APPENDIX D Solutions to Problem Set 4 1. èproblem 3.4.3 in textè èaè Consider an inænite-interval problem,,1 éxé+1, for which hèxè ; for xé0 uèx; 0è = èd.1è,hè,xè ; for xé0 u t èx; 0è = 0 Show that the solution of u tt, c 2 u xx =0 satisfying these initial conditions also solves the following semi-inænite problem: ænd uèx; tè satisfying u tt, c 2 u xx =0, x 2 è0; +1è, with initial conditions uèx; 0è = hèxè, u t èx; 0è = 0, and with æxed end condition uè0;tè = 0. ëhere hèxè isany given function, with hè0è = 0ë. Sketch the solution for the case where hèxè = 1 2, æ x, 3 æ for 1 éxé2, hèxè = 0 elsewhere. 2 èbè Use a similar idea to explain how you could use èd.2è uèx; tè = 1 2 ëuèx + ct; 0è + uèx, ct; 0èë + 1 2c Z x+ct x,ct u t è;0èd to solveany ænite interval problem in which uè0;tè=uèl; tè = 0 for all t, with uèx; 0è = hèxè and u t èx; 0è = 0 for 0 éxél.ëwe take hè0è = hèlè = 0.ë ècè Reconsider parts èaè and èbè for situations in which u t èx; 0è is prescribed, with uèx; 0è = 0. Sketch the solution for a simple case. èaè Equation èd.2è gives the unique solution of the wave equation in the region,1 éxé+1,0été+1, in terms of its Cauchy data along the x-axis. If we use Eq. èd.1è to extend a Cauchy problem on the positive x-axis to the entire x-axis, then by restricting Eq. èd.2è to the region è0; +1èæè0; +1èwe obtain a solution of the wave equation satisfying the boundary conditions uèx; 0è = hèxè, u t èx; 0è = 0, for all x 2 è0; +1è. We only have tocheck that if the boundary condition uè0;tè = 0 is also satisæed. From èd.2è we have èd.3è uèx; 0è = 1 2 ëuèx; 0è + uèx; 0èë + 1 2c = 1 2 hèxè = 0 R x x u tè;0èd èbè Our problem now is to deæne an extension Hèxè of the function hèxè deæned on è0;lè to the entire x-axis so that Eq. è8.22è can be used to write down the solution of the wave equation in the region è0;lèæè0; +1è satisfying èd.4è uèx; 0è = hèxè u t èx; 0è = 0 uè0;tè = 0 uèl; tè = 0 114
1. èproblem 3.4.3 IN TEXTè 115 The validity of the ærst two boundary conditions will be automatic èsince the restriction of our extension must give us exactly what we started withè. Setting uèx; 0è = Hèxè, u t èx; 0è = 0 we obtain from Eq. è8.22è uèx; tè = 1 ëhèx + ctè+hèx, ctèë 2 The third boundary condition in èd.4è thus leads to 0=uè0;tè= 1 2 ëhèctè+hè,ctèë : This will be satisæed automatically if we extend hèxè is such a way that the new function Hèxè is odd with respect to reæections about x =0. The last boundary condition thus leads to 0=uèl; tè = 1 ëhèl + ctè+hèl, ctèë : 2 This will be satisæed automatically if we extend hèxè issuchaway that the new function Hèxè is odd with respect to reæections about x = l. We thus deæne Hèxè as follows: 8 é gèxè = é: hè2l + xè ;,2l éxé,l,hè,xè ;,l éxé0 hèxè ; 0 éxél,hèx, lè ; léxé2l Hèxè =g èx, 4nlè ; 4nl, 2l éxé4nl +2l ; n 2 Z : ècè If instead we had boundary conditions of the form èd.5è uèx; 0è = 0 ; 0 éxél èd.6è u t èx; 0è = pèxè ; 0 éxél èd.7è uè0;tè = 0 èd.8è uèl; tè = 0 we would seek to extend the deænition of pèxè to the entire x-axis so that the last two boundary conditions are satisæed automatically. Wewould thus need to deæne P èxè such that èd.9è 0 = uè0;tè= 1 2c Z ct,ct P èèd èd.10è 0 = uèl; tè = 1 2c Z l+ct l,ct P èèd automatically. To accomplish this we can simply extend pèxè in such a way that it is periodic with period 4l and antisymmetric with respect to reæections about x = 0 and x = l.
