LAPLACE EQUATION If a diffusion or wave problem is stationary (time independent), the pde reduces to the Laplace equation u = u =, an archetype of second order elliptic pde. = 2 is call the Laplacian or del-square operator. In two dimensions, the expression of in rectangular and polar coordinates are = 2 x + 2 2 y 2 and = 2 r + 1 2 r r + 1 2 r 2 θ 2. In three dimensions, the expressions of and in spherical coordinates are and = r ˆr + 1 r = 2 r 2 + 2 r θ ˆθ + 1 r sin θ φ ˆφ r + 1 r 2 sinθ θ sin θ θ + 1 r 2 sin 2 θ The inhomogeneous version of Laplace s equation u = f, is called Poisson s equation. 2 φ 2. 1
A BVP involving Laplace or Poisson s equation is to solve the pde in a domain D with a condition on the boundary of D (to be represented by D). u = f u = h in D or u n = h u or n + au = h on D A solution of the Laplace equation is called a harmonic function. Harmonic functions Maximum principle Theorem: Let D be a path connected bounded open set (in two or three dimensions). Let u be a harmonic function in D that is continuous on D = D D. Then the maximum and the minimum values of u are on D and nowhere inside (unless u = constant). Proof: Here we prove the weak form of the theorem, that the maximum and minimum values occur on the boundary, i.e. max u = max u, for the two dimensional case. D D 2
Let v(x) = u(x) + ɛ x 2 where ɛ >. Then v = u + ɛ (x 2 + y 2 ) = 4ɛ > in D. However, v xx +v yy at an interior maximum. Therefore, v has no interior maximum, and max D v = max D v. ( ) Now, suppose that the maximum of u is at an interior point x and u(x ) max D u = δ >. Then v(x ) max v = u(x ) + ɛ x 2 max D (u) ɛl2 D (u + ɛ x 2 ) u(x ) + ɛ x 2 max D where L is the greatest distance from D to the origin. This expression can be written as v(x ) max D δ ɛ(l 2 x 2 ) > δ 2 > v u(x ) max D (u) ɛ(l2 x 2 ) = which can be realized for small enough ɛ. Thus, there is a contradiction with Eq. ( ). For this reason, δ can only be, and the maximum of u is on D. The minimum of u is located at where the maximum of w = u lies. w is also a harmonic function. Therefore, it has to be on D. 3
Uniqueness of the Dirichlet problem Suppose u 1, u 2 are two solutions to the problem u = f u = h Then u 1 = u 2. in D on D. Proof: Let w = u 1 u 2. w is harmonic and w = on the boundary. Since both the maximum and minimum values of w are on the boundary, w = everywhere on D. Stability of the Dirichlet problem Suppose u 1, u 2 are two solutions to the two problems: u = f in D u 1 = h 1 and u 2 = h 2 on D. Then max u 1 u 2 max h 1 h 2. Proof: max u 1 u 2 = max(u 1 u 2 ) or max(u 2 u 1 ) max h 1 h 2. 4
Harmonic function in special 2D domains The separation of variables method can often be applied to find harmonic functions in special geometries. Rectangular domains Example: Consider the problem: u = on the rectangle D = (, a) (, b) u(, y) = u x (a, y) = u y (x, ) + u(x, ) = u(x, b) = g(x). Steps: With the substitution u(x, y) = X(x) Y (y), the problem is split into two: X = λx X() = X x (a) = and Y = λy Y () + Y () =. X satisfies homogeneous boundary conditions and can be treated by an eigenvalue problem. The x-dependent boundary condition at y = b has to be handled by a series expansion in terms of the X eigenfunctions. The solutions of the mixed type boundary conditions for the X equation are X n = C n sin β n x where β n = (n + 1 2 )π a with n =, 1, 2,... and C n is a constant. 5
The equation for Y is restricted to take the form Y = λ n Y with λ n = β 2 n >. The solutions are Y n (y) = Ae β ny + Be β ny. The boundary condition at y = requires that Aβ n +Bβ n +A+B = B A = 1 β n 1 + β n γ n. Therefore, the sum u(x, y) = A n sin β n x (e βny + γ n e βny ) n= is a harmonic function in D that satisfies all three homogeneous BCs. The remaining BC can be satisfied if g(x) = A n sin β n x (e βnb + γ n e βnb ) n= for < x < a. One can easily verify that a sin β m x sin β n xdx = a 2 δ mn. Thus, the coefficients can be found as A n = 1 (e β nb + γ n e β nb ) 2 a sinβ mx sinβ nx = 1 2 [cos(βm βn)x cos(βm + βn)x] a g(x) sinβ n xdx. 6
Circular domains Example: Consider the following problem in a disk with radius a: Steps: u = u rr + 1 r u r + 1 r 2u θθ =, (use polar coordinates) u = h(θ) on the circle r = a. Let u = R(r)Θ(θ). The problem is split into two: Θ = λθ < θ < 2π with periodic boundary conditions, and r 2 R + rr λr = < r < a with R() < (regularity) and R(a) = h(θ). For Θ, if λ >, the eigenvalues are λ = n 2 (n = 1, 2,...), and the eigenfunctions are Θ(θ) = A cos nθ + B sin nθ. If λ =, Θ(θ) = constant. The equation for R is of Euler type and has solution of the form R(r) = r α. Through substitutions, it yields α(α 1)r α + αr α n 2 r α =, and α = ±n. Thus, for n = 1, 2,... R(r) = Cr n + D r n. 7
