Vibrating-string problem
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1 EE-2020, Spring 2009 p. 1/30 Vibrating-string problem Newton s equation of motion, m u tt = applied forces to the segment (x, x, + x), Net force due to the tension of the string, T Sinθ 2 T Sinθ 1 T[u x (x + δx, t) u x (x, t)], External force, F(x, t), Frictional force against the string, βu t, Restoring force, γu, For a small segment of string, xρu tt = T[u x (x + x, t) u x (x, t)] + xf(x, t) xβu t (x, t) xγu(x, t), u tt = α 2 u xx βu t γu + F(x, t),
2 EE-2020, Spring 2009 p. 2/30 Maxwell s equations with total charge and current Gauss s law for the electric field: E = ρ ǫ 0 S E d A = Q ǫ 0, Gauss s law for magnetism: B = 0 S B d A = 0, ( ) Faraday s law of induction: E = κ t B C E d l = κ t Φ B, Ampére s circuital law: B = κµ 0 (J + ǫ 0 E) t C B d l = κµ 0 (I + ǫ 0 t Φ E)
3 EE-2020, Spring 2009 p. 3/30 Helmholtz wave equations For a source-free medium, ρ = J = 0, 2 ( E) = µ 0 ǫ 0 t 2 E, ( E) 2 2 E = µ 0 ǫ 0 t 2 E. When E = 0, one has wave equation, 2 E = µ 0 ǫ 0 2 t 2 E which has following expression of the solutions, in 1D, with v 2 = 1 µ 0 ǫ 0 = c 2. E = ˆx[f + (z vt) + f (z + vt)], plane wave solutions: E + = E 0 cos(kz ωt), where ω k = c.
4 EE-2020, Spring 2009 p. 4/30 Wave equation The wave equation contains a second-order time derivative, u tt, it requires two initial conditions u(x, t = 0) = f(x), and u t (x, t = 0) = g(x), For electric current along the wire, with the Kirchhoff s laws, i x + C v t + G v = 0, v x + L i t + R i = 0, The transmission-line equations i xx = CL i tt + (CR + GL) i t + GR i, v xx = CL v tt + (CR + GL) v t + GR v, If G = R = 0, and α 2 = 1/CL, i tt = α 2, i xx, v tt = α 2 v xx.
5 EE-2020, Spring 2009 p. 5/30 The D Alembert solution of the wave equation For the diffusion problems (the parabolic case), we solve the bounded case (0 x L) by separation of variables while solve the unbounded case ( < x < ) by the Fourier transform. For the wave problems (the hyperbolic case), we will do the opposite. PDE: u tt = α 2 u xx, < x <, 0 < t < u(x,0) = f(x) ICs:, < x < u t (x, 0) = g(x) Replace (x, t) by new canonical coordinates (ξ, η), i.e. the moving-coordinate, ξ = x + α t η = x α t the PDE becomes u ξη = 0 with the solution of arbitrary functions of ξ or η, i.e. u(ξ, η) = φ(η) + ψ(ξ),
6 EE-2020, Spring 2009 p. 6/30 The D Alembert solution of the wave equation, cont. In the original coordinates x and t, we have u(x, t) = Φ(x α t) + Ψ(x + α t), this is the general solution of the wave equation. Physically it represents the sum of any two moving waves, each moving in opposite direction with the velocity α. Eg. u(x, t) = Sin(x α t), (one right-moving wave) u(x, t) = (x + α t) 2, (one left-moving wave) u(x, t) = Sin(x α t) + (x + α t) 2, (two oppositely moving waves)
7 EE-2020, Spring 2009 p. 7/30 spherical waves spherical wave: U(r) = A r r 0 exp( ik r r 0 ), where k r r 0 = constant, wavefronts resemble sphere surfaces, intensity: I(r) = A 2 r 2,
8 EE-2020, Spring 2009 p. 8/30 ICs Substitute the general solution into the two ICs, ICs: u(x,0) = f(x) u t (x, 0) = g(x), < x < for arbitrary functions φ and ψ, we have φ(x) + ψ(x) = f(x), α φ (x) + α ψ (x) = g(x), then by integrating from x 0 to x, α φ(x) + α ψ(x) = x x 0 g(ξ)d ξ + K where K is an integration constant.
