Wave Equation With Homogeneous Boundary Conditions

Transcription

1 Wave Equation With Homogeneous Boundary Conditions MATH 467 Partial Differential Equations J. Robert Buchanan Department of Mathematics Fall 018

2 Objectives In this lesson we will learn: how to solve the wave equation with homogeneous Dirichlet boundary conditions using separation of variables, how to solve the wave equation with homogeneous Neumann boundary conditions using separation of variables, terms for describing the components of the solution to the wave equation.

3 Initial Boundary Value Problem u tt = c u xx for 0 < x < L and t > 0 u(0, t) = u(l, t) = 0 u(x, 0) = f (x) u t (x, 0) = g(x) Since the PDE is linear and homogeneous and the boundary conditions are homogeneous and of Dirichlet type, the method of separation of variables and the Principle of Superposition apply.

4 Separation of Variables Assume a product solution of the form u(x, t) = X(x)T (t), differentiate and substitute into the wave equation. where λ is a constant. u tt = c u xx X(x)T (t) = c X (x)t (t) T (t) c T (t) = X (x) X(x) = λ This implies the boundary value problem for X(x). X (x) + λx(x) = 0 X(0) = 0 X(L) = 0

5 Eigenvalues and Eigenfunctions The only non-trivial eigenfunctions are X n (x) = sin nπx L corresponding to the eigenvalues λ n = n π L for n N. With these eigenvalues, the implied ODE for function T n (t) has the form T n (t) + c n π L T n (t) = 0 and consequently solution T n (t) = a n cos cnπt L + b n sin cnπt L.

6 Product Solutions Functions of the form u n (x, t) = X n (x)t n (t) = ( a n cos cnπt L + b n sin cnπt ) sin nπx L L for n N solve the wave equation and satisfy the homogeneous Dirichlet boundary conditions. These solutions are called fundamental solutions. By the Principle of Superposition a sum of fundamental solutions will also solve the wave equation and satisfy the homogeneous Dirichlet boundary conditions. u(x, t) = N ( a n cos cnπt L + b n sin cnπt ) sin nπx L L

7 Fundamental Solutions u n (x, t) = ( a n cos cnπt L + b n sin cnπt ) sin nπx L L This solution is known as the nth harmonic. Solution is periodic in t with period L cn. The number of oscillations per π units of time is called the natural frequency and is cnπ L. The number of oscillations per unit time is called the frequency and is cn L. The wavelength of the solution is L n. The intensity of the solution is given by the amplitude a n + b n.

8 First Harmonic u 1 (x, t) = ( a 1 cos cπt L + b 1 sin cπt ) sin πx L L This solution is known as the first harmonic or the fundamental mode. The number of oscillations per π units of time is called the fundamental frequency and is cπ L. First harmonic is periodic in t with period L c. The nth harmonic has a frequency which is n times the fundamental frequency.

9 Initial Displacement We will assume a Fourier series solution to the IBVP. ( u(x, t) = a n cos cnπt L + b n sin cnπt ) sin nπx L L u(x, 0) = a n sin nπx L = f (x) Suppose f (x) can be extended to R as a L periodic, odd function, then f (0) = f (L) = f ( L) and a n = L L 0 f (x) sin nπx L dx, for n N.

10 Initial Velocity If u(x, t) can be differentiated term by term, then u t (x, t) = u t (x, 0) = cnπ L ( a n sin cnπt L cnπ L b n sin nπx L = g(x) + b n cos cnπt ) sin nπx L L Again, if g(x) can be extended to R as a L-periodic, odd function then for n N. cnπ L b n = L b n = L cnπ g(x) sin nπx L dx 0 L 0 g(x) sin nπx L dx

11 Example: Plucked String Find the solution to the following IBVP. u tt = u xx for 0 < x < and t > 0 u(0, t) = u(, t) = 0 x/5 if 0 x 5/, u(x, 0) = 1 if 5/ < x < 15/, 4 x/5 if 15/ x u t (x, 0) = 0

12 Solution (1 of ) The formal solution can be expressed as u(x, t) = ( a n cos nπt + b n sin nπt ) sin nπx. Since u t (x, 0) = 0 then b n = 0 for all n N. a n = = 0 8 n π u(x, 0) sin nπx dx ( sin nπ ) 3nπ + sin 4 4

13 Solution (1 of ) The formal solution can be expressed as u(x, t) = ( a n cos nπt + b n sin nπt ) sin nπx. Since u t (x, 0) = 0 then b n = 0 for all n N. a n = = 0 8 n π u(x, t) = 8 π u(x, 0) sin nπx dx ( sin nπ ) 3nπ + sin 4 4 sin nπ 4 3nπ + sin 4 n cos nπt nπx sin

