VIBRATIONS OF A CIRCULAR MEMBRANE

Transcription

1 VIBRATIONS OF A CIRCULAR MEMBRANE RAM EKSTROM. Solving the wave equation on the isk The ynamics of vibrations of a two-imensional isk D are given by the wave equation..) c 2 u = u tt, together with the bounary conition u = on D an initial conitions ux, y, ) = φx, y) an u t x, y, ) = ψx, y). If D is the isk of raius κ, it is convenient to make a transformation into polar coorinates see 2.). Recasting this PDE into polar coorinates we get:..2) u rr + r u r + r 2 u θθ = c 2 u tt, with bounary conition uκ, θ, t) = an initial conitions ur, θ, ) = φr, θ) an u t r, θ, ) = ψr, θ). Aitionally the isk D = {r, θ): r [, κ], θ [, π]} R 2 is escribe more easily. One approach to solve this partial ifferential equation is via the metho of separation of variables. Assume that u has the form ur, θ, t) = Rr)Θθ)T t), then substituting into the above an iviing by u we have: R R + R rr + Θ r 2 Θ = T..3) c 2 T. The right-han term must be a constant since ifferentiating the right-han sie by r an θ yiels zero an ifferentiating the left-han sie by t yiels zero, hence the erivative of the right-han term is zero an thus must be a constant α C. By a similar argument Θ /Θ is also a constant β C. An so we have a set of orinary) ifferential equations:..4)..5)..6) T + αc 2 T =, Θ + βθ =, R + r R + α βr ) 2 R =. subject to the bounary ata: { Rκ) = an R) <, Θθ) = Θθ + 2π). To solve..5 we see that β = n 2 for n N see 2.2), an so has solution..7) Θ n θ) = A n cosnθ) + B n sinnθ). Thus the raial ifferential equation..6 becomes..8) r 2 R + rr + αr 2 n 2 )R =. Now if α =, the solution is given by Rr) = c r n + c 2 r n.

2 however as Rκ) =, we get that c = c 2 = by the linear inepenence of the exponential) an thus have the unesirable) trivial solution. So assume α. If we let w = αr, then we can write with a slight abuse of notation, Rw) = R αr) so that the ifferential equation becomes..9) w 2 R + wr + w 2 n 2 )R =. This ifferential equation is calle the Bessel ifferential equation. There are several ways to solve the Bessel equation an as w = is a regular singular point we can solve it via Frobenius metho aroun w =. Assume that Rw) = w m c kw k c ) is a solution. Then by inserting Rw) into the ifferential equation we get m + k)m + k )c k w m+k + m + k)c k w m+k + c k 2 w m+k n 2 c k w m+k =. Upon collecting terms this becomes c k [m + k)m + k ) + m + k) n 2 ]w m+k + c k 2 w m+k =. Hence, we require that c m 2 n 2 ) =, so m = ±n. For now, let m = z be a complex number not a negative integer at the en we ll just set z = ±n), then we get k=2 c 2z + ) = an, c k k 2 + 2kz) + c k 2 = for k 2. Thus c =. Hence we have the recurrence relation c k efine by c =, c k = k2z + k) c k 2 for k 2, with initial conition c. The fact that c = forces all c 2t = for t 2, so we only nee to consier even k 2. But it is easily seen by substitution that c 2k = k=2 ) k c 4 k k!k + z)k + z )... z + ). The stanar choice for c is c = 2 z Γz + )) where Γz) is the gamma function), then our solution is written as..) w ) z ) k w ) 2k J z w) = for z C. 2 k!γz + k + ) 2 This is calle the Bessel function of the first kin of orer z an we mention in passing that the raius of convergence for J z is. See 2.3 for the proof that for n N, then J n w) = ) n J n w). As Bessel s equation is a secon orer ifferential equation, we require a secon solution linearly inepenent from J n. A natural choice woul be J n w) however when n an integer, J n w) an J n w) are not linearly inepenent. For noninteger n, J n w) an J n w) are linearly inepenent this can be verifie by computing their Wronskian). To fin a secon linearly inepenent solution in general, we can use the fact that cosnπ) = ) n an so cosnπ)j n w) J n w) is also a solution to Bessel s equation an is ientically zero when 2

