18.303: Introduction to Green s functions and operator inverses

Transcription

1 8.33: Introduction to Green s functions and operator inverses S. G. Johnson October 9, 2 Abstract In analogy with the inverse A of a matri A, we try to construct an analogous inverse  of differential operator Â, and are led to the concept of a Green s function G(, ) [the ( ), matri element of  ]. Although G is a perfectly well-defined function, in trying to construct it and define its properties we are drawn ineorably to something that appears less well defined: ÂG seems to be some kind of function that is infinitely concentrated at a point, a delta function δ( ) that does not have a proper definition as an ordinary function. In these notes, we show how to solve for G in an eample d problem without resorting to delta functions, by a somewhat awkward process that involves taking a limit after solving the PDE. In principle, we could always apply such contortions, but the inconvenience of this approach will lead us in future lectures to a new definition of function in which delta functions are well defined. Review: Matri inverses The inverse A of a matri A is already familiar to you from linear algebra, where A A = AA = I. et s think about what this means. Each column j of A has a special meaning: AA = I, so A (column j of A ) = (column j of I) = e j, where e j is the unit vector (,,...,,,,..., ) T which has a in the j-th row. That is, (column j of A ) = A e j, the solution to Au = f with e j on the right-hand-side. If we are solving Au = f, the statement u = A f can be written as a superposition of columns of A : u = A f = j (column j of A )f j = A j f j e j. Since j f je j = f is just writing f in the basis of the e j s, we can interpret A f as writing f in the basis of the e j unit vectors and then summing (superimposing) the solutions for each e j. Going one step further, and writing: u i = ( A f ) i = j (A ) i,j f j, we can now have a simple physical interpretation of the matri elements of A : (A ) i,j is the effect (solution) at i of a source at j (a right-hand-side e j ), [and column j of A is the effect (solution) everywhere from the source at j]. et us consider a couple of eamples drawn from the finite-difference matrices for a discretized 2 with Dirichlet () boundary conditions. Recall that one interpretation of the problem 2 u = f is that u is a displacement of a stretched string (d) or membrane(2d) and that f is a force density (pressure). Figure (top) shows several columns of A for 2 2 in d, and it is very much like what you might intuitively epect if you pressed on a string at one point. Figure (bottom) shows two column of A for 2 on a square 2d domain (e.g. a stretched square drum head), and again matches the intuition for what happens if you press at one point. In both cases, the columns are, as epected, the solution from a source e j at one point j (and the minimum points in the plot are indeed the points j).

2 6.5.5 columns of A / 3 3 one A column another A column y y.5 Figure : Columns of A where A is a discretized 2 with Dirichlet boundary conditions. Top: several columns in d (Ω = [, ] = [, ]). Bottom: two columns in 2d (Ω = [, ] [, ]). In both d and 2d, the location of the minimum corresponds to the inde of the column: this is the effect of a unit-vector source or force = at that position (and = elsewhere). 2

3 . Inverses in other bases This is not the only way to write down A, of course, because e j are only one possible choice of basis. For eample, suppose that q,..., q m is any orthonormal basis for our vector space, and that u j solves Au j = q j. Then we can write any f = j q j(q j f) and hence the solution u to Au = f by u = A f = j u j(q j f). In other words, we are writing A = j u jq j = j (A q j )q j in terms of this basis. In matri form, if Q is the (unitary) matri whose columns are the q j, then we are simply writing A = (A Q)Q, since QQ = I. [Similarly, if the columns of B are any basis, not necessarily orthonormal, we can write A = (A B)B : we solve any right-hand-side f by a superposition of the solutions for the columns of B with coefficients B f.] For eample, if A is Hermitian (A = A ), then we can choose the q j to be orthonormal eigenvectors of A (Aq j = λ j q j ), in which case u j = q j /λ j and u = A f = j u j(q j f) = j q j(q j f)/λ j. But this is just writing A = QΛ Q in terms of the (hopefully) familiar diagonalization A = QΛQ. Operator inverses Now we would like to consider the inverse  of an operator Â, if such a thing eists. Suppose N(Â) = {} so that Âu = f has a unique solution u (for all f in some suitable space of functions).  is then whatever operator gives the solution u =  f from any f. This must clearly be a linear operator, since if u solves Âu = f and u 2 solves Âu 2 = f 2 then u = αu + βu 2 solves  = αf + βf 2. In fact, we already know how to write down Â, for self-adjoint Â, in terms of an orthonormal basis of eigenfunctions u n :  f = n u n u n, f /λ n. But now we would like to write  in a different way: analogous to the entries (A ) ij above, we would like to write the solution u() as a superposition of effects at from sources at for all (where just means another point, not a derivative): ˆ u() =  f = d d G(, )f( ), () Ω where G(, ) is the Green s function or the fundamental solution to the PDE. We need to figure out what G is (and whether it eists at all), and how to solve for it. This equation is the direct analogue of u i = j (A ) i,j f j for the matri case above, with sums turned into integrals, so this leads us to interpret G(, ) as the matri element ( ),. So, in analogy to the matri case, we hope that G(, ) is the effect at from a source at. But what does this mean for a PDE?. ÂG and a delta function If G eists, one way of getting at its properties is to consider what PDE G(, ) might satisfy (as opposed to the integral equation above). In particular, we can try to apply the equation Âu = f, noting that  is an operator acting on and not on so that it commutes with the Ω dd integral: ˆ Âu = d d [ÂG(, )]f( ) = f(). Ω This is telling us that ÂG(, ) must be some function that, when integrated against any f( ), gives us f(). What must such an ÂG(, ) look like? It must somehow be sharply peaked around so that all of the integral s contributions come from f(), and in particular the integral must give some sort of average of f( ) in an infinitesimal neighborhood of. How could we get such an average? In one dimension for simplicity, consider the function δ () = { < otherwise (2) 3

