 0. (ii) Show that the equipotential lines (the curves where the potential is constant) are circles, and find an expression for their radius in terms of the value of the potential. (i) Substituting r = 1, we get V 0 = λ 2πɛ 0 log r 0 (ii) The equipotential V = V 1 is given by the equation V 1 = λ log r 0 2πɛ 0 r = V 0 λ log r 2πɛ 0 This equation depends only on the radial variable, so the equipotential line V = V 1 is indeed a circle with radius ( r = r 0 exp 2πɛ ) ( 0V 1 = exp 2πɛ ) 0(V 1 V 0 ) λ λ 4.8 Let us forget now about the direction parallel to the wire. The interior of the cylinder then corresponds to the punctured unit disc D. We will consider it as the punctured unit disk in the w-plane. The radial coordinate r in the statement of Problem 5.4 then refers to the modulus of w. 1 This is only true on simply-connected regions of three-dimensional space, but let s obviate this finer point here we haven t even defined what simply-connected means!

8 The Möbius transformation w = T (z) = z i z + i maps the upper half-plane H to the unit disc D, and the point z = i to the origin of the w-plane (see Problem 2.6). (i) Show that the composition W = V T defines a harmonic function on the set H {i} of the z-plane. (ii) Show that W is constant along the real line in the z-plane. Which value does it take? Conclude that W provides a solution of the Laplace equation on H {i} satisfying W H = const. Physically, this is describing the electrostatic potential generated by an infinite wire placed at a constant distance from an infinite plane. (iii) Find equations for the equipotential lines W = const. (i) The transformation w = T (z) takes H {i} to D, and is holomorphic at every point of H {i}. On the other hand, V is harmonic on D. Their composition W = V T is thus harmonic on H {i}. y v i w = T (z) 0 = T (i) u x (ii) The transformation w = T (z) takes the real line in the z-plane to the unit circle in the w-plane. Since V takes the value V 0 on the latter, so does W on the former. (iii) Let us start by finding an explicit expression for W in terms of x and y. Since V only depends on the modulus of w, we just need to calculate Substituting into V, we find w = z i z + i z i = z + i = x2 + (y 1) 2 x 2 + (y + 1) 2 W = V 0 λ log w = V 0 λ log x2 + (y 1) 2 4πɛ 0 4πɛ 0 x 2 + (y + 1) 2 ( ) The equipotential W = W 1 is thus given by the equation W 1 = V 0 λ log x2 + (y 1) 2 4πɛ 0 x 2 + (y + 1) 2

9 which we can rewrite as exp 4πɛ 0(V 0 W 1 ) λ = x2 + (y 1) 2 x 2 + (y + 1) 2 Notice that the left-hand side of this last equation is just a (nonzero) constant. Denoting it by α, we have α [ x 2 + (y + 1) 2] = x 2 + (y 1) 2 If α = 1 (equivalently, W 1 = V 0 ), this equation reduces to y = 0 the boundary of our problem, where we already knew that W takes the value V 0. On the other hand, if α 1, we get ( x 2 + y α + 1 ) 2 = α 1 ( ) α α 1 Thus the equipotential lines (other than the boundary itself) are circles centered on the y-axis. v i u Fluid flow Under certain conditions, the velocity field V of a fluid can be treated much in the same way as the electric field of the last few problems. We need to assume that the flow is stationary, irrotational and incompressible. In other words, that the following equations are satisfied: V t = 0, V = 0, V = 0. As in the case of the electric field, we can write 2 V as the gradient of some function φ usually referred to as the velocity potential which satisfies the Laplace equation: φ = ( φ) = V = 0. We can now attack problems with translational symmetry along an axis with Complex Analysis techniques. 2 Once again, on simply-connected domains; once again, don t worry about it for now.

