7 Green s Functions and Nonhomogeneous Problems

Transcription

1 7 Green s Functions nd Nonhomogeneous Problems The young theoreticl physicists of genertion or two erlier subscribed to the belief tht: If you hven t done something importnt by ge 3, you never will. Obviously, they were unfmilir with the history of George Green, the miller of Nottinghm. Julin Schwinger ( The wve eqution, het eqution, nd plce s eqution re typicl homogeneous prtil differentil equtions. They cn be written in the form u(x, where is differentil opertor. For exmple, these equtions cn be written s ( 2 t 2 c2 2 u, ( t k 2 u, 2 u. (7.1 In this chpter we will explore solutions of nonhomogeneous prtil differentil equtions, u(x f (x, by seeking out the so-clled Green s function. The history of the Green s function dtes bck to 1828, when George Green published work in which he sought solutions of Poisson s eqution 2 u f for the electric potentil u defined inside bounded volume with specified boundry conditions on the surfce of the volume. He introduced function now identified s wht Riemnn lter coined the Green s function. In this chpter we will derive the initil vlue Green s function for ordinry differentil equtions. ter in the chpter we will return to boundry vlue Green s functions nd Green s functions for prtil differentil equtions. As simple exmple, consider Poisson s eqution, George Green ( , British mthemticl physicist who hd little forml eduction nd worked s miller nd bker, published An Essy on the Appliction of Mthemticl Anlysis to the Theories of Electricity nd Mgnetism in which he not only introduced wht is now known s Green s function, but he lso introduced potentil theory nd Green s Theorem in his studies of electricity nd mgnetism. Recently his pper ws posted t rxiv.org, rxiv: u(r f (r.

2 226 prtil differentil equtions Ω Ω Figure 7.1: et Poisson s eqution hold inside region Ω bounded by surfce Ω. The Dirc delt function stisfies ˆn δ(r, r, δ(r dv 1. Ω 1 We note tht in the following the volume nd surfce integrls nd differentition using re performed using the r-coordintes. 2 In mny pplictions there is symmetry, G(r, r G(r, r. Then, the result cn be written s u(r G(r, r f (r dv. Ω et Poisson s eqution hold inside region Ω bounded by the surfce Ω s shown in Figure 7.1. This is the nonhomogeneous form of plce s eqution. The nonhomogeneous term, f (r, could represent het source in stedy-stte problem or chrge distribution (source in n electrosttic problem. Now think of the source s point source in which we re interested in the response of the system to this point source. If the point source is locted t point r, then the response to the point source could be felt t points r. We will cll this response G(r, r. The response function would stisfy point source eqution of the form 2 G(r, r δ(r r. Here δ(r r is the Dirc delt function, which we will consider in more detil in Section 9.4. A key property of this generlized function is the sifting property, δ(r r f (r dv f (r. Ω The connection between the Green s function nd the solution to Poisson s eqution cn be found from Green s second identity: φ ψ ψ φ] n ds φ 2 ψ ψ 2 φ] dv. Ω etting φ u(r nd ψ G(r, r, we hve 1 Ω Ω Ω u(r G(r, r G(r, r u(r] n ds ] u(r 2 G(r, r G(r, r 2 u(r dv u(rδ(r r G(r, r f (r ] dv Ω u(r G(r, r f (r dv. (7.2 Ω Solving for u(r, we hve u(r G(r, r f (r dv Ω + u(r G(r, r G(r, r u(r] n ds. (7.3 Ω If both u(r nd G(r, r stisfied Dirichlet conditions, u on Ω, then the lst integrl vnishes nd we re left with 2 u(r Ω G(r, r f (r dv. So, if we know the Green s function, we cn solve the nonhomogeneous differentil eqution. In fct, we cn use the Green s function to solve nonhomogenous boundry vlue nd initil vlue problems. Tht is wht we will see develop in this chpter s we explore nonhomogeneous problems in more detil. We will begin with the serch for Green s functions for ordinry differentil equtions.

3 green s functions nd nonhomogeneous problems Initil Vlue Green s Functions In this section we will investigte the solution of initil vlue problems involving nonhomogeneous differentil equtions using Green s functions. Our gol is to solve the nonhomogeneous differentil eqution subject to the initil conditions (ty (t + b(ty (t + c(ty(t f (t, (7.4 y( y y ( v. Since we re interested in initil vlue problems, we will denote the independent vrible s time vrible, t. Eqution (7.4 cn be written compctly s where is the differentil opertor The solution is formlly given by y] f, (t d2 dt 2 + b(t d dt + c(t. y 1 f ]. The inverse of differentil opertor is n integrl opertor, which we seek to write in the form y(t G(t, τ f (τ dτ. The function G(t, τ is referred to s the kernel of the integrl opertor nd is clled the Green s function. In the lst section we solved nonhomogeneous equtions like (7.4 using the Method of Vrition of Prmeters. etting, G(t, τ is clled Green s function. y p (t c 1 (ty 1 (t + c 2 (ty 2 (t, (7.5 we found tht we hve to solve the system of equtions This system is esily solved to give c 1 (ty 1(t + c 2 (ty 2(t. c 1 (ty 1 (t + f (t c 2 (ty 2 (t q(t. (7.6 c 1 (t f (ty 2 (t (t y 1 (ty 2 (t y 1 (ty 2(t ] c 2 (t f (ty 1 (t (t y 1 (ty 2 (t y 1 (ty 2(t ]. (7.7

4 228 prtil differentil equtions We note tht the denomintor in these expressions involves the Wronskin of the solutions to the homogeneous problem, which is given by the determinnt y W(y 1, y 2 (t 1 (t y 2 (t y 1 (t y 2 (t. When y 1 (t nd y 2 (t re linerly independent, then the Wronskin is not zero nd we re gurnteed solution to the bove system. So, fter n integrtion, we find the prmeters s t c 1 (t c 2 (t t t t 1 f (τy 2 (τ (τw(τ dτ f (τy 1 (τ dτ, (7.8 (τw(τ where t nd t 1 re rbitrry constnts to be determined from the initil conditions. Therefore, the prticulr solution of (7.4 cn be written s t y p (t y 2 (t t 1 f (τy 1 (τ t (τw(τ dτ y 1(t t f (τy 2 (τ dτ. (7.9 (τw(τ We begin with the prticulr solution (7.9 of the nonhomogeneous differentil eqution (7.4. This cn be combined with the generl solution of the homogeneous problem to give the generl solution of the nonhomogeneous differentil eqution: t y p (t c 1 y 1 (t + c 2 y 2 (t + y 2 (t t 1 f (τy 1 (τ t (τw(τ dτ y 1(t t f (τy 2 (τ (τw(τ dτ. (7.1 However, n pproprite choice of t nd t 1 cn be found so tht we need not explicitly write out the solution to the homogeneous problem, c 1 y 1 (t + c 2 y 2 (t. However, setting up the solution in this form will llow us to use t nd t 1 to determine prticulr solutions which stisfies certin homogeneous conditions. In prticulr, we will show tht Eqution (7.1 cn be written in the form t y(t c 1 y 1 (t + c 2 y 2 (t + G(t, τ f (τ dτ, (7.11 where the function G(t, τ will be identified s the Green s function. The gol is to develop the Green s function technique to solve the initil vlue problem (ty (t + b(ty (t + c(ty(t f (t, y( y, y ( v. (7.12 We first note tht we cn solve this initil vlue problem by solving two seprte initil vlue problems. We ssume tht the solution of the homogeneous problem stisfies the originl initil conditions: (ty h (t + b(ty h (t + c(ty h(t, y h ( y, y h ( v. (7.13

