u tt = a 2 u xx u tt = a 2 (u xx + u yy )
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1 10.7 The wave equation
2 10.7 The wave equation O. Costin: 10.7
3 1 This equation describes the propagation of waves through a medium: in one dimension, such as a vibrating string u tt = a 2 u xx
4 1 This equation describes the propagation of waves through a medium: in one dimension, such as a vibrating string u tt = a 2 u xx in two dimensions, such as a vibrating membrane: u tt = a 2 (u xx + u yy )
5 1 This equation describes the propagation of waves through a medium: in one dimension, such as a vibrating string u tt = a 2 u xx in two dimensions, such as a vibrating membrane: u tt = a 2 (u xx + u yy ) in three dimensions, such as vibrating rock in an earthquake: u tt = a 2 (u xx + u yy + u zz )
6 1 This equation describes the propagation of waves through a medium: in one dimension, such as a vibrating string u tt = a 2 u xx in two dimensions, such as a vibrating membrane: u tt = a 2 (u xx + u yy ) in three dimensions, such as vibrating rock in an earthquake: u tt = a 2 (u xx + u yy + u zz ) All three can be solved by separation of variables, but we will only look at one dimension.
7 1 This equation describes the propagation of waves through a medium: in one dimension, such as a vibrating string u tt = a 2 u xx in two dimensions, such as a vibrating membrane: u tt = a 2 (u xx + u yy ) in three dimensions, such as vibrating rock in an earthquake: u tt = a 2 (u xx + u yy + u zz ) All three can be solved by separation of variables, but we will only look at one dimension. u is the amplitude of the wave.
8 2 Note: none of the above include damping. We deal with a no-damping approximation, valid for short time.
9 3 We need sufficient data as (1) boundary conditions
10 3 We need sufficient data as (1) boundary conditions and (2) initial conditions
11 3 We need sufficient data as (1) boundary conditions and (2) initial conditions to have a unique solution of the problem. Vibrating string A vibrating string has its endpoints rigidly attached. (In this picture, = l, u = y.)
12 3 We need sufficient data as (1) boundary conditions and (2) initial conditions to have a unique solution of the problem. Vibrating string A vibrating string has its endpoints rigidly attached. (In this picture, = l, u = y.) Then, we have u(0, t) = 0; u(, t) = 0
13 4 How about initial conditions?
14 4 How about initial conditions? Now we need two, because the equation is second order in time.
15 4 How about initial conditions? Now we need two, because the equation is second order in time. We give: u(x, 0) and u t (x, 0). Clearly, both matter: where the string starts (sometimes with zero initial velocity, e.g., guitar), and its initial velocity (impact excitation, e.g., in a piano).
16 4 How about initial conditions? Now we need two, because the equation is second order in time. We give: u(x, 0) and u t (x, 0). Clearly, both matter: where the string starts (sometimes with zero initial velocity, e.g., guitar), and its initial velocity (impact excitation, e.g., in a piano). In reality, the conditions are some combinations of the above, often not easy to model. Full problem: u(0, t) = 0; u tt = a 2 u xx u(, t) = 0, u(x, 0) = f(x), u t (x, 0) = g(x)
17 4 How about initial conditions? Now we need two, because the equation is second order in time. We give: u(x, 0) and u t (x, 0). Clearly, both matter: where the string starts (sometimes with zero initial velocity, e.g., guitar), and its initial velocity (impact excitation, e.g., in a piano). In reality, the conditions are some combinations of the above, often not easy to model. Full problem: u(0, t) = 0; u tt = a 2 u xx u(, t) = 0, u(x, 0) = f(x), u t (x, 0) = g(x) Here, a 2 = T/ρ depends on the physical setup only: T is the tension (force) in the string, ρ is its density.
