Spotlight on Laplace s Equation

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Vincent Wilkerson
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1 16 Spotlight on Laplace s Equation Reference: Sections 1.1,1.2, and 1.5. Laplace s equation is the undriven, linear, second-order PDE 2 u = (1) We defined diffusivity on page 587. where 2 is the Laplacian operator defined in Section 1.1. Laplace s equation models steady-state temperatures in a body of constant diffusivity. By steady state we mean that the temperature function u does not change with time, although it may vary from point to point within the body. Laplace s equation also models the gravitation and magnetic potentials in empty space, as well as electric potential and the velocity potential of ideal fluids, so the potential equation is another name for PDE (1). The operator 2 has the following forms in various coordinate systems: Rectangular 2 = 2 / x / y / z 2 (2) Polar 2 u = u rr + 1 r u r + 1 r 2 u θθ (3) x z θ φ ρ Warning: Physicists usually interchange the symbols θ and φ. y Cylindrical 2 u = u rr + 1 r u r + 1 r 2 u θθ + u zz (4) Spherical 2 u = 1 ρ 2 (ρ2 u ρ ) ρ + 1 ρ 2 sin φ (sin φ u 1 φ) φ + ρ 2 sin 2 φ u θθ (5) Rectangular coordinates (x, y, z) and spherical coordinates (ρ, θ, φ) are two ways of locating points in 3. These coordinates are related as follows: φ measures the angle from the vector k to the vector = x i + y j + z k, of length ρ = (x 2 + y 2 + z 2 ) 1/2, and θ is the angle from i to the vector r = x i + y j. So we have x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ Laplace s equation has many solutions. For example, u = c 1 e x cos y + c 2 z + c 3 e 4z cos 4x are solutions in rectangular coordinates for all constants c 1, c 2, c 3, while u = c 1 r cos θ + c 2 r 2 sin 2θ are solutions of the two-dimensional Laplace s equation in polar coordinates for all c 1 and c 2. We need boundary conditions on bounded regions to select a unique solution. We will solve representative problems in various coordinate systems and derive the fundamental properties of these solutions. The Dirichlet Problem: Harmonic Functions Suppose that R is a region in 2 or 3, and that h is a piecewise continuous function defined on R, the boundary of R. Then the Dirichlet problem for R with boundary data h is the following boundary value problem:

2 17 Dirichlet Problem We denote R together with its boundary R by R. Let h be in C ( R) for a region R. Find a function u in C 2 (R) such that 2 u = in R, and u is in C (R ) with u = h on R. Write the Dirichlet problem for R with boundary data h as (PDE) 2 u = in R (BC) u = h on R A solution u of (PDE) is a harmonic function or a potential function. The boundary value problem (6) is the classical Dirichlet problem, and u is a classical solution for that problem. The Dirichlet problem also makes sense when h is only in PC( R), but in that case the boundary condition must be relaxed: if P is a point in R where h is continuous, then (6) lim u(x) = h( p), x x p inside R Steady-State Temperatures in the Unit Disk: Circular Harmonics Suppose that R is the unit disk {(x, y) : x 2 + y 2 1}, and that h is a given piecewise continuous function on R. We will construct a solution of the Dirichlet problem (6) for this region R and the given function h by using the Method of Separation of Variables. If the region R and the equations (PDE) and (BC) are in rectangular coordinates, separation of variables will not provide us with a formal solution to this problem, because R is not composed of level curves in the rectangular coordinate system. From this point of view, if we find a problem that is equivalent to (6) but expressed in terms of polar coordinates, separation of variables would have some chance for success. Using identity (3), we see that problem (6) is equivalent to the problem (PDE) w rr + (1/r)w r + (1/r 2 )w θθ = for θ in, < r < 1 (BC) w(1, θ) = f (θ) for θ in where f is the function h expressed in terms of the polar angle θ and w(r, θ) = u(r cos θ, r sin θ). The function f belongs to PC[ π, π] and we require the function w to be periodic in θ with period 2π and twice continuously differentiable in the strip in the rθ-plane < r < 1, < θ <. To ensure that u is twice continuously differentiable in a region containing the origin, we impose the condition that for any θ in, the limits of all derivatives of w up to second order exist as (r, θ) (, θ ) and are independent of θ. We may interpret problem (7) as the model for steady-state temperatures in a thin homogeneous disk with prescribed edge temperatures whose top and bottom faces are perfectly insulated. We will follow the Method of Separation of Variables to construct a formal solution to problem (7). We first look for all twice continuously differentiable solutions of (7)