5 green s functions nd nonhomogeneous problems 229 We then ssume tht the prticulr solution stisfies the problem (ty p(t + b(ty p(t + c(ty p (t f (t, y p (, y p(. (7.14 Since the differentil eqution is liner, then we know tht y(t y h (t + y p (t is solution of the nonhomogeneous eqution. Also, this solution stisfies the initil conditions: y( y h ( + y p ( y + y, y ( y h ( + y p( v + v. Therefore, we need only focus on finding prticulr solution tht stisfies homogeneous initil conditions. This will be done by finding vlues for t nd t 1 in Eqution (7.9 which stisfy the homogeneous initil conditions, y p ( nd y p(. First, we consider y p (. We hve y p ( y 2 ( t 1 f (τy 1 (τ (τw(τ dτ y 1( t f (τy 2 (τ dτ. (7.15 (τw(τ Here, y 1 (t nd y 2 (t re tken to be ny solutions of the homogeneous differentil eqution. et s ssume tht y 1 ( nd y 2 (. Then, we hve f (τy y p ( y 2 ( 1 (τ dτ (7.16 t 1 (τw(τ We cn force y p ( if we set t 1. Now, we consider y p(. First we differentite the solution nd find tht y p(t y 2 (t t f (τy 1 (τ t (τw(τ dτ y 1 (t f (τy 2 (τ dτ, (7.17 (τw(τ since the contributions from differentiting the integrls will cncel. Evluting this result t t, we hve y p( y 1 ( Assuming tht y 1 (, we cn set t. Thus, we hve found tht t f (τy y p (x y 2 (t 1 (τ (τw(τ dτ y 1(t t ] y1 (τy 2 (t y 1 (ty 2 (τ (τw(τ t t f (τy 2 (τ dτ. (7.18 (τw(τ t f (τy 2 (τ (τw(τ dτ f (τ dτ. (7.19 This result is in the correct form nd we cn identify the temporl, or initil vlue, Green s function. So, the prticulr solution is given s y p (t t G(t, τ f (τ dτ, (7.2

6 23 prtil differentil equtions where the initil vlue Green s function is defined s We summrize G(t, τ y 1(τy 2 (t y 1 (ty 2 (τ. (τw(τ Solution of IVP Using the Green s Function The solution of the initil vlue problem, (ty (t + b(ty (t + c(ty(t f (t, y( y, y ( v, tkes the form where t y(t y h (t + G(t, τ f (τ dτ, (7.21 G(t, τ y 1(τy 2 (t y 1 (ty 2 (τ (τw(τ (7.22 is the Green s function nd y 1, y 2, y h re solutions of the homogeneous eqution stisfying y 1 (, y 2 (, y 1 (, y 2 (, y h( y, y h ( v. Exmple 7.1. Solve the forced oscilltor problem x + x 2 cos t, x( 4, x (. We first solve the homogeneous problem with nonhomogeneous initil conditions: x h + x h, x h ( 4, x h (. The solution is esily seen to be x h (t 4 cos t. Next, we construct the Green s function. We need two linerly independent solutions, y 1 (x, y 2 (x, to the homogeneous differentil eqution stisfying different homogeneous conditions, y 1 ( nd y 2 (. The simplest solutions re y 1 (t sin t nd y 2 (t cos t. The Wronskin is found s W(t y 1 (ty 2 (t y 1 (ty 2(t sin 2 t cos 2 t 1. Since (t 1 in this problem, we compute the Green s function, G(t, τ y 1(τy 2 (t y 1 (ty 2 (τ (τw(τ sin t cos τ sin τ cos t sin(t τ. (7.23 Note tht the Green s function depends on t τ. While this is useful in some contexts, we will use the expnded form when crrying out the integrtion. We cn now determine the prticulr solution of the nonhomogeneous differentil eqution. We hve x p (t t G(t, τ f (τ dτ

7 green s functions nd nonhomogeneous problems 231 t 2 sin t (sin t cos τ sin τ cos t (2 cos τ dτ t t cos 2 τdτ 2 cos t sin τ cos τdτ τ 2 sin t ] t ] 1 t sin 2τ 2 cos t 2 2 sin2 τ t sin t. (7.24 Therefore, the solution of the nonhomogeneous problem is the sum of the solution of the homogeneous problem nd this prticulr solution: x(t 4 cos t + t sin t. 7.2 Boundry Vlue Green s Functions We solved nonhomogeneous initil vlue problems in Section 7.1 using Green s function. In this section we will extend this method to the solution of nonhomogeneous boundry vlue problems using boundry vlue Green s function. Recll tht the gol is to solve the nonhomogeneous differentil eqution y] f, x b, where is differentil opertor nd y(x stisfies boundry conditions t x nd x b.. The solution is formlly given by y 1 f ]. The inverse of differentil opertor is n integrl opertor, which we seek to write in the form y(x b G(x, ξ f (ξ dξ. The function G(x, ξ is referred to s the kernel of the integrl opertor nd is clled the Green s function. We will consider boundry vlue problems in Sturm-iouville form, d dx ( p(x dy(x dx + q(xy(x f (x, < x < b, (7.25 with fixed vlues of y(x t the boundry, y( nd y(b. However, the generl theory works for other forms of homogeneous boundry conditions. We seek the Green s function by first solving the nonhomogeneous differentil eqution using the Method of Vrition of Prmeters. Recll this method from Section B.3.3. We ssume prticulr solution of the form y p (x c 1 (xy 1 (x + c 2 (xy 2 (x, which is formed from two linerly independent solution of the homogeneous problem, y i (x, i 1, 2. We hd found tht the coefficient functions stisfy the equtions c 1 (xy 1(x + c 2 (xy 2(x c 1 (xy 1 (x + c 2 (xy 2 (x f (x p(x. (7.26

8 232 prtil differentil equtions Solving this system, we obtin c 1 (x f y 2 pw(y 1, y 2, c 1 (x f y 1 pw(y 1, y 2, where W(y 1, y 2 y 1 y 2 y 1 y 2 is the Wronskin. Integrting these forms nd inserting the results bck into the prticulr solution, we find x y(x y 2 (x x 1 f (ξy 1 (ξ x p(ξw(ξ dξ y 1(x x f (ξy 2 (ξ p(ξw(ξ dξ, where x nd x 1 re to be determined using the boundry vlues. In prticulr, we will seek x nd x 1 so tht the solution to the boundry vlue problem cn be written s single integrl involving Green s function. Note tht we cn bsorb the solution to the homogeneous problem, y h (x, into the integrls with n pproprite choice of limits on the integrls. We now look to stisfy the conditions y( nd y(b. First we use solutions of the homogeneous differentil eqution tht stisfy y 1 (, y 2 (b nd y 1 (b, y 2 (. Evluting y(x t x, we hve y( y 2 ( y 2 ( x 1 x 1 f (ξy 1 (ξ p(ξw(ξ dξ y 1( x f (ξy 2 (ξ p(ξw(ξ dξ f (ξy 1 (ξ dξ. (7.27 p(ξw(ξ We cn stisfy the condition t x if we choose x 1. Similrly, t x b we find tht b y(b y 2 (b y 1 (b x 1 b f (ξy 1 (ξ p(ξw(ξ dξ y 1(b x b x f (ξy 2 (ξ p(ξw(ξ dξ f (ξy 2 (ξ dξ. (7.28 p(ξw(ξ The generl solution of the boundry vlue problem. This expression vnishes for x b. So, we hve found tht the solution tkes the form x y(x y 2 (x f (ξy 1 (ξ x p(ξw(ξ dξ y 1(x b f (ξy 2 (ξ dξ. (7.29 p(ξw(ξ This solution cn be written in compct form just like we hd done for the initil vlue problem in Section 7.1. We seek Green s function so tht the solution cn be written s single integrl. We cn move the functions of x under the integrl. Also, since < x < b, we cn flip the limits in the second integrl. This gives y(x x f (ξy 1 (ξy 2 (x p(ξw(ξ b dξ + x f (ξy 1 (xy 2 (ξ p(ξw(ξ dξ. (7.3 This result cn now be written in compct form:

