MA 201: Partial Differential Equations Lecture - 10

Transcription

1 MA 201: Partia Differentia Equations Lecture - 10 Separation of Variabes, One dimensiona Wave Equation

2 Initia Boundary Vaue Probem (IBVP) Reca: A physica probem governed by a PDE may contain both boundary as we as initia conditions. Equations that describe the behaviour of the soution on the boundary of the region under consideration are caed boundary conditions. Equations that describe the phenomenon at the initia vaues of the soution at time t = 0 are caed initia conditions. When a physica probem seeks for soutions fufiing both initia and boundary conditions, the probem is caed a initia boundary vaue probems (IBVP).

3 Boundary conditions Three types of boundary conditions: Dirichet conditions: When the condition refers to vaues of the function on the boundary. (e.g. u(x, t) = 0 for x = 0, 1; t > 0) Neumann conditions: When the condition refers to vaues of norma (to the boundary) derivatives of the function on the boundary. (e.g. u t (x, t) = 0 for x = 0, 1; t > 0) Robin conditions: When the condition is a inear combination of both the vaue of the function and its norma derivatives on the boundary. (e.g. αu(x, t) + βu t (x, t) = 0, α 0, β 0, for x = 0, 1; t > 0).

4 Separation of variabes method for PDE A convenient and powerfu soution technique of partia differentia equations. The basic idea of this method is that the soution is assumed to consist of products of two or more functions, each function being a function of one independent variabe ony. The number of functions invoved depends on the number of independent variabes. For exampe to sove an equation for u = u(x, y) assume a soution in the form u(x, y) = X (x)y (y), where X is a function of x ony and Y is a function of y ony. Substituting this soution in the given equation we wi have a pair of ODEs.

5 Vibrating string and the wave equation The equation under consideration is u tt = c 2 u xx, (1) which is a inear second-order homogeneous equation. Here c represents the veocity of the wave. The above equation has been derived without considering any externa force acting on the string. In case of additiona transverse forces: Let F (x, t) denote the amount of force per unit ength acting in the u-direction. Then the equation of forced vibrations of the string is u tt = c 2 u xx + F (x, t). (2) ρ

6 Vibrating string and the wave equation (Contd.) Two specia cases of (2) particuary generate interest. 1. When the externa force is due to the gravitationa acceeration g ony (consider a string oriented horizontay), the equation becomes u tt = c 2 u xx g. (3) 2. When the externa force is due to the resistance which is proportiona to the instantaneous veocity (a string vibrating in a fuid), the equation becomes u tt + 2ku t = c 2 u xx, (4) where k is a positive constant.

7 IBVP for Vibrating string with no externa forces We take up equations of the type (1) ony and find the soutions. The IBVP under consideration consists of the foowing: The governing equation: u tt = c 2 u xx. (5) The boundary conditions for a t > 0: u(0, t) = 0, (6a) u(, t) = 0. (6b) The initia conditions for 0 < x < are u(x, 0) = φ(x), (7a) u t (x, 0) = ψ(x). (7b)

8 IBVP for Vibrating string with no externa forces (Contd.) The boundary conditions state that the ends of the string are hed fixed for a time, whie the initia conditions give the initia shape of the string φ(x) and its initia veocity ψ(x). Assume a soution of the form u(x, t) = X (x)t (t), (8) where X (x) is function of x aone and T (t) is a function of t aone. Substituting (8) in equation (5) XT = c 2 X T. (9)

9 IBVP for Vibrating string with no externa forces (Contd.) Separating the variabes X X = T c 2 T. Here the eft side is a function of x and the right side is a function of t. The equaity wi hod ony if both are equa to a constant, say, k. We get two differentia equations as foows: X kx = 0, (10a) T c 2 kt = 0. (10b)

10 IBVP for Vibrating string with no externa forces (Contd.) Since k is any constant, it can be zero, or it can be positive, or it can be negative. We consider a the possibiities and examine what vaue(s) of k ead to a feasibe soution.

11 IBVP for Vibrating string with no externa forces (Contd.) Case I: k = 0. In this case the equations (10) reduce to X = 0, and T = 0 giving rise to soutions X (x) = Ax + B, T (t) = Ct + D. But the soution u(x, t) = X (x)t (t) is a trivia soution if it has to satisfy the conditions (6): If T (t) is not identicay zero, then u(0, t) = 0 gives B = 0 and u(, t) = 0 gives A = 0. This case of k = 0 is rejected since it gives rise to trivia soutions ony.

12 IBVP for Vibrating string with no externa forces (Contd.) Case II: k > 0, et k = λ 2. In this case the equations (10) reduce to the equations X λ 2 X = 0, and T c 2 λ 2 T = 0 Giving rise to soutions X (x) = Ae λx + Be λx, T (t) = Ce cλt + De cλt. Therefore u(x, t) = (Ae λx + Be λx )(Ce cλt + De cλt ).

13 IBVP for Vibrating string with no externa forces (Contd.) Using boundary condition (6a), A + B = 0, i.e. B = A. boundary condition (6b), Using (Ae λ + Be λ )(Ce cλt + De cλt ) = 0. The t part of the soution cannot be zero as it wi ead to a trivia soution. Then we must have A(e λ e λ ) = 0 which wi require either A = 0 (not possibe since that wi aso give B = 0 giving us trivia soution) or λ = 0, which is possibe ony when k = 0. k > 0 aso gives rise to trivia soution: k > 0 is aso rejected.

