Consider the 2nd order liner inhomogeneous ODE Green s functions d 2 u 2 + k(t)du + p(t)u(t) = f(t). Of course, in prctice we ll only del with the two prticulr types of 2nd order ODEs we discussed lst week, but let me keep the discussion more generl, since it works for ny 2nd order liner ODE. We wnt to find u(t) for ll t >, given the initil conditions u(t = ) = u, du t= = v. Let us ssume tht the two linerly-independent solutions u (t), u 2 (t) of the homogeneous eqution re known (we ve discussed wht these re for the specil kinds of equtions we will need to solve). Then, we know tht the generl solution of the inhomogeneous eqution is: u(t) = C u (t) + C 2 u 2 (t) + u p (t) where u p (t) is ny prticulr solution of the inhomogeneous eqution. After we find u p (t), we cn use the initil conditions to find C nd C 2. Two wys to find u p (t) tht we ve discussed in clss re guessing nd the vrition of prmeters. There is nothing wrong with either, except tht every time new f(t) is given, we hve to try nother guess or go into ll the work required by the vrition of prmeters we hve to redo the whole clcultion gin to get the new u p. This becomes quite difficult, especilly if f(t) is not simple function. The ide behind the Green s function is to find generl expression tht tells us wht u p (t) is for ny f(t) tht we cre to use. We still need to do is one clcultion (to find the Green s function), but once we hve it, we cn find u p (t) for ny f(t) without much further work. Before lunching into how this works, let me point out tht sometimes this solution is shown s sort of mth trick relted to certin specil wy to do vrition of prmeters for instnce, this is how it s presented in the 6th edition of the textbook, pges 2-23. I wnt to try to do bit of better job in explining why this works, nd how should we think bout the mening of this solution, s physicists. This should help us figure out how to generlize this ide, becuse we will use it lter for PDEs. The min ide is to decompose f(t) s sum of simple functions, for which we know the prticulr solutions. Remember tht if f(t) = f (t)+f 2 (t), then u p (t) = u p (t)+u p2 (t), where u p (t) is the prticulr solution for f (t), etc. Of course, this would be true if we broke f(t) in ny number of pieces, so long s we could find the corresponding u p (t) for ech piece. Now, the wy we do this breking is with the Dirc function: remember tht we cn write: f(t) = dτδ(t τ)f(τ) which mens tht f(t) is sum of short kicks (described by the δ-function), nd so tht the kick pplied t τ hs the strength f(τ). Of course, sum here is relly n integrl, becuse the time τ is continuous vrible. The reson why the integrl is from to is tht I m only interested in times t in this sorts of problems we don t cre wht hppened in the pst, the question is lwys wht hppens fter the initil moment t =. So, the ide is tht if we cn find the prticulr solution for kick δ(t τ), i.e. kick pplied t time τ, then we re done we need only sum over the prticulr contributions from ll the kicks tht contributed to our f(t). The Green s function G(t, τ) is the solution of the inhomogeneous eqution d 2 G(t, τ) (t, τ) 2 + k(t) + p(t)g(t, τ) = δ(t τ) () In other words, it tells us wht is prticulr solution is we pply single kick of strength t the time t = τ exctly wht we need. It hs two rguments becuse, of course, the solution will be different if we kick t different times τ, so we need to keep trck of τ s well. Finlly, G is not just ny prticulr solution of this inhomogeneous eqution, but we will ASK tht it stisfies the initil conditions G(t =, τ) =, t= =. Note the very specil form of these initil conditions they look like the ones for the generl solution, but they re both HOMOGENEOUS (i.e., = )! The reson for this is tht I don t wnt to hve to reclculte the Green s function every time I chnge the vlues of u, v s you ll see below, the homog. prt of the solution cn tke cre of u nd v.