9 green s functions nd nonhomogeneous problems 233 Boundry Vlue Green s Function The solution of the boundry vlue problem ( d p(x dy(x + q(xy(x f (x, dx dx < x < b, y(, y(b. (7.31 tkes the form y(x b G(x, ξ f (ξ dξ, (7.32 where the Green s function is the piecewise defined function G(x, ξ y 1 (ξy 2 (x pw, ξ x, y 1 (xy 2 (ξ pw, x ξ b, (7.33 where y 1 (x nd y 2 (x re solutions of the homogeneous problem stisfying y 1 (, y 2 (b nd y 1 (b, y 2 (. The Green s function stisfies severl properties, which we will explore further in the next section. For exmple, the Green s function stisfies the boundry conditions t x nd x b. Thus, G(, ξ y 1(y 2 (ξ pw, G(b, ξ y 1(ξy 2 (b pw. Also, the Green s function is symmetric in its rguments. Interchnging the rguments gives G(ξ, x But creful look t the originl form shows tht y 1 (xy 2 (ξ pw, x ξ, y 1 (ξy 2 (x pw. ξ x b, (7.34 G(x, ξ G(ξ, x. We will mke use of these properties in the next section to quickly determine the Green s functions for other boundry vlue problems. Exmple 7.2. Solve the boundry vlue problem y x 2, y( y(1 using the boundry vlue Green s function. We first solve the homogeneous eqution, y. After two integrtions, we hve y(x Ax + B, for A nd B constnts to be determined. We need one solution stisfying y 1 ( Thus, y 1 ( B.

10 234 prtil differentil equtions So, we cn pick y 1 (x x, since A is rbitrry. The other solution hs to stisfy y 2 (1. So, y 2 (1 A + B. This cn be solved for B A. Agin, A is rbitrry nd we will choose A 1. Thus, y 2 (x 1 x. For this problem p(x 1. Thus, for y 1 (x x nd y 2 (x 1 x, p(xw(x y 1 (xy 2 (x y 1 (xy 2(x x( 1 1(1 x 1. Note tht p(xw(x is constnt, s it should be. Now we construct the Green s function. We hve G(x, ξ { ξ(1 x, ξ x, x(1 ξ, x ξ 1. (7.35 Notice the symmetry between the two brnches of the Green s function. Also, the Green s function stisfies homogeneous boundry conditions: G(, ξ, from the lower brnch, nd G(1, ξ, from the upper brnch. Finlly, we insert the Green s function into the integrl form of the solution nd evlute the integrl. y(x 1 1 x G(x, ξ f (ξ dξ G(x, ξξ 2 dξ (1 x 1 ξ(1 xξ 2 dξ x x ξ 4 (1 x 4 ξ 3 dξ x ] x 1 x ξ 3 x 3 ξ4 4 x(1 ξξ 2 dξ (ξ 2 ξ 3 dξ 1 4 (1 xx x( x(4x3 3x (x4 x. (7.36 Checking the nswer, we cn esily verify tht y x 2, y(, nd y( Properties of Green s Functions ] 1 x We hve noted some properties of Green s functions in the lst section. In this section we will elborte on some of these properties s tool for quickly constructing Green s functions for boundry vlue problems. We list five bsic properties: 1. Differentil Eqution: The ( boundry vlue Green s function stisfies the differentil eqution + q(xg(x, ξ, x ξ. x p(x G(x,ξ x

11 green s functions nd nonhomogeneous problems 235 This is esily estblished. For x < ξ we re on the second brnch nd G(x, ξ is proportionl to y 1 (x. Thus, since y 1 (x is solution of the homogeneous eqution, then so is G(x, ξ. For x > ξ we re on the first brnch nd G(x, ξ is proportionl to y 2 (x. So, once gin G(x, ξ is solution of the homogeneous problem. 2. Boundry Conditions: In the exmple in the lst section we hd seen tht G(, ξ nd G(b, ξ. For exmple, for x we re on the second brnch nd G(x, ξ is proportionl to y 1 (x. Thus, whtever condition y 1 (x stisfies, G(x, ξ will stisfy. A similr sttement cn be mde for x b. 3. Symmetry or Reciprocity: G(x, ξ G(ξ, x We hd shown this reciprocity property in the lst section. 4. Continuity of G t x ξ: G(ξ +, ξ G(ξ, ξ Here we define ξ ± through the limits of function s x pproches ξ from bove or below. In prticulr, G(ξ +, x lim x ξ G(x, ξ, x > ξ, G(ξ, x lim x ξ G(x, ξ, x < ξ. Setting x ξ in both brnches, we hve y 1 (ξy 2 (ξ pw y 1(ξy 2 (ξ pw. Therefore, we hve estblished the continuity of G(x, ξ between the two brnches t x ξ. 5. Jump Discontinuity of G x t x ξ: G(ξ +, ξ x G(ξ, ξ x 1 p(ξ This cse is not s obvious. We first compute the derivtives by noting which brnch is involved nd then evlute the derivtives nd subtrct them. Thus, we hve G(ξ +, ξ x G(ξ, ξ x 1 pw y 1(ξy 2 (ξ + 1 pw y 1 (ξy 2(ξ y 1 (ξy 2(ξ y 1 (ξy 2 (ξ p(ξ(y 1 (ξy 2 (ξ y 1 (ξy 2(ξ 1 p(ξ. (7.37 Here is summry of the properties of the boundry vlue Green s function bsed upon the previous solution.

12 236 prtil differentil equtions 1. Differentil Eqution: ( x p(x G(x,ξ x Properties of the Green s Function + q(xg(x, ξ, x ξ 2. Boundry Conditions: Whtever conditions y 1 (x nd y 2 (x stisfy, G(x, ξ will stisfy. 3. Symmetry or Reciprocity: G(x, ξ G(ξ, x 4. Continuity of G t x ξ: G(ξ +, ξ G(ξ, ξ 5. Jump Discontinuity of G x G(ξ +, ξ x t x ξ: G(ξ, ξ x 1 p(ξ We now show how knowledge of these properties llows one to quickly construct Green s function with n exmple. Exmple 7.3. Construct the Green s function for the problem with ω. y + ω 2 y f (x, < x < 1, y( y(1, I. Find solutions to the homogeneous eqution. A generl solution to the homogeneous eqution is given s Thus, for x ξ, II. Boundry Conditions. y h (x c 1 sin ωx + c 2 cos ωx. G(x, ξ c 1 (ξ sin ωx + c 2 (ξ cos ωx. First, we hve G(, ξ for x ξ. So, G(, ξ c 2 (ξ cos ωx. So, G(x, ξ c 1 (ξ sin ωx, x ξ. Second, we hve G(1, ξ for ξ x 1. So, G(1, ξ c 1 (ξ sin ω + c 2 (ξ cos ω. A solution cn be chosen with c 2 (ξ c 1 (ξ tn ω. This gives G(x, ξ c 1 (ξ sin ωx c 1 (ξ tn ω cos ωx.

