The Wave Equation on a Disk

Transcription

1 R. C. Trinity University Partial Differential Equations April 3, 214

2 The vibrating circular membrane Goal: Model the motion of an elastic membrane stretched over a circular frame of radius a. Set-up: Center the membrane at the origin in the xy-plane and let u(r,θ,t) = deflection of membrane from equilibrium at polar position (r,θ) and time t. Under ideal assumptions: ( u tt = c 2 2 u = c 2 u rr + 1 r u r + 1 ) r 2u θθ, < r < a, < θ < 2π, t >, u(a,θ,t) =, θ 2π, t >. y u= 2 Δ 2 u = c u tt a x

3 Separation of variables Setting u(r,θ,t) = R(r)Θ(θ)T(t) leads to the separated boundary value problems r 2 R +rr + ( λ 2 r 2 µ 2) R =, R() finite, R(a) =, Θ +µ 2 Θ =, Θ 2π-periodic, T +c 2 λ 2 T =. We have already seen that the second yields Θ(θ) = Θ m (θ) = Acos(mθ)+Bsin(mθ), µ = m N. For each such µ, we have also seen that the solution to the first is R(r) = J m (λr), where J m is the Bessel function of the first kind of order m.

4 We still have one more boundary condition: R(a) = J m (λa) = λa = α mn, where α mn is the nth positive zero of J m. This means that λ = λ mn = α mn a, and hence R(r) = R mn (r) = J m (λ mn r), m N, n N. Returning to T, we finally find that T(t) = T mn (t) = C cos(cλ mn t)+dsin(cλ mn t).

5 Normal modes Multiplying our results together gives the separated solutions J m (λ mn r)(acos(mθ)+bsin(mθ))(c cos(cλ mn t)+dsin(cλ mn t)). For convenience we split these in two and write u mn (r,θ,t) = J m (λ mn r)(a mn cos(mθ)+b mn sin(mθ))cos(cλ mn t), u mn(r,θ,t) = J m (λ mn r)(a mncos(mθ)+b mnsin(mθ))sin(cλ mn t). Note that, up to scaling, rotation and a phase shift in time, these have the form u(r,θ,t) = J m (λ mn r) cos(mθ) cos(cλ mn t).

6 Remarks Let s compare with the normal modes of the rectangular membrane problem: u(x,y,t) = sin(µ m x)sin(ν n y)cos(λ mn t) The functions R mn (r) are the polar analogs of X m (x) = sin(µ m x), Y n (x) = sin(ν n y). The numbers λ mn = αmn a are analogous to µ m = mπ a and ν n = nπ b. Moral: We have (essentially) replaced sine by J m and the zeros of sine by those of J m.

7 Initial conditions and superposition In order to completely determine the shape of the membrane at any time we must specify the initial conditions u(r,θ,) = f(r,θ), r a, θ 2π (shape), u t (r,θ,) = g(r,θ), r a, θ 2π (velocity). In order to meet these conditions we use superposition to build the general solution u(r,θ,t) = m= n=1 + J m (λ mn r)(a mn cos(mθ)+b mn sin(mθ))cos(cλ mn t) }{{} u mn(r,θ,t) m= n=1 J m (λ mn r)(amn cos(mθ)+b mn }{{ sin(mθ))sin(cλ mnt). } umn (r,θ,t)

8 Othogonality of Bessel functions We will see later that the functions R mn (r) = J m (λ mn r) are orthogonal relative to the weighted inner product That is, f,g = a f(r)g(r)r dr. R mn,r mk = a In addition, it can also be shown that R mn,r mn = J m (λ mn r)j m (λ mk r)r dr = if n k. a J 2 m(λ mn r)r dr = a2 2 J2 m+1(α mn ).

9 Using the orthogonality relations for Bessel and trigonometric functions, one obtains: Theorem The functions φ mn (r,θ) = J m (λ mn r)cos(mθ), ψ mn (r,θ) = J m (λ mn r)sin(mθ), (m N, n N) form a (complete) orthogonal set of functions relative to the inner product f,g = 2π a f(r,θ)g(r,θ)r dr dθ. That is, φ mn,φ jk = ψ mn,ψ jk = for (m,n) (j,k) and φ mn,ψ jk = for all (m,n) and (j,k).

