MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 17 doc

TIME-INDEPENDENT GREENS FUNCTIONS 593 with the initial conditions z(0) = k(0) = 0, C1 and C2 in Equation (19.147) is zero, hence the solution will be written as where we have defined (19.156) One can easily check that s(t) satisfies the differential equation d'z(t) dz(t) 2 dt2 + 2E- + woz(t) = &-at. dt (19.157) For weak damping the solution reduces to As expected, in the t + 00 limit this becomes FO sin [wot - 71 z(t) = - wo JW' 19.1.12 Green's Function for the Helmholtz Equation in Three Dimensions The Helmholtz equation in three dimensions is given as We now look for a Green's function satisfying + ko G(?, F') = S(7 - 7'). ("' 2, (19.159) (19.160) We multiply the first equation by G(?,7') and the second by $(7) and then subtract, and integrate the result over the volume V to get - ///v F(?)G(7, 7')d3?;t. (19.161) 594 GREEN’S FUNCTIONS Using Green’s theorem JJJ, (FV~G - GV~F) d3F = JJ (PVG - GVF) . iids, (19.162) S where S is a closed surface enclosing the volume V with the outward unit normal ii, we obtain Interchanging the primed and the unprimed variables and assuming that the Green’s function is symmetric in anticipation of the corresponding boundary conditions to be imposed later, we obtain the following remarkable formula: Boundary conditions: The most frequently used boundary conditions are: i) Dirichlet boundary conditions, where G is zero on the boundary. ii) Neumann boundary conditions, where the normal gradient of G on the surface is zero: TG.61 =O. boundary iii) General boundary conditions: = 0, boundary TG + ??(?“)GI where T(F’) is a function of the boundary point 7‘. surface term in the above equation vanishes, thus giving For any one of these cases, the Green’s function is symmetric and the (19.165) 19.1.13 Green’s Functions in Three Dimensions with a Discrete Spectrum Consider the inhomogeneous equation H@(?;’) = F(?;‘), (19.166) TIME-INDEPENDENT GREEN 5 FUNCTIONS 595 where H is a linear differential operator. H has a complete set of orthonormal eigenfunctions, {$A(?)), which are determined by the eigenvalue equation where X stands for the eigenvalues and the eigenfunctioris satisfy the homoge- neous boundary conditions given in the previous section. We need a Green's function satisfying the equation HG(7, ?') = S(? - 7'). (19.167) Expanding @(?) and F(?) in terms of this complete set of eigenfunctions we write (19.168) @(-i i> = Ex axdx(?;') F(?) = Ex cx$x(?") where the expansion coefficients are (19.169) ax = JJj", 43?)@(?)d3? cx = JJJ, $1(?)F(?)d3? (19.170) ax = Using this ax and the explicit form of cx in Elpation (19.169) we can write Q(7) as CA X Substituting these into Equation (19.166) we obtain This gives the Green's function as (19.172) This Green's function can easily be generalized to the equation (H -A,) @(?) = F(?), (19.173) for the operator (H - A,) as (19.174) 596 GREENS FUNCTlONS We now find the Green’s function for the three-dimensional Helmholtz equation (“2 + k;) $(?) = F(?) in a rectangular region bounded by six planes: x=O, z=a y=O, y=b z=o, z=c (19.175) i and with the homogeneous Dirichlet boundary conditions. The corresponding eigenvalue equation is T241mn(T) + k12,,4lmn(T) = 0- The normalized eigenfunctions are easily obtained as where the eigenvalues are L2x2 m2x2 n2r2 klmn = - + - + - a2 b2 c2 ’ (1, m, n = positive integer). (19.177) Using these eigenfunctions [Eq. (19.176)] we can now write the Green’s func- tion as (19.178) 19.1.14 Green’s function for the Laplace operator V2 satisfies Green’s Function for the Laplace Operator Inside a