13 green s functions nd nonhomogeneous problems 237 This cn be simplified by fctoring out the c 1 (ξ nd plcing the remining terms over common denomintor. The result is G(x, ξ c 1(ξ sin ωx cos ω sin ω cos ωx] cos ω c 1(ξ sin ω(1 x. (7.38 cos ω Since the coefficient is rbitrry t this point, s cn write the result s G(x, ξ d 1 (ξ sin ω(1 x, ξ x 1. We note tht we could hve strted with y 2 (x sin ω(1 x s one of the linerly independent solutions of the homogeneous problem in nticiption tht y 2 (x stisfies the second boundry condition. III. Symmetry or Reciprocity We now impose tht G(x, ξ G(ξ, x. To this point we hve tht G(x, ξ { c 1 (ξ sin ωx, x ξ, d 1 (ξ sin ω(1 x, ξ x 1. We cn mke the brnches symmetric by picking the right forms for c 1 (ξ nd d 1 (ξ. We choose c 1 (ξ C sin ω(1 ξ nd d 1 (ξ C sin ωξ. Then, G(x, ξ { C sin ω(1 ξ sin ωx, x ξ, C sin ω(1 x sin ωξ, ξ x 1. Now the Green s function is symmetric nd we still hve to determine the constnt C. We note tht we could hve gotten to this point using the Method of Vrition of Prmeters result where C 1 pw. IV. Continuity of G(x, ξ We lredy hve continuity by virtue of the symmetry imposed in the lst step. V. Jump Discontinuity in x G(x, ξ. We still need to determine C. We cn do this using the jump discontinuity in the derivtive: G(ξ +, ξ x G(ξ, ξ x 1 p(ξ. For this problem p(x 1. Inserting the Green s function, we hve 1 G(ξ+, ξ x G(ξ, ξ x x C sin ω(1 x sin ωξ] xξ x C sin ω(1 ξ sin ωx] xξ ωc cos ω(1 ξ sin ωξ ωc sin ω(1 ξ cos ωξ ωc sin ω(ξ + 1 ξ ωc sin ω. (7.39

14 238 prtil differentil equtions Therefore, C 1 ω sin ω. Finlly, we hve the Green s function: sin ω(1 ξ sin ωx, x ξ, G(x, ξ ω sin ω sin ω(1 x sin ωξ, ξ x 1. ω sin ω (7.4 It is instructive to compre this result to the Vrition of Prmeters result. Exmple 7.4. Use the Method of Vrition of Prmeters to solve y + ω 2 y f (x, < x < 1, y( y(1, ω. We hve the functions y 1 (x sin ωx nd y 2 (x sin ω(1 x s the solutions of the homogeneous eqution stisfying y 1 ( nd y 2 (1. We need to compute pw: p(xw(x y 1 (xy 2 (x y 1 (xy 2(x ω sin ωx cos ω(1 x ω cos ωx sin ω(1 x ω sin ω (7.41 Inserting this result into the Vrition of Prmeters result for the Green s function leds to the sme Green s function s bove The Differentil Eqution for the Green s Function H(x 1 Figure 7.2: The Heviside step function, H(x. x As we progress in the book we will develop more generl theory of Green s functions for ordinry nd prtil differentil equtions. Much of this theory relies on understnding tht the Green s function relly is the system response function to point source. This begins with reclling tht the boundry vlue Green s function stisfies homogeneous differentil eqution for x ξ, x ( p(x G(x, ξ + q(xg(x, ξ, x ξ. (7.42 x For x ξ, we hve seen tht the derivtive hs jump in its vlue. This is similr to the step, or Heviside, function, { 1, x >, H(x, x <. This function is shown in Figure 7.2 nd we see tht the derivtive of the step function is zero everywhere except t the jump, or discontinuity. At the jump, there is n infinite slope, though techniclly, we hve lerned tht

15 green s functions nd nonhomogeneous problems 239 there is no derivtive t this point. We will try to remedy this sitution by introducing the Dirc delt function, δ(x d dx H(x. We will show tht the Green s function stisfies the differentil eqution ( G(x, ξ p(x + q(xg(x, ξ δ(x ξ. (7.43 x x However, we will first indicte why this knowledge is useful for the generl theory of solving differentil equtions using Green s functions. As noted, the Green s function stisfies the differentil eqution ( G(x, ξ p(x + q(xg(x, ξ δ(x ξ (7.44 x x nd stisfies homogeneous conditions. We will use the Green s function to solve the nonhomogeneous eqution ( d p(x dy(x + q(xy(x f (x. (7.45 dx dx The Dirc delt function is described in more detil in Section 9.4. The key property we will need here is the sifting property, b for < ξ < b. f (xδ(x ξ dx f (ξ These equtions cn be written in the more compct forms y] f (x G] δ(x ξ. (7.46 Using these equtions, we cn determine the solution, y(x, in terms of the Green s function. Multiplying the first eqution by G(x, ξ, the second eqution by y(x, nd then subtrcting, we hve Gy] yg] f (xg(x, ξ δ(x ξy(x. Now, integrte both sides from x to x b. The left hnd side becomes b f (xg(x, ξ δ(x ξy(x] dx b f (xg(x, ξ dx y(ξ. Using Green s Identity from Section 4.2.2, the right side is b (Gy] yg] dx ( p(x G(x, ξy (x y(x G x Combining these results nd rerrnging, we obtin ] xb (x, ξ. x Recll tht Green s identity is given by b (uv vu dx p(uv vu ] b. The generl solution in terms of the boundry vlue Green s function with corresponding surfce terms. y(ξ b f (xg(x, ξ dx ( p(x y(x G ] xb x (x, ξ G(x, ξy (x. (7.47 x

16 24 prtil differentil equtions We will refer to the extr terms in the solution, ( S(b, ξ S(, ξ p(x y(x G x (x, ξ G(x, ξy (x s the boundry, or surfce, terms. Thus, y(ξ b f (xg(x, ξ dx S(b, ξ S(, ξ]. ] xb, x The result in Eqution (7.47 is the key eqution in determining the solution of nonhomogeneous boundry vlue problem. The prticulr set of boundry conditions in the problem will dictte wht conditions G(x, ξ hs to stisfy. For exmple, if we hve the boundry conditions y( nd y(b, then the boundry terms yield b y(ξ f (xg(x, ξ dx p(b + b p( ( y(b G ( y( G x (, ξ G(, ξy ( ] x (b, ξ G(b, ξy (b ] f (xg(x, ξ dx + p(bg(b, ξy (b p(g(, ξy (. (7.48 The right hnd side will only vnish if G(x, ξ lso stisfies these homogeneous boundry conditions. This then leves us with the solution y(ξ b f (xg(x, ξ dx. We should rewrite this s function of x. So, we replce ξ with x nd x with ξ. This gives y(x b f (ξg(ξ, x dξ. However, this is not yet in the desirble form. The rguments of the Green s function re reversed. But, in this cse G(x, ξ is symmetric in its rguments. So, we cn simply switch the rguments getting the desired result. We cn now see tht the theory works for other boundry conditions. If we hd y (, then the y( G x (, ξ term in the boundry terms could be mde to vnish if we set G x (, ξ. So, this confirms tht other boundry vlue problems cn be posed besides the one elborted upon in the chpter so fr. We cn even dpt this theory to nonhomogeneous boundry conditions. We first rewrite Eqution (7.47 s b ( y(x G(x, ξ f (ξ dξ p(ξ y(ξ G ] ξb ξ (x, ξ G(x, ξy (ξ. ξ (7.49 et s consider the boundry conditions y( α nd y (b β. We lso ssume tht G(x, ξ stisfies homogeneous boundry conditions, G(, ξ, G (b, ξ. ξ

