Lucio Demeio Dipartimento di Ingegneria Industriale e delle Scienze Matematiche

Transcription

1 Scuola di Dottorato THE WAVE EQUATION Lucio Demeio Dipartimento di Ingegneria Industriale e delle Scienze Matematiche Lucio Demeio - DIISM wave equation 1 / 44

2 1 The Vibrating String Equation 2 Second order PDEs 3 The D Alembert solution 4 The Klein-Gordon and the telegrapher s equations 5 Solutions on the real line (Fourier Transforms) 6 Finite domains 7 Non homogeneous problems and resonances Lucio Demeio - DIISM wave equation 2 / 44

3 The vibrating string equation Lucio Demeio - DIISM wave equation 3 / 44

4 The vibrating string equation m= ρ x T (x+ x) β α T ( x) u( x+ x,t) u( x,t) x x+ x Lucio Demeio - DIISM wave equation 4 / 44

5 The vibrating string equation m= ρ x T (x+ x) β α T ( x) u( x+ x,t) u( x,t) x x+ x Newton s law for m along x and y gives 0 = T (x + x) cos β T (x) cos α No horizontal displacement T (x + x) cos β = T (x) cos α T 0 m 2 u = T (x + x) sin β T (x) sin α (1) t2 Lucio Demeio - DIISM wave equation 4 / 44

6 The vibrating string equation With m = ρ x and by substituting from (1): 2 u t 2 = T 0 tan β tan α ρ x = v 2 u x u (x + x, t) x (x, t) v 2 2 u x x 2 2 u t 2 v2 2 u x 2 = 0 (2) which is called wave equation (in 1 D). Lucio Demeio - DIISM wave equation 5 / 44

7 The vibrating string equation With m = ρ x and by substituting from (1): 2 u t 2 = T 0 tan β tan α ρ x = v 2 u x u (x + x, t) x (x, t) v 2 2 u x x 2 2 u t 2 v2 2 u x 2 = 0 (2) which is called wave equation (in 1 D). Generalizing to 2 or 3 dimensions: 2 u t 2 v2 u = 0 The wave equation describes virtually all wave phenomena: sound waves, light waves, waves in fluids, gases and plasmas, waves in solids, etc. Lucio Demeio - DIISM wave equation 5 / 44

8 Quasilinear second order PDEs Lucio Demeio - DIISM wave equation 6 / 44

9 Quasilinear second order PDEs General form in 2 variables The wave equation is a particular case of a second-order PDE; the most general form is a 2 u x + 2 b 2 u 2 x y + c 2 u y = d (3) 2 where a = a(x, y, u, u/ x, u/ y), and so for b, c and d (quasilinearity). Lucio Demeio - DIISM wave equation 7 / 44

10 Quasilinear second order PDEs General form in 2 variables The wave equation is a particular case of a second-order PDE; the most general form is a 2 u x + 2 b 2 u 2 x y + c 2 u y = d (3) 2 where a = a(x, y, u, u/ x, u/ y), and so for b, c and d (quasilinearity). Boundary Value and Initial Value Problems A differential equation possesses a family of solutions, and auxiliary conditions need to be specified in order to assure uniqueness of the solution. There are two sorts of auxiliary conditions: Lucio Demeio - DIISM wave equation 7 / 44

11 Quasilinear second order PDEs General form in 2 variables The wave equation is a particular case of a second-order PDE; the most general form is a 2 u x + 2 b 2 u 2 x y + c 2 u y = d (3) 2 where a = a(x, y, u, u/ x, u/ y), and so for b, c and d (quasilinearity). Boundary Value and Initial Value Problems A differential equation possesses a family of solutions, and auxiliary conditions need to be specified in order to assure uniqueness of the solution. There are two sorts of auxiliary conditions: Initial Value (Cauchy) Problem: when the auxiliary conditions are imposed on the function and its derivatives along a line belonging to the domain of the independent variables; Lucio Demeio - DIISM wave equation 7 / 44

12 Quasilinear second order PDEs General form in 2 variables The wave equation is a particular case of a second-order PDE; the most general form is a 2 u x + 2 b 2 u 2 x y + c 2 u y = d (3) 2 where a = a(x, y, u, u/ x, u/ y), and so for b, c and d (quasilinearity). Boundary Value and Initial Value Problems A differential equation possesses a family of solutions, and auxiliary conditions need to be specified in order to assure uniqueness of the solution. There are two sorts of auxiliary conditions: Initial Value (Cauchy) Problem: when the auxiliary conditions are imposed on the function and its derivatives along a line belonging to the domain of the independent variables; Boundary Value Problem: the solution is sought for in a domain Ω R n and the auxiliary conditions are imposed on the function (or its derivatives) on the boundary Ω of the domain. Lucio Demeio - DIISM wave equation 7 / 44