Once we hve this prticulr solution, we know tht for ny rbitrry sequence of kicks tht mkes up f(t), the prticulr solution must be: u p (t) = dτg(t, τ)f(τ) (if you re not quite sure bout this, plug this u p in the ODE nd check tht indeed it stisfies Eq. ()). Becuse of the initil conditions stisfied by G, it follows tht u p () =, dup t= =. So the generl solution of our eqution is: where C, C 2 must be chosen such tht: u(t) = C u (t) + C 2 u 2 (t) + dτg(t, τ)f(τ) u = u(t = ) = C u () + C 2 u 2 () + u p () = C u () + C 2 u 2 () v = du du t= = C du 2 t= + C 2 t= + du p du t= = C du 2 t= + C 2 t= This explins why we chose those initil conditions for G(t, τ) this wy we cn djust C nd C 2 to tke cre of the ctul initil conditions, nd we don t need to reclculte G nd therefore u p if we chnge the initil conditions ll tht is needed is to djust C, C 2 ccordingly. In summry, once we know G(t, τ), we only need to do the integrl dτg(t, τ)f(τ) for the function f(t) of interest, nd lso find C nd C 2, nd we re done. And this works for ny f(t), ll we need is to do one integrl. Before clculting G(t, τ), let s see wht generl conditions it must stisfy.. G(t, τ) must be continuous t ll t, becuse we expect u(t), nd therefore u p (t) to be continuous t ll times. For exmple, if this eqution comes from pplying Newton s second lw, then u describes the loction of some object. Clerly, then, it must chnge continuously in time. In prticulr, G(t, τ) must be continuous when we pply the kick: G(t = τ, τ) = G(t = τ +, τ) where τ ± re times infinitesimlly close to τ (just before nd just fter). 2. However, since we pply this singulr kick t t = τ, we expect something discontinuous to hppen to G there. Indeed, it turns out tht the derivtive is discontinuous: t=τ+ t=τ = (2) in other words, the derivtive of G(t, τ) hs jump of precisely t t = τ when the kick is pplied. This comes directly from Eq. () if we integrte it from τ ɛ to τ + ɛ, nd we let ɛ. Wht we get is: [ d 2 ] G(t, τ) (t, τ) τ+ɛ 2 + k(t) + p(t)g(t, τ) = δ(t τ) = (the second equlity is just the vlue of the integrl on the rhs, see delt functions). On the lhs, we hve three terms. Let me tke them from the end. When ɛ, p(t)g(t, τ) becuse the integrnd is continuous function, nd we re shrinking the integrtion intervl to zero. Similrly, fter integrting by prts, we find: (t, τ) k(t) = [k(t)g(t, τ)] t=τ+ɛ t= dk G(t, τ) = becuse gin we re deling only with continuous functions in the limit where the integrtion intervl goes to zero. Finlly: d2 G(t, τ) 2 = t=τ+ɛ t= = t=τ+ t=τ 2
nd Eq. (2) follows directly. 3. As I sid, we sk for the simplest possible initil conditions G(t =, τ) =, t= =. The reson for this is tht we do not wnt the Green s function to depend on the initil conditions u, v of the eqution if this ws the cse, then nytime we chnged the initil conditions we would hve to reclculte G for the new initil conditions. Asking tht G(t =, τ) =, t= = mens tht G(t, τ) lwys hs the sme expression, nd we let C nd C 2 djust so s to tke cre of u, v. So let s find G(t, τ). Let us consider first the intervl < t < τ. Becuse in this intervl δ(t τ) =, here the eqution for G is: d 2 G(t, τ) (t, τ) 2 + k(t) + p(t)g(t, τ) = in other words here G is solution of the homogeneous eqution, so it must be of the generl form: if < t < τ, theng(t, τ) = u (t) + 2 u 2 (t) Similrly, for t > τ, gin δ(t τ) = nd the eqution becomes homogeneous, so we must hve: if t > τ, theng(t, τ) = b u (t) + b 2 u 2 (t) All tht is left is to find, 2, b, b 2 nd we re done. For this, we use the 4 conditions we hve. Let s strt with the initil conditions. The time t = < τ, so we must hve: nd G(t =, τ) = u () + 2 u 2 () = (t =, τ) = du () + du 2 2 () = du Becuse u, u 2 re linerly independent solutions, their Wronskin W (t) = u 2 du u 2 for ny time, therefore lso for t =. As result, the