9 EE-2020, Spring 2009 p. 9/30 D Alembert solution The solutions for φ and ψ are φ(x) = 1 2 f(x) 1 2α ψ(x) = 1 2 f(x) + 1 2α x x 0 g(ξ)d ξ, x x 0 g(ξ)d ξ, The D Alembert solution, u(x, t) = 1 1 [f(x α, t) + f(x + α t)] + 2 2α x+α t x α t g(ξ)d ξ.
10 EE-2020, Spring 2009 p. 10/30 Examples of the D Alembert solution 1. Motion of an initial Sine wave, PDE: u tt = α 2 u xx, < x <, 0 < t < u(x,0) = Sin(x) ICs:, < x < u t (x, 0) = 0 The solution u(x, t) = 1 [Sin(x α t) + Sin(x + α t)] Initial velocity given ICs: u(x,0) = 0 u t (x,0) = Sin(x), < x < The solution u(x, t) = 1 2α x+α t x α t Sin(ξ)d ξ = 1 [Cos(x + α t) Cos(x α t)]. 2α
11 EE-2020, Spring 2009 p. 11/30 The finite vibrating string (standing waves) PDE: u tt = α 2 u xx, 0 < x < L, 0 < t < u(0, t) = 0 BCs: u(l, t) = 0, 0 < t < u(x,0) = f(x) ICs:, 0 x L u t (x, 0) = g(x) Applying separation of variables u(x, t) = X(x) T(t), Substituting this expression into the wave equation, we have two ODEs T α 2 λt = 0 X λx = 0
12 EE-2020, Spring 2009 p. 12/30 The finite vibrating string (standing waves), cont. Only negative values of λ give feasible (nonzero and bounded) solutions, i.e., λ β 2. u(x, t) = [CSin(βx) + DCos(βx)][ASin(αβt) + BCos(αβt)], With the BCs, u(0, t) = X(0) T(t) = 0, D = 0, u(l, t) = X(L) T(t) = 0, β n = nπ L, n = 1,2,... The fundamental solution u n (x, t) = X n (x) T n (t) = Sin( nπx L )[a nsin( nπαt L ) + b ncos( nπαt L )] = R n Sin( nπx L )Cos[nπα(t δ n) ] L
13 EE-2020, Spring 2009 p. 13/30 The finite vibrating string (standing waves), cont. The solution for wave equation u(x, t) = n=1 Sin( nπx L )[a nsin( nπαt L ) + b ncos( nπαt L )] With ICs, u(x, 0) = u t (x,0) = m=1 m=1 b n Sin( nπx L ) = f(x), a n ( nπα L )Sin(nπx L ) = g(x), Using the orthogonality conditions, L 0 a n = b n = 2 L 2 nπα Sin( mπx L L 0 L 0 nπx )Sin( L )d x = L 2 δ mn, we have g(x)sin( nπx )d x, L f(x)sin( nπx )d x. L
14 EE-2020, Spring 2009 p. 14/30 The finite Fourier transforms Finite Sine transform S [f] = S n = 2 L f(x) = n=1 L 0 S n Sin( nπ L x). f(x) Sin( nπ L x)dx, Finite Cosine transform C [f] = C n = 2 L f(x) = C n=1 L 0 f(x) Cos( nπ L x)dx, C n Cos( nπ L x). Transform of derivatives S [f x x] = [ nπ L ]2 S [f] + 2nπ L 2 [f(0, t) + ( 1)n+1 f(l, t)], C [f x x] = [ nπ L ]2 S [f] 2 L [f x(0, t) + ( 1) n+1 f x (L, t)].
15 EE-2020, Spring 2009 p. 15/30 A nonhomogeneous BVP via the Finite Sine Transform PDE: u tt = u xx + Sin(πx), 0 < x < 1, 0 < t < u(0, t) = 0 BCs: u(1, t) = 0, 0 < t < ICs: u(x,0) = 1 u t (x,0) = 0, 0 x 1 Transform the PDE by using the finite Sine transform for the variable x, S n (t) = S [u], d 2 d t 2 S n(t) = (nπ) 2 S n (t) + 2nπ[u(0, t) + ( 1) n+1 u(1, t)] + D n (t), where D n (t) = S [Sin(πx)] = 1 ; n = 1 0 ; n = 2,3,...
