Physics 116C Solution of inhomogeneous differential equations using Green functions

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1 Physics 6C Solution of inhomogeneous differential equations using Green functions Peter Young November 8, 23 Introduction In this handout we give an introduction to Green function techniques for solving inhomogeneous differential equations. We will concentrate on the simpler case of ordinary differential equations, but will also discuss partial differential equations briefly at the end. For this year, 23, I will not require Secs. 6 and 7 for the exams of homework, but I recommend that you read them for your general interest. 2 Inhomogeneous Equations A homogeneous linear ordinary differential equation can be written Ly(x) =, () where L is an operator involving derivatives with respect to x. Green functions, the topic of this handout, appear when we consider the inhomogeneous equation analogous to Eq. () Ly(x) = f(x), (2) where f(x) is some specified function of x. We require the solution for some specified boundary conditions. The idea of the method is to determine the Green function, G(x,x ), which is given by the solution of this equation with a particular right hand side, namely a delta function, i.e. LG(x,x ) = δ(x x ). (3) In Eq. (3), the differential operators in L act on x, and x is a constant. Once G has been determined, the solution of Eq. (2) can be obtained for any function f(x) from y(x) = G(x,x )f(x )dx, (4) which follows since Ly(x) = L G(x,x )f(x )dx = δ(x x )f(x )dx = f(x), (5) so y(x) satisfies Eq, (2) as required. Note that we used Eq. (3) to obtain the second equality in Eq. (5). We emphasize that the same Green function applies for any f(x), and so it only has to be calculated once for a given differential operator L and boundary conditions.

2 3 Eigenvalues of ODEs We have studied, especially in a long HW problem, second order linear ordinary differential equations which can be written as an eigenvalue problem of the form Ly n (x) = λ n y n (x), (6) where L is the same operator involving derivatives as in Eqs. () and (2), λ n is an eigenvalue and y n (x) is an eigenfunction (which satisfies some specified boundary conditions). Note that Eqs. (6) and (2) have the same left hand side but different right hand sides. In Eq. (2) the right hand side is a specified function of x, whereas in Eq. (6) the right hand side is proportional to the function being determined y(x). The general case that we are interested in is called a Sturm-Liouville problem, for which one can show that the eigenvalues are real, and the eigenfunctions are orthogonal, i.e. b a y n (x)y m (x)dx = δ nm, (7) where a and b are the upper and lower limits of the region where we are solving the problem, and we have also normalized the solutions. A simple example, which we will study in detail, will be 2 y + 4y = λy, (8) in the interval x π, with the boundary conditions y() = y(π) =. This corresponds to L = d2 dx (9) This is just the simple harmonic oscillator equation, and so the solutions are coskx and sinkx. The boundary condition y() = eliminates coskx and the condition y(π) = gives k = n a positive integer. (Note: For n = the solution vanishes and taking n < just gives the same solution as that for the corresponding positive value of n because sin( nx) = sin(nx). Hence we only need consider positive integer n.) The normalized eigenfunctions are therefore 2 y n (x) = sinnx, (n =,2,3, ), () π and the eigenvalues in Eq. (8) are λ n = 4 n2, () since the equation satisfied by y n (x) is y n +n 2 y n =. 4 Expression for the Green functions in terms of eigenfunctions In this section we will obtain an expression for the Green function in terms of the eigenfunctions y n (x) in Eq. (6). The general Sturm-Liouville problem has a weight function w(x) multiplying the eigenvalue on the RHS of Eq. (6) and the same weight function multiplies the integrand shown in the LHS of the orthogonality and normalization condition, Eq. (7). Furthermore the eigenfunctions may be complex, in which case one must take the complex conjugate of either y n or y m in in Eq. (7). Here, to keep the notation simple, we will just consider examples with w(x) = and real eigenfunctions. 2 Different books adopt different sign conventions for the definition of L, and hence of the Green functions. 2

