Existence Theory: Green s Functions
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1 Chapter 5 Existence Theory: Green s Functions In this chapter we describe a method for constructing a Green s Function The method outlined is formal (not rigorous) When we find a solution to a PDE by constructing a Green s function we will have to verify the solution is correct The method we outline provides solutions to PDEs and ODEs in a variety of physical domains 51 General Method for finding a Green s Function Suppose we wish to solve a differential equation of the form (51) Lu = f, B(u Here L (L for linear) could be an ODE or a PDE The function f is given, and B(u) refers to boundary conditions and or initial conditions The zero does not mean we only consider homogeneous conditions To solve the differential equation we need to find L 1 - the solution should be u = L 1 f Moreover, since L involves derivatives, we expect L 1 to be an integral We suppose (51) is to be solved on some physical domain Ω If the PDE involves time, we might want the solution on D := Ω [, T ] Here is an algorithm for finding a Green s function (finding L 1 ) 61
2 62 5 Existence Theory: Green s Functions Let x represent the location at which we want the solution For a PDE with the physical domain a subset of R 3, x would be x, y, z, t for example We will express the solution as an integral with respect to x variables All integration and differentiation is with respect to the prime variables - x is just a fixed observation point RDefine a vector space and inner product on that vector space- typically (u, v D u v dx for some appropriate domain D The vector space could be V = {u u C 2 (D)} We consider a linear (differential) operator on V which we denote by L We need to compute the adjoint of L We have integrating by parts a generalized Green s second identity (cf Equation (81)) (52) (Lu, v D Lu v dx = Boundary Terms + (u, L v) The operator L is called the formal adjoint of L Now comes the hard part We need to solve L v = δ(x x ) So v = v(x, x ) (in practice we need to be able to solve the homogeneous equation L v = to make this work) Then, we have (u, L v D u(x)l v(x, x )dx = D u(x)δ(x x )dx = u(x) Finally, the boundary terms (52) contain u evaluated at either a temporal or spatial boundary If u is known on those boundaries (through B(u ), we do nothing If they are not known, we require v(x, x ) to be zero (x evaluated on the boundary) The solution u to the original PDE comes from (52) Indeed, (Lu, v (f, v Known Boundary Terms + u(x), or u(x D f(x )v(x, x ) dx Known Boundary Terms The function v(x, x ) is called a Green s function and we usually denoted it G(x, x ) Many examples follow
3 51 General Method for finding a Green s Function 63 Example 51 Solve the boundary-value problem (one-dimensional Elliptic PDE) u = f(x), u( 1, u(1 Solution Following the algorithm we set we think of x as being fixed in the interval [, 1] We wish to find u(x) We set V = {u u C 2 [, 1]}, (u, v R 1 u(x)v(x) dx, and Lu = u Integrating by parts twice, we find (53) (Lu, v u v 1 uv 1 + uv dx, where the derivatives are with respect to the primed variable Thus, the formal adjoint operator is L v = v Since this is the same as L, we call the operator formally self adjoint The next step is to solve L v = δ(x x ) Since δ(x x when x 6= x, we have v(x, x ax + b x < x cx + d x > x To get v = δ(x x ) we require v to have a jump discontinuity at x = x To find it, we integrate v = δ(x x ) across x We get 1 = x+ x δ(x x ) dx = x+ x v (x, x ) dx = v (x, x + ) v (x, x ) Taking the limit as x approaches x from