BOUNDARY PROBLEMS IN HIGHER DIMENSIONS. kt = X X = λ, and the series solutions have the form (for λ n 0):

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1 BOUNDARY PROBLEMS IN HIGHER DIMENSIONS Time-space separation To solve the wave and diffusion equations u tt = c u or u t = k u in a bounded domain D with one of the three classical BCs (Dirichlet, Neumann, Robin; all homogeneous) and standard initital conditions (ICs), one can employ the separation of variables method. Let Then u = T(t)X(x, y, z). T c T or T kt = X X = λ, and the series solutions have the form (for λ n 0): u(x, t) = [ A n cos( λ n ct) + B n sin( ] λ n ct) X n (x) n or u(x, t) = n A n e λ nkt X n (x), respectively. X n are eigenfunctions of the eigenvalue problem: X = λx with one of the three classical BCs. (1) 1

2 Eigenvalue problems of on D Rewrite Green s Second Identity as: ( (u v v u)dv = u v ) n v u ds. () n D For any pair of u, v satisfying the classical BCs, the right hand side of this equation is 0. A boundary condition is called symmetric for the operator on D if it can make the right hand side of Eq.() zero for all pairs of functions u, v that satisfy the boundary condition. The classical BCs are symmetric so that the Laplacian satisfies u v = ( u)v. D D D where D (...) is a shorthand for D (...)dv. In terms of the inner product defined by (f, g) = f(x)g(x), D this can be written as (u, v) = ( u, v). Associated with the inner product, the norm can be defined as f = (f,f). The set C n (D) of real (or complex) functions having continuous n th -order derivatives in the domain D is a vector space

3 over the real (or complex) number field. It is also an inner product space and normed space. Similar to the familiar vector spaces R n (or C n ), there are a number of theorems that describe the properties of the eigenvalues and eigenfunctions of the eigenvalue problem (1) associated with the symmetric (self-adjoint) operator. Theorem: Consider any of the problem (1). All the eigenvalues are real. The eigenfunctions can be chosen to be real valued. The eigenfunctions that correspond to distinct eigenvalues are necessarily orthogonal. All the eigenfunctions can be chosen to be real and orthogonal. use self-adjoint property An eigenvalue has multiplicity m if it has m linearly independent eigenfunctions. The eigenfunctions can be orthogonalized by the Graham-Schmidt orthogonalization method. Theorem: All the eigenvalues are positive in the Dirichlet case. All the eigenvalues are positive or zero in the Neumann case, as well as in the Rubin case u/ n + au = 0 (a is a real number ) provided that a 0. use Green s first identity 3

4 Eigenfunction expansions (general Fourier series) A general Fourier series is a series form by all orthogonal eigenfunctions X n of (1) ordered in increasing eigenvalues (i.e. λ 1 λ λ 3...): f(x) = n A n X n (x) with the expansion coefficients defined by A n = (f,x n )/(X n, X n ). The series n A nx n (x) is said to converge in the mean-square sense to f(x) if N f(x) f n (x) 0 as N. D The quantity n=1 f g = (f g, f g) = [ D f g ] 1/ measures the distance between two functions f and g. It is sometimes called the L metric. Theorem: (Least-Square Approximation) Let {X n } be any orthogonal set of functions. Let f <. Let N be a fixed positive integer. Among all possible choices of N constants c 1, c,..., c N, the choice that minimizes 4

5 N f c n X n n=1 is c 1 = A 1,..., c n = A N, where A n = (f,x n )/ X n (1 n N). Proof: Recall that the magnitude (or modulus) of a complex number a is usually written as a (not a ), and a = a a. Denote the error by E N, so that E N = f c n X n = (f,f) (f, c n X n ) ( c n X n, f) + ( c n X n, c m X m ) = f (c n A n + c n A n ) X n + c n X n = ( f ) A n X n + c n A n X n where the sum for n (or m) is over the range 1 to N. The two terms inside the large bracket do not depend on c 1,..., c N. The last term is made 0 by choosing c 1 = A 1,..., c N = A N, the choice that minimizes E N. With the choice c n = A n, N EN = f A n X n 0. n=1 5

6 As this is true for arbitrary N, one gets the inequality: A n X n f. n=1 This is called Bessel s inequality. It is valid as long as f is finite. Theorem: The series A n X n converges to f in the mean-square n=1 sense if and only if A n X n = f. n=1 The equation is called Parseval s equality. Proof: Mean-square convergence means that E N 0 as N. Thus, the equality holds. Definition: The infinite orthogonal set of functions X 1, X,... is called complete if f A n X n = 0 n for all f with f <. Therefore, X 1, X,... is complete if and only if Parseval s equality is true for all f with f <. 6

