Method of Green s Functions

Transcription

1 Method of Green s Functions Linear Partial ifferential Equations Matthew J. Hancock Fall 24 Ref: Haberman, h 9, 11 We introduce another powerful method of solving PEs. First, we need to consider some preliminary definitions and ideas. 1 Preliminary ideas and motivation 1.1 The delta function efinition [elta Function] The δ-function is defined by the following three properties, { δ x =, x,, x =, δ x dx =1 f x δ x a dx = f a where f is continuous at x = a. The last is called the sifting property ofthe δ-function. To make proofs with the δ-function more rigorous, we consider a δ-sequence, that is, a sequence of functions that converge to the δ-function, at least in a pointwise sense. onsider the sequence δ n x = n π e nx2 Note that δ n x dx = 2n e nx2 dx = 2 e z2 dz =erf =1 π π 1

2 efinition [2 elta Function] The 2 δ-function is defined by the following three properties, {, x, y, δ x, y =, x, y =, δ x, y da =1, f x, y δ x a, y b da = f a, b. 1.2 Green s identities Recall that we derived the identity G F + F G da = GF ˆndS 1 for any scalar function G and vector valued function F. Setting F = u gives what is called Green s First Identity, G 2 u + u G da = G u ˆn ds 2 Interchanging G and u and subtracting gives Green s Second Identity, u 2 G G 2 u da = u G G u ˆndS. 3 2 Solution of Laplace and Poisson equation onsider the BVP 2 u = F in, 4 u = f on. Let x, y be a fixed arbitrary point in a 2 domain and let ξ,η beavariable point used for integration. Let r be the distance from x, y toξ,η, r = ξ x 2 +η y 2. onsidering the Green s identities above motivates us to write 2 G = δ ξ x, η y =δ r in, 5 G = on. 2

3 The notation δ r isshortforδ ξ x, η y. Substituting 4 and 5 into Green s second identity 3 gives u x, y GF da = f G ˆndS Rearranging gives u x, y = GF da + f G ˆndS 6 Therefore, if we can find a G that satisfies 5, we can use 6 to find the solution u x, y of the BVP 4. The advantage is that finding the Green s function G depends only on the area and curve, not on F and f. Note: this method can be generalized to 3 domains - see Haberman. 2.1 Finding the Green s function Ref: Haberman To find the Green s function for a 2 domain see Haberman for 3 domains, we first find the simplest function that satisfies 2 v = δ r. Suppose that v x, y is axis-symmetric, that is, v = v r. Then 2 v = v rr + 1 r v r = δ r For r>, Integrating gives v rr + 1 r v r = v = A ln r + B For simplicity, we set B =. To find A, we integrate over a disc of radius ε centered at x, y, ε, 1= δ r da = 2 vda ε ε From the ivergence Theorem, we have 2 vda = v nds ε ε where ε is the boundary of ε, i.e. a circle of circumference 2πε. ombining the previous two equations gives v 1= v nds = A ε r ds = ds =2πA r=ε ε ε ε 3

4 Hence v r = 1 2π ln r This is called the fundamental solution for the Green s function of the Laplacian on 2 domains. For 3 domains, the fundamental solution for the Green s function of the Laplacian is 1/4πr, where r = x ξ 2 +y η 2 +z ζ 2 see Haberman. The Green s function for the Laplacian on 2 domains is defined in terms of the corresponding fundamental solution, G x, y; ξ,η = 1 ln r + h, 2π h is regular, 2 h =, ξ,η, G = ξ,η. The term regular means that h is twice continuously differentiable in ξ,η on. Finding the Green s function G is reduced to finding a 2 function h on that satisfies 2 h = ξ,η, h = 1 ln r ξ,η. 2π The definition of G in terms of h gives the BVP 5 for G. Thus,for2regions, finding the Green s function for the Laplacian reduces to finding h. 2.2 Examples i Full plane = R 2. There are no boundaries so h = will do, and G = 1 2π ln r = 1 4π ln [ ξ x 2 +η y 2] ii Half plane = {x, y :y>}. Ref: Haberman We find G by introducing what is called an image point x, y corresponding to x, y. Let r be the distance from ξ,η tox, y andr the distance from ξ,η to the image point x, y, r = ξ x 2 +η y 2, r = ξ x 2 +η + y 2 We add h = 1 2π ln r = 1 2π ln ξ x 2 +η + y 2 4

5 G2 1/2,2 1/2 ;ξ,η η ξ Figure 1: Plot of the Green s function G x, y; ξ,η for the Laplacian operator in the upper half plane, for x, y = 2, 2. to G to make G = on the boundary. Since the image point x, y isnotin, then h is regular for all points ξ,η, and satisfies Laplace s equation, 2 h = 2 h ξ h η 2 = for ξ,η. Writing things out fully, we have G = 1 2π ln r + h = 1 2π ln r 1 2π ln r = 1 2π ln r r = 1 4π ln ξ x2 +η y 2 ξ x 2 +η + y 2 7 G x, y; ξ,η is plotted in the upper half plane in Figure 1 for x, y = 2, 2. Note that G as ξ,η x, y. Also, notice that G<everywhereand G = on the boundary η =. These are, in fact, general properties of the Green s function. The Green s function G x, y; ξ,η acts like a weighting function for x, y and neighboring points in the plane. The solution u at x, y involves integrals of the weighting G x, y; ξ,η times the boundary condition f ξ,η and forcing function F ξ,η. On the boundary, η =,sothatg =and G n = G η = 1 y η= π ξ x 2 + y 2 The solution of the BVP 6 with F = on the upper half plane can now be 5