13 Quasilinear second order PDEs Initial Value and Boundary Value Problems Cauchy Problem: y O Ω γ x u(x, y) = h(x, y) u (x, y) = φ(x, y) x u (x, y) = ψ(x, y) y On γ y Boundary Value Problem: Ω O Ω x u(x, y) = f(x, y) or u (x, y) = ψ(x, y) n On Ω Lucio Demeio - DIISM wave equation 8 / 44

14 Quasilinear second order PDEs Cauchy problem Existence and uniqueness of the solution starting from assigned values of u, u/ x and u/ y on a smooth curve in the (x, y) plane. Let { x = f(s) y = g(s) be a parametric representation of a curve γ in the (x, y) plane. Lucio Demeio - DIISM wave equation 9 / 44

15 Quasilinear second order PDEs Cauchy problem Existence and uniqueness of the solution starting from assigned values of u, u/ x and u/ y on a smooth curve in the (x, y) plane. Let { x = f(s) y = g(s) be a parametric representation of a curve γ in the (x, y) plane. Cauchy problem We must prescribe u(x(s), y(s)) = h(s) (4) u (x(s), y(s)) = φ(s) x (5) u (x(s), y(s)) = ψ(s) y (6) and find the conditions for existence and uniqueness of the solution. Lucio Demeio - DIISM wave equation 9 / 44

16 Quasilinear second order PDEs Let us differentiate (5) and (6) w.r. to s; by adding these to the differential equation (3) we have the following system: 2 u x 2 f (s) + 2 u x y g (s) = φ (s) (7) 2 u x y f (s) + 2 u y 2 g (s) = ψ (s) (8) a 2 u x b 2 u x y + c 2 u = d, (9) y2 where the partial derivatives are to be considered as unknowns. Lucio Demeio - DIISM wave equation 10 / 44

17 Quasilinear second order PDEs Let us differentiate (5) and (6) w.r. to s; by adding these to the differential equation (3) we have the following system: 2 u x 2 f (s) + 2 u x y g (s) = φ (s) (7) 2 u x y f (s) + 2 u y 2 g (s) = ψ (s) (8) a 2 u x b 2 u x y + c 2 u = d, (9) y2 where the partial derivatives are to be considered as unknowns. Unique solution if the determinant f (s) g (s) 0 = 0 f (s) g (s) a 2 b c = a [g (s)] 2 2 b f (s) g (s) + c [f (s)] 2 0 If = 0, γ is called characteristic curve. Lucio Demeio - DIISM wave equation 10 / 44

18 Classification Lucio Demeio - DIISM wave equation 11 / 44

19 Classification The condition a [g (s)] 2 2 b f (s) g (s) + c [f (s)] 2 = 0 is equivalent to a [ g ] 2 (s) f 2 b g (s) (s) f (s) + c = 0 Lucio Demeio - DIISM wave equation 12 / 44

20 Classification The condition a [g (s)] 2 2 b f (s) g (s) + c [f (s)] 2 = 0 is equivalent to a [ g ] 2 (s) f 2 b g (s) (s) f (s) + c = 0 If b 2 a c > 0 we have g (s) f (s) = b ± b2 a c a (10) and the equation is called hyperbolic. Lucio Demeio - DIISM wave equation 12 / 44

21 Classification The condition a [g (s)] 2 2 b f (s) g (s) + c [f (s)] 2 = 0 is equivalent to a [ g ] 2 (s) f 2 b g (s) (s) f (s) + c = 0 If b 2 a c > 0 we have g (s) f (s) = b ± b2 a c a (10) and the equation is called hyperbolic. If b 2 a c < 0 we have no solutions for g (s)/f (s) and the equation is called elliptic Lucio Demeio - DIISM wave equation 12 / 44

22 Classification The condition a [g (s)] 2 2 b f (s) g (s) + c [f (s)] 2 = 0 is equivalent to a [ g ] 2 (s) f 2 b g (s) (s) f (s) + c = 0 If b 2 a c > 0 we have g (s) f (s) = b ± b2 a c a (10) and the equation is called hyperbolic. If b 2 a c < 0 we have no solutions for g (s)/f (s) and the equation is called elliptic If b 2 a c = 0 we have and the equation is called parabolic. g (s) f (s) = b a Lucio Demeio - DIISM wave equation 12 / 44

