Green s Functions and Distributions

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1 CHAPTER 9 Green s Functions and Distributions 9.1. Boundary Value Problems We would like to study, and solve if possible, boundary value problems such as the following: (1.1) u = f in U u = g on U, where U is open and bounded, with smooth boundary U. Now, we know how to solve the PDE on all space - we write the solution as a convolution with the fundamental solution: (1.2) u(x) = Φ(x y)f(y) dy. The idea of the Green s function is to find a formula like this that also takes into account the boundary conditions. Of course, the trouble is that since Φ(x) is set up to localize at x =, it cannot be expected to do the job in the interior of U at the same time as on the boundary. However, we can modify Φ so that it satisfies homogeneous boundary conditions. This is how the Green s function is set up, and in fact such a function allows us to solve general problems like (1.1), even for nonzero boundary data g. On the way to formulating Green s functions, we outline the theory of distributions (the Dirac delta function is an example of a distribution that is not a function in the usual sense), which is useful in the study of weak solutions of PDE. Let s start with a seemingly simple example, namely problem (1.1) in one dimension, with zero boundary conditions to start with: (1.3) u (x) = f(x) < x < 1 u() =, u(1) =. We solve this problem by integrating twice: u (x) = x f(y) dy + C; u(x) = where C, D are constants of integration. 115 x z f(y) dy dz + Cx + D,

2 GREEN S FUNCTIONS AND DISTRIBUTIONS Now we apply the boundary conditions. The boundary condition u() = lead immediately to D =. But the boundary condition at u(1) = gives an equation for C that is hard to understand. It becomes much more transparent when we change the order of integration. To do so, we note that the double integral occurs over a triangle (see Fig. 15.1). z x y z z y x y Figure 9.1. Area of Integration. Thus, x z f(y) dy dz = x x y f(y) dz dy = x (x y)f(y) dy Consequently, the solution we have so far is expressed as (1.4) u(x) = x The boundary condition at x = 1 becomes so that u(1) = 1 C = (x y)f(y) dy Cx. (1 y)f(y) dy C =, 1 (1 y)f(y) dy.

3 9.1. BOUNDARY VALUE PROBLEMS 117 Substituting back into (1.4) and rearranging terms, we arrive at u(x) = Thus, x (y x)f(y) dy+ 1 (1.5) u(x) = x(1 y)f(y) dy) = 1 G(x, y)f(y) dy, x y(1 x)f(y) dy+ where { y(1 x) if y x (1.6) G(x, y) = x(1 y) if x y. The graph of G(x, y) for fixed x (, 1) is shown in Figure The G(x,y) 1 x x(1 y)f(y) dy. x(1 - x) x 1 y Figure 9.2. The integral kernel G(x, y). function G(x, y) is called an integral kernel. It defines an integral operator G : C[, 1] X on the space of continuous functions on [, 1] that is the inverse of the differential operator L = d2 : X C[, 1], where X = {u dx 2 C[, 1] : u() = = u(1)} : (Gf)(x) = 1 G(x, y)f(y) dy. Note that the domain of L is the subspace of X consisting of differentiable functions. In fact, G is the Green s function for this boundary value problem - note that the formula (1.5) is similar to that of (1.2). Here are some properties of G : [, 1] [, 1] R. 1. G is non-negative: G(x, y) ;

4 GREEN S FUNCTIONS AND DISTRIBUTIONS 2. G is symmetric: G(x, y) = G(y, x). 3. G is continuous. 4. G is differentiable, except on the diagonal x = y. On the diagonal, G y has a jump discontinuity: [ ] G = 1, y x=y where [..] means the jump, or difference between the right limit and left limit. Now 2 G = except on the diagonal, where G y 2 y infinite negative slope. We will write 2 G (x, y) = δ(x y), y2 may be thought to have where δ(x) is the Dirac delta function. δ(x) is a measure that assigns mass one at x =, and zero mass elsewhere. We shall treat δ as a distribution, or generalized function, which leads us to the theory of distributions, a useful framework for considering PDEs, and Green s functions such as G(x, y) Distributions We start by considering smooth functions on R. Let D = Cc (R) denote the space of C functions with compact support; this is the space of test functions. To define the topology of D, we define what it means for a sequence of functions {φ n } to converge in the space. Let φ (j) denote the j-th derivative of φ. We say φ n φ D as n if: (a) There is a compact subset K of R such that supp φ n K for all n, and supp φ K. (b) φ (j) n φ (j) as n, uniformly on K, for each j : sup φ (j) n (x) φ (j) (x), as n. x K Similarly, we may define the space D( ) of test functions on.

