Wave Equation Modelling Solutions

Transcription

1 Wave Equation Modelling Solutions SEECS-NUST December 19, 2017

2 Wave Phenomenon Waves propagate in a pond when we gently touch water in it.

3 Wave Phenomenon Our ear drums are very sensitive to small vibrations in air which enable them to listen sound. Sound waves are longitudinal waves generated from the rarefaction and compression of air molecules. These regions of compression and rarefaction are recorded in the form of vibrations on ear drums.

4 Wave Phenomenon An em-wave is an electromagnetic wave produced by accelerated charges. These are synchronized oscillations of electric and magnetic fields that travel at speed of light. These are always perpendicular to the direction of wave propagation therefore these are transverse waves.

5 Link with second-order PDEs We already know that second order PDEs are classified into three main categories. For example, there are three famous PDEs known for their physical characteristics. Type Definition Example Hyperbolic ac b 2 < 0 Wave Parabolic ac b 2 = 0 Heat Elliptic ac b 2 > 0 Laplace Therefore, wave equation is a representative of PDEs of hyperbolic type. It helps to investigate basic properties of hyperbolic PDEs from the known features of wave equation.

6 Modeling of Wave Equation Objective: To derive the equation of small transverse vibrations of the string. Briefly to model standing waves of a string.

7 Modeling of Wave Equation Objective: To derive the equation of small transverse vibrations of the string. Briefly to model standing waves of a string. Consider an elastic string of length L, and m is the mass of the string per unit length. Let us distort the string and allow it to vibrate and let us call u = u(x, t) as the deflection at any point and any time t > 0, where t, x represent time and distance.

8 Modeling of Wave Equation Physical Assumptions: (1) Mass of the string is homogeneous.

9 Modeling of Wave Equation Physical Assumptions: (1) Mass of the string is homogeneous. (2) String is perfectly elastic and offers no bending.

10 Modeling of Wave Equation Physical Assumptions: (1) Mass of the string is homogeneous. (2) String is perfectly elastic and offers no bending. (3) Tension caused by stretching is large enough to avoid gravitational effects.

11 Modeling of Wave Equation Physical Assumptions: (1) Mass of the string is homogeneous. (2) String is perfectly elastic and offers no bending. (3) Tension caused by stretching is large enough to avoid gravitational effects. (4) String performs small transverse motions.

12 Wave Equation Kinematics: We now consider the transverse motion using kinematics.

13 Wave Equation Kinematics: We now consider the transverse motion using kinematics.

14 Wave Equation Kinematics: We now consider the transverse motion using kinematics. T 1 cos(α) = Horizontal component of tension at P

15 Wave Equation Kinematics: We now consider the transverse motion using kinematics. T 1 cos(α) = Horizontal component of tension at P T 2 cos(β) = Horizontal component of tension at Q

16 Wave Equation Kinematics: We now consider the transverse motion using kinematics. T 1 cos(α) = Horizontal component of tension at P T 2 cos(β) = Horizontal component of tension at Q T 2 sin(β) = Vertical component of tension at Q

17 Wave Equation Kinematics: We now consider the transverse motion using kinematics. T 1 cos(α) = Horizontal component of tension at P T 2 cos(β) = Horizontal component of tension at Q T 2 sin(β) = Vertical component of tension at Q T 1 sin(α) = Vertical component of tension at P

18 Wave Equation Since the string offers no resistance and the tension is tangential thus, T 1 cos(α) = T 2 cos(β) = T = constant because the points move only vertically so the horizontal component of the tension must be a constant.

19 Wave Equation Since the string offers no resistance and the tension is tangential thus, T 1 cos(α) = T 2 cos(β) = T = constant because the points move only vertically so the horizontal component of the tension must be a constant. Newton s second law F = ma, implies

20 Wave Equation Since the string offers no resistance and the tension is tangential thus, T 1 cos(α) = T 2 cos(β) = T = constant because the points move only vertically so the horizontal component of the tension must be a constant. Newton s second law F = ma, implies T 2 sin(β) T 1 sin(α) = mass x acceleration

21 Wave Equation Since the string offers no resistance and the tension is tangential thus, T 1 cos(α) = T 2 cos(β) = T = constant because the points move only vertically so the horizontal component of the tension must be a constant. Newton s second law F = ma, implies T 2 sin(β) T 1 sin(α) = mass x acceleration But we know that the mass and acceleration of small portion of string is

