Week 5 Lectures, Math 6451, Tanveer

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1 Week 5 Lectures, Math 651, Tanveer 1 Separation of variabe method The method of separation of variabe is a suitabe technique for determining soutions to inear PDEs, usuay with constant coefficients, when the domain is bounded in at east one of the independent variabes. We iustrate this procedure for 1-D wave equation and -D heat equation for Dirichet, Neumann and Robin boundary conditions, though the idea is equay appicabe for diffusion equation and Lapace s equation, and other constant coefficient equations. The idea of separation of variabe is first to seek simpe soution to the PDE in the form of a product, each term in the product depending on ony one independent variabe. Soutions are then constrained by boundary conditions. This resuts in a countaby infinite set of soutions. A inear superposition of such soution is aso a soution, because of the inearity of the probem. As we sha find ater, such inear superposition is capabe of describing a reasonabe initia conditions. 1.1 Dirichet Probem Consider with initia condition u tt c u xx = for < x < (1) u(,t) = = u(,t) () u(x,) = φ(x) ; u t (x,) = ψ(x) (3) Reca, there is a representation of this soution by doing odd extension about x = to the interva (, ), and then periodicay extending this probem (with period ), and then using D Aembert representation of soution, as you wi see in ast week notes. Uniqueness foows from appication of energy method, as we saw earier. Here, we are seeking a different form of the same soution. We seek simpe soution to (1), ignoring initia conditions (3) for now, in a product form u(x,t) = X(x)T(t) Pugging it into (1) and dividing it the resuting equation by c X(x)T(t), we obtain T (t) c T(t) = (x) X () X(x) If we set t = 1, for instance, then the eft side of () is a number independent of x. Hence the right side of () cannot depend on x at a. Simiary, if we set x to some specific vaue, the right side of () is a constant; hence the eft side () must be independent of t. Therefore, we concude both eft and right side of each side of () is some constant λ. Therefore, and The boundary conditions () impy that X() = and X() =. T (t)+λc λt(t) = (5) X (x)+λx(x) = (6) 1

2 If λ <, then the soution to (6) that satisfies X() = is given by ( ) X(x) = Csinh λx (7) Since the function sinh does not vanish anywhere except the origin, any nontrivia soution in the form (7) is incapabe of satisfying X() =. Therefore, we must discard the possibiity of λ <. If λ =, then the soution to (6) that satisfies X() = is X(x) = Cx, (8) But, this is not capabe of satisfying X() =, uness C =, which corresponds to the trivia soution. Therefore, we concude that λ. We are ony eft with the possibiity λ = β >. Then, soution to (6) satisfying X() = is given by X(x) = Csinβx (9) In order for the soution to be nontrivia, i.e. C, and yet X() =, we must have sinβ =. Therefore β = nπ, for integer n. We note that sin ( ) ( nπx = sin nπx ). This impies that n < does not generate an independent set of functions from what we obtain for n >. Thus, we can restrict to integer n 1, and obtain ( nπx ) X(x) = Csin CX n (x) (1) Corresponding to β = nπ β n, with n > 1, we may sove (5), with λ = βn, to obtain T(t) = A n cosβ n ct+b n sinβ n ct T n (t) (11) where A n and B n are arbitary constants. Therefore, we obtain a countaby infinite set of separabe soutions to to (1) satisfying boundary conditions (), indexed by integer n, and in the form u(x,t) = X n (x)t n (t). A more genera soution to (1), satisfying Dirichet boundary condition () is of the form u(x,t) = n 1 X n (x)t n (t) = n 1 ( A n cos nπct +B n sin nπct ) sin nπx, (1) assuming that this sum converges. We note that this soution (1) satisfies initia condition u(x,) = n 1 A n sin nπx (13) u t (x,) = n 1 nπc B n sin nπx (1) Therefore, if φ(x) and ψ(x) in (3) is in the form of (13) and (1), the separation of variabe method wi have provided soution to IVP (1)-(3). We wi ater discover from the theory of Fourier Series, that this restriction on φ and ψ is rather mid that a initia conditions of physica interest can be accomodated with soution (1) by suitaby choosing A n and B n, depending on φ and ψ.

