4 Separation of Variables

Transcription

1 4 Separation of Variabes In this chapter we describe a cassica technique for constructing forma soutions to inear boundary vaue probems. The soution of three cassica (paraboic, hyperboic and eiptic) PDE probems are discussed in turn. Our treatment is not intended to be rigorous. A proper mathematica justification for the cass of soutions that resut in these simpe cases is knownas Fourier anaysis. An outine of the genera theory that underpins the appication of the separation of variabes soution technique can be found in the concuding section The Heat Equation The heat equation is the prototypica paraboic PDE (see Section 1.5). We consider the simpest possibe boundary vaue probem in the firstinstance The homogeneous probem A generic probem statement is: find a function u(x, t), representing the temperature at position x and time t in a rod with constant therma conductivity κ>, satisfying with boundary conditions and an initia condition u t κu xx = (x, t) (,) (,τ], (4.1) u(,t)=; u(, t) = t (,τ], (4.2) u(x, ) = u (x) x [,]. (4.3) This is a mode for what happens when a heated rod with a given initia temperature profie u has both ends immersed into ice at t =. Physicay 48

2 4.1 The Heat Equation 49 we woud expect the heat to diffuse out of the two ends and for thetemperature to tend to the ambient temperature of zero as t.twospecificexampes are given beow. Probem u (x) ={ 2 x x x 2 x 1 /2 Probem u (x) =1 x 1 1 Note that, in Probem the initia condition is consistent with the boundary conditions, whereas in Probem there is a jump in the temperature at the two ends x =andx = when t =. The fact that the boundary vaue probem (4.1) (4.3) is we posed can be shown by the foowing energy argument. WemutipythePDEbyu and integrate in space, uu t dx κ We now introduce the energy E, ascaarfunctionoftime: E(t) = uu xx dx =. (4.4) u 2 dx, and note that if u is sufficienty smooth then we can differentiate the integra to give E (t) = d u 2 1 dx = dt t (u2 ) dx =2 uu t dx. (4.5)

3 5 Separation of Variabes Combining (4.4) with (4.5) we find that E (t) =2κ uu xx dx. We now foow the argument used in Theorem 3.2 and integrate the right hand side term by parts to give E (t) = 2κ (u x ) 2 dx. (4.6) {{ The boundary term [uu x ] 1 in (4.6) drops out because u(,t)=andu(1,t)=. This shows that the energy is a monotonicay non-increasing function of time. Thus we deduce that E(t) E() for a t>, which means that u 2 dx = E(t) E() u 2 dx. (4.7) This is a cassica stabiity estimate for a time dependent probem. The soution u is forever bounded by the size of the initia condition, so sma perturbations to u cannot generate arge changes in the soution. This is true even in the ong-time imit τ. The cruciay important point here is that the boundary vaue probem (4.1) (4.3) is inear, which means that we can appy the principe of superposition (see Section 1.3). Our target wi be to represent the soution u(x, t) as a inear combination of the increasingy osciatory sine waves that are iustrated in Figure 4.1. These sine waves (usuay referred to as Fourier modes) are associated with the so-caed Fourier expansion of the initia data, u (x) = n=1 a n sin nπx, (4.8) and the numbers a n are caed the Fourier coefficients. Wewiseethateach of the component sine waves wi correspond to a different particuar soution of the boundary vaue probem. Two important properties of these Fourier modes are isted beow. ➊ The modes are mutuay orthogona with respect to the inner products (a, b) = ab dx: thatis nπx sin sin mπx dx =, n m. ➋ The modes are ineary independent. Moreovertheyformabasisforthe space L 2 (,)ofsquareintegrabefunctionsdefinedon(,).

4 4.1 The Heat Equation 51 sin(π x) sin(2π x) sin(3π x) sin(4π x) sin(8π x) sin(16π x) Figure 4.1. Fourier modes: the scaed anaogues sin( kπx ) ook ike this over [,]. An immediate consequence of ➋ the representation (4.8) is unique as ong as u is square integrabe which is the case for either of the two exampes above. The condition ➊ eads to a characterization of the Fourier coefficients a n. Starting from (4.8) and mutipying by sin mπx and integrating over the interva (,)gives: Thus we deduce that u (x)sin mπx a n = 2 dx = a n n=1 = a m = 2 a m. sin nπx (sin mπx ) 2 dx u (x)sin nπx sin mπx dx dx. (4.9) Let s get back to separation of variabes. Theideaistoookforsoutions of the form u(x, t) =X(x)T (t). (4.1)

5 52 Separation of Variabes Substituting the ansatz (4.1) into (4.1) gives and coecting ike terms gives 1 1 dt κ T {{ dt function of t X dt dt = κd2 X dx 2 T = 1 d 2 X X {{ dx 2. function of x Ceary both sides of this equation must be equa to a constant, µ say, caed the separation constant. Thuswegetapairofordinarydifferentiaequations, one for each variabe: X = µx, (4.11) T = κµt. (4.12) To get boundary conditions for (4.11) we substitute (4.1) into (4.2) and since T (otherwise,theonypossibesoutionisu =)weobtain u(,t) = X()T (t) = u(, t) = X()T (t) = X() =, X() =. (4.13) Combining (4.11) with (4.13) gives a differentia eigenvaue probem: X = µx X() = ; X() = (4.14) where µ is an eigenvaue and X is the associated eigenfunction. Note that the trivia soution X = must be excuded if we are to find a nontrivia soution u. Unike matrix eigenvaue probems, which are finite dimensiona, the probem (4.14) may have infinitey many soutions {µ n,x n.itturnsoutthatony negative eigenvaues µ n < makesense.toseethis,wemutipy(4.11)byx and integrate over (,)togive XX dx + µ X 2 dx =. At first sight this does not seem too hepfu. The trick though is to integrate the first term by parts to remove the second derivative, thus (X ) 2 dx + µ X 2 dx = [XX ].