4 δ () δ()?? + +? Figure 2: eft: A finite localized function δ () from eq. 2: a rectangle of area on [, ]. Right: We would like to take the + limit to obtain a Dirac delta function δ(), an infinitesimally wide and infinitely high spike of area, but this is not well defined according to our ordinary understanding of function. as depicted in figure 2(left): a bo of width and height /, with area. The integral ˆ d δ ( )f( ) = ˆ f( )d is just the average of f( ) in [, + ]. As, this goes to f() (for continuous f), which is precisely the sort of integral we want. That is, we want ÂG(, ) = lim δ ( ) = δ( ), (3) where δ() is a Dirac delta function that is a bo infinitely high and infinitesimally narrow with area around =, as depicted in figure 2(right). Just one problem: this is not a function, at least as we currently understand function, and the definition of δ() here doesn t actually make much rigorous sense. In fact, it is possible to define δ(), and even lim δ (), in a perfectly rigorous fashion, but it requires a new level of abstraction: we will have to redefine our notion of what a function is, replacing it with something called a distribution. We aren t ready for that, yet, though we should see how far we can get with ordinary functions first. In fact, we will see that it is perfectly possible to obtain G as an ordinary function, even though ÂG will not eist in the ordinary sense (G will have some discontinuity or singularity that prevents it from being classically differentiable). But the process is fairly indirect and painful: once we obtain G the hard way, we will appreciate how much nicer it will be to have δ(), even at the price of an etra layer of abstraction..2 G by a limit of superpositions et s consider the same problem from another angle. We saw for matrices that we could interpret the columns of A as the response to unit vectors e j, and A f as representing f in the basis of unit vectors and then superimposing (summing) the solutions for each e j. et s try to do the same thing here, at least approimately. Consider one dimension for simplicity, with a domain Ω = [, ]. et s approimate f() by a piecewise-constant function as shown in figure 3: a sum of N + rectangles of width = /(N + ). Algebraically, that could be written as a sum in terms of precisely the δ () functions from (2) above: N f() f N () = f(n )δ ( n ), (4) n= since δ ( n ) is just a bo of height on [n, (n + ) ]. In the limit N ( ), f N () clearly goes to the eact f() (for continuous f). But we don t want to 4

5 f() f N () N Figure 3: A function f() (left) can be approimated by a piecewise-constant f N () (right) as in eq. (4), where each segment [n, (n + ) ] takes the value f(n ). take that limit yet. Instead, we will find the solution u N to Âu N = f N, and then take the N limit. For finite N, f N is the superposition of a finite number of bo functions that are localized sources. Our strategy, similar to the matri case, will be to solve [analogous to (3)] Âg n () = δ ( n ) (5) for each of these right-hand sides to obtain solutions g n (), to write u N () = n f(n )g n(), and then to take the limit to obtain an integral where / u() = lim u N() = lim f(n )g n () = N n= ˆ G(, )f( )d, G(, ) = lim g / (). (6) We will find that, even though δ does not have a well-defined limit (at least, not according to the familiar definition of function ), g m does have a perfectly well-defined limit G as an ordinary function. 2 Eample: Green s function of d 2 /d 2 As an eample, consider the familiar  = d2 d 2 on [, ] with Dirichlet boundaries u() = u() =. Following the approach above, we want to solve equation (5) and then take the limit to obtain G(, ). That is, we want to solve g n() = δ ( n ) (7) with the boundary conditions g n () = g n () =, and then take the limit (6) of g n to get G. As depicted in figure 4, we will solve this by breaking the solution up into three regions: (I) [, n ); (II) [n, (n + ) ); and (III) [(n + ), ]. In region II, = /, and the most general g m () with this property is the quadratic function 2 + γ + κ for some unknown constants γ and κ to be determined. In I and III, g n =, so the solutions must be straight lines; combined with the boundary conditions at and, g n 2 this gives a solution of the form: α g n () = γ + κ β( ) < n [n, (n + ) ], (8) > (n + ) for constants α, β, γ, κ to be determined. These unknowns are determined by imposing 5