16 A nonhomogeneous BVP via the Finite Sine Transform, Cont. Solve the new initial-value problems for ODE, ODE: ICs: d 2 d t 2 S n(t) + (nπ) 2 S n = S n (0) = d d t S n(0) = 0 4 nπ ; n = 1,3,... 0 ; n = 2,4,... 1 ; n = 1 0 ; n = 2,3,... The solutions are S n (t) = [ 4 π 1 π 2 ]Cos(πt) + ( 1 π )2 ; n = 1,... 0 ; n = 2,4,... 4 Cos(nπt) ; n = 3,5, 7,... nπ the solution u(x, t) of the problem is u(x, t) = {[ 4 π 1 π 2 ]Cos(πt)+( 1 π )2 }Sin[πx]+ 4 π n=1 1 2n + 1 Cos[(2n+1)πt]Sin[(2n+1)πx] EE-2020, Spring 2009 p. 16/30
17 EE-2020, Spring 2009 p. 17/30 Wave equation in polar coordinates PDE: u tt = c 2 [u rr + 1 r u r + 1 r 2 u θθ], 0 < r < 1, 0 < t < BCs: u(r = 1) = 0, 0 < t < u(t = 0) = f(r, θ) ICs:, 0 r 1 u t (t = 0) = g(r, θ) By separation of variables, u(r, θ, t) = U(r, θ) T(t), i.e. 2 U + λ 2 U = 0, Helmholtz equation T + λ 2 c 2 T = 0, Simple harmonic motion, where 2 U = [U rr + 1 r U r + 1 r 2 U θθ]
18 EE-2020, Spring 2009 p. 18/30 Helmholtz eigenvalue problem PDE: 2 U + λ 2 U = 0, BCs: U(r = 1, θ) = 0, By separation of variables again, U(r, θ) = R(r) Θ(θ), i.e. r 2 R + rr + (λ 2 r 2 n 2 )R = 0, Θ + n 2 Θ = 0 Bessel s equation The solutions for the Bessel s equation, J n (λr) and Y n (λr), n-th-order Bessel function of the first and second kinds.
19 EE-2020, Spring 2009 p. 19/30 Helmholtz eigenvalue problem, Cont. For R(r = 0) <, we have R(r) = A J n (λr). The BC, R(1) = 0, we have J n (λ) = 0 λ = k n,, where k nm is the m-th root of J n (λ). The corresponding eigenfunctions U nm (r, θ) U nm = J n (k nm r)[asin(nθ) + BCos(nθ)],
20 Cavity modes EE-2020, Spring 2009 p. 20/30
21 EE-2020, Spring 2009 p. 21/30 Hermite/Laguerre/Ince-Gaussian beams M. A. Bandres, J. Opt. Soc. Am. A, 21, , (2004)
22 EE-2020, Spring 2009 p. 22/30 Helmholtz eigenvalue problem, Cont. For the time equation, we have T + k 2 nm c2 T = 0, Simple harmonic motion, T nm (t) = ASin(k nm ct) + BCos(k nm ct). The total solution of our problem is u(r, θ, t) = J n (k nm r)cos(nθ)[a nm Sin(k nm ct) + B nm Cos(k nm ct)] n=0 m=1
23 EE-2020, Spring 2009 p. 23/30 Wave equation in polar coordinates, θ-independent For the special case independent of θ, i.e. n = 0 u(r, t) = m=1 J 0 (k 0m r)[a m Sin(k 0m ct) + B m Cos(k 0m ct)] In particular, we consider the ICs, ICs: u(t = 0) = f(r) = m=1 A mj 0 (k 0m r) u t (t = 0) = 0 = B m, 0 r 1 Using the orthogonality condition of the Bessel functions, i.e. we have 1 0 r J 0 (k 0m r) J 0 (k 0n r)d r = δ nm 1 2 J2 1 (k 0m), A m = 2 J 2 1 (k0m) 1 0 r f(r) J 0 (k 0m r)d r, m = 1,2,...,
24 EE-2020, Spring 2009 p. 24/30 Laplace s equation 2 u = 0, in Cartesian coordinates: x, y, z 2 u(x, y, z) = u xx + u yy + u zz in Cylindrical coordinates: x = r cos θ, y = r sin θ, z 2 u(r, θ, z) = [u rr + 1 r u r + 1 r 2 u θθ + u zz ] in Spherical coordinates: x = r cos θ sin φ, y = r sin θ sin φ, z = r cos φ 2 u(r, θ, φ) = [u rr + 2 r u r + 1 r 2 u φφ + cot φ r 2 u φ + = 1 r 2 [ r (r2 u r ) + 1 sin φ φ 1 r 2 sin 2 φ u θθ] (sin φ u φ ) + 1 sin 2 φ 2 u θ 2 ]