3 n an integer). For non-integer n, if we put Y n w) = cosnw)j nw) J n w), sinnw) then Y n w) is a solution of Bessel s equation linearly inepenent of J n w). The case for integer n, we can efine Y n by Y n w) = lim Y z w) for z a non-integer complex number converging to n. This z n limit can teiously) be evaluate using L Hospital s rule. Then this Y n is also linearly inepenent from J n again can be verifie by computing its Wronskian which will not be one here). Such Y n are calle the Bessel functions of the secon kin. Thus, J n w) an Y n w) are linearly inepenent solutions to Bessel s equation an thus their linear combination form the general solution to Bessel s equation. If we go back to solving our particular wave equation, we have initial an bounary conitions R) < an Rκ) = respectively. As explaine, our general solution is of the form Rw) = c J n w) + c 2 Y n w) wherein our case n N ). However, a significant property of Bessel functions of the secon kin, Y n, is that lim Y n w) = while J n ) is always finite. As we require w R) to be finite, we must exclue Y n w) from forming our solution that is c 2 = ) an so our solution is now of the form Rw) = c J n w) for c a constant. For the bounary conition, we require the following lemma. Lemma.. For n, then J n has only real zeros. Moreoever, J n an J n have infinitely many positive zeros an can be orere < z n < z n2 <... where z nm is the m-th positive zero of J n. Since we nee Rw) = R ακ) = we have that α is real an positive an must be chosen such that J n ακ) =. Write < λ n < λ n2 <... where λ nm κ is the m-th positive zero of J n. So, α = λ nm for n N an m =, 2, 3,.... Hence our solution for the raial part is..) R nm r) = c J n λ nm r) for c a constant. For..4, as αc 2 > we easily see that ) )..2) T mn t) = cos λnm ct + 2 sin λnm ct, is the general solution. Thus a solution to..2 is u nm r, θ, t) = R nm r)θ n θ)t nm t). These u nm are the normal moes of the vibration. By summing over n an m we obtain the general solution + ur, θ, t) = Rr)Θθ)T t) = J λ m r) [ ) ) ] A m cos λm ct + C m sin λm ct m= J n λ nm r) [ ) A nm cosnθ) + B nm sinnθ)) cos λnm ct n= m= ) +C nm cosnθ) + D nm sinnθ)) sin λnm ct ) ].) Next we must consier the initial conitions of u as well as fin formulas for the coefficients. Recall the initial conitions are ur, θ, ) = φr, θ) an t ur, θ, ) = ψr, θ). Using the full solution we obtain 3

4 ..3)..4) ψr, θ) = φr, θ) = J λ m r)a m + m= c λ m J λ m r)c m + m= J n λ nm r)a nm cosnθ) + B nm sinnθ)), n= m= c λ nm J n λ nm r)c nm cosnθ) + D nm sinnθ)). n= m= The expression of this initial ata makes sense if they can be expresse in this manner. As we will fin later, this is the case provie φ, ψ L 2 D). To fin the coefficients we nee a few lemmas on orthogonality. Lemma.2. In L 2 [, π]), the set {cosnθ): n N } {sinnθ): n N} is orthogonal with respect to the stanar L 2 inner prouct. The above lemma can be prove via the metho of Googling trig ientities. Lemma.3. Let n be fixe, then J n λ nm r)j n λ np r)r r = {, if m p, 2λ nm κ 2 [J n λ nm κ)] 2 + n2 2λ nm J n ) 2, otherwise. Proof. Case: m p. Consier fr) = J n λ nm r) an gr) = J n λ np r) where < p < m. These satisfy the Bessel) ifferential equations r 2 2 f r 2 + r f r + λ nmr 2 n 2 )f =, These ifferential equations can be rewritten as r f ) n2 r r r f = λ nmrf, r 2 2 g r 2 + r g r + λ npr 2 n 2 )g =. r g ) n2 r r r g = λ nprg. Multiplying the first equation by g an the secon equation f an subtracting the resulting secon equation from the first we obtain: r f ) g r g ) f = λ np λ nm )rfg. r r r r Integrating over [, κ] an then integrating by parts yiels r f ) g r g ) f r = λ np λ nm )rfg r, = r r r r r f r g r g r f κ κ = λ np λ nm )rfg r. Now, fκ) = gκ) = an f an g are finite at r = as n ), so the left han sie is zero. Hence, λ np λ nm )rfg r =, = J n λ nm r)j n λ np r)r r =. Case: m = p. Now let hr) = J n λ nm r), then we know that h satisfies r h ) + r r r λ nmr 2 n 2 )h =. 4