7 The following theorem is stated without proof. Theorem: (L Convergence & Completeness) The set of eigenfunctions of a symmetric BVP as defined earlier is complete. The eigenfunction expansion converges to f in the mean-square sense in D provided only that f(x) is any function for which f = f(x) is finite. D Note: Due to Parseval s equality, all the Fourier coefficients A n = 0 f = 0. This does not mean that f 0. f can be non-zero at some points. However, we can write f = 0 to label it as belonging to an equivalence class of functions satisfying f = 0. Expansions for special geometric regions In two or three dimensions, it is often more convenient to write the index n as a double or triple index, one for each coordinate, e.g. (l, m, n). The series can be expressed as a double or triple series. Eigenvalues in a rectangular region Consider the domain D = {0 < x < a, 0 < y < b} on the plane. The eigenvalue problem (1) with Dirichlet BC takes the form 7

8 X xx + X yy = λx in D X = 0 on D. With the substitution X = F(x)G(y), the pde can be written as F /F + G /G = λ, and can be split as F = λ x F F(0) = F(a) = 0, G = λ y G G(0) = G(b) = 0, with λ = λ x + λ y. From previous discussions, it is clear that λ x = (mπ/a) (m = 1,,...) and λ y = (nπ/b) (n = 1,,...), so that ( mπ ) ( nπ ) λ = λ mn = +. a b The corresponding eigenfunction is X mn (x, y) = sin(mπx/a) sin(nπy/b). Example: Consider the case a = b = π for the wave equation u tt = c (u xx + u yy ) for 0 < x < a, 0 < y < b with u(x, y, 0) = 0, u t (x, y, 0) = xy(π x)(π y). B nm = 1 c 16 m + n n 3 m 3 π [( 1)n 1][( 1) m 1] Vibration of a drumhead - a disk region If u represents the small displacement of the drum surface from the equilibrium position (u = 0), the problem can be formulated as 8

9 u tt = c u in D = {x + y < a } u = 0 on D u, u t are given functions at t = 0. Following the separation procedure discussed earlier, one can split the temporal and spatial parts u = T(t)X(x). The spatial function X(x) needs to satisfy the eigenvalue problem: X rr + 1 r X r + 1 r X θθ = λx X = 0 on D. Further separation can be made by writing X = R(r)Θ(θ). An ODE for each spatial coordinate is obtained. For θ, the problem is Θ + αθ = 0 with periodic BC. The solution is Θ(θ) = A n cos nθ + B n sin nθ (n = 0, 1,,...), and α = n. For r, the problem is R rr + 1 r R r + (λ n r)r = 0 R(0) finite, R(a) = 0. With the rescaling ρ = λr, the equation can be written as R ρρ + 1 ρ R ρ + (1 n ρ)r = 0. This is Bessel s differential equation of order n. It s solutions need to be found by the power series method. As n is a non-negative integer, the two independent solutions are 9

10 and J n (ρ) = Y n (ρ) = π ( 1) j (ρ/)n+j j!(n + j)!, j=0 ( γ + ln ρ ) J n (ρ) 1 π 1 π n 1 j= n j=0 n j (n j 1)! j!ρ n j ( 1) j (H j + H n+j ) j!(n + j)! ( ρ ) n+j where γ is called Euler s constant and H j = n. J n and Y n are respectively called Bessel function of order n of the first and second kind. So R(ρ) = c 1 J n (ρ) + c Y n (ρ). From the expression for Y n, it is clear that Y n as ρ 0. In order that R be finite at r = 0, the Y n term needs to be dropped. For fixed n, the separated solution is J n ( λr)(a n cos nθ + B n sinnθ). To satisfy the boundary condition at r = a, it is necessary that λ r be a root of J n, namely J n ( λa) = 0. Asymptotically, 10