3 1. Neumann Condition The same method works for Neumann probem as we. The text iustrates this for 1-D wave equation in section.. We iustrate here for -D heat equation in a rectanguar domain in x = (x,y). So, our probem of interest is satisfying Neumann Boundary condition and initia condition u t κ(u xx +u yy ) = for < x <, < y <,t > (15) u x (,y,t) = = u x (,y,t), u y (x,,t) = = u y (x,,t) (16) u(x,y,) = φ(x,y) (17) Using energy method (see Homework 3 probem), we can prove that any soution satisfying (1)-(3) is unique. Here, we are showing existence. As in the ast subsection, first, weseeksimpe soutionto (15) satisfyingony(16) in the product formu(x,y,t) = X(x)Y(y)T(t). Pugginginto (15)and dividingthe resut byx(x)y(y)t(t), we obtain X (x) X(x) = Y (y) Y(y) + T (t) (18) κt(t) The eft side depends ony on x, whie the right depends on (y,t). It foows that each side of the equation has to be some constant λ. Therefore, X (x) X(x) = λ, impying X (x)+λx(x) = (19) The first set of boundary conditions in (16) impy that X () = and X () =. However, if λ <, the soution satisfying X () = is given by X(x) = Ccosh λx which cannot satisfy X () = for nonzero C, since sinh is nonzero for nonzero argument. Therefore, we are eft with ony possibiity of λ = β. In that case, a nontrivia soution to (19) satisfying X () = is X(x) = cosβx () In order to satisfy X () =, we must have sinβ =, therefore nπ λ = β = λ n (1) and non-trivia soution for X(x) is restricted to X(x) = cos nπx X n (x) for n () Note that n < is excuded from the above set, since cos is an even function and repacing n by n does not generate any new set of soutions. However, unike the Dirichet probem, (1) aows the possibiity of n =. Having determined X(x), we now determine Y(y). For that purpose, it is convenient to rewrite (18) as Y (y) Y(y) = X (x) X(x) + T (t) (3) κt(t) 3

4 and concude that a function of y cannot be equa to a function of x and t uness each is some constant γ. Therefore, Y (y) Y(y) = γ, impying Y (y)+γy(y) = () The second set of boundary conditions in (16) impies that Y () = = Y () (5) The determination of a nontrivia Y(y) mirrors the procedure to determine X(x). It is cear therefore, that γ and is restricted by the requirement for integer m, and Y is restricted to γ = mπ Y(y) = cos mπx γ m (6) Y m (y) (7) Using (3), and reations () and (7), we obtain ( n T π (t)+κ + m π ) T(t) = (8) Therefore, we must have T(t) of the form [ ( n π T(t) = A m,n exp + m π ) ] κt T m,n (t) (9) for some constants A m,n, which can depend on integers m,n. Therefore, using inear superposition of soutions of the form X n (x)y m (y)t mn (t), we obtain a more genera soution satisfying (15) and (16) u(x,y,t) = [ ( n π A m,n exp + m π ) ] κt cos nπx cos mπy (3) m,n This satisfies initia condition u(x,y,) = m,n A m,n cos nπx cos mπy (31) Therefore, if φ(x,y) in (17) is expressibe in the form (31), we wi have a representation of the soution to the Neumann probem, which is known to be unique from using energy method.. From Fourier series theory, which we wi soon review, this is not much of a restriction on φ(x,y). Very genera functions φ in the domain (,) (,) have the representation (31) for suitaby chosen A m,n, which of course depends on φ.

5 1.3 Robin Boundary Condition We continue discussion of method of separation of variabe for the case of Robin boundary condition. We wi iustrate this for the 1-D wave equation (1), though there is a parae for every step for heat equation (see Text, section.3). Our boundary conditions u x (,t) a u(,t) = = u x (,t)+a u(,t) (3) for some constant a, a. This is a specia case of the more genera Neumann condition in n-dimension of the form Ω u +a(x)u = for x Ω n It is not difficut to prove (Homework probem) that if we define ( 1 E = ( u) + 1 ) c u t dx+ 1 au (33) for n-dimensiona wave equation with the above Robin boundary condition, then the energy E is a constant in time. This can be used to prove uniqueness of soution to the Robin probem when a(x) >. We now return to the showing the existence of souton to 1-D wave equation with Robin condition (3) with separation of variabe technique. The initia conditions wi be the same as before for the Dirichet probem and is given by (3). As before when we try simpe soutions of the form X(x)T(t), we obtain Ω X (x)+λx(x) = (3) T (x)+λc T(t) = (35) Consider first the case, when λ = β >. Then, soution to (3) is of the form so that X(x) = Ccosβx+Dsinβx (36) X (x)±ax(x) = (βd ±ac)cosβx+( βc ±ad)sinβx (37) At the eft end, x =, we require that = X () a X() = βd a C (38) So, we can sove for D in terms of C. At the right end x =, we have = (βd +a C)cosβ+( βc +a D)sinβ (39) equations (38) and (39) impy the matrix reation ( ) ( a β C = a cosβ βsinβ βcosβ+a sinβ)( D ) () 5

