Spring 2012 Polar
Where is it optimal to strike a circular drum? Polar
Daniel Bernoulli (1700-1782) - introduced concept of Bessel functions Leonhard Euler (1707-1783) - used Bessel funtions of both zero and integral orders Friedrich Bessel (1784-1846) - generalized the Bessel function Polar
I wish to ascertain the difference one hears between striking a drum at the center and off center. Two Dimensions Polar vs. Striking Off Center Polar
Rectangular Membrane The wave equation in 2 ( u 2 ) t 2 = u c2 x 2 + 2 u T y 2 ; c = ρ t > 0, 0 x L, 0 < y H Polar
Conditions The fixed boundary conditions are with initial conditions u(0, y, t) = 0, u(l, y, t) = 0 u(x, 0, t) = 0, u(x, H, t) = 0 u(x, y, 0) = f (x, y) u t (x, y, 0) = g(x, y) Polar
Separation of Variables Let u(x, y, t) = F (x)g(y)t (t), then F (x)g(y)t (t) = c 2 (F (x)g(y)t (t) + F (x)g (y)t (t)) which leads to thus obtaining two equations 1 T c 2 T = F F + G G = λ T + c 2 λt = 0. (1) Polar F F + G G = λ (2)
Time Dependence T (t) = A cos ωt + B sin ωt, ω = c λ. Frequency of oscillations for the harmonics are ν = ω 2π = c λ. 2π Polar
Spatial Equations F F = G G This leads to two equations λ = µ, µ < 0. F + µf = 0 (3) Polar G + (λ µ)g = 0. (4)
Apply BCs Similarly, F (0) = 0 = A = 0, F (L) = 0 = B = 0 or sin λx L = 0 F n (x) = B n sin nπx ( nπ ) 2 L, λ =, n = 1, 2,.... L G(0) = 0, G(H) = 0 Polar G m (y) = D m sin mπy H ( mπ ) 2, λ µ =, m = 1, 2,.... H
Yields the product solutions u nm (x, y, t) = (A nm cos ω nm t + B nm sin ω nm t) sin nπx L where, ( nπ λ nm = L ω nm = c (nπ Thus the general solutions is u(x, y, t) = n=1 m=1 L ) 2 ( mπ ) 2 + H ) 2 ( mπ ) 2. + H (A nm cos ω nm t+b nm sin ω nm t) sin nπx L mπy sin H ; sin mπy H. Polar
Fourier Coefficients Initial displacement is u(x, y, 0) = f (x, y), thus f (x, y) = n=1 m=1 Rewriting as a single sum gives A nm sin nπx L mπy sin H. f (x, y) = A n (y) sin nπx L, where n=1 A n (y) = A nm sin mπy H. m=1 Polar
Fourier Coefficients The coefficients of Fourier sine series A n (y) = 2 L A nm = 2 H L 0 H 0 f (x, y) sin nπx L dx, A n (y) sin mπy H dy. Results in a double Fourier sine series, A nm = 4 LH H L 0 0 f (x, y) sin nπx L mπy sin H dxdy. Polar
Fourier Coefficients Initial velocity, u t (x, y, 0) = g(x, y), B nm = g(x, y) = n=1 m=1 4 H ω nm LH 0 L 0 1 B nm sin nπx ω nm L g(x, y) sin nπx L mπy sin H. mπy sin H dxdy. Polar
This completes the full solution of the vibrating rectangular membrane problem: u(x, y, t) = n=1 m=1 (A nm cos ω nm t+b nm sin ω nm t) sin nπx L A nm = 4 H LH 0 B nm = 4 ω nm LH L 0 H L 0 ω nm = c f (x, y) sin nπx L mπy sin H dxdy g(x, y) sin nπx mπy sin 0 L H dxdy (nπ ) 2 ( mπ ) 2 + L H mπy sin H, Polar
ϕ nm (x, y) = sin nπx L Nodal lines occur when ϕ nm (x, y) = 0. mπy sin H Polar
sin πx L sin πy H n = m = 1 = 0, x = 0, L; = 0, y = 0, H [3] Polar
[3] Polar n = 3, m = 2 sin 3πx L = 0, x = 0, L 3, 2L 3, L; sin 2πy H = 0, y = 0, H 2, H
Circular Membrane The wave equation in polar coordinates 2 [ u 2 t 2 = u c2 r 2 + 1 u r r + 1 2 ] u r 2 θ 2 ; t > 0, 0 r a, 0 θ 2π Polar [3]
Conditions The fixed boundary conditions are The periodic conditions are with initial conditions u(a, θ, t) = 0 t > 0, 0 θ 2π. u(r, θ, t) = u(r, θ + 2π, t) u θ (r, θ, t) = u θ (r, θ + 2π, t), Polar u(r, θ, 0) = f (r, θ) u t (r, θ, 0) = g(r, θ)
Separation of Variables Let u(r, θ, t) = R(r)Θ(θ)T (t), Θ + n 2 Θ = 0 (5) T + c 2 λ 2 T = 0 (6) r 2 R + rr + (r 2 λ 2 n 2 )R = 0. (7) Polar
Simple Equations The solutions to (5) and (6) are Θ n (θ) = A n cos nθ + B n sin nθ; T (t) = C nm cos ω nm t + D nm sin ω nm t, ω nm = c λ nm. Polar
Radial Equation The solutions to r 2 R + rr + (r 2 λ 2 n 2 )R = 0, are Bessel functions. Polar Figure: A few Bessel Functions with their zeros, z nm [7].