5 Important example of a test function DISTRIBUTIONS 119 Let Ce 1 1 x 2 if x < 1 (2.7) η(x) = otherwise, where C = 1/ 1 x <1 e 1 x 2 dx is chosen so that η(x) dx = 1. To see that η is a test function, we observe that it has continuous derivatives of all orders, even where the definition is split, at x = 1. For example, with n = 1, the derivatives approach zero at x = 1, since every derivative is the product of a rational function and the exponential e 1 1 x 2 ; the exponential dominates the rational function. It is useful to rescale η using a parameter ɛ > : (2.8) η ɛ (x) = 1 ( x ) ɛ n η. ɛ Then: supp η ɛ = {x : x ɛ} and η ɛ (x) dx = 1. The space of distributions D is defined to be the space of continuous linear functionals on D. That is, f D means f : D R, and f has the properties (i) f is linear, i.e., f(aφ 1 +bφ 2 ) = af(φ 1 )+bf(φ 2 ) for each a, b R, φ 1, φ 2 D. (ii) f is continuous, i.e., φ n φ in D implies f(φ n ) f(φ) (as a sequence of numbers). Following the usual custom, we denote f(φ) using the more suggestive notation (f, φ). (Sometimes, f, φ is used, but not in these notes.) Examples of distributions 1. (f 1, φ) = R φ(x) dx (defined because φ is continuous and has compact support). 2. (f 2, φ) = φ(x) dx. 3. (f 3, φ) = φ(). 4. (f 4, φ) = φ ().

6 12 9. GREEN S FUNCTIONS AND DISTRIBUTIONS All of these are distributions, and the last three are related, as we shall see, by differentiation in the sense of distributions. The first two examples are associated with locally integrable functions: Let L loc 1 (Rn ) denote the space of locally integrable (in the sense of Lebesgue) functions on : g L loc 1 ( ) if g(x) dx < for any compact K. Examples. K 1. g(x) = 1 is not integrable on R, but it is in L loc 1 (R). 2. g(x) = x is unbounded on R, but is in L loc 1 (R). CLAIM If g L loc 1 (Rn ), then g defines a distribution f D by (f, φ) = g(x)φ(x) dx. Proof. Linearity of f is clear; continuity follows from the definition. However, not every distribution defines an L loc 1 function. We say a distribution f is regular if it has an L loc 1 representative f : (f, φ) = f(x)φ(x) dx for all φ D( ). Otherwise, the distribution f is called singular. Example f 3 is the Dirac δ-function: (δ, φ) = φ(); we show below that δ is a singular distribution. It will then follow that example f 4 is also singular. However, examples f 1 and f 2 define L loc 1 functions: For f 1 : take f(x) = 1. For f 2 : take f to be the Heaviside function H(x) = { if x < 1 if x Lemma 9.1. δ is a singular distribution. Proof. Suppose δ is regular, so that there is g L loc 1 such that (2.9) (δ, φ) = g(x)φ(x) dx for all φ D( ).

7 9.2. DISTRIBUTIONS 121 Now define a one-parameter family of test functions, modeled on η(x) : φ ɛ e 1 1 x ɛ 2 if x < ɛ (x) = if x ɛ. Now we calculate the effect of δ on φ ɛ in two ways. first, from the definition of δ : (δ, φ ɛ ) = φ ɛ () = 1 e Second, using the assumption (2.9): (δ, φ ɛ ) = g(x)φ ɛ (x) dx. But then 1 e = g(x)φ ɛ (x) dx g(x) 1 x ɛ e dx. But the latter integral approaches zero as ɛ zero, providing a contradiction for small enough ɛ. Note that the function φ ɛ used in the proof of the lemma is a mulitple (ɛ n /C to be precise) of η ɛ, but this multiple provides the very different behaviors of the two functions as ɛ. The function η ɛ is called a mollifier because it can be used to smooth rough functions. It does this through the convolution product: Recall that the convolution of two square integrable functions (i.e., in L 2 ( )) f and g is defined as g f(x) = g(x y)f(y) dy. Example: Compute g f for the functions { 1 if < x < 1 f(x) = g(x) = otherwise; and graph the convolution product g f. { x if x < 1 otherwise, Remark Convolution is symmetric: g f = f g, as can be seen by changing variables in the integral. Here is how the mollifier does its smoothing: Let f C( ), and define, for ɛ >, f ɛ (x) = η ɛ f(x). Then f ɛ C, and f ɛ (x) f(x) for all x, as ɛ.