only solution of those two equtions is = 2 =. Nice nd simple. So we find tht G(t, τ) = if t < τ. Now we use conditions nd 2 to find G(t, τ) for t > τ. First: becuse the functions u, u 2 re continuous. Also, G(t = τ, τ) = G(t = τ +, τ) = b u (τ) + b 2 u 2 (τ) t=τ+ du t=τ = b (τ) + b du 2 2 (τ) = Since the Wronskin is gin gurnteed to be non-zero, the solution of this system of coupled equtions is: b = u 2(τ) W (τ) ; b 2 = u (τ) W (τ) So the conclusion is tht the Green s function for this problem is: { if < t < τ G(t, τ) = if τ < t u (τ)u 2(t) u 2(τ)u (t) W (τ) nd we bsiclly know it if we know u nd u 2 (which we need to clculte in ny event). Let me mke some comments.. As I sid, in the textbook this formul is derived s specil cse of the vrition of prmeters solution, nd then is clled Green s function. There is nothing wrong with tht derivtion s such, except it is not very cler how to extend tht procedure to equtions with boundry conditions (which is wht we will do next). The derivtion I hve here is much more generl nd we will go through precisely the sme steps to find the Green s functions if we re given boundry conditions, insted of initil conditions. Of course, in tht cse the solution for G will chnge, but we cn find it just s esily. 2. The formul we derived here is quite esy to understnd if we think in physicl terms. Suppose tht u(t) is the loction of some object of mss m =, nd the ODE is Newton s second lw: mybe there is some drg (the term 3
proportionl to du/) nd mybe some elstic force (the term proportionl to u) nd so f(t) describes whtever other externl force is pplied t time t. Then, s discussed, G(t, τ) will be the loction of the object if it strts t the origin nd t rest (see initil conditions for G), nd if we kick it with f(t) δ(t τ) t the time τ. Obviously, before we kick it the object will remin t rest t the origin, which explins why G(t, τ) = if t < τ. After the kick, the object will move s described by G(t, τ), t > τ. For t τ +, i.e. just when it strts to move, it is still t the origin where it ws ll the time until t = τ tht s condition. But becuse it ws kicked with this very short but intense force, its speed jumps from to (tht s condition 2 remember tht pplying force chnges the momentum of n object. For this unit of kick-force we pply, the speed increses by one unit s well). Wht the method does, then, is to sy tht ny externl force cn be thought of s sequence of kicks with vrious strengths. Becuse the eqution is liner, to find the whole solution we simply need to sum the contributions due to ech kick (superposition principle). We will spend some time prcticing this in clss, nd compring it ginst the other two methods. After you re comfortble with this, we will derive together the Green s function for ODEs where we re given boundry conditions think vrible x (spce) insted of t (time), nd tht we re interested in finite region of spce x b (for exmple, we might wnt to know the temperture of rod which is locted between nd b). Then, we might be given the temperture t ech end: u(x = ) = T L, u(x = b) = T R. Conditions like this, which specify the vlue of the unknown, re known s Dirichlet conditions. Or we could be told wht the derivtives of u re t the ends of the rod (s we ll see in bit, this derivtive is proportionl to how much het flows into/out of the rod, nd mybe tht s wht we control, not the tempertures s such for instnce, for n isolted end, no het cn flow out nd the derivtive is zero). Such conditions re known s von Neumnn conditions: du x= = h L, du x=b = h R. Or we could hve mixed bg, where we know u t one end nd du/ t the other one. To ech of these situtions will correspond different Green s function; but once we know tht Green s function, we cn solve ny inhomogeneous ODE with tht type of boundry conditions without ny further