18 5 Separation of variables in u tt = a 2 u xx u(x, t) = X(x)T(t)
19 5 Separation of variables in u tt = a 2 u xx u(x, t) = X(x)T(t) X(x)T (t) = a 2 X (x)t(t)
20 5 Separation of variables in u tt = a 2 u xx u(x, t) = X(x)T(t) X(x)T (t) = a 2 X (x)t(t) (1) T (t) a 2 T(t) = X (x) X(x)
21 5 Separation of variables in u tt = a 2 u xx u(x, t) = X(x)T(t) X(x)T (t) = a 2 X (x)t(t) (1) T (t) a 2 T(t) = X (x) = λ X(x) Thus the pair of ODEs is: X (x) + λx(x) = 0; X(0) = X() = 0
22 5 Separation of variables in u tt = a 2 u xx u(x, t) = X(x)T(t) X(x)T (t) = a 2 X (x)t(t) (1) T (t) a 2 T(t) = X (x) X(x) = λ Thus the pair of ODEs is: X (x) + λx(x) = 0; X(0) = X() = 0 (2) (an eigenvalue problem). T (t) + λa 2 T(t) = 0;
23 5 Separation of variables in u tt = a 2 u xx u(x, t) = X(x)T(t) X(x)T (t) = a 2 X (x)t(t) (1) T (t) a 2 T(t) = X (x) X(x) = λ Thus the pair of ODEs is: X (x) + λx(x) = 0; X(0) = X() = 0 (2) (an eigenvalue problem). T (t) + λa 2 T(t) = 0; no conditions yet
24 6 X (x) + λx(x) = 0; X(0) = X() = 0
25 6 X (x) + λx(x) = 0; X(0) = X() = 0 (3) We have studied exactly this eigenvalue problem. Its solutions are: λ n = n 2 π 2 / 2 ; X n = c n sin nπx How about T?
26 6 X (x) + λx(x) = 0; X(0) = X() = 0 (3) We have studied exactly this eigenvalue problem. Its solutions are: λ n = n 2 π 2 / 2 ; X n = c n sin nπx How about T? T (t) + λ n a 2 T(t) = 0
27 6 X (x) + λx(x) = 0; X(0) = X() = 0 (3) We have studied exactly this eigenvalue problem. Its solutions are: λ n = n 2 π 2 / 2 ; X n = c n sin nπx How about T? T (t) + λ n a 2 T(t) = 0 T (t) + n 2 π 2 a 2 / 2 T(t) = 0 T(t) = A n sin nπat + B n cos nπat
28 7 Example: nonzero initial displacement f(x), zero initial velocity (g(x) = 0). In this case u t (x, 0) = 0;
29 7 Example: nonzero initial displacement f(x), zero initial velocity (g(x) = 0). In this case u t (x, 0) = 0; thus T (0)X(x) = 0;
30 7 Example: nonzero initial displacement f(x), zero initial velocity (g(x) = 0). In this case Then, u t (x, 0) = 0; thus T (0)X(x) = 0; T (0) = 0 = A n X(x)T(t) = c n sin nπx nπat cos
31 7 Example: nonzero initial displacement f(x), zero initial velocity (g(x) = 0). In this case u t (x, 0) = 0; thus T (0)X(x) = 0; T (0) = 0 = A n Then, X(x)T(t) = c n sin nπx General solution should be nπat cos u(x, t) = n=1 c n sin nπx nπat cos
32 7 Example: nonzero initial displacement f(x), zero initial velocity (g(x) = 0). In this case u t (x, 0) = 0; thus T (0)X(x) = 0; T (0) = 0 = A n Then, X(x)T(t) = c n sin nπx General solution should be nπat cos u(x, t) = n=1 c n sin nπx nπat cos
33 8 The initial condition. u(x, t) = n=1 c n sin nπx nπat cos
34 8 The initial condition. u(x, t) = n=1 c n sin nπx nπat cos u(x, 0) = f(x) = n=1 c n sin nπx
35 8 The initial condition. u(x, t) = n=1 c n sin nπx nπat cos u(x, 0) = f(x) = which is again a sine-series. n=1 c n sin nπx Thus we have to odd-extend f and then calculate c n from the usual sine-series formula c n = 2 0 f(x) sin nπx dx
36 9 u(x, t) = n=1 c n sin nπx nπat cos
37 9 u(x, t) = c n = 2 n=1 0 c n sin nπx nπat cos f(x) sin nπx dx
38 9 u(x, t) = n=1 c n sin nπx nπat cos c n = 2 f(x) sin nπx 0 dx u is an infinite sum of terms (modes) of the form sin nπx nπat cos
39 9 u(x, t) = n=1 c n sin nπx nπat cos c n = 2 f(x) sin nπx 0 dx u is an infinite sum of terms (modes) of the form sin nπx nπat cos In t, this is periodic with frequency nπa.
40 9 u(x, t) = n=1 c n sin nπx nπat cos c n = 2 f(x) sin nπx 0 dx u is an infinite sum of terms (modes) of the form sin nπx nπat cos In t, this is periodic with frequency nπa. Each such mode has a periodic x behavior too, with space frequency nπ.