below, and x + approaches x from above, we conclude c a = 1 We also require v to be continuous Moreover, since we do not know u on the boundary (x =, x = 1), we require v(x, v(x, 1 This removes the unknown boundary terms in (53) This leads to the equations, respectively, c a = 1, ax + b = cx + d, b =, c + d = Solving, we find a = x 1, b =, c = x, and d = x Thus, the Green s function is G(x, x (x 1)x x < x (x 1)x x > x, and the solution follows from (53) We set G(x, x v(x, x ) Then u(x = x f(x )G(x, x ) dx dg(x, 1) + u(1) (x 1)x f(x ) dx + x dx u() dg(x, ) dx (x 1)xf(x ) dx (x 1) In order the check the solution we first recall Leibniz s Formula from Chapter One Its proof can be found in Chapter One, 135
4 64 5 Existence Theory: Green s Functions Theorem 52 (Leibniz s Formula) Suppose f and f/ x are continuous on D to R, where D := {(x, x ) : a x b, c x d} If α and β 2 are functions which are differentiable on the interval [c, d] with values in [a, b], then F defined by F (x is differentiable for each x [c, d] and F (x β(x) α(x) f(x, x ) x β(x) α(x) f(x, x ) dx, dx + f(x, β(x)) β (x) f(x, α(x)) α (x) Proof See Theorem 111 in Chapter One Now we can check our answer Theorem 53 Suppose f C r ([, 1]) Then solves u(x x Moreover, u C r+2 ([, 1]) (x 1)x f(x ) dx + u = f(x), u( 1, u(1 x (x 1)xf(x ) dx (x 1) Proof We easily check that u( 1 and u(1 Applying Leibniz s formula we get u (x x x f(x ) dx + (x 1)xf(x) + Applying the fundamental theorem of calculus, we get u (x xf(x) (x 1)f(x f(x) x (x 1)f(x ) dx (x 1)xf(x) 1 Note that we need f to be at least continuous to check the solution The next example is probably familiar Example 54 Suppose p and q are continuous on an interval containing t = Solve the ODE u + p(t)u = q(t) u( u Solution Set Lu = u + pu We wish to find L 1 To do this we first find the adjoint operator L Let t be a fixed observation point Set, for some T > t, (54) (Lu, v = Lu v dt (u + pu)v dt = uv T + u( v + pv) dt
5 51 General Method for finding a Green s Function 65 The the adjoint operator is therefore L v = v +pv We must solve L v = δ(t t ) The ODE L v = is easy enough to solve since it is separable The solution is v = c exp( R t p) To eliminate the unknown boundary term u(t )v(t ), we require v to vanish for t > t (this eliminates T as well) Thus, the Green s function is of the form ( v(t, t ce t p t < t t > t We need some kind of jump discontinuity across t = t to get L v = δ To find it we formally integrate L v = δ from just below t to slightly above We find 1 = t+ t δ(t t t+ t = (v(t, t + ) v(t, t )) + ( v (t, t ) + p(t )v(t, t )) dt t+ t p(t )v(t, t ) dt Taking the limit as t approaches t from below, and t + approaches t from above the integral vanishes, and we conclude v should have a jump discontinuity across t In particular we require ( ce t p 1, or Thus, the Green s function is G(t, t ( c = e t p Returning to (54), we find the solution to be e ( t p t p) t < t t > t u(t (Lu, v) uv T = e t t p e t p q(t ) dt + u e t p in agreement with the integrating factor approach in elementary ODEs Example 55 Solve the second-order initial-value problem u 2u + u = f(t), u( u, u ( u Solution Set Lu = u 2u + u We first compute the adjoint Let t be fixed We have, by integrating by parts twice with T > t (55) Lu v dt = v(u 2u + u) dt = vu T uv T 2uv T + u(v + 2v + v) dt