3 18 (PDE) that have the form R(r) (θ). Substitute R(r) (θ) into (PDE) and separate the variables to obtain the ODEs λ =, r 2 R + r R + λr = (8) where λ is the separation constant. We want the solution R(r) (θ) to be periodic in θ and smooth across θ = π. So impose the conditions ( π) = (π), ( π) = (π). This leads to the Sturm Liouville problem with periodic boundary conditions λ =, ( π) = θ (π), ( π) = (π) We considered a similar problem in Example and found that the problem has a nontrivial solution if and only if for n =, 1, 2,, λ = λ n = n 2. The solutions of this Sturm Liouville problem are (θ) = C, n (θ) = A n cos nθ + B n sin nθ, n = 1, 2,... where C is any nonzero real number and A n and B n are arbitrary real numbers, not both zero. Replace λ by λ n in the other separated ODE in (8) to obtain the ODE r 2 R + r R n 2 R =, n =, 1, 2,... Euler s equation For each n, this is an Euler equation with the general solution appears in Problem 6 in Section 3.7. R(r) = Ar n + Br n when n = 1, 2,... (9a) R(r) = A + B ln r when n = (9b) Since R must be well behaved as r +, take B = and so R n = r n, n =, 1, 2,... Look for a formal solution to problem (7) in the form w(r, θ) = A 2 + r n ( A n cos nθ + B n sin nθ) (1) Compute the constants A n and B n by imposing the boundary condition n=1 f (θ) = A 2 + ( A n cos nθ + B n sin nθ) for π θ π n=1 Since {1, cos x, sin x,...} is a basis for PC[ π, π], A n = 1 π B n = 1 π π π π π f (θ) cos nθ dθ, n =, 1, 2,... f (θ) sin nθ dθ, n = 1, 2,... The terms of the series (1) are circular harmonics. Solution (1) with coefficients (11) is only a formal solution to the Dirichlet problem since we have said nothing as yet about convergence properties. Although we (11)

4 19 C (R) is the class of all functions that have derivatives of all orders on R. will not show it, if the boundary function h is piecewise continuous on R, then the function w defined by formulas (1) and (11) belongs to C (R) and satisfies the PDE in (8). Steady-state temperature functions have the same strong smoothness properties as the time-dependent solutions of the diffusion equation in Section 1.4. If, in addition, h is continuous and piecewise smooth on R, then formulas (1) and (11) define a classical solution of problem (7). If the disk has radius r > rather than radius 1, then (1) and (11) define the solution of the corresponding Dirichlet problem when we replace r in (1) xby r/r. Properties of Harmonic Functions Harmonic functions have a number of distinctive properties. For simplicity, we only state results for planar regions. THEOREM 1 Mean Value Property Suppose that u(x, y) is a solution of Laplace s equation in a planar region R. Suppose also that (x, y ) is a point of R, and D is a disk of radius r centered at (x, y ) and lying entirely in R. Then u(x, y ) = 1 2π 2π u(x + r cos θ, y + r sin θ) dθ (12) That is, the value of a harmonic function at a point is the average of its values around any circle centered at the point. Proof. Figure 1 illustrates the geometrical setup of the theorem. To show that (12) is valid, suppose that the polar coordinates r, θ are relative to the point (x, y ). Then, using solution formula (1) with r replaced by r/r, we get By (11), the coefficient A is 1 π π π and we have verified inequality (13). u(x, y ) = w(, θ) = A 2 u(x + r cos θ, y + r sin θ) dθ Harmonic functions also satisfy a Maximum Principle. THEOREM 2 Maximum Principle for Harmonic Functions If u is a nonconstant harmonic function, then u cannot attain its maximum (or its minimum) value inside a region R of 2.