Apply BCs u(a, θ, t) = 0 for t > 0 and 0 θ 2π = R(a) = 0. Since we expect solutions to be finite at the center, therefore R(r) = CJ n ( λr). Polar
Apply BCs Since, R(a) = 0, J n ( λa) = 0. Listing a few of the mth zeros of J n z nm n = 0 n = 1 n = 2 n = 3 m = 1 2.4048 3.8317 5.1356 6.3802 m = 2 5.5201 7.0156 8.1472 9.7610 m = 3 8.6537 10.173 11.620 13.015 m = 4 11.792 13.324 14.796 16.223 Polar Table: Approximate location of the zeros of Bessel functions of the first kind, J n (z nm ) = 0 [7].
Denoting that z nm is the mth zero of J n (x), then, J n ( λa) = 0, tells us that λa = znm λ nm = ( znm Substituting this into R(r) gives a ) 2. R(r) = J n ( znm a r ). Polar
General Solution u(r, θ, t) = J n (λ nm r) [A nm cos nθ + B nm sin nθ] cos ω nm t n=0 m=1 +J n (λ nm r) [C nm cos nθ + D nm sin nθ] sin ω nm t. Polar
The circular membrane harmonics are given by ( ) z nm r Φ(r, θ) = cos(nθ)j n. a Nodal curves occur when cos nθ = 0 or J n ( z nmr a ) = 0. Polar Figure: Vibrational Modes of a Circular Membrane with dashed nodal curves [3].
Nodal Circles We wish to find values of r such that z nmr a Bessel function. Thus, z nm r a = z nj, j m r = z nj z nm a, r a. is a zero of the Polar
[3] For m = 2, we have two circles, r = a and r = z n1 z n2 for each n. We will need to calculate r for each n = 0, 1, 2 by using the table. For n = 0, For n = 1, and for n = 2, r = z n1 z n2 = 2.4048 5.5201 a 0.4356a. r = 3.8317 7.0156 a 0.5462a, r = 5.1356 8.1472 a 0.6304a. Polar
General Solution u(r, θ, t) = J n (λ nm r) [A nm cos nθ + B nm sin nθ] cos ω nm t n=0 m=1 +J n (λ nm r) [C nm cos nθ + D nm sin nθ] sin ω nm t Polar
Initial displacement, u(r, θ, 0) = f (r, θ), u(r, θ, 0) = J nm (λ nm r) [A nm cos nθ + B nm sin nθ]. n=0 m=1 Let a n (r) = A nm J n (λ nm r) and b n (r) = B nm J n (λ nm r). m=1 m=1 Rewriting the equation gives u(r, θ, 0) = [a n (r) cos nθ + b n (r) sin nθ], where n=0 a n (r) = 1 π 2π 0 f (r, θ) cos nθdθ, Polar b n (r) = 1 π 2π 0 f (r, θ) sin nθdθ.
Since a n (r) = Similarly, A nm = B nm = m=1 A nm J n (λ nm r), 2 a 2 [J n+1 (z n )] 2 2 a 2 [J n+1 (z n )] 2 a 0 a 0 a n (r)j n (λ nm r)dr. b n (r)j n (λ nm r)dr Polar
Intitial velocity, u t (r, θ, 0) = g(r, θ), C nm = D nm = 2 ω nm a 2 [J n+1 (z n )] 2 2 ω nm a 2 [J n+1 (z n )] 2 a 0 a 0 c n (r)j n (λ nm r)dr, d n (r)j n (λ nm r)dr, Polar
Axisymmetric Solution The solution independent of θ, u(r, t) = m=1 ( z0m r ) J 0m [α m cos ω 0m t + β m sin ω 0m t] a. Notice for n 1, each vibrational mode has nodal curves that pass through the center so none of these modes can be excited. Polar
Conclusion Where is it optimal to strike a drum? Off Center Polar
Acknowledgements Polar Dr. Russel Herman, Allison Martin, Navid Sharifian, and my friends and family!
A. Bruder. Two mathematical models for the tympanic membrane. pages 157 166, 2005. http://mtbi.asu.edu/files/two_mathematical_ Models_for_the_Tympanic_Membrane.pdf. R. Herman. An introduction to fourier and complex analysis with applications to the spectral analysis of signals. http://people.uncw.edu/hermanr/mat367/ FCABook/Book2010/FCA_Main.pdf, March 2012. R. Herman. An introduction to mathematical physics via oscillations. Polar http://people.uncw.edu/hermanr/phy311/ MathPhysBook/MathPhys_Main.pdf, March 2012. M. Kac. Can one hear the shape of a drum? The American Mathematical Monthly, 73:1 23, 1966.
http://mathdl.maa.org/images/upload_library/ 22/Ford/MarkKac.pdf. O. Shipilova. Differential equations lecture vi, February 2008. https://noppa.lut.fi/noppa/opintojakso/ bm20a2101/luennot/lecture_4.pdf. A. Wazwaz. Partial Differential Equation and Solitary Wave Theory. Higher Education, Beijing, 2009. D. Young. An introduction to partial differential equations in the undergraduate curriculum. https://http: //www.math.hmc.edu/~ajb/pcmi/lecture1.pdf. [4] [6] [3] [2] [1] [5] [7] Polar