6 66 5 Existence Theory: Green s Functions The adjoint it L v = v +2v +v Again we must solve L v = δ(t t ) Moreover, as in the previous initial-data problem, to make the unknown boundary terms vanish at t = T, we require v(t, t for t > t In this case we expect v(t, t ae t + bt e t t < t t > t Integrating L v = δ(t t ) gives 1 = t + v (t, t + ) v (t, t ) + 2 v(t, t + ) v(t, t ) + v(t, t ) dt t We take the limit as t + t from above and t t from below The integral vanishes, and we require v to be continuous and have the appropriate jump This requires ae t + bte t = ( ae t + be t bte t 1 Solving, we find a = te t and b = e t The Green s function is G(t, t (t t )e t t t < t t > t From (56) we conclude the solution is u(t = t G(t, t )dt + G(t, )u () G t (t, )u() 2G(t, )u() (t t )e t t f(t )dt + u te t + u (1 t)e t For fun suppose f(t δ(t 1) with u( 1 and u ( 1 Then our solution becomes e u(t t t < 1 (t 1)e t 1 + e t t > 1 If we recall the Heaviside function u c (t) introduced in elementary ODEs, we may write this u(t (t 1)e t 1 u 1 (t) + e t in agreement with Laplace Transform methods Sometimes it is possible, or more convenient, to find solutions to Lu = rather than L v = Suppose our original problem was L v = f with boundary conditions Then u would be the Green s function for L Not surprisingly the two Green s functions are related Theorem 56 Suppose G and G satisfy LG = δ(x x ) and L G = δ(x x ) Moreover, suppose (LG, G (G, L G) (G and G satisfy appropriate boundary conditions) Then G(x, x G (x, x)
7 51 General Method for finding a Green s Function 67 Proof The proof is formal (assumes properties of delta functions are valid) Set u = G (x 1, x ), and v = G(x 2, x ) Then (Lu, v δ(x 1 x )G(x 2, x ) dx = G(x 2, x 1 ) Similarly, (Lu, v (u, L G G (x 1, x )δ(x 2 x ) dx = G (x 1, x 2 ) The result follows Example 57 Derive the Variations of Parameters formula found in elementary texts on ODEs Specifically, solve y + p(t)y + q(t)y = f(t) y( y, y ( y Solution Set Lu = u + pu + qu As in other cases v(t )(u + pu + qu) dt = v u T vu T + pvu T + (v (pv) + qv)u(t ) dt Thus, L v = v (pv) + qv with boundary conditions v(t, t for t > t We could solve this Or we could solve LG (t, t δ(t t ) with G (t, = G t (t, With this choice (v, Lu (L v, u) and the above theorem applies That is, G(t, t G (t, t) We will find G Let us suppose y 1 and y 2 are two linear independent solutions to Lu = Then the Green s function has the form G (t, t c1 y 1 (t ) + c 2 y 2 (t ) t < t c 3 y 1 (t ) + c 4 y 2 (t ) t > t The boundary conditions imply (they are all at t = ) y1 () y 2 () c1 y 1 () y 2 ( c 2 The determinant is W (y 1, y 2 )( y 1 ()y 2 () y 1 ()y 2() and is called Wronskian If y 1 and y 2 are linear independent, the Wronskian is never zero Therefore, c 1 = c 2 = To find the other constants we continuity and the jump discontinuity That is, = c 3 y 3 (t) + c 4 y 4 (t) and 1 = t + t u (t ) + p(t )u (t ) + q(t )u(t ) dt The second term presents some difficulty With some extra assumptions on p we have t + t t + t t + t p(t )u dt = (p(t )u) dt p (t )u dt
8 68 5 Existence Theory: Green s Functions If we assume p is continuous, the first term is zero because pu is continuous, and the second is zero (as t t + ) because the integral itself converges to zero Thus, the jump discontinuity is 1 = u (t, t + ) u (t, t ) = c 3 y 1 (t) + c 4y 2 (t) Putting these together, we require y1 (t) y 2 (t) c3 y 1 (t) y 2 (t) c 4 = 1 This is equivalent to 1 y 2 (t) y 2 (t) W (y 1, y 2 )(t) y 1 (t) y = 1(t) 1 Thus, the Green s function is ( By our theorem G (t, t G(t, t c3 c 4 t < t y2(t)y1(t ) W (y + y1(t)y2(t ) 1,y 2)(t) W (y 1,y 2)(t) t > t ( y1(t )y 2(t) W (y 1,y 2)(t ) y2(t )y 1(t) W (y 1,y 2)(t ) t > t t < t The solution is t y 2 (t ) t u(t y 1 (t) W (y 1, y 2 )(t ) f(t ) dt y 1 (t ) + y 2 (t) W (y 1, y 2 )(t ) f(t ) dt in agreement with our result from elementary ODEs Example 58 Solve the boundary-value problem u u = f(x) u ( a, u(1 b, Solution Set Lu = u u We first compute the adjoint Let x (, 1) be fixed We have, by integrating by parts twice, (56) Lu v dx = v(u u ) dx = vu 1 u(v + v) 1 + u(v + v ) dx Thus, the adjoint is L v = v + v We need to solve v + v = δ(x x ) For x 6= x, L v = implies v(x, x c 1 + c 2 e x Thus, the Green s function has the form v(x, x c1 + c 2 e x x < x c 3 + c 4 e x x > x