Sphere q2G( 7, ?’) = S( ?, ?’). (19.179) Using spherical polar coordinates this can be written as T2G(?, 7’) = ~ S(r - r’) S(cm e - cos8’)6(4 - 4’) (19.180) r ’2 TIME-INDEPENDENT GREEN’S FUNCTIONS 597 where we have used the completeness relation of the spherical harmonics. For the Green’s function inside a sphere, we use the boundary conditions G(0,T’) = finite, (19.182) G(u, 7’) = 0. (19.183) In spherical polar coordinates we can separate the angular part and write the Green’s function as (19.184) 1=0 m=-l We now substitute Equation (19.184) into Equation (19.181) to find the dif- ferential equation that gl(r, r’) satisfies as 1 gl(r,r’) = -6(r - r’). r ‘2 1 d2 1(1+ 1) - - [Tgl (T, ?-’)I - 7 r dr2 A general solution of the homogeneous equation can be obtained by trying a solution as ~r‘ + clr-(l+l). (19.185) (19.186) (19.187) We can now construct the radial part of the Green’s function for the inside of a sphere by finding the appropriate u and the u solutions as r < r‘, r >r’. (19.188) Now the complete Green’s function can be written into Equation (19.184). by substituting this result 19.1.15 Green’s Functions for the Helmholtz Equation for All Space-Poisson and Schrdinger Equations We now consider the operator Ho=q2+X (19.189) in the continuum limit. Using this operator we can write the following differ- ential equation: Ho@(?;’) = F’(?). (19.190) 598 GREENS FUNCTIONS + * + Let us assume that the Fourier transforms @( k ) and F( k ) of @(?) and F(F) exists: (19.191) (19.192) Taking the Fourier transform of Equation (19,190) we get Using the Green's theorem [Eq. Equation (19.193) as (19.162)] we can write the first term in (19.194) where S is a surface with an outward unit normal 2 enclosing the volume V. We now take our region of integration as a sphere of radius R and consider the limit R -+ 00. In this limit the surface term becomes where 6 = ^er and dR = sin0dOd4. If the function *I(?) goes to zero suffi- ciently rapidly as 14 -+ 00, that is, when @(?) goes to zero faster than :, then the surface term vanishes, thus leaving us with in Equation (19.194). Consequently, Equation (19.193) becomes "i F( T) *(k)= (42 + A)' (19.196) (19.197) In this equation we have to treat the cases X > 0 and X 5 0 separately. TIME-INDEPENDENT GREEN’S FUNCTIONS 5% Casel: XSO: In this case we can write X = -K~; thus the denominator (k2 + K~) in -4 F(2) *(k)= k2 + K2 (19.198) never vanishes. Taking the inverse Fourier transform of this, we write the general solution of Equation (19.190) as where [(?) denotes the solution of the homogeneous equation Hot(?) = (T2 - K~) [(T) = 0. (19.200) Defining a Green’s function G(?;‘,?:”) as we can express the general solution of Equation (19.190) as @(?) = [(F) + /// G(?,7:”)F(?:”)d3?, (19.202) The integral in the Green’s function can be evaluated by using complex contour integral techniques. Taking the k vector as V + we write (19.203) (19.204) where d3x = k2 sin BdkdBd4. We can take the #I and 0 integrals imme diately, thus obtaining (19.205) 600 GREEN’S FUNCTIONS Using Jordan’s lemma (Section 13.7) we can show that the integral over the circle in the upper half complex k-plane goes to zero as the radius goes to infinity; thus we obtain I as keikI-‘-+ 2~2 residues of { k2 + K2 ’ } (19.206) 2T I=. 