For λ =, the solution is R = C + D ln r. Solutions (r n and ln r) that are infinite at r = are to be rejected. Summing up the rest, one has u = 1 2 A + r n (A n cos nθ + B n sin nθ). To satisfy the BC at r = a, it is necessary that h(θ) = 1 2 A + a n (A n cos nθ + B n sinnθ). This is the full Fourier series expansion for h. The coefficients are A n = 1 πa n B n = 1 πa n 2π 2π h(φ) cos nφdφ, h(φ) sinnφ dφ. Poisson s formula Now, substitute the expressions of A, A n, and B n into the series for u above to obtain u(r, θ) = + 2π r n πa n h(φ) dφ 2π 2π h(φ)(cos nφ cos nθ + sin nφ sin nθ)dφ 8
= 2π h(φ) [ 1 + 2 ] ( r ) n cos n(θ φ) a dφ 2π. The terms in the rectangular bracket can be summed as ( r ) n ( r ) n 1 + e in(θ φ) + e in(θ φ) a a = 1+ rei(θ φ) re i(θ φ) a re i(θ φ)+ a re = a 2 r 2 i(θ φ) a 2 2ar cos(θ φ) + r 2. Thus, u(r, θ) = 1 2π 2π a 2 r 2 a 2 2ar cos(θ φ) + r 2 h(φ)dφ. This formula is known as Poisson s formula. It expresses any harmonic function inside a circle in terms of its boundary values. Poisson s formula can be written in a more geometrical way. Let x be the position vector of the point having polar coordinates (r, θ), x be a point on the circle, and ds = adφ be the differential of arc length on the circle. The denominator of the integrand can be recognized as (x x ) (x x ) = x 2 2x x + x 2 = r 2 2ar cos(θ φ) + a 2. The formula can be written as u(x) = 1 2πa x =a a 2 x 2 x x 2 u(x ) ds. 9
Mean value property Let u be a harmonic function in a disk D, continuous in its closure D. Then the value of u at the center of D equals the average of u on its circumference. Proof: In the above equation, choose x to be at the center of the circle, then u() = 1 u(x ) ds. 2πa x =a Strong form of maximum principle in 2D If the maximum or minimum value occurs at an interior point, the harmonic function is a constant. Proof: Let u(x) be harmonic in the open set D. Suppose that the maximum value M of u on D occurs at an interior point x M D. Let D D be any open disk centered at x M, then u(x) = M x D. 1
Steps: Let C be the circle centered entered at x M and passing through x. Because of the mean value property, u(x M ) = M is equal to the mean value of u on C. Since M u, 1 M u dφ = 1 (M u)dφ = M 1 udφ 2π 2π 2π = u = M on C, including the point x. Similar argument can be applied to any other point where u has the maximum value. One can show the following: Let D, D be two open disks in D and D D φ. If u(x) = M in D, then u(x) = M x D. Steps: Let d = the distance between the two intersecting points of the circular boundaries of D and D. Let < ɛ < d. Use ɛ as the radius for a number of 2 smaller disks to connect the centers of D and D so that consecutive disk centers are separated by a distance less than ɛ. Then the value of u at the center of D can be shown to be equal to M. As D is assumed to be path connected, any two points can be connected by a finite string of intersecting open disks. Therefore, u = M x D. 11
Differentiability Let u be a harmonic function in any open set D of the plane. Then u(x, y) possesses all partial derivatives of all orders in D. This is a straightforward consequence of Poisson s formula (vector form). Other types of circular domains Example: Consider, for example, an infinite region between the two angles θ =, θ = β, and exterior of the circle r = a. Solve u = with boundary conditions Steps: u(r, ) = u(r, β) =, u (a, θ) = h(θ). r Separation of variables splits the problem into two parts: (i) Θ + λθ = Θ() = Θ(β) = This is a Dirichlet eigenvalue problem. The solutions are ( ) 2 nπ λ = Θ(θ) = sin nπθ n = 1, 2,... β β 12
(ii) r 2 R +rr λr = R (a) = h(θ), R( ) finite. This is the same ODE problem as the one discussed earlier. The solutions are of the form R(r) = r α with α = ± λ = ±nπ/β. The positive exponents are rejected due to the condition at infinity. The following series makes a solution u(r, θ) = A n r nπ/β sin nπθ β. The boundary condition at r = a requires that nπ h(θ) = A n β a 1 nπ/β sin nπθ β. So, A n = a 1+nπ/β 2 nπ β h(θ) sin nπθ β dθ. Other domains, for example annuli and exterior of a circle, can be handled similarly. Problem Set 9 13