22 Wave Equation Since the string offers no resistance and the tension is tangential thus, T 1 cos(α) = T 2 cos(β) = T = constant because the points move only vertically so the horizontal component of the tension must be a constant. Newton s second law F = ma, implies T 2 sin(β) T 1 sin(α) = mass x acceleration But we know that the mass and acceleration of small portion of string is mass = mass of the portion = ρ x acceleration = 2 u t 2

23 Wave Equation Since the string offers no resistance and the tension is tangential thus, T 1 cos(α) = T 2 cos(β) = T = constant because the points move only vertically so the horizontal component of the tension must be a constant. Newton s second law F = ma, implies T 2 sin(β) T 1 sin(α) = mass x acceleration But we know that the mass and acceleration of small portion of string is Therefore we obtain mass = mass of the portion = ρ x acceleration = 2 u t 2 T 2 sin(β) T 1 sin(α) = ρ x 2 u t 2

24 Wave Equation Dividing the above equation by T, so T 2 sin(β) T T 1 sin(α) T = ρ x 2 u t 2 T

25 Wave Equation Dividing the above equation by T, so T 2 sin(β) T T 1 sin(α) T = ρ x 2 u t 2 T T 2 sin(β) T 2 cos(β) T 1 sin(α) T 1 cos(α) = ρ xu tt T

26 Wave Equation Dividing the above equation by T, so T 2 sin(β) T T 1 sin(α) T = ρ x 2 u t 2 T T 2 sin(β) T 2 cos(β) T 1 sin(α) T 1 cos(α) = ρ xu tt T tan(β) tan(α) = ρ xu tt T Therefore, we arrive at the left hand side of the wave equation. Further note that

27 Wave Equation Dividing the above equation by T, so T 2 sin(β) T T 1 sin(α) T = ρ x 2 u t 2 T T 2 sin(β) T 2 cos(β) T 1 sin(α) T 1 cos(α) = ρ xu tt T tan(β) tan(α) = ρ xu tt T Therefore, we arrive at the left hand side of the wave equation. Further note that tan(α) = slope of the curve u(x, t) at x = u x x

28 Wave Equation Dividing the above equation by T, so T 2 sin(β) T T 1 sin(α) T = ρ x 2 u t 2 T T 2 sin(β) T 2 cos(β) T 1 sin(α) T 1 cos(α) = ρ xu tt T tan(β) tan(α) = ρ xu tt T Therefore, we arrive at the left hand side of the wave equation. Further note that tan(α) = slope of the curve u(x, t) at x = u x x tan(β) = slope of the curve u(x, t) at x+ x = u x x+ x

29 Wave Equation By placing these values into the last equation we get

30 Wave Equation By placing these values into the last equation we get ( u x ) (x+ x) ( u x ) x x = ρ T u tt

31 Wave Equation By placing these values into the last equation we get ( u x ) (x+ x) ( u x ) x x = ρ T u tt u x (x + x) u x (x) lim = ρ x 0 x T u tt

32 Wave Equation By placing these values into the last equation we get ( u x ) (x+ x) ( u x ) x x = ρ T u tt u x (x + x) u x (x) lim = ρ x 0 x T u tt u xx = ρ T u xx

33 Wave Equation By placing these values into the last equation we get ( u x ) (x+ x) ( u x ) x x = ρ T u tt u x (x + x) u x (x) lim = ρ x 0 x T u tt u xx = ρ T u xx Thus, we arrive at a very important equation = u tt = c 2 u xx, c 2 = T ρ

34 Solution of wave equation Solution In order to obtain general solutions of wave equation we apply following steps.

35 Solution of wave equation Solution In order to obtain general solutions of wave equation we apply following steps. Step-1: Separation of variables.

36 Solution of wave equation Solution In order to obtain general solutions of wave equation we apply following steps. Step-1: Separation of variables. Step-2: Solution of ODE s from Step-1.

37 Solution of wave equation Solution In order to obtain general solutions of wave equation we apply following steps. Step-1: Separation of variables. Step-2: Solution of ODE s from Step-1. Step-3: Use Fourier Series to construct a general solution.

38 Solution of Wave Equation Step-1: Count the number of independent variables and assume

39 Solution of Wave Equation Step-1: Count the number of independent variables and assume u(x, t) = F (x)g(t)

40 Solution of Wave Equation Step-1: Count the number of independent variables and assume u(x, t) = F (x)g(t) = u t = F (x) G t = F (x) G = u t = F G = u tt = F G

41 Solution of Wave Equation Step-1: Count the number of independent variables and assume u(x, t) = F (x)g(t) = u t = F (x) G t = F (x) G = u t = F G = u tt = F G Similarly, u xx = GF.