5 Multiplying by 2rh we can collect erivatives to get r [ r h ) ] 2 + λ nmr 2 n 2 )h 2 = 2λ nm rh 2. r Integrating over [, κ] we have 2λ nm h 2 r r = r h ) 2 + λ nmr 2 n 2 )h 2 κ r, = [J n λ nm r)] 2 r r = κ2 Jn ) 2 λ nm κ) + n2 J n )) 2. 2λ nm r 2λ nm If we require n >, then the last term vanishes since J n ) = for integer n >. Also note that this integral is nonzero. Now, we will completely fin a formula for A nm an just simply state the formulas for the remaining coefficients as they are all obtaine in a similar manner. To fin a formula for A nm let s use φr, θ) = J n λ nm r)a nm cosnθ)) + J n λ nm r)b nm sinnθ)) n= m= n= m= If we multiply both sies by J n λ nm r) cosnθ)r an integrate over D π π φr, θ)j n λ nm r) cosnθ)r θr = J n λ nm r)a nm cosnθ) + B nm sinnθ)))j n λ nm r) cosnθ)r θr. n= m= For sake of clarity, replace the inices in the summation n an m with q an p. To interchange the sum an integral we orinarily woul require certain conitions of the convergence of the series to hol true. However, to be shown later, we are in actuality exploiting the continuity of a certain inner prouct in a certain Hilbert space this is sometimes calle Fourier s trick ). Bearing this in min, we interchange the sum an the integral = q= p= π J q λ qp r)a qp cosqθ) + B qp sinqθ))j n λ nm r) cosnθ)r θr. By orthogonality all the terms vanish except for when q = n an p = m, so all our work simplifies to = A nm J n ) π ) λ nm r) 2 r r cos 2 nθ) θ. Write ξ nm = J n λ nm r) 2 r r, C n = π cos2 nθ) θ an S n = π sin2 nθ) θ, then our complete formula for A nm becomes..5) A nm = φr, θ)j n λ nm r) cosnθ)r µ, n N. C n ξ nm D 5

6 Similarly the remaining coefficients are B nm = φr, θ)j n λ nm r) sinnθ)r µ, n N S n ξ nm D C nm = C n ξ nm c ψr, θ)j n λ nm r) cosnθ)r µ, n N λ nm D D nm = S n ξ nm c ψr, θ)j n λ nm r) sinnθ)r µ, n N. λ nm D These coefficients can be foun using a particular inner prouct an orthonormal basis. In orer to illustrate this we state the following theorem an lemmas. Theorem.4. Let B be an orthonormal system in a Hilbert space H. Then the following are equivalent. ) The orthonormal system B is a Hilbert basis. 2) For all v H such that v, b = for all b H the equality v = hols true. 3) For all v H, v amits the Fourier expansion v = b B v, b b. Lemma.5 Full Fourier series). The set {cosnθ): n N } {sinnθ): n N} is an orthonormal basis for the Hilbert space L 2 [, π]) with the stanar L 2 inner prouct. Lemma.6 Fourier-Bessel series). Let n be fixe. The set {J n λ nm r)} m N is an orthogonal basis for the Hilbert space L 2 [, κ]) with respect to the weighte inner prouct f, g = Now we can prove the following theorem. rfr)gr) r. Theorem.7. The Hilbert space L 2 D) where D = {r, θ): r [, κ], θ [, π]} is the isk has the set J = {J n λ nm r) cosnθ): n N, m N} {J n λ nm r) sinnθ): n N, m N} as an orthogonal basis an with inner prouct, given by f, g = rfg µ. D Proof. The inner prouct given is simply the stanar L 2 R 2 ) inner prouct in polar coorinates or really the weighte inner prouct with weight wr, θ) = r), so it is an inner prouct. To show that J is an orthogonal basis first note that by first applying Lemma.2 an then Lemma.3 we obtain orthogonality. Note that J L 2 D) an consier fr, θ) L 2 D) an suppose it is orthogonal to each element of J with respect to, ). Hence, fr, θ), J n λ nm r) cosnθ) = fr, θ), J n λ nm r) sinnθ) = So in particular, the functions π π π fr, θ) cosnθ) θ, fr, θ)j n λ nm r) cosnθ)r θr =, fr, θ)j n λ nm r) sinnθ)r θr =. π fr, θ) sinnθ) θ, are in L 2 [, κ]) which can be verifie by applying the Cauchy-Schwarz inequality an using f L 2 D)) an are orthogonal to all {J n λ nm r)} n,m N. However, as {J n λ nm r)} n,m N is an 6