23 Classification Characteristics Thus, hyperbolic equations possess two families of (real) characteristic curves in the (x, y) plane; Lucio Demeio - DIISM wave equation 13 / 44

24 Classification Characteristics Thus, hyperbolic equations possess two families of (real) characteristic curves in the (x, y) plane; parabolic equations have one family; Lucio Demeio - DIISM wave equation 13 / 44

25 Classification Characteristics Thus, hyperbolic equations possess two families of (real) characteristic curves in the (x, y) plane; parabolic equations have one family; elliptic equations have none (or have complex characteristics); Lucio Demeio - DIISM wave equation 13 / 44

26 Classification Characteristics Thus, hyperbolic equations possess two families of (real) characteristic curves in the (x, y) plane; parabolic equations have one family; elliptic equations have none (or have complex characteristics); If a, b and c depend only on x and y (linear equation), equation (10) can be written as dy dx = b ± b2 a c λ ±(x, y) (11) a and it becomes a set of two differential equations for the characteristic curves. Lucio Demeio - DIISM wave equation 13 / 44

27 Classification Characteristics Thus, hyperbolic equations possess two families of (real) characteristic curves in the (x, y) plane; parabolic equations have one family; elliptic equations have none (or have complex characteristics); If a, b and c depend only on x and y (linear equation), equation (10) can be written as dy dx = b ± b2 a c λ ±(x, y) (11) a and it becomes a set of two differential equations for the characteristic curves. Role of the characteristic curves In time dependent problems, one can say, qualitatively, that the characteristics are the lines along which solutions are transported ; Lucio Demeio - DIISM wave equation 13 / 44

28 Classification Characteristics Thus, hyperbolic equations possess two families of (real) characteristic curves in the (x, y) plane; parabolic equations have one family; elliptic equations have none (or have complex characteristics); If a, b and c depend only on x and y (linear equation), equation (10) can be written as dy dx = b ± b2 a c λ ±(x, y) (11) a and it becomes a set of two differential equations for the characteristic curves. Role of the characteristic curves In time dependent problems, one can say, qualitatively, that the characteristics are the lines along which solutions are transported ; The presence of two families of characteristics in hyperbolic equations corresponds to counter-propagating wave forms; Lucio Demeio - DIISM wave equation 13 / 44

29 Classification Role of the characteristic curves the presence of one family of characteristics in parabolic equations corresponds to relaxation towards a statistical equilibrium (e.g., in gases); Lucio Demeio - DIISM wave equation 14 / 44

30 Classification Role of the characteristic curves the presence of one family of characteristics in parabolic equations corresponds to relaxation towards a statistical equilibrium (e.g., in gases); elliptic equations, which do not possess characteristics, describe static problems (e.g., static electric fields due to a charge distribution, steady-state temperature distributions, linear elasticity problems and more). Lucio Demeio - DIISM wave equation 14 / 44

31 Classification Role of the characteristic curves the presence of one family of characteristics in parabolic equations corresponds to relaxation towards a statistical equilibrium (e.g., in gases); elliptic equations, which do not possess characteristics, describe static problems (e.g., static electric fields due to a charge distribution, steady-state temperature distributions, linear elasticity problems and more). Prototypical equations Wave equation: Heat equation: 2 u t 2 v2 2 u x 2 = 0 b 2 a c = v 2 > 0, hyperbolic equation u t κ 2 u x 2 = 0 b 2 a c = 0, parabolic equation Laplace equation: 2 u x u y 2 = 0 b 2 a c = 1 < 0, elliptic equation Lucio Demeio - DIISM wave equation 14 / 44

32 Classification Canonical form Lu = a 2 u x b 2 u x y + c 2 u y 2 Lucio Demeio - DIISM wave equation 15 / 44

33 Classification Canonical form Lu = a 2 u x b 2 u x y + c 2 u y 2 Polynomial form (a, b, c constants for simplicity) p L (λ) = aλ b λ + c = a(λ λ 1 ) (λ λ 2 ) and, by analogy, ( ) ( ) Lu = a x λ 1 y x λ 2 u y Lucio Demeio - DIISM wave equation 15 / 44

34 Classification Reduction to canonical form (after linear transformation to (ξ, η)): (λ 1 λ 2 R) : Lu 2 u ξ η or also Lu 2 u ξ 2 2 u η 2 (λ 1 = λ 2 R) : Lu 2 u η 2 (λ 1 = λ 2 C) : Lu 2 u ξ u η 2 hyperbolic case parabolic case elliptic case Lucio Demeio - DIISM wave equation 16 / 44