9 51 General Method for finding a Green s Function 69 We need to find the jump discontinuities We do this by integrating L v = δ(x x ) near x We get x + 1 = (v + v ) dx x = (v (x, x + ) v (x, x )) + (v(x, x + ) v(x, x )) This suggests requiring v to have a jump discontinuity of one across x and v to be continuous across x That is, we require c 2 e x c 4 e x = 1, c 1 + c 2 e x = c 3 + c 4 e x, respectively The unknown boundary terms in (57) must be eliminated This requires v () + v( and v(1 The first requires c 2 + (c 1 + c 2 or c 1 = Putting the remainder together we need e x 1 e x e x e x 1 e 1 c 2 c 3 c 4 = 1 The determinant is 1/e x+1 and the inverse is easily found to be e ex e e 1 1 e e e The middle column gives c 2, c 3 and c 4, and the Green s function is v(x, x (e x e)e x x < x 1 e 1 x x > x The solution is obtained from (57) and is u(x = x v(x, x )f(x ) dx + v(x, )u () + (v (x, 1) + v(x, 1))u(1) (e x e)e x f(x ) dx + x (1 e 1 x )f(x ) dx + (e x e)a + b Alternatively, we could have reduced the number of unknowns by looking for v in the form v(x, x c 1 + c 2 e x x < x c 3 + c 4 (e x e 1 ) x > x The boundary conditions imply c 1 = c 3 = and we would have to invert a 2 2 matrix to solve for c 2 and c 4 That matrix is e x e x c2 1 e x e 1 e x = c 4 The determinant is e 1 x, and so c2 = e 1+x e 1 e x e x e x e x c 4 1 Thus, c 2 = e x e and c 4 = e Plugging back in we arrive at the same Green s function as before - only with far fewer calculations
10 7 5 Existence Theory: Green s Functions 52 Modified Green s Functions Sometimes a differential operator Lu does not have an inverse This happens when the homogeneous equation, Lu = has a nontrivial solution-the null space of L is nontrivial (a square matrix would not be invertible for the same reason) For example, the boundary-value problem u + u =, u( = u(π), has the nontrivial solution u = sin x We expect the solution to Lu = f, u( u(π not be unique since we can add any multiple of sin x to a solution and obtain another solution In this case the procedure outlined in the previous section breaks down When a boundary-value problem has a nontrivial homogeneous solution, Lu = f may not have a solution A similar situation occurs in linear algebra Indeed, recall (a shortened version) Theorem 59 (The Fundamental Theorem of Linear Algebra) If A is an m n matrix, then R(A N(A T ) In other words Ax = f has a solution if and only if f is perpendicular to the null space of the adjoint (transpose) There is an interesting analogy for differential equations (loosely stated here) Theorem 51 (Fredholm Alternative Theorem) Either the boundary-value problem Lu = f with boundary conditions has a unique solution, or Lu = (same boundary conditions) has a nontrivial solution In this case Lu = f has a solution if and only if (f, v Here L v = with v 6= If, as above, we try to solve L v = δ(x x ), we may not be able to if L v = has a nontrivial solution In fact, according to the Fredholm Alternative Theorem, applied to L, L v = f has a solution if and only if (u h, f where Lu h = To get a solution we modify G slightly to make the right side perpendicular to nontrivial solutions to Lu = and try again In particular, we require L G = δ(x x ) u h (x)u h (x ), with (u h, u h 1 and Lu h = Now, the right side of L G satisfies δ(x x ) u h (x)u h (x ), u h (x δ(x x )u h (x )dx u h (x)u h (x )u h (x )dx = u h (x) u h (x) = and so it is perpendicular to the null space of L The Fredholm Alternative Theorem applies and we expect L G = δ(x x ) u h (x)u h (x ) to have a solution