14 Solution ( of ) Let f (x) be the odd, 0 periodic extension of u(x, 0), then f (x) = 8 π sin nπ 4 3nπ + sin 4 n sin nπx

15 Solution ( of ) Let f (x) be the odd, 0 periodic extension of u(x, 0), then f (x) = 8 π f (x + t) = 8 π f (x t) = 8 π sin nπ 4 sin nπ 4 sin nπ 4 3nπ + sin 4 n sin nπx 3nπ + sin 4 nπ(x + t) n sin 3nπ + sin 4 nπ(x t) n sin

16 Solution ( of ) Let f (x) be the odd, 0 periodic extension of u(x, 0), then f (x) = 8 π f (x + t) = 8 π f (x t) = 8 π f (x + t) + f (x t) = 16 π sin nπ 4 sin nπ 4 sin nπ 4 sin nπ 4 3nπ + sin 4 n sin nπx u(x, t) = 1 (f (x + t) + f (x t)) 3nπ + sin 4 nπ(x + t) n sin 3nπ + sin 4 nπ(x t) n sin 3nπ + sin 4 n cos nπt nπx sin = u(x, t)

17 0 Periodic Extension of Initial Displacement f o (x) x

18 Shifting Initial Displacement Horizontally f o (x) x

19 Adding Shifts of Initial Displacement u(x,t) x

20 Check by Differentiation Consider the following IBVP. u tt = u xx for 0 < x < and t > 0 u(0, t) = u(, t) = 0 x/5 if 0 x 5/, u(x, 0) = 1 if 5/ < x < 15/, 4 x/5 if 15/ x u t (x, 0) = 0 Show by direct differentiation that u(x, t) = 1 (f (x + t) + f (x t)) solves the IBVP when f is the odd, 0 periodic extension of u(x, 0).

21 Solution u(x, t) = 1 (f (x + t) + f (x t)) u xx = 1 ( f (x + t) + f (x t) ) u tt = 1 ( f (x + t) + f (x t) )

22 Solution u(x, t) = 1 (f (x + t) + f (x t)) u xx = 1 ( f (x + t) + f (x t) ) u tt = 1 ( f (x + t) + f (x t) ) u tt = u xx

23 Solution u(x, t) = 1 (f (x + t) + f (x t)) u xx = 1 ( f (x + t) + f (x t) ) u tt = 1 ( f (x + t) + f (x t) ) u tt = u xx Initial displacement: u(x, 0) = 1 (f (x) + f (x)) = f (x)

24 Solution u(x, t) = 1 (f (x + t) + f (x t)) u xx = 1 ( f (x + t) + f (x t) ) u tt = 1 ( f (x + t) + f (x t) ) u tt = u xx Initial displacement: u(x, 0) = 1 (f (x) + f (x)) = f (x) Initial velocity: u t (x, 0) = 1 ( f (x) f (x) ) = 0

25 Example: Struck String Find the solution to the following IBVP. u tt = u xx for 0 < x < and t > 0 u(0, t) = u(, t) = 0 u(x, 0) = 0 { u t (x, 0) = x/5 if 0 x 5, x/5 if 5 < x

26 Solution (1 of 3) The formal solution can be expressed as u(x, t) = ( a n cos nπt + b n sin nπt ) sin nπx. Since u(x, 0) = 0 then a n = 0 for all n N. b n = nπ 0 = 80 n 3 π 3 sin nπ u t (x, 0) sin nπx dx

27 Solution (1 of 3) The formal solution can be expressed as u(x, t) = ( a n cos nπt + b n sin nπt ) sin nπx. Since u(x, 0) = 0 then a n = 0 for all n N. b n = nπ = 80 u(x, t) = 80 π 3 0 n 3 π 3 sin nπ sin nπ n 3 u t (x, 0) sin nπx dx sin nπt sin nπx

28 Solution ( of 3) Let g(x) be the odd, 0 periodic extension of u t (x, 0), then g(x) = 8 π sin nπ n sin nπx. Define G(x) = term. x 0 g(s) ds and integrate the Fourier series term by G(x) = 8 π = 8 π = 80 π 3 = 80 π 3 sin nπ n x 0 sin nπs ds sin nπ ( n 1 cos nπx nπ sin nπ ( n 3 1 cos nπx sin nπ n 3 80 π 3 ) sin nπ n 3 ) cos nπx