5 11 y D r R (x, y ) FIGURE 1 Geometry for the Mean Value Property. x Proof. Suppose that u attains its maximum in R at some interior point P of R. Then this maximum value is the average of the values of u around the edge of any disk centered at P and lying in R (Theorem 1). However, u is continuous, so u(p) cannot be both an average and a maximum unless the values of u are constant. A similar argument applies to the minimum. In other words, if R is a region, and u is continuous on R, then a nonconstant harmonic function u only attains extreme values on the boundary of R. A central question is whether a problem has exactly one solution, and whether the solution changes continuously with the data. If it does, the problem is well-posed. The existence of a solution for an arbitrary Dirichlet problem is not easy to show, although for regions of simple shape, we can construct solutions by the Method of Separation of Variables (e.g., the Dirichlet problem solved above). We apply the Maximum Principle to derive the other two aspects of a well-posed Dirichlet problem (i.e., with no more than one solution and continuity in the data). THEOREM 3 A region in 2 is bounded if a rectangle can contain it. Uniqueness and Continuity The Dirichlet problem (6) for a bounded region R in 2 has no more than one continuous solution u in R if the boundary data h are continuous on R. Moreover, the solution (if it exists) varies continuously with respect to the data. Proof. Let s show first that a solution is continuous in the data. Suppose that 2 u = 2 v = in R while u = h and v = h + ε on R, where h and ε are continuous on R. We will show that min ε(p) v u max ε(p) (13) P on R P on R for all values of v(q) u(q), Q in R. If ε(p) is small, then v is close to

6 111 u and we have continuity in the data. Now to show inequality (13), let w = v u, so that 2 w = 2 v 2 u = in R. By the Maximum Principle (and the corresponding Minimum Principle), min w(p) w(q) max w(p) P on R P on R for all w(q), Q in R. We see that w = v u = h + ε h = ε on R, and we are done. Next, we use the inequality (13) to show that problem (6) has no more than one solution. Suppose that u 1 and u 2 are solutions of problem (6). Then U = u 1 u 2 satisfies (PDE) in R but with U = on R. By the same argument as above, U(Q) for all Q in R. So U =, and u 1 = u 2, and the Dirichlet problem (6) has no more than one solution. Looking Back From the results above we see that harmonic functions have strong and distinctive properties. We can interpret these properties in terms of steady-state temperatures within a thin plate R whose top and bottom faces are perfectly insulated and edge temperatures prescribed. PROBLEMS Plate Temperatures. Use the Method of Separation of Variables to solve these IBVPs. 1. Show that the formal solution of the steady-state temperature problem in the rectangular plate R: < x < L, < y < M, modeled by is (PDE) u xx + u yy =, in R (BC) 1 u(, y) = α(y), y M (BC) 2 u(l, y) =, y M (BC) 3 u(x, ) =, x L (BC) 4 u(x, M) =, x L u α (x, y) = where sinh α = 1 2 (eα e α ) and A n sinh(nπl/m) = n=1 A n sin nπy M sinh nπ (L x) M α, sin(nπy/m) sin(nπy/m) 2 = 2 M M α(y) sin(nπy/m) dy 2. Show that the formal solution of the problem in R with general boundary temperatures (PDE) u xx + u yy = in R (BC) 1 u(, y) = α(y), y M (BC) 2 u(l, y) = β(y), y M (BC) 3 u(x, ) = γ(x), x L (BC) 4 u(x, M) = δ(x), x L

7 112 is u = u α + u β + u γ + u δ, where u α solves the same problem but with β(y) = γ(x) = δ(x) = and we define u β, u γ, u δ analogously. 3. Insulated Edge Repeat Problem 1 with the insulation condition u x (, y) = α(y), y M replacing (BC) Perfect Insulation Repeat Problem 1 with (BC) 3 and (BC) 4 replaced by the perfect insulation conditions. Temperatures in a Ring. u y (x, ) = = u y (x, M), x L 5. A thin ring R = {(r, θ) : < ρ < r < R, π θ π} has insulated faces. Find the steady temperatures in R if we maintain the boundaries at the temperatures f (θ) and g(θ) at r = ρ and r = R, respectively. [Hint: proceed as in the text for a disk, but keep the second terms in formulas (9a), (9b). Then f (θ) = A + B ln ρ + n ρ n [A n cos nθ + B n sin nθ] + n ρ n [C n cos nθ + D n sin nθ] and a similar expression with f replaced by g, and ρ by R.] Temperatures in a Disk; Maximum and Minimum Temperatures. 6. Solve the Dirichlet problem for steady-state temperatures in the unit disk with the edge temperatures f (θ) as below. Then find the maximum and minimum temperatures in the disk: (a) f (θ) = 3 sin θ (b) f (θ) = θ + π for π θ < ; f (θ) = θ + π for θ π. (c) f (θ) = for π < θ < ; f (θ) = 1 for < θ π.

21 Laplace s Equation and Harmonic Functions

21 Laplace s Equation and Harmonic Functions 2 Laplace s Equation and Harmonic Functions 2. Introductory Remarks on the Laplacian operator Given a domain Ω R d, then 2 u = div(grad u) = in Ω () is Laplace s equation defined in Ω. If d = 2, in cartesian