trouble. How to derive the proper Green s functions for such ODEs with boundry conditions (=BC) is the topic of the in-clss ctivity 6. I will summrize here the results most of you rrived t by the end of the clss. I hope tht those of you who hve not quite finished will try to do so t home, before looking t this solution. So we wnt to find the Green s function G(x, ξ) tht will llow us to find prticulr solution of the form u p (x) = dξg(x, ξ)f(ξ) for the ODE d2 u + k(x) du 2 + p(x)u = f(x) with the BC u(x = ) = u, u(x = b) = u b. The first step is to find wht eqution must be stisfied by G(x, ξ). Since dup(x) = for the second derivtive, putting this into the ODE nd grouping the terms leds to: [ d 2 ] G(x, ξ) (x, ξ) dξf(ξ) 2 + k(x) + p(x)g(x, ξ) = f(x) = dξ (x,ξ) dξf(ξ)δ(x ξ) f(ξ) nd similrly where the second eqution is true for ny x [, b] (see Dirc functions). For this to hold for ny f(x), it follows tht we must hve: d 2 G(x, ξ) (x, ξ) 2 + k(x) + p(x)g(x, ξ) = δ(x ξ) In other words, just like we found for G(t, τ), the Green s function stisfies precisely the sme ODE s u(x), but the inhomogeneous prt is δ-function. This should not be surprising, since if we chnged x t, this would be identicl to the time-dependent ODE, so nturlly we should get the sme result. 2. If this ws Poisson eqution, then we know tht u(x) would be the electric potentil t x due to distribution of chrges described by f(x). Since G stisfies the sme eqution, it follows tht G(x, ξ) is the electric potentil t x due to distribution of chrge δ(x ξ) i.e. unit chrge locted t ξ. 4
So we cn see the logic here: if we cn figure out the potentil due to ny point chrge (i.e., G) then using the superposition principle, the totl potentil due to ll chrges will be the sum (integrl, relly) of contributions from ech chrge tht s precisely the reltionship between u p nd G. 3. The generl solution of the ODE will be: To find C, C 2, I need to use the BC: u(x) = C u (x) + C 2 u 2 (x) + u(x = ) = u = C u () + C 2 u 2 () + u(x = b) = u b = C u (b) + C 2 u 2 (b) + dξg(x, ξ)f(ξ) dξg(x =, ξ)f(ξ) dξg(x = b, ξ)f(ξ) The simplest choice (since it is independent of u, u b, so it won t chnge if we chnge those vlues) is to sk: G(x =, ξ) = ; G(x = b, ξ) = in which cse the integrls vnish, nd we must solve u = C u () + C 2 u 2 (), u b = C u (b) + C 2 u 2 (b) which is esy enough. So we sk for these simplest possible BC for G. Note tht if insted of u(x = ) = x, we were given t tht end the von Neumnn condition du x= = v, then we would need to solve: v = C du x= + C 2 du 2 x= + dξ (x, ξ) x= f(ξ) So t this end, it now mkes sense to sk tht (x,ξ) x= = s the simplest possible choice. In conclusion, we will sk tht G stisfies the sme kind of BC like u, but lwys homogeneous (i.e., equl to ): If u is given t n end, we sk G = t tht end. If du is given t n end, we sk tht = there. We cn tret similrly mixed conditions if we re given the vlue of αu + β du t n end, we sk tht αg + β = t tht end. This mens tht even if the ODE is unchnged, the expression for G will be different for different BC. In terms of Poisson eqution, G = mens the electric potentil creted by the point chrge is zero t the boundry, while = mens tht the electric field is zero t tht boundry. So the two solutions will nturlly be different, depending on wht we sk to hppen t the boundry. 