35 The D Alembert solution Lucio Demeio - DIISM wave equation 17 / 44

36 Back to the wave equation Characteristics From equation (11) with a = 1, b = 0, c = v 2 dy dx = ±v x + v t = ξ x v t = η Lucio Demeio - DIISM wave equation 18 / 44

37 Back to the wave equation Characteristics From equation (11) with a = 1, b = 0, c = v 2 dy dx = ±v x + v t = ξ x v t = η x+v t= ξ x-v t= η Lucio Demeio - DIISM wave equation 18 / 44

38 Back to the wave equation Characteristics From equation (11) with a = 1, b = 0, c = v 2 dy dx = ±v x + v t = ξ x v t = η x+v t= ξ x-v t= η Canonical Form The transformation (x, t) (ξ, η) leads to: 2 u t 2 v2 2 u x 2 = 0 = 2 U ξ η = 0 Lucio Demeio - DIISM wave equation 18 / 44

39 Back to the wave equation Characteristics From equation (11) with a = 1, b = 0, c = v 2 dy dx = ±v x + v t = ξ x v t = η x+v t= ξ x-v t= η Canonical Form The transformation (x, t) (ξ, η) leads to: 2 u t 2 v2 2 u x 2 = 0 = 2 U ξ η = 0 General Solution U(ξ, η) = F (ξ) + G(η) (12) u(x, t) = F (x + v t) + G(x v t) (13) Lucio Demeio - DIISM wave equation 18 / 44

40 D Alembert s form of the solution Initial value problem We prescribe the initial data u(x, 0) = h(x) (14) u (x, 0) = ψ(x) t (15) Lucio Demeio - DIISM wave equation 19 / 44

41 D Alembert s form of the solution Initial value problem We prescribe the initial data u(x, 0) = h(x) (14) u (x, 0) = ψ(x) t (15) Equations (12) and (13) give: F (x) + G(x) = h(x) (16) v [F (x) G (x)] = ψ(x) (17) Lucio Demeio - DIISM wave equation 19 / 44

42 D Alembert s form of the solution Initial value problem We prescribe the initial data u(x, 0) = h(x) (14) u (x, 0) = ψ(x) t (15) Equations (12) and (13) give: F (x) + G(x) = h(x) (16) v [F (x) G (x)] = ψ(x) (17) and, by differentiating (16): F (x) = v h (x) + ψ(x) 2 v G (x) = v h (x) ψ(x) 2 v (18) (19) Lucio Demeio - DIISM wave equation 19 / 44

43 D Alembert s form of the solution After integration F (x) = h(x) 2 G(x) = h(x) v 1 2 v x 0 x 0 ψ(λ)dλ (20) ψ(λ)dλ (21) Lucio Demeio - DIISM wave equation 20 / 44

44 D Alembert s form of the solution After integration F (x) = h(x) 2 G(x) = h(x) v 1 2 v x 0 x 0 ψ(λ)dλ (20) ψ(λ)dλ (21) We finally obtain the d Alembert solution of the Initial Value Problem: u(x, t) = F (x + v t) + G(x v t) = = 1 1 [h(x + v t) + h(x v t)] v x+v t x v t ψ(λ)dλ (22) Lucio Demeio - DIISM wave equation 20 / 44

45 D Alembert s form of the solution After integration F (x) = h(x) 2 G(x) = h(x) v 1 2 v x 0 x 0 ψ(λ)dλ (20) ψ(λ)dλ (21) We finally obtain the d Alembert solution of the Initial Value Problem: u(x, t) = F (x + v t) + G(x v t) = = 1 1 [h(x + v t) + h(x v t)] v x+v t x v t ψ(λ)dλ (22) Works well on infinite domains; on a finite domain, we must use it piecewise and take reflections into account. Lucio Demeio - DIISM wave equation 20 / 44

46 Dependence domain and Influence cone Domain of dependence From the d Alembert form (22), we see that u(x, t) depends upon the values of the functions h and ψ in the interval [x v t, x + v t] on the x axis. This is called domain of dependence. Lucio Demeio - DIISM wave equation 21 / 44

47 Dependence domain and Influence cone Domain of dependence From the d Alembert form (22), we see that u(x, t) depends upon the values of the functions h and ψ in the interval [x v t, x + v t] on the x axis. This is called domain of dependence. The meaning is that the solution at position x and time t depends upon the initial data within [x v t, x + v t] and not outside this interval. Lucio Demeio - DIISM wave equation 21 / 44