42 Solution of Wave Equation Step-1: Count the number of independent variables and assume u(x, t) = F (x)g(t) = u t = F (x) G t = F (x) G = u t = F G = u tt = F G Similarly, u xx = GF. Place these values in the wave equation. F G = c 2 GF = G G = c2 F F

43 Solution of Wave Equation Step-1: Count the number of independent variables and assume u(x, t) = F (x)g(t) = u t = F (x) G t = F (x) G = u t = F G = u tt = F G Similarly, u xx = GF. Place these values in the wave equation. F G = c 2 GF = G G = c2 F Note that left hand side is only a function of t whereas right hand side is of x, which means variables are separated. F

44 Solution of Wave Equation Step-1: Now LHS(t)=RHS(x) is only possible if and only if

45 Solution of Wave Equation Step-1: Now LHS(t)=RHS(x) is only possible if and only if LHS = RHS = k = G c 2 G = F F = k

46 Solution of Wave Equation Step-1: Now LHS(t)=RHS(x) is only possible if and only if LHS = RHS = k = G c 2 G = F F = k Therefore we obtain systems of two ODE s F kf = 0 ODE 1 G kc 2 G = 0 ODE 2 Remark: So the problem of solving the wave equation (PDE) is transformed into the problem of finding solutions to the above system of ODEs. This is the standard way of solving a partial differential equation.

47 Solution of Wave Equation In order to solve this system we now need to find the boundary conditions F kf = 0 ODE 1 G kc 2 G = 0 ODE 2 Step-2: Check Boundary conditions u = F (x)g(t)

48 Solution of Wave Equation In order to solve this system we now need to find the boundary conditions F kf = 0 ODE 1 G kc 2 G = 0 ODE 2 Step-2: Check Boundary conditions u = F (x)g(t) Since, u(0, t) = 0 and u(l, t) = 0 F (0)G = 0 and F (L)G = 0

49 Solution of Wave Equation In order to solve this system we now need to find the boundary conditions F kf = 0 ODE 1 G kc 2 G = 0 ODE 2 Step-2: Check Boundary conditions u = F (x)g(t) Since, As, u(0, t) = 0 and u(l, t) = 0 F (0)G = 0 and F (L)G = 0 G 0 F (0) = 0 = F (L)

50 Solution of Wave Equation In order to solve this system we now need to find the boundary conditions F kf = 0 ODE 1 G kc 2 G = 0 ODE 2 Step-2: Check Boundary conditions u = F (x)g(t) Since, u(0, t) = 0 and u(l, t) = 0 F (0)G = 0 and F (L)G = 0 As, G 0 F (0) = 0 = F (L) So we have boundary conditions for ODE-1 F kf = 0, F (0) = 0, F (L) = 0

51 Solution of Wave Equation Our next task is to obtain general solution of F kf = 0, F (0) = 0, F (L) = 0. Since there are two boundary conditions so there will be two independent solutions of above ODE. Note that it depends on the sign of k, whether it is positive, negative or zero. Therefore we consider each of these cases in detail and rule out those which are not feasible solutions.

52 Solution of Wave Equation Case-1 (k > 0, k = µ 2 )

53 Solution of Wave Equation Case-1 (k > 0, k = µ 2 ) F = Ae µx + Be µx F (0) = A + B = 0 F (L) = Ae µl + Be µl = 0

54 Solution of Wave Equation Case-1 (k > 0, k = µ 2 ) F = Ae µx + Be µx F (0) = A + B = 0 F (L) = Ae µl + Be µl = 0 Since, both e µx and e µx are independent so, A = 0 = B F = 0 trivial solution u = 0.

55 Solution of Wave Equation Case-1 (k > 0, k = µ 2 ) F = Ae µx + Be µx F (0) = A + B = 0 F (L) = Ae µl + Be µl = 0 Since, both e µx and e µx are independent so, A = 0 = B F = 0 trivial solution u = 0. Case-2 (k = 0) If k = 0, F = 0 F = Ax + B, then F (0) = B = 0 F (L) = AL + B = 0 A = 0, B = 0 F = 0 trivial sol. u = 0

56 Solution of Wave Equation So we discard the above two cases as they lead to trivial solutions and solve the last case Case-3 (k < 0, k = p 2 )

57 Solution of Wave Equation So we discard the above two cases as they lead to trivial solutions and solve the last case Case-3 (k < 0, k = p 2 ) F + p 2 F = 0 F = A cos px + B sin px