217-?;tq k>O Using this in (19.201) we obtain the Green’s function as (19.207) (19.208) (19.209) To complete the solution [Eq. (19.202)] we also need [(F), which is easily obtained as 7 (19.210) [(T) = Coe*~lXef?2Ve*~3~ IE2 = K4 + 6; + K;. Because this solution diverges for Irl + co, for a finite solution every- where we set CO = 0 and write the general solution as In this solution if F(7’) goes to zero sufficiently rapidly as lr’l -+ (x, or if F(?) is zero beyond some lr’l = ro, we see that for large r, q(7) decreases exponentially as (19.212) This is consistent with the neglect of the surface term in our derivation in Equation (19.195). ePnr r @(?) f c Example 19.8. Green’s function for the Poisson equation: Using the above Green’s function [Eq. (19.209)] with K = 0, we can now convert the Poisson equation V%$(T+) = -4Tp(T+) (19.213) into an integral equation. In this case X = 0; thus the solution is given as 4(7) = -4~/// G(7, ?;f’)p(?’)d3F’, (19.214) V TIME-INDEPENDENT GREEN'S FUNCTIONS 601 where (19.215) Example 19.9. Green's function for the Schrodinger equation-E < 0: Another application for Green's function [Eq. independent Schrodinger equation: (19.209)] is the time- For central potentials and bound states (E < 0) K~ is given as 2m PI K2 = -__ fi2 . Thus the solution of Equation (19.216) can be written as (19.216) (19.217) This is also the integral equation version of the time-independent Shrdinger equation for bound states. CaseII: X>O: In this case the denominator in the definition [Eq. (19.197)] of $(c) has zeroes at k = kfi. To eliminate this problem we add a small imaginary piece ZE to the X values as X=(q*i&), E>O. (19.219) Substituting this in Equation (19.197) we get (19.220) which is now well defined everywhere on the real k-axis. Taking the inverse Fourier transform of this we get We can now evaluate this integral in the complex k-plane using the complex contour integral theorems and take the limit as E + 0 to obtain the final result. Because our integrand has poles at k = (q f ZE), 602 GREEN5 FUNCTlONS we use the Cauchy integral theorem. However, as before, we first take the 8 and q3 integrals to write 1’ - (k - q F is) (k + q f ZE) (19.222) For the first integral we close the contour in the upper half complex k-plane and get 03 ?‘I I. (19.223) lcdk - - 7rie+iq] r’- -+-+ s, (k - q3= iE) (k + q&iiE) Similarly, for the second integral we close our contour in the lower half complex k-plane to get (19.224) Combining these we obtain the Green’s function and the solution as The choice of the f sign is very important. In the limit as /?“I -+ 00 this solution behaves as or e- iqr K(*)+E(?’f)+C-, (19.228) where C is a constant independent of r, but it could depend on 6 and 4. The f signs physically correspond to incoming and outgoing waves. We now look at the solutions of the homogeneous equation: (“2 + 9’) t(7) = 0, (19.229) [...]... function satisfying the singlepoint boundary conditions Y ( ~ o = YO and Y’(ZO) = Y& ) Hint: First write the Green’s function as G(z, z’) = Ayl (z) + Bya(z), z > x’, where yl(z) and y2(2) are two linearly independent solutions of Ly(z) = 0 Because the Green’s function is continuous a t 2 = Z’ and its derivative has a discontinuity of magnitude l / p ( z ) at 1c = J:’, find the constants A , B , C, and I),... operator independent of T We again assume that H has a complete set of orthonormal eigenfunctions satisfying H+n(7) = An& (*) 7 (19.314) where A are the eigenvalues We expand *(7,7) terms of this complete , in and orthonormal set as *(T+,T) = CAn(T)+nF), (19.315) n where the coefficients A,(T) carry the Equation (19.313) we obtain 7 dependence Substituting this in (19.316) Because 4 are linearly independent,... G2(7, 7’) and G2(?, 7;’’,~, (19.354) 7’,7, both 7’) satisfy (19.355) Guided by our experience in G I , to find the Green’s function we start by introducing a discontinuity in either G2 or G2 as ~ ~ T( ’ 7 , , G,(T, + ,~ = Gc stands for G2 or G2, ?;fl,T,T’)e (7- 7’) (19.356) while the subscript R will be explained later Oper- 7 ’ 7’) ating on GR(7,, r , with (19.357) 622 GREEN’S FUNCTIONS Since Gz(?”,... 