14 IBVP for Vibrating string with no externa forces (Contd.) Case III: k < 0, et k = λ 2. to the equations In this case equations (10) reduce X + λ 2 X = 0 and T + c 2 λ 2 T = 0 giving rise to soutions X (x) = A cos λx + B sin λx, T (t) = C cos(cλt) + D cos(cλt). Hence u(x, t) = (A cos λx + B sin λx)(c cos(cλt) + D sin(cλt).

15 IBVP for Vibrating string with no externa forces (Contd.) Using boundary condition (6a) A = 0. Using boundary condition (6b), B sin λ = 0. have B 0 as that wi ead to a trivia soution. sin λ = 0 Hence we must which gives us λ n = nπ, n = 1, 2, 3,...

16 IBVP for Vibrating string with no externa forces (Contd.) These λ n s are caed eigenvaues. Note that corresponding to each n there wi be an eigenvaue. The function corresponding to each eigenvaue is caed an eigenfunction. The eigenfunction corresponding to the eigenvaue nπ wi be u n (x, t) = sin nπx [ A n cos nπct + B n sin nπct ]. (11)

17 IBVP for Vibrating string with no externa forces (Contd.) Since the wave equation is inear, therefore the superposition of a the inear soutions is aso a soution of equation (5). Hence the soution exists in the foowing form: u(x, t) = = u n (x, t) n=1 n=1 sin nπx [ A n cos nπct + B n sin nπct ], (12) where A n and B n are the coefficients (Fourier coefficients) to be determined by using the initia conditions (7).

18 IBVP for Vibrating string with no externa forces (Contd.) Using the initia condition (7a) φ(x) = n=1 A n sin nπx. (13) This series can be recognized as the haf-range sine expansion of a period function φ(x) defined in the range (0, ). A n can be obtained by mutipying equation (13) by sin nπx and integrating with respect to x from 0 to. Therefore A n = 2 0 φ(x) sin nπx dx, n = 1, 2, 3,... (14)

19 IBVP for Vibrating string with no externa forces (Contd.) To use the initia condition (7b) we need to differentiate (12) w.r.t. t to get u t (x, t) = Then Simiary n=1 sin nπx ψ(x) = ( nπc n=1 ) [ A n sin nπct B n nπc B n = 2 ψ(x) sin nπx nπc 0 sin nπx. + B n cos nπct ]. dx, n = 1, 2, 3,... (15) Hence (12) is the soution of the given IBVP with the coefficients A n and B n, respectivey, given by (14) and (15).

20 IBVP for Vibrating string (Contd.) Figure: Fundamenta mode of a vibrating string

21 IBVP for Vibrating string with gravity(contd.) Figure: Second norma mode of a vibrating string

22 IBVP for Vibrating string (Contd.) Figure: Third norma mode of a vibrating string

23 IBVP for Vibrating string (Contd.) Figure: Fourth norma mode of a vibrating string

24 IBVP for Vibrating string (Contd.) Figure: A puse traveing through a string with fixed endpoints as modeed by the wave equation

25 We seek to find the dispacement of the string at any position and at any time the externa force is due to the gravitationa acceeration g ony (consider a string oriented horizontay). u tt = c 2 u xx g. (16) Boundary condition (for t > 0) and initia conditions (0 < x < ) are: u(0, t) = 0, (17a) u(, t) = 0, (17b) u(x, 0) = φ(x), (18a) u t (x, 0) = ψ(x). (18b)

26 (Contd.) Due to the presence of the term g in equation (16), the direct appication of the method of separation of variabes wi not work. We intend to convert the given IBVP into two probems: one IBVP and one BVP, former resembing the probem we soved previousy and the other taking care of the nonhomogeneous term in the governing equation. Seek a soution in the form: u(x, t) = v(x, t) + h(x), (19) where h(x) is an unknown function of x aone.

27 (Contd.) Now substituting (19) into (16) v tt = c 2 [v xx + h (x)] g, (20) subject to v(0, t) + h(0) = 0, (21a) v(, t) + h() = 0, (21b) and v(x, 0) + h(x) = φ(x), (22a) v t (x, 0) = ψ(x). (22b)

28 (Contd.) Equations (20)-(22) can be convenienty spit into two probems: Probem I: v tt = c 2 v xx, v(0, t) = 0 = v(, t), v(x, 0) = φ(x) h(x), v t (x, 0) = ψ(x). Probem II: c 2 h (x) = g, h(0) = 0 = h().

29 (Contd.) The soution of Probem I is known to us from the previous probem, which is: v(x, t) = n=1 sin nπx [ A n cos( nπct where A n and B n are given, respectivey, by ) + B n sin( nπct ] ), (23) A n = 2 B n = 2 nπc [φ(x) h(x)] sin( nπx ) dx, n = 1, 2, 3,... (24) 0 0 ψ(x) sin( nπx ) dx, n = 1, 2, 3,... (25)

30 (Contd.) The soution for Probem II can be easiy found by integrating h (x) twice h(x) = gx 2 + Ax + B. 2c2 Upon using the boundary conditions we get B = 0 and A = g/(2c 2 ) and hence ( x)x h(x) = g 2c 2. (26) Hence the soution u(x, t) for our IBVP is given by the sum of (23) and (26).