3 We assume that the solution y(x) of the inhomogeneous equation, Eq. (2), can be written as a linear combination of the eigenfunctions y n (x), obtained with the same boundary conditions, i.e. y(x) = n c n y n (x), (2) for some choice of the constants c n. Substituting into Eq. (2) gives f(x) = Ly(x) = n c n Ly n (x) = n c n λ n y n (x). (3) To determine the c n we multiply by one of the eigenfunctions, y m (x) say, and integrate use the orthogonality of the eigenfunctions, Eq. (7). This gives b a f(x)y m (x)dx = n Substituting for c n into Eq. (2) gives y(x) = n λ n b c n λ n y n (x)y m (x)dx = c m λ m. (4) b which can be written in the form of Eq. (4) with a G(x,x ) = n a y n (x )f(x )dx y n (x), (5) λ n y n (x)y n (x ). (6) Note that G(x,x ) is a symmetric function of x and x which is a quite general result. Furthermore, it only depends on the eigenfunctions of the corresponding homogeneous equation, i.e. on the boundary conditions and L. It is independent of f(x) and so can be computed once and for all, and then applied to any f(x) just by doing the integral in Eq. (4). 5 Other boundary conditions We have chosen our Green function to have the boundary conditions of the desired solution. Suppose it was easier to find the Green function for some other boundary condition. How can we incorporate this solution into a Green function for the actual boundary condition? We have a Green function G, say, which satisfies boundary condition. This satisfies the equation LG (x,x ) = δ(x x ). (7) We also have a Green function G 2 for boundary condition 2 which satisfies the same equation, Subtracting these last two equations we get LG 2 (x,x ) = δ(x x ). (8) L [ G (x,x ) G 2 (x,x ) ] =, (9) so the difference, G G 2, satisfies the corresponding homogeneous equation. In other words the general expression for the Green function is G(x,x ) = G p (x,x )+F(x,x ), (2) where G p (x,x ) is a particular solution of the inhomogeneous equation, Eq. (3), while F(x,x ) is a solution of corresponding homogeneous equation chosen so that G satisfies the desired boundary conditions. In fact, you have already know that the general solution of an inhomogeneous differential equation is the sum of a particular solution of the inhomogeneous equation, plus a solution of the homogeneous equation chosen to satisfy the boundary conditions. Equation (2) is an example of this. 3

4 6 A simple example As an example of the use of Green functions let us determine the solution of the inhomogeneous equation corresponding to the homogeneous equation in Eq. (8), i.e. y + 4y = f(x), with y() = y(π) =. (2) First we will evaluate the solution by elementary means for two choices of f(x) (i) f(x) = sin2x, (ii) f(x) = x/2. (22) We will then obtain the solutions for these cases from the Green function determined according to Eq. (6). In the elementary approach, one writes the solution of Eq. (2) as a combination of a complementary function y c (x) (the solution with f(x) = ) and the particular integral y p (x) (a particular solution with f(x) included). Since, for f(x) =, the equation is the simple harmonic oscillator equation, y c (x) is given by y c (x) = Acos(x/2)+Bsin(x/2). (23) For the particular integral, we assume that y p (x) is of a similar form to f(x). We now determine the solution for the two choices of f(x) in Eq. (2). (i) For f(x) = sin2x we try y p (x) = Ccos2x+Dsin2x and substituting gives ( 4+/4)D =, and C =. This gives y(x) = y c (x)+y p (x) = 4 sin2x+acos(x/2)+bsin(x/2). (24) 5 The boundary conditions are y() = y(π) =, which gives A = B =. Hence y(x) = 4 sin2x, for f(x) = sin2x. (25) 5 (ii) Similarly for f(x) = x we try y p (x) = C +Dx and substituting into Eq. (2) gives C =,D = 2, and so y(x) = y c (x)+y p (x) = 2x+Acos(x/2)+Bsin(x/2). (26) The boundary conditions, y() = y(π) =, give A =,B = 2π. Hence This is plotted in the figure below y(x) = 2x 2πsin(x/2), for f(x) = x/2. (27) 4