6 δ ( n ) I II III n (n+) Figure 4: When solving g n() = δ ( n ), we will divide the solution into three regions: regions I and III, where the right-hand side is zero (and g n is a line), and region II where the right-hand side is / (and g n is a quadratic). continuity conditions at the interfaces between the regions, to make sure that the solutions match up with one another. Equation (7) tells us that both g n and g n must be continuous in order to obtain a finite piecewise-continuous g n. Thus, we obtain four equations for our four unknowns: αn = n 2 + γn + κ, 2 α = n + γ, β( [n + ] ) = (n + ) 2 + γ(n + ) + κ, 2 β = (n + ) + γ. After straightforward but tedious algebra, letting = n, one can obtain the solutions: β = + 2 α = β, γ = + α, κ = 2 2. The resulting function g m () is plotted for three values of in figure 5(left). It looks much like what one might epect: linear functions in regions I and III that are smoothly patched together by a quadratic-shaped kink in region II. Physically, this can be interpreted as the shape of a stretched string when you press on it with a localized pressure δ ( ). We are now in a position to take the limit, keeping = n fied. Region II disappears, and we are left with β {( G(, ) =, in region III and α ) < = ( ) ) {( < ( ), in region I: a pleasingly symmetrical function [whose symmetry G(, ) = G(, ) is called reciprocity and, we will eventually show, derives from the self-adjointness of Â]. This function is plotted in figure 5(right), and is continuous but with a discontinuity in its slope at =. [Indeed, it looks just like what we got from the discrete aplacian in figure (top), ecept with a sign flip since we are looking at d 2 /d 2.] Physically, this intuitively corresponds to what happens when you press on a stretched string at one point. Thus, we have obtained the Green s function G(, ), which is indeed a perfectly welldefined, ordinary function. As promised in the previous section, ÂG does not eist according to our ordinary definition of derivative, not even piecewise, since G is discontinuous: the second derivative is infinity at. Correspondingly, δ does not have a well-defined limit 6

7 .25 =.5.25 =.5 g() for = =.25 =.75 eact G(, ).2.5. =.25 = (=) (=) Figure 5: Green s function G(, ) for  = 2 / 2 on [, ] with Dirichlet boundary conditions, for =.25,.5, and.75. eft: approimate Green s function, using our finite-width δ () as the right-hand-side. Right: eact Green s function, from limit. as an ordinary function. So, even though the right-hand side of Âg n = δ ( n ) did not have a well-defined limit, the solution g n did, and indeed we can then write the solution u for any f eactly by our integral () of G. If only there were an easier way to obtain G Deriving the slope discontinuity of G Why is G/ discontinuous at =? It is easy to see this just by integrating both sides of equation (7) for g n (): ˆ (n+) n g n() d = g n(n ) g n(n + ) = ˆ (n+) n δ () d =. Thus, the slope g drops by when going from region I to region II. The limit + of these epressions (the integrals) is perfectly well defined (even though the limit of the integrands is not), and gives G G =, = + = so the slope G/ must drop discontinuously by at =. In fact, we can use this to easily solve for G(, ). For < (region { I) and for > (region III) we must have ÂG = and hence G is a line: G(, α < ) = β( ) >. We have two unknown constants α and β, but we have two equations: continuity of G, α = β( ); and our slope condition α ( β) =. Combining these, one can quickly find that α = and β = as above! 3 To infinity... and beyond! It is perfectly possible to work with Green s functions and  in this way, putting finite δ -like sources on the right-hand sides and only taking the + limit or its equivalent after solving the PDE. Physically, this corresponds to putting in a source in a small but non-infinitesimal region (e.g. a force in a small region for a stretched string, or a small ball of charge for electrostatics), and then taking the limit of the solution to obtain the limiting behavior as the source becomes infinitely concentrated at a point (e.g. to obtain the displacement for a string pressed at one point, or the potential for a point charge). However, this is all terribly cumbersome and complicated. 7

8 It would be so much simpler if we could have an infinitely concentrated source at a point to start with. For eample, when you study freshman mechanics or electromagnetism, you start with point masses and point charges and only later do you consider etended distributions of mass or charge. But to do that in a rigorous way seems problematic because δ() = lim δ () apparently does not eist as a function in any sense that you have learned it does not have a value for every. It turns out that the problem is not with δ(), the problem is with the definition of a function surprisingly, classical functions don t really correspond to what we want in many practical circumstances, and they are often not quite what we want to deal with for PDEs. A clue is provided in section 2., where we saw that even though our integrands might not always have well-defined limits, our integrals did. If we can only define a function in terms of its integrals rather than by its values at points, that might fi our problems. In fact, we can do something very much like this, which leads to the concept of a distribution (or generalized function) or a weak solution. At the cost of an increase in the level of abstraction, we can completely eliminate the tortured contortions that were required above (and only get worse in higher dimensions) to avoid delta functions. 8