6 4.1 The Heat Equation 53 The right-hand side term is zero because of the boundary conditions X() = and X() =sothatweareeftwithtwonon-negativetermswhichsumto zero (X ) 2 dx +µ {{ X 2 dx =, (4.15) {{ > X thus µ<. We write µ = ω 2 and note that the genera soution of (4.11) is given by X(x) =A cos(ωx)+b sin(ωx). (4.16) The eft end condition X() = impies that A =. Therightendcondition X() =givessin(ω) =eadingtothenontriviasoutionsω = nπ for a positive integers n. Wededucethat (4.14) hasaninfinitenumberof eigenfunctions X n (x) =sin nπx, n =1, 2,... (4.17) and corresponding eigenvaues ( nπ µ n = ) 2, n =1, 2,... (4.18) Note that the eigenfunctions are the Fourier modes in (4.8). Returning to (4.12), for each distinct n we have that T n κµ n T n =. This is a inear first order ordinary differentia equation with integrating factor e κµnt.ithasthesoution T n (t) =e κµnt = e nπ κ( )2 t (4.19) when the constant of integration is set to be equa to one. Combining (4.17) with (4.19) gives an infinite famiy of separated soutions u n (x, t) =X n (x) T n (t) =e nπ κ( )2t sin nπx, n =1, 2,... It is easiy checked that each u n satisfies the heat equation (4.1) and the zero boundary condition (4.2). Thus, using the principe of superposition we can construct a genera soution to (4.1) (4.2) by taking a inear combination of these soutions nπ κ( u(x, t) = c n e )2t sin nπx. (4.2) n=1

7 54 Separation of Variabes Finay if we put t =in(4.2)werecover(4.8)andweidentifythecoefficients c n as being the Fourier coefficients of the initia condition u (x). Thus from (4.9) we deduce that u(x, ) = u = n=1 c n sin nπx c n = 2 u (x)sin nπx dx. (4.21) Remark 4.1. From (4.21) it seems that we can sove any probem for which u (x) ispiecewise continuous. Theonyissueistomakesurethatthecoefficients c n in (4.2) are a finite. 1 Probem temperature x Figure 4.2. Evoution of soution u(x, t) to Exampe Exampe u (x) ={ 2 x x x 2 x. Note that u (x) issymmetricwithrespecttotheverticainethroughthe midpoint x = 2,andsoc n = u sin nπx dx =foratheanti-symmetric, even-numbered waves n =2, 4,... as shown in Figure 4.1. For the odd numbered waves n = 1, 3, 5,... the characterization (4.9) gives c n = 2 u sin nπx dx = 4 /2 u sin nπx dx = 8 2 /2 x sin nπx dx

8 4.1 The Heat Equation 55 = 8 [ ] /2 1 nπx π 2 sin = 8 { 1 n2 π 2 1 2, 1 3 2, 1 5 2, 1 7 2, 1 9 2,.... Note that the Fourier coefficients are decreasing ike 1/n 2 in this case. Substituting c n into (4.2) gives the soution u(x, t) = 8 π 2 { e κπ2 2 t sin πx 1 9 9κπ2 e 2 t sin 3πx κπ2 e 2 t sin 5πx... Two important observations are that a the Fourier modes components of the soution decay exponentiay in time, with the highy osciating waves decaying faster than ow frequency ones. Thus, utimatey, u ast which is consistent with the physics. Aso, the bigger the conductivity constant κ is, the faster the decay. The evoution of the temperature profie with time is shown in Figure The numerica soution of this exampe probem wi be discussed in Chapter 6. Exampe c n = 2 sin nπx dx = 2 π u (x) =1, <x<1. [ 1 ] nπx cos = 4 { 1 n π 1, 2, 1 3, 4, 1 5, 6,.... Since the Fourier coefficients are proportiona to 1/n rather that 1/n 2 the high frequency waves are more significant in this exampe. Substituting the c n into (4.2) gives the soution u(x, t) = 4 π { e κπ2 2 t sin πx κπ2 e 2 t sin 3πx κπ2 e 2 t sin 5πx +... Remark 4.2. Our Fourier series soution (4.2) (4.21) is often referred to as a forma soution. Stricty speaking, to be sure that such an infinite series soution is vaid we woud need to show two extra things. First, that the series converges to a we-defined imit function, and second, that the imit function soves the boundary vaue probem (4.1) (4.3). Such technica considerations are beyond the scope of this book. Further detais are discussed in Chapter 9 of the foowing textbook. Asak Tvieto & Ragner Winther, Introduction to Partia Differentia Equations: A Computationa Approach, Springer, The soution can be visuaized using the matab function fourier heat, forexampe, using the caing sequence [u,m] = fourier heat(8,64,1,.3);