11 J n (ρ) (ρ πρ cos nπ π ) + O(ρ 3/ ). 4 For each fixed n, J n has an infinite number of roots. The roots ρ nm = λ nm a need to be counted by a different index m, with the arrangement 0 < λ n1 < λ n <... < λ nm <... m = 1,,... The solution for the wave equation can then be written as u(r, θ, t) = (C 0m cos λ 0m ct + D 0m sin λ 0m ct)j 0 ( λ 0m r) m=1 + m,n=1 (C nm cos λ nm ct + D nm sin λ nm ct) J n ( λ nm r)(a nm cos nθ + B nm sinnθ). The initial conditions require that u(r, θ, 0) = φ(r, θ) = C 0m J 0 ( λ 0m r) and + + m,n=1 m=1 C nm J n ( λ nm r)(a nm cos nθ + B nm sin nθ) u t (r, θ, 0) = ψ(r, θ) = m,n=1 D 0m λ0m cj 0 ( λ 0m r) m=1 D nm λnm cj n ( λ nm r)(a nm cos nθ + B nm sinnθ). 11

12 Using the formulas a 0 J n ( λ nm r)j n ( λ np r)rdr = 0 for m p and a [J n ( λ nm r)] rdr = 1 a [J n( λ nm a)], 0 one can find the coefficients through the standard procedure. Note that rdr comes from rdrdθ. Vibrations in a solid ball The Dirichlet eigenvalue problem for a ball with radius a is X rr + r X r + 1 [ 1 r sin θ (sin θx θ) θ + 1 ] sin θ X φφ = λx X(a, θ, φ) = 0. Separating the angular coordinates from the radial coordinate X = Y (θ, φ) R(r), one gets the split problems 1 sin θ (sin θy θ) θ + 1 sin θ Y φφ + αy = 0 and R rr + r R r + (λ α ) R = 0 R(0) finite, R(a) = 0. r The Y equation can be further split with Y (θ, φ) = p(θ)q(φ). 1

13 The equation for q is q φφ + βq = 0, The eigenfunctions are q(φ) = A cosmφ + B sin mφ so that β = m (m = 0, 1,,...). with periodic BC on [ π, π]. The p equation is ( 1 d sin θ d ) ) sin θdθ dθ p + (α m sin p = 0 θ with the conditions p finite at θ = 0, π. With the substitution χ = cos θ, the equation can be cast in the from ( d (1 χ ) d ) ) dχ dχ p + (α m p = 0 1 χ with p finite at χ = ±1. The singular behavior of the equation at χ = ±1 is due to the degeneracy of the coordinate system at the poles and is the reason for the special BC. This equation is known as the associated Legendre equation. It also has to be handled by the power series method. The eigenvalues are α = l(l + 1) where l is an integer m, and the eigenfunctions are of the form P m l (χ) = ( 1)m l l! (1 χ m/ dl+m ) 1) l, dχ l+m(χ 13

14 called the associated Legendre functions. Example: P 0 0 (x) = 1 P 0 1 (x) = x, P 1 1 (x) = (1 x ) 1/ P 0 (x) = 1 (3x 1), P 1 (x) = 3x(1 x ) 1/, P (x) = 3(1 x ) These functions satisfy the orthogonal relationship 1 1 P m l (χ)p m l (χ) dχ = 0 if l l, and 1 [Pl m (χ)] dχ = 1 (l + m)! (l + 1) (l m)!. Combining the θ and φ eigenfunctions, one gets the spherical harmonics Y m l (θ, φ) = P m (cos θ) e imφ l where l = 0, 1,,... and m = l,..., 0,..., l, The sines and cosines have been replaced by the complex exponentials. These functions satisfy the orthogonal relationship π π 0 0 Y m l (θ, φ)y m l (θ, φ) sinθdθdφ = δ ll δ mm 4π (l + m)! l + 1(l m)!. Given α = l(l + 1), the R equation can be written as R rr + ( ) r R l(l + 1) r + λ R = 0. r 14

15 It is similar to the Bessel equation. The difference is in the factor r (instead of 1 r ) in front of the derivative R r. It is possible to convert the equation to the Bessel form with a substitution w(r) = rr(r), i.e. R(r) = r 1/ w(r) and a rescaling ρ = λ r. The result is w ρρ + 1 ) ρ w ρ + (1 s w = 0. ρ where s = l(l + 1) = l + 1. As s is not an integer, the two independent solutions are ( 1) j ( ρ ) j±s J ±s (ρ) = Γ(j + 1)Γ(j + s + 1) where Γ(s) = j=0 0 e t t s 1 dt is the Γ function which has the property Γ(s + 1) = sγ(s). If s is an integer Γ(s + 1) = s!. Since J (l+ 1 ) is singular at ρ = 0, only the solution J l+ 1 should be retained. As ρ, J s again has the asymptotic formula J s (ρ) (ρ πρ cos sπ π ) + O(ρ 3/ ). 4 This formula turns out to be exact when s = ±1/ so that J 1/ = πρ sin ρ and J 1/ = cos ρ. πρ The asymptotic formula shows that J s has infinitely many roots. If the index j (= 1,,...) is used to count the roots 15