29 Solution (3 of 3) G(x + t) = G(x t) = x+t 0 x t 0 g(s) ds = 80 π 3 g(s) ds = 80 π 3 sin nπ n 3 80 π 3 sin nπ n 3 80 π 3 sin nπ nπ(x + t) n 3 cos sin nπ nπ(x t) n 3 cos

30 Solution (3 of 3) G(x + t) = G(x t) = x+t 0 x t 0 g(s) ds = 80 π 3 g(s) ds = 80 π 3 sin nπ n 3 80 π 3 sin nπ n 3 80 π 3 sin nπ nπ(x + t) n 3 cos sin nπ nπ(x t) n 3 cos Subtract the two equations. G(x + t) G(x t) = 80 1 x+t x t x+t x t π 3 g(s) ds = 160 π 3 g(s) ds = 80 π 3 sin nπ n 3 sin nπ n 3 sin nπ n 3 ( nπ(x t) cos cos sin nπt sin nπt sin nπx sin nπx = u(x, t) ) nπ(x + t)

31 Illustration u(x,t) u(x,0) u(x,1) u(x,) u(x,3) u(x,4) x

32 Check by Differentiation Consider the following IBVP. u tt = u xx for 0 < x < and t > 0 u(0, t) = u(, t) = 0 u(x, 0) = 0 { u t (x, 0) = Show by direct differentiation that u(x, t) = 1 x/5 if 0 x 5, x/5 if 5 < x x+t x t g(s) ds solves the IBVP when g is the odd, 0 periodic extension of u t (x, 0).

33 Solution x+t u(x, t) = 1 x t g(s) ds u t (x, t) = 1 (g(x + t) + g(x t)) u tt (x, t) = 1 ( g (x + t) g (x t) ) u x (x, t) = 1 (g(x + t) g(x t)) u xx (x, t) = 1 ( g (x + t) g (x t) ) u tt = u xx

34 Solution x+t u(x, t) = 1 x t g(s) ds u t (x, t) = 1 (g(x + t) + g(x t)) u tt (x, t) = 1 ( g (x + t) g (x t) ) u x (x, t) = 1 (g(x + t) g(x t)) u xx (x, t) = 1 ( g (x + t) g (x t) ) u tt = u xx When t = 0, u(x, 0) = 1 x x g(s) ds = 0 if g is continuous at x.

35 Solution x+t u(x, t) = 1 x t g(s) ds u t (x, t) = 1 (g(x + t) + g(x t)) u tt (x, t) = 1 ( g (x + t) g (x t) ) u x (x, t) = 1 (g(x + t) g(x t)) u xx (x, t) = 1 ( g (x + t) g (x t) ) u tt = u xx When t = 0, u(x, 0) = 1 When t = 0, x x g(s) ds = 0 if g is continuous at x. u t (x, 0) = 1 (g(x) + g(x)) = g(x).

36 Combination Suppose u(x, t) and v(x, t) solve the respective IBVPs for 0 < x < L and t > 0: u tt = c u xx u(0, t) = u(l, t) = 0 u(x, 0) = f (x) u t (x, 0) = 0 v tt = c v xx v(0, t) = v(l, t) = 0 v(x, 0) = 0 v t (x, 0) = g(x) Question: what IBVP would w(x, t) = u(x, t) + v(x, t) solve?

37 Combination Suppose u(x, t) and v(x, t) solve the respective IBVPs for 0 < x < L and t > 0: u tt = c u xx u(0, t) = u(l, t) = 0 u(x, 0) = f (x) u t (x, 0) = 0 v tt = c v xx v(0, t) = v(l, t) = 0 v(x, 0) = 0 v t (x, 0) = g(x) Question: what IBVP would w(x, t) = u(x, t) + v(x, t) solve? w tt = c w xx for 0 < x < L and t > 0 w(0, t) = w(l, t) = 0 w(x, 0) = f (x) w t (x, 0) = g(x)

38 Example Find the solution to the IBVP: u tt = u xx for 0 < x < 1 and t > 0 u(0, t) = u(1, t) = 0 u(x, 0) = sin(πx) u t (x, 0) = sin(πx)

39 Solution Let f (x) = sin(πx) and g(x) = sin(πx) which both odd functions and periodic. u(x, t) = 1 (sin(π(x + t)) + sin(π(x t))) + 1 = sin(πx) cos(πt) + 1 sin(πx) sin(πt) π x+t x t sin(πs) ds

40 Graph u(x,t) x u(x,0) u(x,1/6) u(x,1/3) u(x,1/) -1.0

41 Homework Read Section 5.1 Exercises: 1 5