6 Condition for an non-trivia soution impies that the determinant of the coefficient matrix shoud be, impying (β a a )tanβ = (a +a )β (1) uness cosβ =, which is consistent with (38) and (39) ony when β = a a. Any root β > of the transcendenta equation (1) woud give us an eigenvaue λ = β. Determining reation between C and D in (), it foows that the corresponding non-trivia soution X(x), caed an eigen-function, is ( X(x) = C cosβx+ a ) β sinβx () for nonzero C. Finding β expicity from the transcendenta equation (1) is not possibe. However, it is convenient to rewrite it in the form tanα = (a +a 1 )α α a a, where α = β (3) Case 1: a,a > (Radiating boundary conditions) We now graphicay pot each side of (3), as shown in Figure 1, for the case a >, a >, physicay corresponding to radiation at both ends. The intersection of the two graphs determine eigenvaues {λ n } n= as shown in the Figure. 6 x Figure 1: Pot of eft and right side of (3) as function of α. The α vaues of the n-th intersection point, starting from n =, determine eigenvaues λ n = β n = α n. Since the n-th intersection point has α (nπ,(n + 1)π) and for arge n approaches nπ, it foows n π < λ n < (n+1) π, () im λ n n π n = (5) 6

7 The arge n-imit corresponds to setting a = a =, i.e. we get the same eigen vaues as for the Neumann probem Case : a <,a >, but a +a > Physicay, this corresponds to absorption at x =, radiation at x =,but more radiation than absorption. In this case we pot both eft and right sides of (3), as shown in Figures or 3, depending on the reative sizes of a and a. The intersection points determine a discrete set of α n, which are the α vaues of the intersection point. This determine eigenvaues λ n = βn = α n. Note that once again reations () and (5) ( hod. However, in the case corresponding to Figure 3, there is no eigenvaue λ in the interva, π ). 6 x Figure : Pot of eft and right side of (3) as function of α, when a <. a >, a +a >, when a +a > a a. There is an eigenvaue in the interva (, π ony if the rationa curve crosses the first branch of the tangent curve.since the rationa curve has a singe maximum, this corssing can happen ony if the sope of the rationa curve is greater than the sope of the tangent curve at the origin. This ( eads to the condition a +a > a a. This is the condition that there is an eigenvaue λ, π ). Non-positive eigenvaues: If λ = γ <, then the soution to (3) is ) X(x) = Ccoshγx+Dsinhγx. (6) Satisfying the Robin boundary conditions at x = and x =, in the same way as before, we 7

8 x Figure 3: Pot of eft and right side of (3) as function of α, when a <. a >, a +a > and a +a < a a. find that in this case the eigenvaue equation is tanhγ = (a +a )γ γ +a a (7) So, we ook for intersection of the two graphs on the eft and right of (7); if there is such an intersection, the corresponding eigenfunction is X(x) = coshγx+ a γ sinhγx (8) Severa different cases are possibe. First case, is when both a,a >, i.e. there is radition at both ends. In that case, the two curves cannot intersect, since one is positive and the other is negative, and there can be no γ satisfiying (7). The second case is a < and a >, with a +a >, i.e. more radiation than absorption. Additionay, if we have a +a < a a then the sope of the tanh curve at the origin is and it exceeds the sope of the rationa curve, which is a+a a a. In this case, there is an intersection (see Figure ); otherwise there is no intersection. It is to be noted that for this case the missing eigenvaue λ has now shifted to a negative eigenvaue. As an exercise, you can show that a zero eigenvaue are possibe if and ony a +a = a a. 8

9 1.5 1 x Figure : Pot of eft and right side of (7) as function of γ, when a <. a >, a +a > and a +a < a a. 1. Summary for Robin BC for 1-D Wave equation A genera soution obtained from separation of variabe is of the form u(x,t) = n T n (t)x n (x) (9) where for λ n > } } T n (t) = A n cos{ λn ct +B n sin{ λn ct (5) and for λ n < (from (35) ] T n (t) = A n cosh[ λn ct ] +B n sinh[ λn ct (51) where eigenvaue λ n is determined as described graphicay in the ast subsection, and corresponding eigen fuctions are determined by () for λ n > and (8) for λ n <. The initia conditions satisfied by (9) is u(x,) = n A n X n (x) ; u t (x,) = n λn cb n X n (x) (5) We wi earn ater from genera Sturm-Liouvie theory that the set of eigen functions {X n } n is compete, meaning that (), with appropriate choice of A n and B n is abe to represent any φ and ψ, with some minima restrictions. 9