58 Solution of Wave Equation So we discard the above two cases as they lead to trivial solutions and solve the last case Case-3 (k < 0, k = p 2 ) F + p 2 F = 0 F = A cos px + B sin px As, F (0) = 0 A = 0 F (L) = 0 B sin pl = 0

59 Solution of Wave Equation So we discard the above two cases as they lead to trivial solutions and solve the last case Case-3 (k < 0, k = p 2 ) F + p 2 F = 0 F = A cos px + B sin px As, F (0) = 0 A = 0 F (L) = 0 B sin pl = 0 If B = 0 then F = 0 so B 0, hence

60 Solution of Wave Equation So we discard the above two cases as they lead to trivial solutions and solve the last case Case-3 (k < 0, k = p 2 ) F + p 2 F = 0 F = A cos px + B sin px As, F (0) = 0 A = 0 F (L) = 0 B sin pl = 0 If B = 0 then F = 0 so B 0, hence sin pl = 0 pl = nπ, n = 1, 2, 3,... p = nπ L

61 Solution of Wave Equation Setting B = 1, without loss of generality we obtain ( nπ ) F n = sin L x n = 1, 2,...

62 Solution of Wave Equation Setting B = 1, without loss of generality we obtain ( nπ ) F n = sin L x n = 1, 2,... Notice that we obtain n solutions of a second-order ODE. Does it contradict anything?

63 Solution of Wave Equation Setting B = 1, without loss of generality we obtain ( nπ ) F n = sin L x n = 1, 2,... Notice that we obtain n solutions of a second-order ODE. Does it contradict anything? We carry out same steps for ODE-2 and obtain G kc 2 G = 0 G + λ n 2 G = 0, G n = B n cos(λ n t) + B n sin(λ n t),

64 Solution of Wave Equation Setting B = 1, without loss of generality we obtain ( nπ ) F n = sin L x n = 1, 2,... Notice that we obtain n solutions of a second-order ODE. Does it contradict anything? We carry out same steps for ODE-2 and obtain G kc 2 G = 0 G + λ n 2 G = 0, G n = B n cos(λ n t) + B n sin(λ n t), where k = p 2 = ( nπ L )2, λ n = cp = cnπ L

65 Solution of Wave Equation Setting B = 1, without loss of generality we obtain ( nπ ) F n = sin L x n = 1, 2,... Notice that we obtain n solutions of a second-order ODE. Does it contradict anything? We carry out same steps for ODE-2 and obtain G kc 2 G = 0 G + λ n 2 G = 0, G n = B n cos(λ n t) + B n sin(λ n t), where k = p 2 = ( nπ L )2, λ n = cp = cnπ L So the original supposition u = FG, becomes u n (x, t) = F n (x)g n (t) = (B n cos(λ n t) + B n sin(λ n t)) sin ( nπx ) L

66 Solution of Wave Equation What do we do with this plethora of solutions of the wave equation for each value of n?

67 Solution of Wave Equation What do we do with this plethora of solutions of the wave equation for each value of n? There is a famous principle that can help us deal with these solutions.

68 Solution of Wave Equation What do we do with this plethora of solutions of the wave equation for each value of n? There is a famous principle that can help us deal with these solutions. Superposition Principle: If u 1 (x) and u 2 (x) are two solutions of a linear PDE then the general solution, u(x) = c 1 u 1 (x) + c 2 u 1 (x) is also a solution of that PDE.

69 Solution of Wave Equation What do we do with this plethora of solutions of the wave equation for each value of n? There is a famous principle that can help us deal with these solutions. Superposition Principle: If u 1 (x) and u 2 (x) are two solutions of a linear PDE then the general solution, u(x) = c 1 u 1 (x) + c 2 u 1 (x) is also a solution of that PDE. In general, u(x) = is also a solution of the PDE. c n u n (x) n=1

70 Solution of Wave Equation Therefore we obtain the general solution of wave equation using superposition principle. Step 3: u(x, t) = n=1 u n (x, t) = (B n cos(λ n t) + B n sin(λ n t))(sin(λ n x)) n=1 Such functions u n, are called eigenfunctions and λ n, are corresponding eigenvalues. The sequence (λ 1, λ 2,...) is a spectrum. In the last we find out the coefficients B n and B n.