48 Dependence domain and Influence cone Domain of dependence From the d Alembert form (22), we see that u(x, t) depends upon the values of the functions h and ψ in the interval [x v t, x + v t] on the x axis. This is called domain of dependence. The meaning is that the solution at position x and time t depends upon the initial data within [x v t, x + v t] and not outside this interval. Cone of influence We also see that the initial condition at x = x 0 influences the solution at later times within the cone (a triangle in a plane) delimited by the surfaces x 0 v t = 0 and x 0 + v t = 0. This is called cone of influence. Lucio Demeio - DIISM wave equation 21 / 44

49 Dependence domain and Influence cone Domain of dependence From the d Alembert form (22), we see that u(x, t) depends upon the values of the functions h and ψ in the interval [x v t, x + v t] on the x axis. This is called domain of dependence. The meaning is that the solution at position x and time t depends upon the initial data within [x v t, x + v t] and not outside this interval. Cone of influence We also see that the initial condition at x = x 0 influences the solution at later times within the cone (a triangle in a plane) delimited by the surfaces x 0 v t = 0 and x 0 + v t = 0. This is called cone of influence. These concepts are consistent with the experience of a finite propagation speed. Lucio Demeio - DIISM wave equation 21 / 44

50 Dependence domain and Influence cone t ( x,t) x- vt x+vt Domain of dependence x Lucio Demeio - DIISM wave equation 22 / 44

51 Dependence domain and Influence cone t ( x,t) x - vt 0 0 x +vt x- vt x+vt Domain of dependence x 0 Cone of influence x Lucio Demeio - DIISM wave equation 22 / 44

52 The Klein-Gordon and the telegrapher s equations Lucio Demeio - DIISM wave equation 23 / 44

53 The Klein-Gordon and the telegrapher s equations The Klein-Gordon equation If the vibrating string is subjected to a uniformly distributed elastic force, we obtain the Klein-Gordon equation: 2 u t 2 v2 2 u + γu = 0 (23) x2 where γ is a constant (e.g., proportional to the elastic constant). u( x,t) x Lucio Demeio - DIISM wave equation 24 / 44

54 The Klein-Gordon and the telegrapher s equations The telegrapher s equation Propagation of waves rarely happens without dissipation; when taken into account, we obtain the telegrapher s equation (or transmission line equation): where λ is a damping constant. 2 u t 2 v2 2 u x λ u + γu = 0 (24) t u( x,t) u( x,t) x x Lucio Demeio - DIISM wave equation 25 / 44

55 Solutions on the real line (Fourier Transforms) Lucio Demeio - DIISM wave equation 26 / 44

56 Solutions by Fourier Transforms The wave equation Assume that u(x, t) can be written as a Fourier integral (which means that u(x, t) 0 fast enough as x ± ) u(x, t) = û(k, t) e i k x dk û(k, t) = 1 2π and substitute in the wave equation. One obtains 2 û t 2 + ω(k)2 û = 0 with ω(k) = k v. The general solution is given by û(k, t) = A(k) e iωt + B(k)e iωt u(x, t) e i k x dx Inversion of the Fourier transform gives back the d Alembert solution. Lucio Demeio - DIISM wave equation 27 / 44

57 Solutions by Fourier Transforms The wave equation Assume that u(x, t) can be written as a Fourier integral (which means that u(x, t) 0 fast enough as x ± ) u(x, t) = û(k, t) e i k x dk û(k, t) = 1 2π and substitute in the wave equation. One obtains 2 û t 2 + ω(k)2 û = 0 with ω(k) = k v. The general solution is given by û(k, t) = A(k) e iωt + B(k)e iωt u(x, t) e i k x dx Inversion of the Fourier transform gives back the d Alembert solution. However, the Fourier form says something very important: the solution of the wave equation can be written as a superposition of plave waves with constant group velocity v g = dω/dk = v, so there is no dispersion. Lucio Demeio - DIISM wave equation 27 / 44

58 Solutions by Fourier Transforms The Klein-Gordon Equation Again, let u(x, t) = û(k, t) e i k x dk û(k, t) = 1 2π and substitute in the Klein-Gordon equation. One obtains 2 û t 2 + ω(k)2 û = 0 with ω(k) = k 2 v 2 + γ. The general solution is again given by û(k, t) = A(k) e iωt + B(k)e iωt u(x, t) e i k x dx Lucio Demeio - DIISM wave equation 28 / 44