8 GREEN S FUNCTIONS AND DISTRIBUTIONS Note that f ɛ is defined and is C for f L loc 1. In that case, it takes a little measure theory (the Lebesgue dominated convergence theorem) to prove that f ɛ f almost everywhere (see [Evans, Appendix]). Convergence of distributions Let {f k } k=1 D ( ) be a sequence of distributions, and let f D ( ). We say if Examples. 1. Let n = 1. Then, for φ D(R), f k f in D in the sense of distributions (f k, φ) (f, φ) as k, for all φ D( ) (f k, φ) = f k (x) = = R 1 2k 1 2k { k if x 1 2k otherwise f k (x)φ(x) dx = φ(x) dx φ() as k = (δ, φ). 1 2k 1 2k Thus, f k δ in the sense of distribitions as k. kφ(x) dx 2. (Exercise): Prove that η ɛ δ in the sense of distributions, as ɛ Distributional derivatives. Let f C 1 ( ). Then f defines a distribution f D : (f, φ) = f(x)φ(x) dx. But f x i C( ) also defines a distribution: ( ) f, φ = x i integrating by parts over the support of φ. f x i (x)φ(x) dx = f(x) φ (x) dx x i ( = f, φ ), x i

9 9.2. DISTRIBUTIONS 123 This calculation suggests that for any distribution f we define its derivative, or distributional derivative f x i to be a distribution given by ( ) ( f, φ = f, φ ), for all φ D. x i x i Then every distribution is differentiable in the sense of distributions, hence has derivatives of all orders. It is not hard to show that f x i is indeed a distribution, by checking directly that it is continuous and linear. Examples with n = Consider the Heaviside function H(x). As we saw earlier, H is in L loc, and it acts on test functions by (H, φ) = φ(x) dx. Thus, for any test function φ, Therefore, 1 ( H (x), φ ) = (H, φ ) = φ (x) dx = φ() = (δ, φ), H = δ. 2. We can also differentiate the δ function directly from the definition of derivative: ( δ, φ ) = (δ, φ ) = φ () 3. Let f(x) = x. Then 1 if x < f (x) = 1 if x > is an L loc 1 function, hence defines a distribution. In fact, f = 2H 1, so f = 2δ. Here, we use the fact that differentiation is a linear operation on distributions, just as it is on differentiable functions. Here are two further properties, or more precisely, definitions: 1. Translation by y. If f D, we define f( + y) = f y D by (f y, φ) = (f, φ y ),

10 GREEN S FUNCTIONS AND DISTRIBUTIONS where φ y D is the test function defined by φ y (x) = φ(x y). To see that this makes sense, we simply check that it is consistent for f L loc 1 : (f y, φ) = f(x + y)φ(x) dx = f(z)φ(z y) dz = (f, φ y ). An example where this is useful is the δ function. We often write δ(x y) to mean the distribution δ y, and it is sometimes helpful to leave x in the arguments of both the distribution and the test function: (δ(x y), φ(x)) = φ(y). 2. Multiplication by a C function. Let c C, f D. Then we define cf D by (cf, φ) = (f, cφ), noting that cφ is a test function if φ is. Again, this definition is motivated by the case of a regular distribution f, in which case the formula makes sense as integrals. As an example of this definition, consider c(x) = x, f = δ. Then (xδ(x), φ(x)) = (δ(x), xφ(x)) =, where again we have left the argument x in the calculation for clarity. Since δ = H, we see that y(x) = H(x) is a solution of the differential equation x dy dx =. In fact, the general distributional solution of this equation (which is singular at x = ) is y(x) = ah(x) + b, for arbitrary constants a, b. Thus, we have a two parameter family of solutions of a first order equation Example. Here is a quick application to conservation laws: Consider the scalar conservation law (2.1) u t + f(u) x =, in which f : R R is a given C 1 function. We can interpret equation (2.1) in the sense of distributions, and in particular say what it means for a discontinuous function to be a solution. The equation simply states that if u D (R 2 ), and f(u) can be interpreted as a distribution, then the combination of distributional derivatives on the left hand side of the equation should be zero on every test function. But this implies for a test function φ(x, t) : (u, φ t ) + (f(u), φ x ) =.