41 9 u(x, t) = n=1 c n sin nπx nπat cos c n = 2 f(x) sin nπx 0 dx u is an infinite sum of terms (modes) of the form sin nπx nπat cos In t, this is periodic with frequency nπa. Each such mode has a periodic x behavior too, with space frequency nπ. The higher the space frequency, the higher the time frequency. Furthermore, the time frequencies are integer multiples of the
42 10 first one (that is, the one with n = 1), πa.
43 10 πa first one (that is, the one with n = 1),. This first one is the fundamental frequency, and the higher ones are harmonics of it.
44 11
45 12 Example: u(x, 0) = f(x) = { x/10; 0 x 10 (30 x)/20; 10 < x < 30 t = K x K0.4 K0.6 K0.8 Curve 1
46 x t
47 x t
48 14 Actual waveform of a guitar string vibration at fixed x
49 15 Other initial conditions. Suppose now we are given u tt = a 2 u xx u(0, t) = 0; u(, t) = 0, u(x, 0) = 0, u t (x, 0) = g(x)
50 15 Other initial conditions. Suppose now we are given u tt = a 2 u xx u(0, t) = 0; u(, t) = 0, u(x, 0) = 0, u t (x, 0) = g(x) Such as the string of a piano. Now the eigenvalue problem is X (x) + λx(x) = 0; X(0) = X() = 0
51 15 Other initial conditions. Suppose now we are given u(0, t) = 0; Such as the string of a piano. Now the eigenvalue problem is Thus X n (x) = c n sin nπx u tt = a 2 u xx u(, t) = 0, u(x, 0) = 0, u t (x, 0) = g(x) X (x) + λx(x) = 0; X(0) = X() = 0 (4) T (t) + λa 2 T(t) = 0;
52 15 Other initial conditions. Suppose now we are given u(0, t) = 0; Such as the string of a piano. Now the eigenvalue problem is Thus X n (x) = c n sin nπx u tt = a 2 u xx u(, t) = 0, u(x, 0) = 0, u t (x, 0) = g(x) X (x) + λx(x) = 0; X(0) = X() = 0 (4) T (t) + λa 2 T(t) = 0; T(0) = 0
53 16 and then T n (t) = sin nπat X n T n = c n sin nπx nπat sin
54 16 and then T n (t) = sin nπat X n T n = c n sin nπx u(x, t) = n=1 c n sin nπx nπat sin nπat sin
55 16 and then T n (t) = sin nπat X n T n = c n sin nπx u(x, t) = u t = n=1 n=1 c n nπa c n sin nπx nπx sin nπat sin nπat sin nπat cos
56 16 and then T n (t) = sin nπat X n T n = c n sin nπx u(x, t) = n=1 nπa u t = c n n=1 u t (0) = n=1 c n sin nπx nπx sin c n nπa nπat sin nπat sin nπat cos nπx sin
57 16 and then again a sine series. T n (t) = sin nπat X n T n = c n sin nπx u(x, t) = n=1 nπa u t = c n n=1 u t (0) = n=1 c n sin nπx nπx sin c n nπa nπat sin nπat sin nπat cos nπx sin
58 17 General initial conditions. u tt = a 2 u xx u(0, t) = 0; u(, t) = 0, u(x, 0) = f(x), u t (x, 0) = g(x)
59 17 General initial conditions. u tt = a 2 u xx u(0, t) = 0; u(, t) = 0, u(x, 0) = f(x), u t (x, 0) = g(x) The general solution is u(x, t) = F(x, t) + G(x, t), where F tt = a 2 F xx F(0, t) = 0; F(, t) = 0, F(x, 0) = f(x), F t (x, 0) = 0
60 17 General initial conditions. u tt = a 2 u xx u(0, t) = 0; u(, t) = 0, u(x, 0) = f(x), u t (x, 0) = g(x) The general solution is u(x, t) = F(x, t) + G(x, t), where F tt = a 2 F xx F(0, t) = 0; F(, t) = 0, F(x, 0) = f(x), F t (x, 0) = 0 G(0, t) = 0; G tt = a 2 G xx G(, t) = 0, G(x, 0) = 0, G t (x, 0) = g(x)
61 17 General initial conditions. u tt = a 2 u xx u(0, t) = 0; u(, t) = 0, u(x, 0) = f(x), u t (x, 0) = g(x) The general solution is u(x, t) = F(x, t) + G(x, t), where F tt = a 2 F xx F(0, t) = 0; F(, t) = 0, F(x, 0) = f(x), F t (x, 0) = 0 G tt = a 2 G xx G(0, t) = 0; G(, t) = 0, G(x, 0) = 0, G t (x, 0) = g(x) (check!)
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