59 Solutions by Fourier Transforms The Klein-Gordon Equation Again, let u(x, t) = û(k, t) e i k x dk û(k, t) = 1 2π and substitute in the Klein-Gordon equation. One obtains 2 û t 2 + ω(k)2 û = 0 with ω(k) = k 2 v 2 + γ. The general solution is again given by û(k, t) = A(k) e iωt + B(k)e iωt u(x, t) e i k x dx Now the group velocity v g = dω/dk is not constant and there is dispersion: waves with different k s propagate at different speeds and the shape (signal) acquires a distorsion in time. Lucio Demeio - DIISM wave equation 28 / 44

60 Solutions by Fourier Transforms The telegrapher s equation Again, with u(x, t) = and γ = 0, one obtains û(k, t) e i k x dk û(k, t) = 1 2π 2 û t 2 + 2λ û t + ω(k)2 û = 0 with ω(k) = k v. The general solution is again given by u(x, t) e i k x dx û(k, t) = e λt [ A 1 (k) e Ωt + A 2 (k) e Ωt] for k λ v = e λt [ B 1 (k) e iνt + B 2 (k)e iνt] for k > λ v where Ω = λ 2 k 2 v 2 and ν = k 2 v 2 λ 2. Lucio Demeio - DIISM wave equation 29 / 44

61 The solution shows both dispersion and diffusion. Lucio Demeio - DIISM wave equation 29 / 44 Solutions by Fourier Transforms The telegrapher s equation Again, with u(x, t) = and γ = 0, one obtains û(k, t) e i k x dk û(k, t) = 1 2π 2 û t 2 + 2λ û t + ω(k)2 û = 0 with ω(k) = k v. The general solution is again given by u(x, t) e i k x dx û(k, t) = e λt [ A 1 (k) e Ωt + A 2 (k) e Ωt] for k λ v = e λt [ B 1 (k) e iνt + B 2 (k)e iνt] for k > λ v where Ω = λ 2 k 2 v 2 and ν = k 2 v 2 λ 2.

62 An example Initial condition We compare solutions of the wave equation, the Klein-Gordon equation and the telegrapher s equation with initial condition u(x, 0) = e x2 /2 u (x, 0) = 0 t (black line: wave eq., red line: Klein-Gordon, blue line: telegr.) Lucio Demeio - DIISM wave equation 30 / 44

63 An example Initial condition We compare solutions of the wave equation, the Klein-Gordon equation and the telegrapher s equation with initial condition u(x, 0) = e x2 /2 u (x, 0) = 0 t (black line: wave eq., red line: Klein-Gordon, blue line: telegr.) Lucio Demeio - DIISM wave equation 30 / 44

64 The wave equation on a finite domain Lucio Demeio - DIISM wave equation 31 / 44

65 The wave equation on a finite domain Vector space with scalar product Consider now the wave equation in the interval 0 x l. On a finite domain, we must impose two boundary conditions (the equation is of 2 nd order in x). Let u(0, t) = u(l, t) = 0 (25) be the homogeneous Dirichlet boundary conditions and consider the linear operator L u = 2 u x 2 defined on all twice differentiable functions on [0, l] with boundary conditions (25). Equipped with the scalar product (u, v) = l 0 u(x) v(x) dx it becomes a vector space with scalar product (pre-hilbert space) and it is easily seen that L is a self-adjoint operator w.r. to this scalar product. Lucio Demeio - DIISM wave equation 32 / 44

66 The wave equation on a finite domain Eigenfunctions Then, L has a set of real eigenvalues, kn 2 = (nπ/l) 2, for n = 1, 2,... and real orthogonal eigenfunctions φ n (x), given by 2 φ n (x) = l sin k n x (26) which form an orthonormal basis for the vector space, that is (φ n, φ m ) = l 0 φ n (x) φ m (x) dx = δ mn (27) Lucio Demeio - DIISM wave equation 33 / 44

67 The wave equation on a finite domain Eigenfunctions Then, L has a set of real eigenvalues, kn 2 = (nπ/l) 2, for n = 1, 2,... and real orthogonal eigenfunctions φ n (x), given by 2 φ n (x) = l sin k n x (26) which form an orthonormal basis for the vector space, that is (φ n, φ m ) = l 0 φ n (x) φ m (x) dx = δ mn (27) Eigenfunction expansion Then, any function of this vector space can be expressed as 2 u(x, t) = c n (t) φ n (x) = c n (t) l sin k n x (28) n=1 Lucio Demeio - DIISM wave equation 33 / 44 n=1