19.3 sin knx sin knx’ k: n Show that the Green’s function for the differential operator with the boundary conditions y(0) = 0 and y(L) = 0 is given as G(x, = d) 1 ko sin koL { sin h x s i n h ( x ’ - L ) x < 2’ sin b x ’ sin ~co(x L ) x - > x’ Show that this is equivalent to the eigenvalue expansion 2 C O0 G ( z , d )= L n=l n7rx n7rx‘ sin sin ~ ~ L g -(n7r/L)2 ‘ PROBLEMS 627 19.4 Single-point boundary... and then t o 72 as (from now on we use j d 3 T instead of sljd3’?;’ ) > TI 610 GREEN'S FUNCTlONS (19.271) Using the definition o propagators [Eq (19.268)] we can also write this as f (19.272) Using the orthogonality relation Equation (19.272) becomes (19.273) (19.274) Using this in Equation (19.271) we obtain the propagator, G1(7'),7')", T ~ , T o ) , that takes us from TO to 7 2 in a single step in. .. obtaining the Green’s function as G(z,z’) = CYI (z) + Dy2(~), J: < z’, where W[yl(z),92(z)]is the Wronskian defined as W[91,y2] = ylya - y2y/i Now impose the single-point boundary conditions G(z0,d) 0 and G’(z0,z’) = 0 = to show that C = D = 0 Finally show that the differential equation () YZ =4 b ) with the singlepoint boundary conditions ~(zo) 90 and ~ ’ ( z o ) = y is = ; equivalent t o the integral... limit, L -+00, the difference between two neighboring eigenvalues becomes infinitesimally small; thus we may replace the summation with an integral as (19.284) 612 GREENS FUNCTlONS This gives us the propagator as Completing the square: Zk(z- 2 ’ )- Ic2T = -7 (19.286) and defining we can write GI(2, z’, r ) as (19.287) This integral can be taken easily, thus giving us the propagator as 1 r-r‘)’ G l ( z ,... function is now given as (19.383) + In this solution A stands for advanced times: that is, t’ = t 1- 7 1In 7 ’ /c other words, whatever “happens” a t the source point shows its effect a t the field point before its happening by the amount of time 1- 71which is 7 ’, /c again equal t o the amount of time that light takes t o travel from the source to the field point In summary, in advanced solutions effects... Am are the eigenvalues and dm are the eigenfunctions We now write the solution of Equation (19.243) as @ ( ~ ,= x A m ( r ) 4 m ( T ) , r) (19.249) m where the time dependence is carried in the expansion coefficients Am(r) Operating on @(T,T) H and remembering that H is independent of 7, with we obtain m I m Using Equation (19.250) and the time derivative of Equation (19.249) in Equation (19.243) we... from the and @(?,O) are given, we write initial conditions Assuming that e(?’,O) and @(7,0) Z C J X n [ a n - b n ] & ( T ) = n (19.321) TIME-DEPENDEN T GREEN’S FUNCTIONS Using the orthogonality relation [Eq (19.256)] of &(?) tions between a, and b, as [a, + b,] = 1 617 we obtain two rela- (b3;i;)’)@(?;f’,0)d3?;f’ (19.322) and [a, - b,] = - /-2 b p + ’ ) i ~ ( ? ’ , O ) d ~ ? ’ ( 6 (19.323) Solving these . surface enclosing the volume V with the outward unit normal ii, we obtain Interchanging the primed and the unprimed variables and assuming that the Green’s function is symmetric in anticipation. (19.216) (19. 217) This is also the integral equation version of the time-independent Shrdinger equation for bound states. CaseII: X>O: In this case the denominator in the definition [Eq Am(r). Operating on @(T,T) with H and remembering that H is independent of 7, we obtain m I m m Using Equation (19.250) and the time derivative of Equation (19.249) in Equa- tion