17 green s functions nd nonhomogeneous problems 241 in both x nd ξ since the Green s function is symmetric in its vribles. Then, we need only focus on the boundry terms to exmine the effect on the solution. We hve S(b, x S(, x Therefore, we hve the solution y(x b ( p(b p( ] ξ (x, b G(x, by (b y(b G ( y( G ξ (x, G(x, y ( ] βp(bg(x, b αp( G (x,. (7.5 ξ G(x, ξ f (ξ dξ + βp(bg(x, b + αp( G (x,. (7.51 ξ This solution stisfies the nonhomogeneous boundry conditions. Generl solution stisfying the nonhomogeneous boundry conditions y( α nd y (b β. Here the Green s function stisfies homogeneous boundry conditions, G(, ξ, ξ (b, ξ G. Exmple 7.5. Solve y x 2, y( 1, y(1 2 using the boundry vlue Green s function. This is modifiction of Exmple 7.2. We cn use the boundry vlue Green s function tht we found in tht problem, G(x, ξ { ξ(1 x, ξ x, x(1 ξ, x ξ 1. (7.52 We insert the Green s function into the generl solution (7.51 nd use the given boundry conditions to obtin 1 y(x G(x, ξξ 2 dξ y(ξ G ] ξ1 ξ (x, ξ G(x, ξy (ξ is x x4 1 (x 1ξ 3 dξ + x (x 1x x(1 x4 4 x(ξ 1ξ 2 dξ + y( G ξ x(1 x3 3 ξ + (x 1 2x G (x, y(1 (x, 1 ξ x 1. (7.53 Of course, this problem cn be solved by direct integrtion. The generl solution y(x x c 1x + c 2. Inserting this solution into ech boundry condition yields the sme result. We hve seen how the introduction of the Dirc delt function in the differentil eqution stisfied by the Green s function, Eqution (7.44, cn led to the solution of boundry vlue problems. The Dirc delt function lso ids in the interprettion of the Green s function. We note tht the Green s function is solution of n eqution in which the nonhomogeneous function is δ(x ξ. Note tht if we multiply the delt function by f (ξ nd integrte, we obtin δ(x ξ f (ξ dξ f (x. The Green s function stisfies delt function forced differentil eqution.

18 242 prtil differentil equtions Derivtion of the jump condition for the Green s function. We cn view the delt function s unit impulse t x ξ which cn be used to build f (x s sum of impulses of different strengths, f (ξ. Thus, the Green s function is the response to the impulse s governed by the differentil eqution nd given boundry conditions. In prticulr, the delt function forced eqution cn be used to derive the jump condition. We begin with the eqution in the form ( G(x, ξ p(x + q(xg(x, ξ δ(x ξ. (7.54 x x Now, integrte both sides from ξ ɛ to ξ + ɛ nd tke the limit s ɛ. Then, ξ+ɛ ( ] G(x, ξ ξ+ɛ lim p(x + q(xg(x, ξ dx lim δ(x ξ dx ɛ ξ ɛ x x ɛ ξ ɛ 1. (7.55 Since the q(x term is continuous, the limit s ɛ of tht term vnishes. Using the Fundmentl Theorem of Clculus, we then hve ] G(x, ξ ξ+ɛ lim p(x 1. (7.56 ɛ x ξ ɛ This is the jump condition tht we hve been using! Series Representtions of Green s Functions There re times tht it might not be so simple to find the Green s function in the simple closed form tht we hve seen so fr. However, there is method for determining the Green s functions of Sturm-iouville boundry vlue problems in the form of n eigenfunction expnsion. We will finish our discussion of Green s functions for ordinry differentil equtions by showing how one obtins such series representtions. (Note tht we re relly just repeting the steps towrds developing eigenfunction expnsion which we hd seen in Section 4.3. We will mke use of the complete set of eigenfunctions of the differentil opertor,, stisfying the homogeneous boundry conditions: φ n ] λ n σφ n, n 1, 2,... We wnt to find the prticulr solution y stisfying y] f nd homogeneous boundry conditions. We ssume tht y(x n φ n (x. Inserting this into the differentil eqution, we obtin y] n φ n ] λ n n σφ n f.

19 green s functions nd nonhomogeneous problems 243 This hs resulted in the generlized Fourier expnsion f (x c n σφ n (x with coefficients c n λ n n. We hve seen how to compute these coefficients erlier in section 4.3. We multiply both sides by φ k (x nd integrte. Using the orthogonlity of the eigenfunctions, b φ n (xφ k (xσ(x dx N k δ nk, one obtins the expnsion coefficients (if λ k k ( f, φ k N k λ k, where ( f, φ k b f (xφ k(x dx. As before, we cn rerrnge the solution to obtin the Green s function. Nmely, we hve y(x ( f, φ n b φ n (x N n λ n φ n (xφ n (ξ N n λ n } {{ } G(x,ξ f (ξ dξ Therefore, we hve found the Green s function s n expnsion in the eigenfunctions: φ n (xφ n (ξ G(x, ξ. (7.57 λ n N n We will conclude this discussion with n exmple. We will solve this problem three different wys in order to summrize the methods we hve used in the text. Green s function s n expnsion in the eigenfunctions. Exmple 7.6. Solve y + 4y x 2, x (, 1, y( y(1 using the Green s function eigenfunction expnsion. Exmple using the Green s function eigenfunction expnsion. The Green s function for this problem cn be constructed firly quickly for this problem once the eigenvlue problem is solved. The eigenvlue problem is φ (x + 4φ(x λφ(x, where φ( nd φ(1. The generl solution is obtined by rewriting the eqution s φ (x + k 2 φ(x, where k λ.

20 244 prtil differentil equtions Solutions stisfying the boundry condition t x re of the form Forcing φ(1 gives So, the eigenvlues re nd the eigenfunctions re φ(x A sin kx. A sin k k nπ, k 1, 2, λ n n 2 π 2 4, n 1, 2,... φ n sin nπx, n 1, 2,.... We lso need the normliztion constnt, N n. We hve tht N n φ n 2 1 sin 2 nπx 1 2. We cn now construct the Green s function for this problem using Eqution (7.57. sin nπx sin nπξ G(x, ξ 2 (4 n 2 π 2. (7.58 Using this Green s function, the solution of the boundry vlue problem becomes y(x G(x, ξ f (ξ dξ ( sin nπx sin nπξ 2 (4 n 2 π 2 ξ 2 dξ sin nπx (4 n 2 π 2 sin nπx (4 n 2 π 2 1 ξ 2 sin nπξ dξ (2 n 2 π 2 ( 1 n ] 2 n 3 π 3 (7.59 We cn compre this solution to the one we would obtin if we did not employ Green s functions directly. The eigenfunction expnsion method for solving boundry vlue problems, which we sw erlier is demonstrted in the next exmple. Exmple 7.7. Solve y + 4y x 2, x (, 1, y( y(1 Exmple using the eigenfunction expnsion method. using the eigenfunction expnsion method. We ssume tht the solution of this problem is in the form y(x c n φ n (x.