68 The wave equation on a finite domain Solution of the wave equation By substituting (28) into the wave equation (2) we obtain { cn(t) φ n(x) v 2 c n(t) φ n(x) } = 0 n=1 { cn(t) φ n(x) + v 2 c n(t) L φ } n(x) = 0 n=1 { cn(t) φ n(x) + v 2 c n(t) kn 2 φ } n(x) = 0 n=1 { cn(t) + ωn 2 c } n(t) φ n(x) = 0 with ω n = v k n n=1 By taking the scalar product with φ m we obtain c m(t) + ωm 2 c m(t) = 0 whose general solution is c m(t) = A m cos ω mt + B m sin ω mt Lucio Demeio - DIISM wave equation 34 / 44

69 The wave equation on a finite domain Solution of the wave equation The general solution of the wave equation (2) can then be written as a superposition of all eigenfunctions as u(x, t) = {A n cos ω nt + B n sin ω nt} φ n(x) (29) n=1 Each of the terms in the sum is called mode of vibration or free vibrations. The general solution is thus a linear superposition of the vibration modes. Lucio Demeio - DIISM wave equation 35 / 44

70 The wave equation on a finite domain Solution of the wave equation The general solution of the wave equation (2) can then be written as a superposition of all eigenfunctions as u(x, t) = {A n cos ω nt + B n sin ω nt} φ n(x) (29) n=1 Each of the terms in the sum is called mode of vibration or free vibrations. The general solution is thus a linear superposition of the vibration modes. Initial conditions The wave equation must be accompanied by the initial conditions (14)-(15): u(x, 0) = h(x) u (x, 0) = ψ(x) t from which the coefficents A n and B n can be determined: A n = l 0 h(x) φ n(x) dx B n = 1 ω n l 0 ψ(x) φ n(x) dx Lucio Demeio - DIISM wave equation 35 / 44

71 The wave equation on a finite domain The Klein-Gordon equation and the telegrapher s equation By following similar steps, we may write the general solutions of the Klein-Gordon equation (23) and the telegrapher s equation (24): u(x, t) = {A n cos ω n t + B n sin ω n t} φ n (x) (K.-G.) (30) n=1 u(x, t) = e λt n=1 {A n cos ν n t + B n sin ν n t} φ n (x) (telegr.) (31) where the frequencies now are ω n = v 2 k 2 n + γ and ν n = v 2 k 2 n λ 2. We again observe dispersion in the Klein-Gordon equation and both dispersion and diffusion in the telegrapher s equation. Lucio Demeio - DIISM wave equation 36 / 44

72 The wave equation on a finite domain The Klein-Gordon equation and the telegrapher s equation By following similar steps, we may write the general solutions of the Klein-Gordon equation (23) and the telegrapher s equation (24): u(x, t) = {A n cos ω n t + B n sin ω n t} φ n (x) (K.-G.) (30) n=1 u(x, t) = e λt n=1 {A n cos ν n t + B n sin ν n t} φ n (x) (telegr.) (31) where the frequencies now are ω n = v 2 k 2 n + γ and ν n = v 2 k 2 n λ 2. We again observe dispersion in the Klein-Gordon equation and both dispersion and diffusion in the telegrapher s equation. Numerical simulations FiniteDomain.W, FiniteDomain.KG, FiniteDomain.tele, Lucio Demeio - DIISM wave equation 36 / 44

73 Non homogeneous problems Lucio Demeio - DIISM wave equation 37 / 44

74 Non homogeneous problems Generalities Non homogeneous terms arise whenever nonhomogeneous boundary conditions or external forces are present. As customary in these cases, the solution is a sum of the general solution of the homogeneous problem and a particular solution of the complete equation. Lucio Demeio - DIISM wave equation 38 / 44

75 Non homogeneous problems Generalities Non homogeneous terms arise whenever nonhomogeneous boundary conditions or external forces are present. As customary in these cases, the solution is a sum of the general solution of the homogeneous problem and a particular solution of the complete equation. Nonhomogeneous boundary conditions 2 u t 2 u 2 v2 x = 0 2 u(0, t) = 0 u(l, t) = A General solution: u(x, t) = u p(x) + ũ(x, t) with u p(x) and ũ such that v 2 u p = 0 u p(0) = 0 stationary solution u p(l) = A which gives u p(x) = A x/l. 2 ũ t 2 ũ 2 v2 x = 0 2 ũ(0, t) = ũ(l, t) = 0 Lucio Demeio - DIISM wave equation 38 / 44