4. Besides the two boundry conditions discussed t the previous point, we lso hve mtching conditions for x = ξ where we plced the point chrge. First, G must be continuous becuse we expect u nd therefore u p to be continuous, so: G(x = ξ, ξ) = G(x = ξ +, ξ) If this ws Poisson eqution, it would just men tht we expect the electric potentil to be continuous t ll points, including where we plced the chrge. The second condition is tht the derivtive is discontinuous: x=ξ + x=ξ = For mthemticl derivtion of this equlity, you just need to follow the sme steps we used for G(t, τ) fter ll, since the ODEs re identicl nd the only chnge is t x, τ ξ, then this condition (which is hs nothing to do with the boundry conditions or initil conditions) should sty the sme. Physiclly, if G is n electric potentil, then its derivtive is (minus) the electric field. So this eqution sys tht the electric field chnges discontinuously from one side to the other side of the chrge. But this is very resonble, becuse we know tht the electric field points towrds the chrge if the chrge is negtive, nd wy if it is positive. If we re in the lter cse, then E < to the left of the chrge, i.e. for x < ξ; nd E > to the right of the chrge, for x > ξ. So indeed, the electric field chnges discontinuously. The fct tht the difference is precisely is becuse we used unit chrge. 5
5. Now we consider the ODE: d 2 u 2 du x = f(x). This ODE is of Cuchy-Euler form therefore the solutions re power lws, which fter some work re found to be u (x) =, u 2 (x) = x 2. We wnt to find G(x, ξ) for x [, 3]. The eqution stisfied by G is (see ): d 2 G 2 = δ(x ξ) x We ll solve this piecewise. If x < ξ, then δ(x ξ) = (there is no chrge in this intervl) nd the eqution of G becomes: d 2 G 2 x = which is hom. ODE nd so hs the generl solution G(x, ξ) = + 2 x 2. If ξ < x 3, the eqution gin becomes homogeneous, so here we must hve G(x, ξ) = b + b 2 x 2. All tht s left to do is to find the 4 coefficients, 2, b, b 2 nd we re done. For this we need 4 equtions. We know tht we hve to sk for boundry conditions G(x =, ξ) =, G(x = 3, ξ) = (see discussion t 3), so this gives 2 equtions. We lso hve the 2 mtching conditions, so this should work out. At the left boundry: G(x =, ξ) = + 2 = G(x, ξ) = (x 2 ) At the right boundry: G(x = 3, ξ) = b + b 2 9 = G(x, ξ) = b(x 2 9) So now we re down to 2 unknowns,, nd b. We hve 2 more conditions. Continuity of G t x = ξ mens nd the jump in the derivtive mens tht: nd fter some work we find: (ξ 2 ) = b(ξ 2 9) 2bξ 2ξ = = ξ2 9 6ξ ; b = ξ2 6ξ. So to conclude, the Green s function for this ODE with this type of BC is: G(x, ξ) = { ξ 2 9 6ξ (x2 ), if x < ξ ξ 2 6ξ (x2 9), if ξ < x 3 6. Since here G(x, ξ) hs the mening of the electric potentil t point x creted by chrge plced t ξ, it is resonble tht it is non-zero on both sides of the chrge. 7. Using the Green s function, we cn now immeditely get the prticulr solution: u p (x) = dξf(ξ)g(x, ξ) = x In the first integrl x > ξ, in the second x < ξ, so we hve: u p (x) = x where we re given f(ξ) = 6ξ. With this: u p (x) = (x 2 9) dξf(ξ) ξ2 6ξ (x2 9) + x dξf(ξ)g(x, ξ) + dξ(ξ 2 ) + (x 2 ) [ x = (x 2 3 ] [ 27 x 9) (x ) + (x 2 3 ) 3 3 6 x x dξf(ξ)g(x, ξ) dξf(ξ) ξ2 9 6ξ (x2 ) x ] 9(3 x) dξ(ξ 2 9) =... = 6 3 x3 + 2 52 3 x2
fter simplifying. This looks resonble guessing would suggest something proportionl to x 3 for u p, nd the lst two terms re prt of the homogenenous solution (whtever is needed to mke sure u p sttisfies homogeneous BC). To get the full solution u(x) = C + C 2 x 2 + u p (x) we use the BC for u: u() =, u(3) = 9. Now you see how nice it is tht we sked tht u p () =, u p (3) = (nd therefore tht G goes to zero t both ends) becuse we don t hve to bother with it. All we hve left is C + C 2 =, C + 9C 2 = 9 C =, C 2 =, nd u(x) = x 2 + u p (x) = 6 3 x3 + 2 49 3 x2 for the u p we found bove. Of course, for this simple function, you could hve guess u p quite esily (try!)... but the dvntge here is tht you cn use this sme G for ny other f(x) in this eqution, s long s the BC sty of the sme type (specify u t both ends). Extr: In this cse, we should choose x=3 = t the right boundry. Since here G(x, ξ) = b + b 2 x 2 2b 2 3 = b 2 = which mkes the clcultion even simpler thn before. The other 3 equtions re the sme, but of course the overll solution will chnge. 7