2. èproblem 3.4.4 IN TEXTè 116 2. èproblem 3.4.4 in textè Consider the ëwhip-cracking" problem: èd.11è in the region xé0, té0. tt, c 2 xx = 0 èx; 0è = 0 t èx; 0è = 0 è0;tè = æètè è0; 0è = 0 ; We know from the discussion in Lecture 9 that èd.12è is the general solution to the wave equation èd.13è The boundary conditions in èd.11è imply èd.14è èx; tè =æ èx + ctè+æ èx, ctè tt, c 2 xx =0 : æèxè+æèxè = 0 cæ 0 èxè, cæ 0 èxè = 0 æèctè, æè,ctè = æètè : The equation tells us that æèxè =,æèxè. Making this substitution, we get from the second equation that so 2cæ 0 èxè =0 æèxè =K : It would seem that this trivializes everything; however, the ærst two conditions in èd.14è are only imposed only for xé0. We are therefore free to adjust the functions æèxè and æèxè in the region where xé0. The third equation, in fact, tells us that æècxè+æè,cxè =æèxè which we can satisfy by extending the deænition of æèxè toxé0 x æè,xè =,K + æ ; 8 xé0 : c It is not necessary to extend the domain of æèxè toxé0, since in the expression èd.12è for èx,tè, the argument ofæ is always positive. Thus, we take K ; é0 æèè = K ; é0 èd.15è, æ æ, c, K ; é0 æèè =,K ; é0 Thus,, æ æ, ; é0 c æè;è=æèè+æèè = 0 ; é0 and so the solution of èd.11è is, æ æ t, x ;, ct é 0 c èd.16è èx; tè = 0 ; x, ct é 0 :
3. èproblem 3.12.1 IN TEXTè 117 3. èproblem 3.12.1 in textè èaè Let uèx; tè satisfy the equation u tt = c 2 u xx ; c = costant ; in some region of the èx; tè plane. Show that the quantity èu t, cu x è is constant along each straight line deæned by x, ct = constant, and that èu t + cu x è is constant along each straight line of the form x + ct = constant. These straight lines are called characteristics; we will refer to typical members of the two families as C + and C, curves, respectively; thus èx, ct = constantè isac + curve. Set èd.17è Along a C + curve wehave èd.18è and so along such a curve èd.19è Diæerentiating + with respect to t we obtain èd.20è èd.21è èd.22è + èx; tè =u t èx; tè, cu x èx; tè : x = k 1 + ct + èx; tè = + ètè =u t èk 1 + ct; tè, cu x èk 1 + ct; tè : d + dt = cu tx + u tt, c 2 u xx, cu tx = u tt, c 2 u xx = 0 since u satisæes the wave equation. Therefore, + is constant along any curve of the form èd.18è. Similarly, if we set èd.23è Then along the curve èd.24è we have and so èd.25è èd.26è èd.27è, èx; tè =u t èx; tè+cu x èx; tè : x = k 2, ct, èx; tè =, ètè =u t èk 2, ct; tè+cu x èk 2, ct; tè d, dt =,cu tx + u tt, c 2 u xx + cu tx = u tt, c 2 u xx = 0 Thus,, is constant along any curve of the form èd.24è. èbè Let uèx; 0è and u t èx; 0è be prescribed for all values of x between,1 and +1, and let èx o ;t o è be some point in the èx; tè plane, with t o é 0. Draw the C + and C, curves through èx o ;t o è and let A and B denote, respectively, their intercepts with the x-axis. Use the properties of C + and C, derived in part èaè to determine u t èx o ;t o è in terms of initial data at points èa; 0è and èb;0è. Using a similar technique to obtain u t èx o ;è with 0 éét o, determine u èx o ;t o èbyintegration with respect to, and compare with Equation è8.22è. Observe that this ëmethod of characteristics" again shows that u èx o ;t o è depends only on that part of the initial data between points èa; 0è and èb;0è. Let èd.28è k æ = x o æ ct o
3. èproblem 3.12.1 IN TEXTè 118 and set èd.29è c æ = æ èx; tè 2 R 2 j x æ ct = k æ æ From part èaè we know that èd.30è + = u t èx; tè, cu x èx; tè, = u t èx; tè+cu x èx; tè are, respectively, constant along the lines c + and c,. At the point èa; 0è where the line c + intersects the x-axis we have èd.31è + = u t èa; 0è, cu x èa; 0è and so the constant + is completely determined by the Cauchy data at the point èa; 0è. Similarly, at the point èb;0è where the line c, intersects the x-axis we have èd.32è, = u t èb;0è + cu x èb;0è and so the constant, is completely determined by the Cauchy data at the point èb;0è. Using èd.31è and èd.32è we can rewrite equations èd.30è as èd.33è u t èa; 0è, cu x èa; 0è = u t èx o ;t o è, cu x èx o ;t o è u t èb;0è + cu x èb;0è = u t èx o ;t o è+cu x èx o ;t o è Adding the second equation to the ærst and then dividing by 2we obtain èd.34è u t èx o ;t o è= 1 2 èu t èa; 0è + u t èb;0è, cu x èa; 0è + cu x èb;0èè : We can be a even more explicit than this. For the value of A is precisely k + = x o, ct o, and the value of B is precisely k, = x o + ct o.thus, èd.35è u t èx o ;t o è = 2 1 èu t èx o, ct o ; 0è + u t èx o + ct o ; 0èè + c 2 è,u x èx o, ct o ; 0è + u x èx o + ct o ; 0èè This equation is perfectly valid for any choice of x o and t o, and so we can write èd.36è u t èx o ;tè = 1 2 èu t èx o, ct; 0è + u t èx o + ct; 0èè + c 2 è,u x èx o, ct; 0è + u x èx o + ct; 0èè Integrating both sides with respect to t from 0 to t o we obtain èd.37è u èx o ;t o è, u èx o ; 0è = 1 2 R to R to 0 u t èx o, ct; 0è dt + 1 2 0 u t èx o + ct; 0è dt R to R 0 u to x èx o, ct; 0è dt + c 2 0 u x èx o + ct; 0è dt, c 2
3. èproblem 3.12.1 IN TEXTè 119 If we make achange of variables = x o, ct in the ærst and third integrals and a change of variables = x o + ct in the second and fourth integrals, the èd.37è becomes èd.38è èd.39è èd.40è èd.41è èd.42è èd.43è èd.44è èd.45è or u èx o ;t o è, u èx o ; 0è =, 1 2c which is precisely Equation è8.22è. = 1 2c = 1 2c + 1 2 Z,cto Z,cto Z +cto,cto u t è;0è d + 1 2c u x è;0è d + 1 2 u t è;0è d + 1 2 u è;0èæ æ,cto Z +cto,cto u t è;0è d Z +cto Z +cto + 1 2 u è;0èæ æ +cto u t è;0è d u x è;0è d + 1 2 èu èx o, ct o è+u èx o + ct o èè + u èx o ; 0è u èx o ;t o è= 1 2 èu èx o, ct o è+u èx o + ct o èè + 1 2c Z +cto,cto u t è;0è d