25 EE-2020, Spring 2009 p. 25/30 Laplace equation in spherical coordinates PDE: 2 u = 1 r 2 [ u (r2 r r ) + 1 sin φ φ u(r = R, φ) = f(φ) BCs: lim r u(r, φ) = 0 (sin φ u)] = 0, φ By separation of variables, u(r, φ) = R(r) Φ(φ), i.e. r 2 d2 R d r 2 + 2r d R = k R n(n + 1)R, d r 1 d sin φ d φ (sin φ d Φ d φ ) = kφ, Euler-Cauchy equation Assume R(r) = r a, then a(a 1) + 2a n(n + 1) = 0 = (a n)(a + n + 1), with roots a = n and a = n 1. The solutions for the Euler-Cauchy equation are R n (r) = A n r n + B n 1 r n+1,
26 EE-2020, Spring 2009 p. 26/30 Legendre polynomials Set w = cos φ, then sin 2 φ = 1 w 2, then d d φ = sin φ d d w, We have the Legendre s equation for Φ(φ), d d w [(1 w2 ) d Φ ] + n(n + 1)Φ, d w = (1 w 2 ) d2 Φ d w 2 2w d Φ + n(n + 1)Φ = 0. d w For integer n = 0,1,2,..., the solutions are the Legendre polynomials Φ = P n (w) = P n (cos φ), P 0 (w) = 1, P 1 (w) = w, P 2 (w) = 1 2 (3w2 1), P 3 (w) = 1 2 (5w3 3w), P n (w) = 1 2 n n! d n d w n [(w2 1) n ]
27 EE-2020, Spring 2009 p. 27/30 Laplace equation in spherical coordinates, Cont. The fundamental solutions for the Laplace equation is u n (r, φ) = [A n r n + B n 1 r n+1 ] P n(cos φ), Interior problem: potential within the sphere, u(r, φ) = n=0 A n r n P n (cos φ), BC: u(r, φ) = n=0 A nr n P n (cos φ) = f(φ), The orthogonality condition for the Legendre polynomials 1 1 P n (w) P m (w)d w = 2 2n + 1 δ mn then the coefficients are A n = 2n + 1 2R n π 0 f(φ) P n (cos φ) sin φd φ.
28 EE-2020, Spring 2009 p. 28/30 Dimensionless problem, heat equation PDE: u t = α 2 u xx, 0 < x < L, 0 < t < u(0, t) = T 1 BCs:, 0 < t < u(l, t) = T 2 IC: u(x,0) = sin(πx/l), 0 x L Define U(x, t) = u(x,t) T 1 T 2 T 1, PDE: U t = α 2 U xx, 0 < x < L, 0 < t < U(0, t) = 0 BCs: U(L, t) = 1, 0 < t < IC: U(x,0) = sin(πx/l) T 1 T 2 T 1, 0 x L
29 EE-2020, Spring 2009 p. 29/30 Dimensionless problem, heat equation, Cont. Define dimensionless-space variable ξ = x/l, PDE: U t = (α/l) 2 U ξξ, 0 < ξ < 1, 0 < t < U(0, t) = 0 BCs: U(1, t) = 1, 0 < t < IC: U(ξ,0) = sin(πξ) T 1 T 2 T 1, 0 ξ 1 Define dimensionless-time variable τ = c t = [α/l] 2 t, PDE: U τ = U ξξ, 0 < ξ < 1, 0 < τ < U(0, τ) = 0 BCs: U(1, τ) = 1, 0 < τ < IC: U(ξ, 0) = sin(πξ) T 1 T 2 T 1 = φ(ξ), 0 ξ 1 And the solution to the original problem is u(x, t) = T 1 + (T 2 T 1 )U(x/l, α 2 t/l 2 )
30 EE-2020, Spring 2009 p. 30/30 Normalization of coefficients Nonlinear Schrödinger equation: set η = az and τ = bt, then NLSE becomes then by choosing i U z + D 2 U 2 t 2 + χ 3 U 2 U = 0 ia U η + D 2b 2 2 U τ 2 + χ 3 U 2 U = 0 a = χ 3, b = χ3 2D 2, NLSE is reduced to scaleless form, i U η U τ 2 + U 2 U = 0
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