21 green s functions nd nonhomogeneous problems 245 Inserting this solution into the differentil eqution y] x 2, gives ] x 2 c n sin nπx d 2 ] c n sin nπx + 4 sin nπx dx2 c n 4 n 2 π 2 ] sin nπx (7.6 This is Fourier sine series expnsion of f (x x 2 on, 1]. Nmely, f (x b n sin nπx. In order to determine the c n s in Eqution (7.6, we will need the Fourier sine series expnsion of x 2 on, 1]. Thus, we need to compute 1 b n 2 x 2 sin nπx 1 (2 n 2 π 2 ( 1 n ] 2 2 n 3 π 3, n 1, 2,.... (7.61 The resulting Fourier sine series is x 2 (2 n 2 2 π 2 ( 1 n ] 2 n 3 π 3 sin nπx. Inserting this expnsion in Eqution (7.6, we find (2 n 2 2 π 2 ( 1 n ] 2 n 3 π 3 sin nπx c n 4 n 2 π 2 ] sin nπx. Due to the liner independence of the eigenfunctions, we cn solve for the unknown coefficients to obtin c n 2 (2 n2 π 2 ( 1 n 2 (4 n 2 π 2 n 3 π 3. Therefore, the solution using the eigenfunction expnsion method is y(x 2 c n φ n (x sin nπx (4 n 2 π 2 (2 n 2 π 2 ( 1 n 2 n 3 π 3 ]. (7.62 We note tht the solution in this exmple is the sme solution s we hd obtined using the Green s function obtined in series form in the previous exmple. One remining question is the following: Is there closed form for the Green s function nd the solution to this problem? The nswer is yes! Exmple 7.8. Find the closed form Green s function for the problem y + 4y x 2, x (, 1, y( y(1

22 246 prtil differentil equtions Using the closed form Green s function. nd use it to obtin closed form solution to this boundry vlue problem. We note tht the differentil opertor is specil cse of the exmple done in section 7.2. Nmely, we pick ω 2. The Green s function ws lredy found in tht section. For this specil cse, we hve sin 2(1 ξ sin 2x, x ξ, G(x, ξ 2 sin 2 sin 2(1 x sin 2ξ, ξ x 1. 2 sin 2 (7.63 Using this Green s function, the solution to the boundry vlue problem is redily computed y(x 1 x G(x, ξ f (ξ dξ sin 2(1 x sin 2ξ 1 ξ 2 sin 2(ξ 1 sin 2x dξ + ξ 2 dξ 2 sin 2 x 2 sin 2 1 ] x 2 sin 2 + (1 cos 2 x sin 2 + sin x cos x(1 + cos 2. 4 sin 2 1 ] x 2 sin sin 2 x sin 1 cos sin x cos x cos sin 2 1 ] x 2 sin sin x cos 1(sin x sin 1 + cos x cos 1. 8 sin 1 cos 1 x2 sin x cos(1 x. ( sin 1 In Figure 7.3 we show plot of this solution long with the first five terms of the series solution. The series solution converges quickly to the closed form solution. Figure 7.3: Plots of the exct solution to Exmple 7.6 with the first five terms of the series solution x As one lst check, we solve the boundry vlue problem directly, s we hd done in the lst chpter. Exmple 7.9. Solve directly: y + 4y x 2, x (, 1, y( y(1. Direct solution of the boundry vlue problem. The problem hs the generl solution y(x c 1 cos 2x + c 2 sin 2x + y p (x,

23 green s functions nd nonhomogeneous problems 247 where y p is prticulr solution of the nonhomogeneous differentil eqution. Using the Method of Undetermined Coefficients, we ssume solution of the form y p (x Ax 2 + Bx + C. Inserting this guess into the nonhomogeneous eqution, we hve 2A + 4(Ax 2 + Bx + C x 2, Thus, B, 4A 1 nd 2A + 4C. The solution of this system is A 1 4, B, C 1 8. So, the generl solution of the nonhomogeneous differentil eqution is y(x c 1 cos 2x + c 2 sin 2x + x We next determine the rbitrry constnts using the boundry conditions. We hve y( c y(1 c 1 cos 2 + c 2 sin (7.65 Thus, c nd c cos 2 sin 2 Inserting these constnts into the solution we find the sme solution s before. y(x cos 2x cos 2 ] sin 2x + x2 sin (cos 2x 1 sin 2 sin 2x(1 + cos 2 8 sin 2. + x2 4 ( 2 sin2 x2 sin 1 cos 1 sin 2x(2 cos sin 1 cos 1 (sin2 x sin 1 + sin x cos x(cos 1 4 sin 1 x2 4 + x2 4 + x2 4 sin x cos(1 x. ( sin The Generlized Green s Function When solving u f using eigenfuction expnsions, there cn be problem when there re zero eigenvlues. Recll from Section 4.3 the

24 248 prtil differentil equtions solution of this problem is given by y(x c n c n φ n (x, b f (xφ n(x dx b λ m φ2 n(xσ(x dx. (7.67 Here the eigenfunctions, φ n (x, stisfy the eigenvlue problem φ n (x λ n σ(xφ n (x, x, b] The Fredholm Alterntive. subject to given homogeneous boundry conditions. Note tht if λ m for some vlue of n m, then c m is undefined. However, if we require ( f, φ m b f (xφ n (x dx, then there is no problem. This is form of the Fredholm Alterntive. Nmely, if λ n for some n, then there is no solution unless f, φ m ; i.e., f is orthogonl to φ n. In this cse, n will be rbitrry nd there re n infinite number of solutions. Exmple 7.1. u f (x, u (, u (. The eigenfunctions stisfy φ n(x λ n φ n (x, φ n(, φ n(. There re the usul solutions, φ n (x cos nπx ( nπ 2, λ n, n 1, 2,.... However, when λ n, φ (x. So, φ (x Ax + B. The boundry conditions re stisfied if A. So, we cn tke φ (x 1. Therefore, there exists n eigenfunction corresponding to zero eigenvlue. Thus, in order to hve solution, we hve to require f (x dx. Exmple u + π 2 u β + 2x, u(, u(1. In this problem we check to see if there is n eigenfunctions with zero eigenvlue. The eigenvlue problem is φ + π 2 φ, φ(, φ(1. A solution stisfying this problem is esily founds s φ(x sin πx. Therefore, there is zero eigenvlue. For solution to exist, we need to require 1 (β + 2x sin πx dx β π cos πx π x cos πx 1 π 2 sin πx ] 1 2 (β + 1. (7.68 π Thus, either β 1 or there re no solutions.

25 green s functions nd nonhomogeneous problems 249 Recll the series representtion of the Green s function for Sturm-iouville problem in Eqution (7.57, G(x, ξ φ n (xφ n (ξ. (7.69 λ n N n We see tht if there is zero eigenvlue, then we lso cn run into trouble s one of the terms in the series is undefined. Recll tht the Green s function stisfies the differentil eqution G(x, ξ δ(x ξ, x, ξ, b] nd stisfies some pproprite set of boundry conditions. Using the bove nlysis, if there is zero eigenvlue, then φ h (x. In order for solution to exist to the Green s function differentil eqution, then f (x δ(x ξ nd we hve to require ( f, φ h b φ h (xδ(x ξ dx φ h (ξ, for nd ξ, b]. Therefore, the Green s function does not exist. We cn fix this problem by introducing modified Green s function. et s consider modified differentil eqution, G M (x, ξ δ(x ξ + cφ h (x for some constnt c. Now, the orthogonlity condition becomes ( f, φ h b φ h (ξ + c φ h (xδ(x ξ + cφ h (x] dx b φh 2 (x dx. (7.7 Thus, we cn choose c φ h(ξ b φ2 h (x dx Using the modified Green s function, we cn obtin solutions to u f. We begin with Green s identity from Section 4.2.2, given by b etting v G M, we hve b (G M u] ug M ] dx (uv vu dx p(uv vu ] b. ( p(x G M (x, ξu (x u(x G M x ] xb (x, ξ. x Applying homogeneous boundry conditions, the right hnd side vnishes. Then we hve u(ξ b b b (G M (x, ξu(x] u(xg M (x, ξ] dx (G M (x, ξ f (x u(xδ(x ξ + cφ h (x] dx b G M (x, ξ f (x dx c u(xφ h (x dx. (7.71