71 Solution of Wave Equation The solution is u(x, t) = (B n cos(λ n t) + B n sin(λ n t))(sin nπx L )), n=1

72 Solution of Wave Equation The solution is u(x, t) = (B n cos(λ n t) + B n sin(λ n t))(sin nπx L )), n=1 where the initial condition u(x, 0) = f (x) implies B n cos(λ n (0)) sin( nπx L ) = f (x) B n sin( nπx L ) = f (x) n=1 n=1 B n = 2 L L 0 f (x) sin ( nπx ) dx L

73 Solution of Wave Equation The other initial condition u t (x, 0) = g(x), can be used by first differentiating solution u, w.r.t. t u t (x, t) = ( λ n B n sin(λ n t) + B n cos(λ n t))(sin nπx L ) n=1 u t (x, 0) = Bnλ n cos(λ n 0) sin( nπx L ) = g(x) λ n B n = 2 L n=1 L Bn = 2 L L cnπ B n = 2 nπc 0 L 0 g(x) sin( nπx L )dx L 0 g(x) sin nπx L dx g(x) sin nπx L dx

74 Solution of Wave Equation Thus, u(x, t) gives the general solution of the wave equation, which is the sum of infinitely many functions in the form of a Fourier Series. u(x, t) = (B n cos(λ n t) + B n sin(λ n t))(sin nπx L ), n=1 where λ n = cp = cnπ/l, and B n = 2 L L B n = 2 nπc f (x) sin( nπx L )dx 0 L 0 g(x) sin( nπx L )dx

75 Particular Case In order to comprehend the solution we assume L = π, so λ n = nc, therefore u(x, t) = (B n cos(nct) + B n sin(nct)) sin (nx), n=1 We further assume that f (x) = 0 and g(x) = 1, then we obtain B n = 0, Bn = 2 nπc = 1 πc π 0 ( 2 n sin(nx)dx ) 2, for, n = odd Thus the solution becomes u(x, t) = 4 ( ) 1 2 sin(nct) sin (nx), πc n n=1

76 Questions Type There are three type of questions that can be generated from above discussion. Type-1: Obtain the systems of ODEs from separable technique Type-2: Solve a system of ODEs for a given PDEs and obtain its boundary conditions Type-3: Perform a rigorous solution analysis In the next slide I outline examples (without solutions) of each of these types.

77 Questions Type There are three type of questions that can be generated from above discussion. Type-1: Obtain the systems of ODEs from separable technique of u xx + u yy + u zz

78 Questions Type There are three type of questions that can be generated from above discussion. Type-2: Solve a system of ODEs for a given PDEs and obtain its boundary conditions from the given data F = kf + x, G c 2 kg = 0

79 Questions Type There are three type of questions that can be generated from above discussion. Type-3: Assume length of a string is L = π. Suppose the string is initially at rest and given an initial speed of sin xm/sec. Compare the speeds of string at t = 5sec and t = 10sec. Determine if the speed has increased, decreased or remains the same. Give the reason why is it so.

80 Solution of Wave Equation The reference book is Advanced Engineering Mathematics (Erwin Kryszic) Textbook :

81 Heat Equation Problem: If u(x, t) represents heat in a rod of length L then it obeys an equation of the form u t = c 2 u xx where t, x represent time and distance.

82 Heat Equation Problem: If u(x, t) represents heat in a rod of length L then it obeys an equation of the form u t = c 2 u xx where t, x represent time and distance. Data: Boundary Conditions: For the sake of simplicity we assume that both ends of rod are isolated (no heat enter or exit from the ends of rod) then u(0, t) = 0, u(l, t) = 0 Initial Conditions: so Assume that initially rod has some heat u(x, 0) = f (x)

83 Heat Equation Solution: To obtain solution we apply separation of variables technique Step-1: Assume u = F (x)g(t) then u t = F G, u xx = F G and from heat equation G c 2 G = F F = k Prove that k = 0 and k > 0 are not feasible as they lead to trivial solution and for non-trivial solution we must have k = p 2 < 0 F + p 2 F = 0 G + c 2 p 2 F = 0

84 Heat Equation Solution: Step-2: As in the case of wave equation the first ODE yields F (x) = A cos (px) + B sin (px) It can be proved in a similar way that A = 0 and value of B can be assumed to be 1. Now, F n (x) = sin ( nπx ), n = 1, 2,... L G + λ 2 n G = 0, λ n = cnπ L G n = B n e λn2 t Hence, u n (x, t) = B n sin ( nπx 2 t L )e λn

85 Heat Equation Solution: Step-3: Therefore the general solution of heat equation is u(x, t) = u n (x, t) = n=1 n=1 B n sin ( nπx 2 t L )e λn where coefficient B n is determined from the initial condition u(x, 0) = n=1 B n sin ( nπx L ) = f (x) B n = 2 L L 0 f (x) sin ( nπx L )dx