11 52 Modified Green s Functions 71 Example 511 Find the Green s function for differential operator with Neumann boundary conditions given by u = f u ( = u (l) Solution We first note the u = 1 is a nontrivial solution, which we normalize by setting u h = l 1/2 Then (u h, u h 1 Our usual method for finding a Green s function breaks down (verify this) We still need to find the adjoint We have (Lu, v u v l uv l + l uv dx Thus, the formal adjoint operator is L v = v Moreover, since u () and u (l) are known and u is unknown on the boundary, we need to require v ( = v (l) Based on the above discussion we need to solve Thus v has the form L v = δ(x x ) u h (x)u h (x ) v(x, x = δ(x x ) l 1 ( c 1 + c 2 x x 2 2l x < x c 3 + c 4 x x 2 2l x > x The boundary conditions on v imply c 2 = and c 4 1 = Continuity implies c 1 + c 2 x = c 3 + c 4 x With the other conditions, we find c 1 = c 3 + x Thus ( G(x, x c 3 + x x 2 2l x < x c 3 + x x 2 2l x > x Note that the jump discontinuity is automatically satisfied and c 3 remains arbitrary - there are many Green s functions for this problem This is not surprising since the solution to the original problem is not unique We may write the solution as u(x C + x x x 2 2l were C is an arbitrary constant To check the solution, we compute u (x f(x ) dx + x l x f(x ) dx We immediately see u ( as expected However, u (l l f(x ) dx x x 2 2l f(x ) dx The operator L is self adjoint in this case, so the Fredholm Alternative Theorem implies Lu = f, u ( = u (l) if and only if f is perpendicular to the null space
12 72 5 Existence Theory: Green s Functions of L = L The null space is spanned by the vector (function) v = 1, and so, f must be perpendicular to 1 That is, u (L l 1 f(x ) dx = as required Returning to u, we find u (x f(x) as required
13 53 Homework Homework Exercise 51 Construct a Green s function for y = f(x) y( 2y(1) and find the solution Check that your solution is correct Exercise 52 Construct a Green s function for y = f(x) y( = y (1) and find the solution Check that your solution is correct Exercise 53 Solve the following ODE by constructing a Green s function y + y = f(x) y( = y(1) and find the solution Check that your solution is correct Construct a modified Green s function if instead y( y(π Exercise 54 Construct a Green s function for the initial-value problem u u = f(t), u( a, u ( b Using Leibniz Rule, check your answer Exercise 55 Consider the Cauchy Problem a(x, y)u x + b(x, y)u y + c(x, y)u = F (x, y), u(x, f(x), with x R, y > and a, b, c, f, and F given smooth functions Of course we could solve this by the method of characteristics Let s find a Green s function instead! (a) Set Lu = a(x, y)u x + b(x, y)u y + c(x, y)u Integrate vlu over the rectangle L x x L x, y L y By integrating by parts (Calculus II style) find a generalized Green s second identity (b) Look at the boundary terms carefully Find conditions on v to eliminate the unknown terms (c) Take the limit as L x and L y Find the conditions the Green s function must satisfy (c) Represent the solution in terms of the Green s function (d) Find the Green s function in the case a = 1, b = 1, and c =
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