26 25 prtil differentil equtions Noting tht u(x, t c 1 φ h (x + u p (x,, the lst integrl gives b c u(xφ h (x dx φ h (ξ b φ2 h (x dx Therefore, the solution cn be written s u(x b b f (ξg M (x, ξ dξ + c 1 φ h (x. φ 2 h (x dx c 1φ h (ξ. Here we see tht there re n infinite number of solutions when solutions exist. Exmple Use the modified Green s function to solve u + π 2 u 2x 1, u(, u(1. We hve lredy seen tht solution exists for this problem, where we hve set β 1 in Exmple We construct the modified Green s function from the solutions of φ n + π 2 φ n λ n φ n, φ(, φ(1. The generl solutions of this eqution re φ n (x c 1 cos π 2 + λ n x + c 2 sin π 2 + λ n x. Applying the boundry conditions, we hve c 1 nd π 2 + λ n nπ. Thus, the eigenfunctions nd eigenvlues re φ n (x sin nπx, λ n (n 2 1π 2, n 1, 2, 3,.... Note tht λ 1. The modified Green s function stisfies where d 2 dx 2 G M(x, ξ + π 2 G M (x, ξ δ(x ξ + cφ h (x, c φ 1(ξ (x dx 1 φ2 1 sin πξ 1 sin2 πξ, dx 2 sin πξ. (7.72 We need to solve for G M (x, ξ. The modified Green s function stisfies d 2 dx 2 G M(x, ξ + π 2 G M (x, ξ δ(x ξ 2 sin πξ sin πx, nd the boundry conditions G M (, ξ nd G M (1, ξ. We ssume n eigenfunction expnsion, G M (x, ξ c n (ξ sin nπx.

27 green s functions nd nonhomogeneous problems 251 Then, δ(x ξ 2 sin πξ sin πx d2 dx 2 G M(x, ξ + π 2 G M (x, ξ The coefficients re found s λ n c n 2 1 λ n c n (ξ sin nπx (7.73 δ(x ξ 2 sin πξ sin πx] sin nπx dx 2 sin nπξ 2 sin πξδ n1. (7.74 Therefore, c 1 nd c n 2 sin nπξ, for n > 1. We hve found the modified Green s function s G M (x, ξ 2 n2 sin nπx sin nπξ λ n. We cn use this to find the solution. Nmely, we hve (for c 1 u(x (2ξ 1G M (x, ξ dξ n2 n2 n2 sin nπx λ n 1 sin nπx (n 2 1π 2 (2ξ 1 sin nπξ dx 1 ] 1 1 (2ξ 1 cos nπξ + nπ n 2 sin nπξ π2 1 + cos nπ n(n 2 sin nπx. (7.75 1π3 We cn lso solve this problem exctly. The generl solution is given by u(x c 1 sin πx + c 2 cos πx + 2x 1 π 2. Imposing the boundry conditions, we obtin u(x c 1 sin πx + 1 2x 1 cos πx + π2 π 2. Notice tht there re n infinite number of solutions. Choosing c 1, we hve the prticulr solution u(x 1 2x 1 cos πx + π2 π 2. In Figure 7.4 we plot this solution nd tht obtined using the modified Green s function. The result is tht they re in complete greement. 7.3 The Nonhomogeneous Het Eqution Figure 7.4: The solution for Exmple Boundry vlue Green s functions do not only rise in the solution of nonhomogeneous ordinry differentil equtions. They re lso importnt in rriving t the solution of nonhomogeneous prtil differentil equtions. In this section we will show tht this is the cse by turning to the nonhomogeneous het eqution.

28 252 prtil differentil equtions Nonhomogeneous Time Independent Boundry Conditions Consider the nonhomogeneous het eqution with nonhomogeneous boundry conditions: u t ku xx h(x, x, t >, u(, t, u(, t b, u(x, f (x. (7.76 We re interested in finding prticulr solution to this initil-boundry vlue problem. In fct, we cn represent the solution to the generl nonhomogeneous het eqution s the sum of two solutions tht solve different problems. First, we let v(x, t stisfy the homogeneous problem v t kv xx, x, t >, v(, t, v(, t, v(x, g(x, (7.77 The stedy stte solution, w(t, stisfies nonhomogeneous differentil eqution with nonhomogeneous boundry conditions. The trnsient solution, v(t, stisfies the homogeneous het eqution with homogeneous boundry conditions nd stisfies modified initil condition. which hs homogeneous boundry conditions. We will lso need stedy stte solution to the originl problem. A stedy stte solution is one tht stisfies u t. et w(x be the stedy stte solution. It stisfies the problem kw xx h(x, x. w(, t, w(, t b. (7.78 Now consider u(x, t w(x + v(x, t, the sum of the stedy stte solution, w(x, nd the trnsient solution, v(x, t. We first note tht u(x, t stisfies the nonhomogeneous het eqution, u t ku xx (w + v t (w + v xx v t kv xx kw xx h(x. (7.79 The boundry conditions re lso stisfied. Evluting, u(x, t t x nd x, we hve u(, t w( + v(, t, u(, t w( + v(, t b. (7.8 The trnsient solution stisfies v(x, f (x w(x. Finlly, the initil condition gives u(x, w(x + v(x, w(x + g(x. Thus, if we set g(x f (x w(x, then u(x, t w(x + v(x, t will be the solution of the nonhomogeneous boundry vlue problem. We ll redy know how to solve the homogeneous problem to obtin v(x, t. So, we only need to find the stedy stte solution, w(x. There re severl methods we could use to solve Eqution (7.78 for the stedy stte solution. One is the Method of Vrition of Prmeters, which

29 green s functions nd nonhomogeneous problems 253 is closely relted to the Green s function method for boundry vlue problems which we described in the lst severl sections. However, we will just integrte the differentil eqution for the stedy stte solution directly to find the solution. From this solution we will be ble to red off the Green s function. Integrting the stedy stte eqution (7.78 once, yields dw dx 1 x h(z dz + A, k where we hve been creful to include the integrtion constnt, A w (. Integrting gin, we obtin w(x 1 x ( y h(z dz dy + Ax + B, k where second integrtion constnt hs been introduced. This gives the generl solution for Eqution (7.78. The boundry conditions cn now be used to determine the constnts. It is cler tht B for the condition t x to be stisfied. The second condition gives b w( 1 ( y h(z dz dy + A +. k Solving for A, we hve A 1 k ( y h(z dz dy + b. Inserting the integrtion constnts, the solution of the boundry vlue problem for the stedy stte solution is then w(x 1 x ( y h(z dz dy + x ( y h(z dz dy + b k k x +. The stedy stte solution. This is sufficient for n nswer, but it cn be written in more compct form. In fct, we will show tht the solution cn be written in wy tht Green s function cn be identified. First, we rewrite the double integrls s single integrls. We cn do this using integrtion by prts. Consider integrl in the first term of the solution, x ( y I h(z dz dy. Setting u y h(z dz nd dv dy in the stndrd integrtion by prts formul, we obtin x ( y I h(z dz dy y y x h(z dz x x yh(y dy (x yh(y dy. (7.81