11 9.2. DISTRIBUTIONS 125 In this way, we can define distribution solutions of differential equations, even nonlinear equations. To see that this interpretation has some substance, consider a jump discontinuity u if x < st (2.11) u(x, t) = if x > st, u + where s, u ± are constants. Now f(u)(x, t) = f(u ) f(u + ) if x < st if x > st, But then u(x, t) = (u + u )H(x st)+u, f(u)(x, t) = (f(u + ) f(u ))H(x st)+f(u ). Thus, u t = (u + u )H (x st)( s); and H = δ. Equation (2.1) becomes x f(u) = (f(u +) f(u ))H (x st), ( s(u + u ) + (f(u + ) f(u )))δ(x st) =, in the sense of distributions. But then the constant s(u + u ) + f(u + ) f(u ) must be zero: s(u + u ) + f(u + ) f(u ) =. Thus, we have derived the Rankine-Hugoniot condition for shock wave solutions of (2.1). We are almost ready to describe Green s functions, but we first need to review integration by parts in. Consider a bounded open set U, with piecewise smooth boundary U. Consider functions f, g in C 1 (U) C(U), and let ν = (ν 1,..., ν n ) be the unit outward normal on U. Differentiating the product f(x)g(x) : U fg = f g + g f. x i x i x i Now integrate over U and apply the divergence theorem to F(x) = (,...,, fg,,..., ) : div F = fg. After rearranging, we get the formula for integration by x i parts: (2.12) f g dx = fg ν i ds g f dx. x i U U x i

12 GREEN S FUNCTIONS AND DISTRIBUTIONS 9.3. Green s Functions In this section, we first give a framework for Green s functions for a rather general differential operator, and then show how this applies when the operator is the Laplacian General Framework. Consider a linear partial differential operator L (L = for example), that acts on functions u : R, and consider solving a boundary value problem (3.13) Lu = f in U Bu = g on U. Here, f is a given function on U, g and given function on U, and U may be unbounded. Bu represents a linear combination of u and derivatives of u of lower order than the order of the partial differential operator L. Associated with L we have the fundamental solution Φ(x, y) : (3.14) LΦ(x, y) = δ(x y), x, y. In order to see how this works, consider Φ to be locally integrable in y for each x, and let f D( ). Then (3.15) v(x) = Φ(x, y)f(y) dy = (Φ(x, y), f(y)) satisfies Lv = f in : Lv(x) = L(Φ(x, y), f(y)) = (LΦ(x, y), f(y)) = (δ(x y), f(y)) = f(x). That is, the fundamental solution is the key to solving the inhomogeneous PDE. However, the formula (3.15) does not generally satisfy the boundary condition Bu = g. To satisfy the boundary condition, we add a solution w of the homogeneous equation Lw = so that u(x) = v(x) + w(x) satisfies the boundary condition Bu = g. But then w must satisfy the boundary condition Bw = g Bv. Thus, provided we have the fundamental solution Φ(x, y), the boundary value problem (3.13) is reduced to solving the problem Lw = in U; Bw = g Bv on U. The boundary condition for w is a bit clumsy, because it relies on applying the boundary operator to the integral (3.15) and restricting to the boundary. The way to express this more smoothly is to introduce the Green s function for the differential operator L with boundary operator B. For clarity, we consider L and B to have independent variable x U, and we let y U be independent.

13 9.3. GREEN S FUNCTIONS 127 The Green s function G = G(x, y) is defined as the solution of the problem (3.16) for each y U. LG(x, y) = δ y (x), x U BG(x, y) =, x U, We construct G using the fundamental solution, by defining a function φ y (x) : G(x, y) = Φ(x, y) φ y (x). Then φ y (x) satisfies Lφ y =, in U Bφ y = BΦ(, y), in U Green s Functions for the Laplacian. Let U be open and bounded, with piecewise smooth boundary U, and let f, g be continuous on U, U respectively. Consider the Dirichlet boundary value problem for Poisson s equation: (3.17) u = f, in U u = g, in U. The fundamental solution for is a function Φ(x y) of x y, since is translation invariant. (More generally, the fundamental solution for any constant coefficient L will be a function of x y.) The Green s function G(x, y) = Φ(x y) φ y (x) is then specified by φ y (x) = x U φ y (x) = Φ(x y), x U. In particular, φ y is a harmonic function in U. Theorem 9.2. If u C 2 (U) solves (3.17), then G (3.18) u(x) = G(x, y)f(y) dy (x, y)g(y) ds(y). ν y U Remark. Note that the first integral solves the inhomogeneous PDE, with zero boundary condition, while the second integral should be harmonic, and should satisfy the given boundary condition. That it is harmonic follows from the fact that G(x, y) is harmonic in x U for each y U, so that the normal derivative with respect to y is also harmonic. To verify the boundary condition, we could try to show that the normal derivative acts U