16 ρ lj = λ lj a of J l+ 1, the eigenfunctions associated with the eigenvalue λ lj can be expressed as X lmj (r,θ, φ) = J l+ 1 ( λ lj r) r Y m l (θ, φ) l m l. Therefore, the eigenvalue λ lj has multiplicity l + 1. The series solution of the wave equation has the form u(r, θ, φ, t) l = l=0 m= l (A lmj cos λ lj ct + B lmj sin λ lj ct) X lmj. j=1 Note that Y m l = Yl m. To ensure that u is real, the coefficients satisfy A l( m)j = A lmj (same for the B coefficients). Example: Consider a firecracker set off at the center of a spherical enclosure with hard walls. What kind of sound waves will be generated? Suppose that this can be model by the following mathematical problem u tt = c u in D = {x + y + z < a } { 1 for r < ɛ u(x, 0) = 0 u t (x, 0) = δ ɛ (r) = 0 for r > ɛ. Steps: 16

17 The series solution is of the general form u = (A lmj cos β lj ct + B lmj sinβ lj ct)x lmj lmj where β lj = λ lj and X lmj = 1 J l+1/ (β lj r) P m r l (cos θ)e imφ. First, note that the initial conditions do not depend on θ and φ. So the solution has the same property (the expansion coefficients are determined by the initial conditions). This means that the only non-zero terms are those with l = m = 0. The sum is therefore only over j = 1,... For this reason, only the j index will be displayed. Second, u(x, 0) = 0 = j A jx j A j = 0. It is only necessary to determine B j from the condition δ ɛ u t (x, 0) = j B jβ j cx j, as B j = (δ ɛ, X j ) β j c X j = a 0 β j c δ ɛ (r)(1/ r)j 1/ (β j r)r dr a 0 (1/r)J 1/ (β jr)r dr. J 1/ has a very nice connection with the sine function, J 1/ (β j r) = πβ j r sin β jr. As β j a has to be a root of J 1/ (β j r), it is clear from the sine function that β j a = jπ with j = 1,,... The 17 (3)

18 integral in the numerator of Eq.(3) becomes ɛ sin(β j r)rdr = 0 πβ j 1 πβ j βj [sin(β j ɛ) (β j ɛ) cos(β j ɛ)]. The integral in the denominator is a sin (β j r)dr = β j a 0 πβ j πβj = a jπ. Therefore, πβ j B j = and the solution is a c(jπ) 3[sin(β jɛ) (β j ɛ) cos(β j ɛ)], u = a cπ 3 [sin(β j ɛ) (β j ɛ) cos(β j ɛ)] rj 3 j=1 ( ) ( jπct jπr sin sin a a ). Problem Set 11 18

19 Justification for separation of variables Given that the set of eigenfunctions {v m } of in an interval (a, b) is complete, one can prove the following. Theorem: Let D = D 1 D be a rectangle in the xy plane. The set of products {v m (x)w n (y) : v m is an eigenfunction in D 1, w n is an eigenfunction in D, m, n = 1,,...} is a complete set of eigenfunctions for with the given boundary conditions. Proof: Note that each product is an eigenfunction since (v m w n ) = (α m + β n )v m w n. It is also clear that the eigenfunctions of are mutually orthogonal. Suppose that there were an eigenfunction u(x, y) of D other than these products. Then, for some λ, u = λu. If λ were different from every one of the α m + β n, then u is orthogonal to all the products v m w n. So 0 = (u, v m w n ) = [ u(x, y)v m (x)dx]w n (y)dy. By the completeness of w n, u(x, y)v m (x) = 0 y. By the completeness of v m, u(x, y) = 0 (x, y). It is not a valid eigenfunction. 19

20 If λ = α m + β n for some m, n (this could be true for one or several such pairs), consider the difference φ = u c mn v m w n where the sum is over all the m, n pairs for which λ = α m + β n and c mn = (u, v m w n )/ v m w n. As φ is then orthogonal to all the v m w n, it is again 0 according to the earlier agrument. So, u is no more than a linear combination of the old products. 0