30 254 prtil differentil equtions Thus, the double integrl hs now collpsed to single integrl. Replcing the integrl in the solution, the stedy stte solution becomes w(x 1 k x (x yh(y dy + x ( yh(y dy + b k x +. We cn mke further simplifiction by combining these integrls. This cn be done if the integrtion rnge,, ], in the second integrl is split into two pieces,, x] nd x, ]. Writing the second integrl s two integrls over these subintervls, we obtin w(x 1 k + x k x (x yh(y dy + x k Next, we rewrite the integrnds, w(x 1 k + 1 k x x x x ( yh(y dy ( yh(y dy + b x +. (7.82 (x y h(y dy + 1 k x( y x x( y h(y dy h(y dy + b x +. (7.83 It cn now be seen how we cn combine the first two integrls: w(x 1 k x y( x h(y dy + 1 k x x( y h(y dy + b x +. The resulting integrls now tke on similr form nd this solution cn be written compctly s w(x G(x, y 1 b h(y] dy + k x +, The Green s function for the stedy stte problem. where x( y, x y, G(x, y y( x, y x, is the Green s function for this problem. The full solution to the originl problem cn be found by dding to this stedy stte solution solution of the homogeneous problem, u t ku xx, x, t >, u(, t, u(, t, u(x, f (x w(x. (7.84 Exmple Solve the nonhomogeneous problem, u t u xx 1, x 1, t >, u(, t 2, u(1, t, u(x, 2x(1 x. (7.85

31 green s functions nd nonhomogeneous problems 255 In this problem we hve rod initilly t temperture of u(x, 2x(1 x. The ends of the rod re mintined t fixed tempertures nd the br is continully heted t constnt temperture, represented by the source term, 1. First, we find the stedy stte temperture, w(x, stisfying Using the generl solution, we hve w xx 1, x 1. w(, t 2, w(1, t. (7.86 w(x 1 1G(x, y dy 2x + 2, where G(x, y { x(1 y, x y, y(1 x, y x 1, we compute the solution x 1 w(x 1y(1 x dy + 1x(1 y dy 2x + 2 x 5(x x 2 2x + 2, 2 15x 5x 2. (7.87 Checking this solution, it stisfies both the stedy stte eqution nd boundry conditions. The trnsient solution stisfies v t v xx, x 1, t >, v(, t, v(1, t, v(x, x(1 x 1. (7.88 Recll, tht we hve determined the solution of this problem s v(x, t b n e n2 π 2t sin nπx, where the Fourier sine coefficients re given in terms of the initil temperture distribution, 1 b n 2 x(1 x 1] sin nπx dx, n 1, 2,.... Therefore, the full solution is u(x, t b n e n2 π 2t sin nπx x 5x 2. Note tht for lrge t, the trnsient solution tends to zero nd we re left with the stedy stte solution s expected.

32 256 prtil differentil equtions Time Dependent Boundry Conditions In the lst section we solved problems with time independent boundry conditions using equilibrium solutions stisfying the stedy stte het eqution snd nonhomogeneous boundry conditions. When the boundry conditions re time dependent, we cn lso convert the problem to n uxiliry problem with homogeneous boundry conditions. Consider the problem u t ku xx h(x, x, t >, u(, t (t, u(, t b(t, t >, u(x, f (x, x. (7.89 We define u(x, t v(x, t + w(x, t, where w(x, t is modified form of the stedy stte solution from the lst section, Noting tht w(x, t (t + b(t (t x. we find tht v(x, t is solution of the problem v t kv xx h(x ȧ(t + u t v t + ȧ + ḃ ȧ x, u xx v xx, (7.9 ] ḃ(t ȧ(t x, x, t >, v(, t, v(, t, t >, ] b( ( v(x, f (x ( + x, x. (7.91 Thus, we hve converted the originl problem into nonhomogeneous het eqution with homogeneous boundry conditions nd new source term nd new initil condition. Exmple Solve the problem u t u xx x, x 1, t >, u(, t 2, u(, t t, t > u(x, 3 sin 2πx + 2(1 x, x 1. (7.92 We first define u(x, t v(x, t (t 2x. Then, v(x, t stisfies the problem v t v xx, x 1, t >, v(, t, v(, t, t >, v(x, 3 sin 2πx, x 1. (7.93

33 green s functions nd nonhomogeneous problems 257 This problem is esily solved. The generl solution is given by v(x, t b n sin nπxe n2 π 2t. We cn see tht the Fourier coefficients ll vnish except for b 2. This gives v(x, t 3 sin 2πxe 4π2t nd, therefore, we hve found the solution u(x, t 3 sin 2πxe 4π2t (t 2x. 7.4 Green s Functions for 1D Prtil Differentil Equtions In Section 7.1 we encountered the initil vlue Green s function for initil vlue problems for ordinry differentil equtions. In tht cse we were ble to express the solution of the differentil eqution y] f in the form y(t G(t, τ f (τ dτ, where the Green s function G(t, τ ws used to hndle the nonhomogeneous term in the differentil eqution. In similr spirit, we cn introduce Green s functions of different types to hndle nonhomogeneous terms, nonhomogeneous boundry conditions, or nonhomogeneous initil conditions. Occsionlly, we will stop nd rerrnge the solutions of different problems nd recst the solution nd identify the Green s function for the problem. In this section we will rewrite the solutions of the het eqution nd wve eqution on finite intervl to obtin n initil vlue Green;s function. Assuming homogeneous boundry conditions nd homogeneous differentil opertor, we cn write the solution of the het eqution in the form u(x, t G(x, ξ; t, t f (ξ dξ. where u(x, t f (x, nd the solution of the wve eqution s u(x, t G c (x, ξ, t, t f (ξ dξ + G s (x, ξ, t, t g(ξ dξ. where u(x, t f (x nd u t (x, t g(x. The functions G(x, ξ; t, t, G(x, ξ; t, t, nd G(x, ξ; t, t re initil vlue Green s functions nd we will need to explore some more methods before we cn discuss the properties of these functions. For exmple, see Section.] We will now turn to showing tht for the solutions of the one dimensionl het nd wve equtions with fixed, homogeneous boundry conditions, we cn construct the prticulr Green s functions Het Eqution In Section 3.5 we obtined the solution to the one dimensionl het eqution on finite intervl stisfying homogeneous Dirichlet conditions, u t ku xx, < t, x,

34 258 prtil differentil equtions u(x, f (x, < x <, u(, t, t >, u(, t, t >. (7.94 The solution we found ws the Fourier sine series u(x, t b n e λnkt sin nπx, where ( nπ 2 λ n nd the Fourier sine coefficients re given in terms of the initil temperture distribution, b n 2 f (x sin nπx dx, n 1, 2,.... Inserting the coefficients b n into the solution, we hve ( 2 u(x, t f (ξ sin nπξ dξ Interchnging the sum nd integrtion, we obtin ( 2 u(x, t sin nπx sin nπξ This solution is of the form u(x, t e λ nkt sin nπx. eλ nkt G(x, ξ; t, f (ξ dξ. f (ξ dξ. Here the function G(x, ξ; t, is the initil vlue Green s function for the het eqution in the form G(x, ξ; t, 2 sin nπx nπξ sin eλnkt. which involves sum over eigenfunctions of the sptil eigenvlue problem, X n (x sin nπx Wve Eqution The solution of the one dimensionl wve eqution (2.2, ws found s u tt c 2 u xx, < t, x, u(, t, u(,, t >, u(x, f (x, u t (x, g(x, < x <, (7.95 u(x, t A n cos nπct + B n sin nπct ] sin nπx.