76 Non homogeneous problems The solution then is: u(x, t) = A x l + {A n cos ω n t + B n sin ω n t} φ n (x) n=1 where now the coefficients A n and B n are given by A n = l 0 [ h(x) A x ] φ n (x) dx B n = 1 l ω n l 0 ψ(x) φ n (x) dx Lucio Demeio - DIISM wave equation 39 / 44

77 Non homogeneous problems The solution then is: u(x, t) = A x l + {A n cos ω n t + B n sin ω n t} φ n (x) n=1 where now the coefficients A n and B n are given by A n = l 0 [ h(x) A x ] φ n (x) dx B n = 1 l ω n l 0 ψ(x) φ n (x) dx Numerical simulations FiniteDomain.WNH Lucio Demeio - DIISM wave equation 39 / 44

78 Non homogeneous problems External loads, e.g. gravity 2 u t 2 v2 2 u x 2 + k = 0 u(0, t) = 0 u(l, t) = A General solution: u(x, t) = u p (x) + ũ(x, t) with u p (x) and ũ such that v 2 u p + k = 0 stationary solution u p (0) = 0 u p (l) = 0 which gives u p (x) = k x (x l) 2 v2 2 ũ t 2 v2 2 ũ x 2 = 0 ũ(0, t) = ũ(l, t) = 0 Lucio Demeio - DIISM wave equation 40 / 44

79 Non homogeneous problems The solution then is: u(x, t) = k x 2 v 2 (x l) + {A n cos ω n t + B n sin ω n t} φ n (x) n=1 where now the coefficients A n and B n are given by A n = l 0 [ h(x) k x ] (x l) 2 v2 φ n (x) dx B n = 1 ω n l 0 ψ(x) φ n (x) dx Lucio Demeio - DIISM wave equation 41 / 44

80 Non homogeneous problems The solution then is: u(x, t) = k x 2 v 2 (x l) + {A n cos ω n t + B n sin ω n t} φ n (x) n=1 where now the coefficients A n and B n are given by A n = l 0 [ h(x) k x ] (x l) 2 v2 φ n (x) dx B n = 1 ω n l 0 ψ(x) φ n (x) dx Numerical simulations FiniteDomain.WNH Lucio Demeio - DIISM wave equation 41 / 44

81 Non homogeneous problems Exteral excitations and resonances Resonance phenomena occur when harmonic external sources (forcing terms) or harmonic boundary conditions are present: Sources: 2 u t 2 u 2 v2 = f(x, t) x2 u(0, t) = 0 u(l, t) = 0 Non hom. b.c.: 2 u t 2 u 2 v2 x = 0 2 u(0, t) = 0 u(l, t) = f(t) Consider the case of harmonic boundary conditions with f(t) = A 0 sin µt. The unknown function can be written as u(x, t) = A 0 x l sin µt + ũ(x, t) Then, ũ obeys the equation (looks like a problem with source) 2 ũ t 2 ũ 2 v2 x = x 2 A0 l µ2 sin µt (32) ũ(0, t) = ũ(l, t) = 0 Lucio Demeio - DIISM wave equation 42 / 44

82 Non homogeneous problems External excitation We expand ũ in the usual eigenfunctions: ũ(x, t) = After substituting into equation (32): n=1 c n(t) φ n(x). n=1 { cn(t) + ω 2 n c n(t) } φ n(x) = A 0 x l µ2 sin µt with ω n = v k n and, by taking scalar products with the φ m s, c m(t) + ω 2 m c m(t) = A 0 γ m l µ 2 sin µt with γ m = (x, φ m(x)) whose general solution is c m(t) = A m cos ω mt + B m sin ω mt + A0 l γ m µ 2 sin µt ωm 2 µ 2 Lucio Demeio - DIISM wave equation 43 / 44

83 Non homogeneous problems Solution of the wave equation The general solution of the wave equation (2) can then be written as a superposition of all eigenfunctions as u(x, t) = A 0 x l sin µt + n=1 { A n cos ω nt + B n sin ω nt + A0 l } γ n µ 2 sin µt φ ωn 2 µ 2 n(x) With the initial conditions u(x, 0) = h(x) we have for the coefficients A n and B n: u (x, 0) = ψ(x) t A n = l 0 B n = 1 ω n h(x) φ n(x) dx { l [ 0 ψ(x) A 0 µ x l Numerical simulations: FiniteDomain.WNH ] A0 µ γn φ n(x) dx l (1 + µ2 ω 2 n µ 2 )} Lucio Demeio - DIISM wave equation 44 / 44