14 GREEN S FUNCTIONS AND DISTRIBUTIONS like a δ function as x approaches the boundary. In fact, the proof uses the divergence theorem to recover the boundary data. Proof of Theorem 9.2. Recall that G(x, y) has a singularity at y = x, because the fundamental solution has. As with our solution of Poisson s equation on all of space, using the fundamental solution, we exclude a small ball around x U, by integrating on the domain V ɛ = U B(x, ɛ). Let u C 2 (U) (not necessarily a solution). Then, using Green s identity, ( ) ( u(y) Φ(x y) Φ(x y) u(y) dy = u Φ Φ u ) ds(y). ν y ν y V ɛ Now Φ(x y) is harmonic away from y = x, so the first term on the left hand side is zero. On the right hand side, there are two terms, but there are also two parts of the boundary, namely U and B(x, ɛ). Let I ɛ = B(x,ɛ) V ɛ u(y) Φ ν y (x y) ds(y). As for the solution of Poisson s equation, I ɛ u(x) as ɛ. (See 8.2.) Similarly, let J ɛ = B(x,ɛ) Φ(x y) u ν y (y) ds(y). Now u is continuously differentiable, and Φ(x y) is constant on B(x, ɛ) (proportional to ln ɛ for n = 2, and to ɛ (n 2) for n > 2). Consequently, J ɛ as ɛ. Letting ɛ, we obtain Φ(x y) u(y) dy = u(x)+ U U ( u Φ (x y) Φ(x y) u ) ds(y). ν y ν y Similarly, since φ x is harmonic in U, has no singularity at x = y, and φ x (y) = Φ(x y) for y U, we have φ x u dy = (u φx Φ(x y) u ) ds(y). ν y ν y U U Subtracting, and using G(x, y) = Φ(x y) φ x (y), we obtain (3.19) G(x, y) u(y) dy = u(x) + u(y) G (x, y) ds(y). ν y (The other terms cancel.) U Finally, if u is a solution of (3.17), then we have proved the theorem. U

15 9.3. GREEN S FUNCTIONS The Method of Images. In some examples, we can use the socalled Method of Images to construct the Green s function. The idea is that Φ(x y) represents a field (more precisely, a potential; the study of solutions of Poisson s equation is sometimes referred to as Potential Theory) induced by a singularity at x = y, which induces a field on the boundary U. Suppose we can construct a point x outside U such that the function Φ(C( x y)) exactly cancels Φ(x y) on the boundary U, for some scale factor C, possibly depending on x. Then the new field is harmonic with respect to y U (since the Laplacian is invariant under scaling and translation by a constant), so we can set φ x (y) = Φ(C( x y)). The point x is the image of x that gives the technique its name. Here is how it works in two examples: (i) U is a half space, and (ii) U is a ball. (i) Let U = {x : x n > }. For x = (x 1,..., x n ) U, define the image x = (x 1,..., x n 1, x n ), and let G(x, y) = Φ(x y) Φ( x y), x n, y n. Since Φ( x y) is harmonic in U (with respect to y, for x U), we see that G(x, y) is the Green s function if we can show G(x, y) = for y n =, x n >. But for y n =, we have y x = n 1 (y j x j ) 2 + x 2 n = y x, j=1 so that Φ(x y) = Φ( x y). (See Figure ) (ii) Now consider U = B(, 1), the unit ball in. Here, the image x includes a scaling, in addition to reflection through the boundary. (See Figure ) We define x = x, and let x 2 G(x, y) = Φ(x y) Φ( x ( x y)). (Note that x = 1/ x.) To prove that G(x, y) = for x U, y U, we need to show that x y = x x y, x = x, y = 1. x 2 Proof. x 2 x y 2 = x x x y 2 = x 2 2y x + 1 = x y 2.

16 13 9. GREEN S FUNCTIONS AND DISTRIBUTIONS xn U x y x,. x 1 n-1 ~ x Figure 9.3. Green s function for a half plane. ~ x 1 x y Figure 9.4. Green s function for the unit ball.

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