5 Now lets work out the Green function, which is given by Eq. (6). The eigenfunction are given by () and the eigenvalues are given by Eq. () so we have G(x,x ) = 2 π sinnx sinnx n=. (28) 4 n2 We can now substitute this into Eq. (4) to solve Eq. (2) for the two choices of f(x) in Eq. (22). (i) First of all for f(x) = sin2x Eq. (4) becomes y(x) = 2 ( π ) sinnx sinnx sin2x dx = 2 π 4 n2 π n= n= sinnx 4 n2 π sinnx sin2x dx. (29) Because of the orthogonality of the sinnx in the interval from to π only the n = 2 term contributes, and the integral for this case is π/2. Hence the solution is in agreement with Eq. (25). y(x) = sin2x 4 22 = 4 5 sin2x. (3) (ii) Now we consider the case of f(x) = x/2. Substituting Eq, (28) into Eq. (4) gives y(x) = π n= sinnx 4 n2 π x sinnx dx. (3) There is no longer any orthogonality to simplify things and we just have to do the integral: π x sinnx dx = [ x cosnx ] π π cosnx + dx (32) n n = πcosnπ [ ] sinnx π + n n 2 (33) = ( ) nπ n. (34) 5

6 For the particular case of n = the integral is zero. Hence the solution is y(x) = ( ) n+ sinnx n( 4 n2 ). (35) n= This may not look the same as Eq. (27), but it is in fact, as one can check by expanding 2x 2πsinx/2 as a Fourier sine series in the interval to π, i.e. one writes 2x 2πsinx/2 = c n sinnx, (36) n= and determines the c n from the usual Fourier integral. Using the Green function may seem to be a complicated way to proceed, especially for our second choice of f(x). However, you should realize that the elementary derivation of the solution may not be so simple in other cases, and you should note that the Green function applies to all possible choices of the function on the RHS, f(x). Furthermore, we will see in the next section that one can often get a closed form expression for G, rather than an infinite series. It is then much easier to find a closed form expression for the solution. 7 Closed form expression for the Green function In many useful cases, one can obtain a closed form expression for the Green function by starting with the defining equation, Eq. (3). We will illustrate this for the example in the previous section for which Eq. (3) is G + 4 G = δ(x x ). (37) Remember that x is fixed (and lies between and π) while x is a variable, and the derivatives are with respect to x. We solve this equation separately in the two regions (i) x < x, and (ii) x < x π. In each region separately the equation is G +(/4)G =, for which the solutions are G(x,x ) = Acos(x/2)+Bsin(x/2), (38) where A and B will depend on x. Since y() =, we require G(,x ) = (recall Eq. (4)) and so, for the solution in the region x < x, the cosine is eliminated. Similarly G(π,x ) = and so, for the region x < x π, the sine is eliminated. Hence the solution is G(x,x ) = { Bsin(x/2) ( x < x ), Acos(x/2) (x < x π). (39) How do we determine the two coefficients A and B? We get one relation between them by requiring that the solution is continuous at x = x, i.e. the limit as x x from below is equal to the limit as x x from above. This gives Bsin(x /2) = Acos(x /2). (4) The second relation between A and B is obtained by integrating Eq. (37) from x = x ǫ to x +ǫ, and taking the limit ǫ, which gives so lim ǫ [ ] dg x +ǫ dx x ǫ + 4 lim ǫ lim ǫ ( dg dx x +ǫ x ǫ x +ǫ G(x,x )dx = lim dg dx x ǫ ǫ x +ǫ x ǫ δ(x x )dx (4) ) + =. (42) 6