6 written as, from 6, u x, y = f G ˆndS = y f ξ π ξ x 2 + y dξ, 2 which is the same as we found from the Fourier Transform, on page 13 of fourtran.pdf. iii Upper right quarter plane = {x, y : x>, y >}. Weusetheimage points x, y, x, y and x, y, G = 1 2π ln ξ x 2 +η y 2 1 2π ln ξ x 2 +η + y π ln ξ + x 2 +η y π ln ξ + x 2 +η + y 2 For ξ,η = the boundary, either ξ =orη =, and in either case, G =. Thus G = on the boundary of. Also, the second, third and fourth terms on the r.h.s. are regular for ξ,η, and hence the Laplacian 2 = 2 / ξ / η 2 of each of these terms is zero. The Laplacian of the first term is δ r. Hence 2 G = δ r. Thus 8 is the Green s function in the upper half plane. For ξ,η = the boundary, f G ˆndS = f,η G ξ dη + f ξ, G ξ= η dξ η= = f,η G ξ dη f ξ, G ξ= η dξ η= Note that G 4yxη ξ = ξ= π x 2 +y + η 2 x 2 +y η 2 G 4yxξ η = η= π x ξ 2 + y 2 x + ξ 2 + y 2 The solution of the BVP 6 with F = on the upper right quarter plane and boundary condition u = f can now be written as, from 6, u x, y = f G ˆndS = 4yx π + 4yx π ηf,η x2 +y + η 2 x 2 +y η 2dη ξf ξ, x ξ 2 + y 2 x + ξ 2 + y 2dξ iv Unit disc = {x, y :x 2 + y 2 1} Haberman By some simple geometry, for each point x, y, choosing the image point x,y alongthesame 6

7 ray as x, y and a distance 1/ x 2 + y 2 away from the origin guarantees that r/r is constant along the circumference of the circle, where r = ξ x 2 +η y 2, r = ξ x 2 +η y 2. To see this, we draw a diagram similar to Figure on p 431 of Haberman. Using the law of cosines, we obtain r 2 = ρ 2 + ρ 2 2ρ ρ cos r 2 = ρ ρ 2 ρ θ 2 ρ cos θ where ρ = x 2 + y 2, ρ = ξ 2 + η 2 and θ, θ are the angles the rays x, y andξ,η make with the horizontal. Note that for ξ,η on the circumference ξ 2 +η 2 = ρ 2 =1, we have r 2 1+ρ2 2ρ cos r = = ρ θ 2, ρ =1. 1 cos ρ 2 ρ θ Thus the Green s function for the Laplacian on the 2 disc is G ξ,η; x, y = 1 2π ln r r ρ = 1 ρ2 + ρ2 2ρ ρ cos 4π ln ρ 2 ρ ρ ρ cos Note that G ˆn = G ρ = 1 1 ρ 2 ρ=1 2π 1+ρ 2 2ρ cos Thus, the solution to the BVP 5 on the unit circle is in polar coordinates, u ρ, θ = 1 2π 2π + 1 2π 4π 1 ρ 2 f θ d θ 1+ρ 2 2ρ cos 1 ln ρ2 + ρ 2 2ρ ρ cos F ρ 2 ρ ρ ρ cos The solution to Laplace s equation is found be setting F =, u ρ, θ = 1 2π 2π 1 ρ 2 1+ρ 2 2ρ cos f θ d θ This is called the Poisson integral formula for the unit disk. ρ, θ ρd ρdθ 7

8 2.3 onformal mapping and the Green s function onformal mapping allows us to extend the number of 2 regions for which Green s functions of the Laplacian 2 u can be found. We use complex notation, and let α = x + iy be a fixed point in and let z = ξ + iη be a variable point in what we re integrating over. If is simply connected a definition from complex analysis, then by the Riemann Mapping Theorem, there is a conformal map w z analytic and one-to-one from into the unit disk, which maps α to the origin, w α = and the boundary of to the unit circle, w z =1forz and w z < 1 for z /. The Greens function G is then given by G = 1 ln w z 2π To see this, we need a few results from complex analysis. First, note that for z, w z =sothatg =. Also,sincew z is1-1, w z > forz α. Thus, we can write w z =z α n H z whereh z is analytic and nonzero in. Since w z is1-1, w z > on. Thusn = 1. Hence and where r = z α = w z =z α H z G = 1 2π ln r + h ξ x 2 +η y 2 h = 1 ln H z 2π Since H z is analytic and nonzero in, then1/2πlnh z isanalyticin and hence its real part is harmonic, i.e. h = R 1/2πlnH z satisfies 2 h =in. Thus by our definition above, G is the Green s function of the Laplacian on. Example 1. The half plane = {x, y :y>}. The analytic function w z = z α z α maps the upper half plane onto the unit disc, where asterisks denote the complex conjugate. Note that w α = and along the boundary of, z = x, whichis equidistant from α and α,sothat wz = 1. Points in the upper half plane y > are closer to α = x + iy, also in the upper half plane, than to α = x iy, inthe lower half plane. Thus for z /, w z = z α / z α < 1. The Green s function is G = 1 1 z α ln w z = ln 2π 2π z α = 1 2π ln r r which is the same as we derived before, Eq. 7. 8

9 3 Solution to other equations by Green s function The method of Green s functions can be used to solve other equations, in 2 and 3. For instance, for a 2 region, the problem 2 u + u = F in, u = f on, has the fundamental solution 1 4 Y r where Y r is the Bessel function of order zero of the second kind. The problem 2 u u = F in, u = f on, has fundamental solution 1 2π K r where K r is the modified Bessel function of order zero of the second kind. The Green s function method can also be used to solve time-dependent problems, such as the Wave Equation and the Heat Equation see hapter 11, Haberman. 9