10 4.9 Consider the stationary flow V = V 0 û (here û is the unit vector pointing in the positive u-direction) in the upper-half plane H = {(u, v) R 2 v > 0}. (i) Find a velocity potential φ that gives this velocity field. (ii) For your choice of φ, find a harmonic conjugate ψ. Show that the velocity field above is tangent to the level curves ψ = const. In particular ψ takes a constant value along the boundary of H. (i) We need to solve the equations φ u = V 0, φ v = 0. The second of these say that the potential is independent of the v-coordinate; the first, that it is a linear function of u. One solution is φ = V 0 u. (ii) The function V 0 w is holomorphic on H, and its real part coincides with φ. Thus, ψ = Im(V 0 w) = V 0 v is a harmonic conjugate to φ throughout H. Its level curves are horizontal lines to which V is obviously tangent. The situation in this last problem is paradigmatical. From a physical point of view, the velocity field should be parallel to an impermeable boundary a component perpendicular to it would signal that there is a flux of fluid through that boundary. A potential for such a velocity field will then never be constant along the boundary; a harmonic conjugate to it, though, always will. A strategy for finding the velocity field of a fluid near an impermeable boundary is then the following. 1) Find a solution of the Laplace equation ψ = 0 on a domain Ω, subject to the boundary condition ψ Ω = const (if the boundary consists of different connected components, the constant value of ψ may be different in different components). 2) Find a harmonic conjugate to ψ on Ω, and let the velocity potential φ be minus that function (recall Problem 4.6). 3) Take the gradient of that velocity potential. The easiest way of visualizing this velocity field, though, is to simply look at the level curves of ψ, which coincide with the integral lines of the vector field V = φ (try it in the problem above!). We will now find the velocity field of a fluid around a cylinder (see Figure 1). The first observation is that the picture is symmetric with respect to reflection across the x-axis. In particular, the velocity field along the the x-axis (when x < 1 and x > 1, of course) is horizontal, and so it is enough to solve the problem in the domain { Ω = H (x, y) R 2 } x 2 + y 2 > 1 ; that is, the upper-half of Figure Show that the transformation w = f(z) = z + 1/z takes the boundary of Ω to the real axis, and Ω to the upper half-plane H. Notice that f(z) is holomorphic outside of z = 0 so, in particular, it is holomorphic on Ω.

11 Figure 1: Fluid flow around a cylinder Hint: consider separately the three pieces that make up the boundary of Ω: the real axis to the left of z = 1, the upper half of the unit circle (where z = e iθ with 0 θ π), and the real axis to the right of z = 1. Restricted to real values greater than 1, the function z z + 1/z takes on positive real values. It is moreover always increasing (considered as real-valued function of a real variable, its derivative is 1 1/z 2, which is positive on R >1 ). Hence it maps R >1 to R >2. Essentially the same argument shows that R <1 is mapped to R <2. The semicircular part of the boundary of Ω consists of complex numbers of the form z = e iθ, with θ ranging from 0 to π. The images of these points are f(e iθ ) = e iθ + e iθ = 2 cos θ, which traverse the real interval [ 2, 2] in the image. The transformation w = f(w) is conformal at every point of Ω (except for z = ±1). Since traversing this curve from left to right leaves Ω to the left, the image of Ω should also be to the left of the image of Ω. In other words, f(ω) = H Consider the function F (w) = ψ iφ, where φ and ψ are the functions you constructed in Problem 4.9. According to Problem 4.2, F (w) is a holomorphic function on the upper-half plane. Composing F f then gives you a holomorphic function on Ω. The velocity field we are after is then Im(F f), and its integral lines are given by the level curves of Re(F f). Find equations for the latter, and plot a few of them (probably aided by some computer algebra system; e.g., Wolfram Alpha) to see that you are reproducing Figure 1). With ψ = V 0 v and φ = V 0 u, we have F (z) = iv 0 w, and ( Re(F f) = Re [ iv 0 z + 1 )] ( = V 0 y z y x 2 + y 2 ) ( = V 0 r 1 ) sin θ (3) r Here are a few of the level curves of this function. The velocity field of the fluid is tangent to these curves.

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