7 Hence dg/dx has a discontinuity of at x = x, i.e. A 2 sin(x /2) B 2 cos(x /2) =. (43) Solving Eqs. (4) and (43) gives Substituting into Eq. (39) gives B = 2cos(x /2), (44) A = 2sin(x /2). (45) G(x,x ) = { 2cos(x /2)sin(x/2) ( x < x ), 2sin(x /2)cos(x/2) (x < x π). (46) A sketch of the solution is shown in the figure below. The discontinuity in slope at x = x (I took x = 3π/4) is clearly seen. It is instructive to rewrite Eq. (46) in terms of x <, the smaller of x and x, and x >, the larger of x and x. One has G(x,x ) = 2sin(x < /2)cos(x > /2), (47) irrespective of which is larger, which shows that G is symmetric under interchange of x and x as noted earlier. We now apply the closed form expression for G in Eq. (46) to solve our simple example, Eq. (2), with the two choices for f(x) shown in Eq. (22). (i) For f(x) = sin2x, Eqs. (4) and (46) give y(x) = 2cos(x/2) x sin2x sin(x /2)dx 2sin(x/2) Using formulae for sines and cosines of sums of angles and integrating gives ( sin(3x/2) y(x) = 2cos(x/2) sin(5x/2) ) ( cos(3x/2) 2sin(x/2) ( ) ( ) = 2 3 sin2x +2 5 sin2x π x sin2x cos(x /2)dx. (48) + cos(5x/2) ) (49) 5 (5) = 4 5 sin2x, (5) 7

8 where we again used formulae for sums and differences of angles. This result is in agreement with Eq. (25). (ii) For f(x) = x/2, Eqs. (4) and (46) give y(x) = cos(x/2) Integrating by parts gives x x sin(x /2)dx sin(x/2) π x x cos(x /2)dx (52) y(x) = cos(x/2)( 2xcos(x/2)+4sin(x/2)) sin(x/2)(2π 4cos(x/2) 2xsin(x/2)), (53) = 2x 2πsin(x/2), which agrees with Eq. (27). Note that we obtained the closed form result explicitly, as opposed to the method in Sec. 4 where the solution was obtained as an infinite series, Eq. (35). In general, to find a closed form expression for G(x,x ) we note from Eq. (3) that, for x x, it satisfies the homogeneous equation (54) LG(x,x ) =, (x x ). (55) We solve this equation separately for x < x and x > x, subject to the required boundary conditions, and match the solutions at x = x with the following two conditions (i) G(x,x ) is continuous at x = x, i.e. (ii) The derivative dg/dx has a discontinuity of at x = x, i.e. 8 Summary and PDE s lim x x +ǫ G(x,x ) = lim x x ǫ G(x,x ). (56) dg(x,x ) dg(x,x ) lim lim =. (57) x x +ǫ dx x x ǫ dx We have shown how to solve linear, inhomogeneous, ordinary differential equations by using Green functions. These can be represented in terms of eigenfunctions, see Sec. 4, and in many cases can alternatively be evaluated in closed form, see Secs. 6 and 7. The advantage of the Green function approach is that the Green function only needs to be computed once for a given differential operator L and boundary conditions, and this result can then be used to solve for any function f(x) on the RHS of Eq. (2) by using Eq. (4). So far we have used Green function techniques for ordinary differential equations. They can also be used, in a very similar manner, for partial differential equations. In particular, if we consider Poisson s equation, 2 y( r) = f( r), (58) the solution can be written as where the Green function satisfies y( r) = d r G( r, r )f( r ), (59) 2 G( r, r ) = δ (3)( r r ), (6) 8

9 in which the derivatives are with respect to the components of r, and the three-dimensional delta function is a product of three Dirac delta functions The Green function for this problem is given by δ (3)( r r ) = δ(x x )δ(y y )δ(z z ). (6) G( r, r ) = 4π r r +F( r, r ), (62) where F( r, r ) is a solution of Laplace s equation chosen so satisfy the boundary conditions, see e.g. Boas 3.8. To obtain in the /4π r r term in Eq. (62) one can argue physically that Eq. (6) is the equation for the potential at r due to a point charge at r and we know that this is given by Coulomb s law. From Eq. (62) it follows that the solution of Poisson s equation, Eq. (58) is y( r) = d r f( r ) 4π r r +H( r), (63) where H( r) is a solution of Laplace s equation chosen to satisfy the boundary conditions. While the advantages of Green functions may not be readily apparent from the simple examples presented here, they are used in many advanced applications in physics. 9