7 orthogonal basis for L 2 [, κ]) by Lemma.6, each integral must be zero. Note that for fixe r, it is easy to see that fr, θ) L 2 [, π]). It then follows that fr, θ) is orthogonal to each sinnθ) an cosnθ) for each r [, κ]. As {sinnθ)} n N {cosnθ)} n N is an orthonormal basis of L 2 [, π]) by Lemma.5, we must have fr, θ) = for all r [, κ]. Thus, f = an so it follows by Theorem.4 that J is an orthogonal basis of L 2 D). Hence, if we assume that φ L 2 D), then by Theorem.4 it has the Fourier expansion φ = φ,j n λ nmr) cosnθ) n= m= J n λ nmr) cosnθ) J n λ nm r) cosnθ) + φ,j n λ nmr) sinnθ) n= m= J n λ nmr) sinnθ) J n λ nm r) sinnθ). Equating coefficients we fin..6) A nm = φ, J n λ nm r) cosnθ) J n λ nm r) cosnθ), B nm = φ, J n λ nm r) sinnθ) J n λ nm r) sinnθ). This agrees with our previous erivation in..5 an the remaining coefficients have similar expressions. 2. Appenix 2.. Conversion of the 2D) Laplacian into polar coorinates. By elementary mathematics x = r cosθ), y = r sinθ). Hence, by the chain rule, the partial ifferential for x is x = r r x + θ θ x. Using r = x 2 + y 2 an θ = arctany/x), we have r x becomes, θ = cosθ) an x = sinθ)/r. So the above Similarly, x = cosθ) r sinθ) r θ. y = sinθ) r + cosθ) r θ. Squaring both operators, we obtain the secon partials 2 x 2 = cos2 θ) 2 2 cosθ) sinθ) r2 r rθ + sin2 θ) 2 r 2 θ 2 + sin2 θ) r r 2 y 2 = sin2 θ) 2 2 cosθ) sinθ) 2 + r2 r rθ + cos2 θ) 2 r 2 θ 2 + cos2 θ) r r Then aing we get the 2D Laplacian in polar coorinates. 2 = 2 x y 2 = 2 r 2 + r 7 2 r r 2 θ cosθ) sinθ) r 2 2 cosθ) sinθ) r 2 θ, θ.

8 2.2. Proof for the fact that β N. Assume for now that β R, we seek to fin a general solution to 2.2.) Θ + βθ =, subject to the perioic bounary conition Θθ) = Θθ + 2π) which naturally implies Θ θ) = Θ θ + 2π). Suppose that β <, then we obtain a solution that is a linear combination of sinh an cosh with real argument) which is clearly not perioic in 2π. Now suppose β >, we see that the general solution is of the form ) ) *) Θθ) = v cos βθ + v 2 sin βθ, for v, v 2 C. Then using the perioic bounary conitions we get Θ) Θ2π) = v cos 2π )) β v 2 sin 2π ) β =, Θ ) Θ 2π) = v sin 2π ) β + v 2 cos 2π )) β =. ) v Writing this as a matrix an realizing that for all vectors v = C v 2, we have that 2 ) cos 2π β sin 2π β ) ) sin 2π β ) cos 2π β ) v =. Hence the eterminant is zero, simplifying the eterminant we fin cos 2π β ) =, thus β = n Z. Quickly checking, the case for *) that β = is also vali. Hence β Z an thus β N if β R. Now suppose that β C. Then our solution becomes Θθ) = v e βθ + v 2 e βθ, where β = x + iy. By evaluating a system of equations in a similar manner to the above, we fin that x =, hence β R. So, in conclusion β N Proof that for n N, J n w) = ) n J n w). The gamma function Γz) has poles at all negative integer z. So w ) n ) k w ) 2k J n w) = 2 k!γ n + k + ) 2 w ) n ) k w ) 2k =. 2 k!γ n + k + ) 2 k=n Make the substitution k = m + n, then w ) n ) m+n w ) 2m+n) w ) n = = 2 m + n)!γm + ) 2 2 m= References m= ) m+n w ) 2m Γm + n + )m! 2 = ) n J n w). [] Walter Strauss. Partial Differential Equations: An Introuction. John Wiley & Sons, Inc., 28. [2] Lawrence C. Evans. Partial Differential Equations. American Mathematical Society, 2. [3] Markel Epele García. Bessel Functions an Equations of Mathematical Physics. Euskal Herriko Unibertsitatea, 25. [4] Yum-Tong Siu. Bessel Functions an Vibrating Circular Membrane. Harvar University, 26. 8