10 Imposing the initial conditions Setting t = in the general solution gives f(r,θ) = u(r,θ,) = J m (λ mn r)(a mn cos(mθ)+b mn sin(mθ)) = g(r,θ) = u t (r,θ,) = = m= n=1 m= n=1 m= n=1 m= n=1 (a mn φ mn (r,θ)+b mn ψ mn (r,θ)), which are called Fourier-Bessel expansions. cλ mn J m (λ mn r)(amn cos(mθ)+b mn sin(mθ)) (cλ mn amn φ mn(r,θ)+cλ mn bmn ψ mn(r,θ)),

11 Integral formulae for a mn and b mn Using orthogonality, the usual argument gives a mn = f,φ mn φ mn,φ mn = 2π a 2π a f(r,θ)j m (λ mn r)cos(mθ)r dr dθ. Jm 2 (λ mnr) cos 2 (mθ)r dr dθ for m, n 1. Using the complementary orthogonality relation, the integral in the denominator is equal to 2π cos 2 (mθ)dθ a πa 2 J1 2(α n) if m =, Jm 2 (λ mnr)r dr = πa 2 2 J2 m+1(α mn ) if m 1.

12 We finally find that a n = a mn = 1 πa 2 J 2 1 (α n) 2π a 2 πa 2 J 2 m+1 (α mn) 2π a and likewise (using ψ mn instead) b mn = for m,n N. 2 πa 2 J 2 m+1 (α mn) f(r,θ)j (λ n r)r dr dθ, 2π a f(r,θ)j m (λ mn r) cos(mθ)r dr dθ, f(r,θ)j m (λ mn r) sin(mθ)r dr dθ,

13 Integral formulae for a mn and b mn The same reasoning using the Fourier-Bessel expansion of g(x,y) yields a n = a mn = b mn = for m,n N. 1 πcα n aj 2 1 (α n) 2 πcα mn aj 2 m+1 (α mn) 2 πcα mn aj 2 m+1 (α mn) 2π a 2π a 2π a g(r,θ)j (λ n r)r dr dθ, g(r,θ)j m (λ mn r) cos(mθ)r dr dθ, g(r,θ)j m (λ mn r) sin(mθ)r dr dθ, This (almost) completes the statement of the general solution to the vibrating circular membrane problem!

14 Remark Since cos = 1 and sin = we have m= n=1 J m (λ mn r)(a mn cos(mθ)+b mn sin(mθ))cos(cλ mn t) = a n J (λ n r)cos(cλ n t) + (as above) n=1 } {{ } m= m=1 n=1 Note that there are really no b n coefficients. This is the true form of the first series in the solution. Analogous comments hold for the second series.

15 Remark If f(r,θ) = f(r) (i.e. f is radially symmetric), then for m a mn = ( ) = ( ) 2π a a dr f(r)j m (λ mn r) cos(mθ)r dr dθ 2π cos(mθ) dθ =, } {{ } and b mn =, too. That is, there are only a n terms. Likewise, if g is radially symmetric, then for m and there are only a n terms. a mn = b mn =,

16 Example Solve the vibrating membrane problem with a = c = 1 and initial conditions f(r,θ) = 1 r 4, g(r,θ) =. Because g(r,θ) =, we immediately find that a mn = b mn = for all m and n. Because f is radially symmetric, we only need to compute a n. Since a = 1, λ mn = α mn, so 1 a n = πj1 2(α n) = 2 J 2 1 (α n) 2π 1 1 f(r)j (α n r)r drdθ (1 r 4 )J (α n r)r dr } {{ } substitute x=α n r

17 2 αn = (1 x4 α 2 n J2 1 (α n) = 2 α 2 n J2 1 (α n) αn α 4 n xj (x)dx } {{ } A ) J (x)x dx 1 αn α 4 n x 5 J (x)dx. } {{ } B According to earlier results αn α n A = xj (x)dx = xj 1 (x) = α n J 1 (α n ), αn B = x 5 J (x)dx = x 5 J 1 (x) 4x 4 J 2 (x)+8x 3 α n J 3 (x) = α 5 nj 1 (α n ) 4α 4 nj 2 (α n )+8α 3 nj 3 (α n ).

18 It follows that a n = so that finally u(r,θ,t) = 2 α 2 n J2 1 (α n) n=1 ( A 1 ) α 4 B n = 8(α nj 2 (α n ) 2J 3 (α n )) α 3 n J2 1 (α, n) 8(α n J 2 (α n ) 2J 3 (α n )) α 3 n J2 1 (α J (α n r) cos(α n t). n) Remark: This solution can easily be implemented in Maple, since the command BesselJZeros(m,n) will compute α mn numerically.