10 Separation of variabe in Circuar Geometry Separation of variabe is not restricted to rectanguar geometry. The method of soution provides soution to either Lapace, heat or wave equation and other reated equations in circuar (in - D) and spherica geometries (in 3-D). We iustrate this for Lapace s equation in a circe with Dirichet boundary condition..1 Lapace s equation in a circe Consider the foowing -D probem: u = ; for x < 1 (53) with Dirichet boundary condition on x = 1. It is convenient in this case to introduce poar coordinates (r, θ). Using expression for Lapacian in poar coordinates, u = u(r, θ) satisfies The Dirichet boundary condition is u = u rr + 1 r u r + 1 r u θθ = for r < 1 (5) We seek a separation of variabe soution to (5) in the form u(1,θ) = f(θ) (55) u(r, θ) = R(r)Θ(θ) (56) Pugging into (5) and dividing the resut by R(r)Θ(θ)/r, we obtain ( Θ (θ) R Θ(θ) = (r) r R(r) + 1 R ) (r) r R(r) (57) Now, since eft and right sides of (57) are functions of θ and r respectivey, it foows that each side is some constant λ. Therefore, Θ(θ), R(r) satisfy: Θ (θ)+λθ(θ) = (58) R (r)+ 1 r R (r) = λ rr(r) (59) Now, since the soution u(r,θ) sought is univaued in the disk of radius 1, it foows that u(r,θ+ π) = u(r,θ). Hence, we must have Θ(θ +π) = Θ(θ) (6) Now, consider the case λ <. Then the soution to (58) is of the form ( Θ(θ) = Aexp ) ( λθ ) λθ +Bexp (61) But this does not satisfy (6) regardess of the vaue of A and B, except for the trivia case (A,B) = (,). Therefore, we rue this out. 1

11 For λ =, we have Θ(θ) = Aθ +a (6) This does not satisfy (6), uness A =. However, note that it does not rue out a nonzero a. Now, comes the case of λ = µ >. Then, the soution to (58) is Then condition (6) for any θ impies Θ(θ) = acosµθ +bsinµθ (63) µ = m > an integer impying λ = λ m m (6) Therefore, Θ(θ) = Θ m (θ) a m cosmθ +b m sinmθ (65) Since m = in this formua incudes the case of λ =, (65) provides a the eigen functions Θ m for λ = λ m = m for integer m. Having determined λ, we go back to (59) to obtain R (r)+ 1 r R (r) m R(r) = (66) r Noting that the equation is homogeneous in r, we ook for soutions of the form R(r) = r α and pug into (66), to discover that nontriva soutions are possibe when α = ±m, i.e R(r) is a inear combination of r m or r m for m >. In the specia case m =, we find R(r) = c + d ogr for constants c and d. However, since the soution u(r,θ) we are seeking is a cassica soution without any singuarities for r < 1, R(r) must be we-behaved in particuar at r =. This rues out r m term for m > and og r for m =. Thus, Taking a inear combination, we obtain for r < 1 R(r) = R m (r) r m for integer m (67) u(r,θ) = a + (a m r m cosmθ +b m r m sinθ) (68) m=1 Assuming that this series converges, the soution is capabe of satisfying Dirichet boundary condition u(1,θ) = f(θ) if f(θ) = a + (a m cosmθ +b m sinmθ) (69) m=1 As we sha see from the theory of Fourier Series, under mid restrictions, quite genera functions f(θ) possess such an expansion, which is caed its Fourier Series. Note aso, an aternate representation of the soution (65) in the form u(r,θ) = a + c m cos(mθ δ m ) (7) m=1 11

12 where a m = c m cosδ m and b m = c m sinδ m. and this can be written as { } u(r,θ) = R C m z m, (71) m= where z = re iθ, RC = a and C m = c m e iδm. The quanitity on the right of (71) is the rea part of an anaytic function of z = x+iy = re iθ and this representation is not surpising since any soution to Lapace s equation in -D, with (x, y) chosen as independent variabes, is aways either the rea or imaginary part of an anaytic function of x+iy. 1