The 1-D Wave Equation 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 1 1-D Wave Equation : Physical derivation Reference: Guenther & Lee §1.2, Myint-U & Debnath §2.1-2.4 [Oct. 3, 2006] We consider a string of length l with ends fixed, and rest state coinciding with x-axis. The string is plucked into oscillation. Let u (x, t) be the position of the string at time t. Assumptions: 1. Small oscillations, i.e. the displacement u (x, t) is small compared to the length l. (a) Points move vertically. In general, we don’t know that points on the string move vertically. By assuming the oscillations are small, we assume the points move vert ically. (b) Slope of tangent to the string is small everywhere, i.e. |ux (x, t)| ≪ 1, so stretching of the string is negligible R l p (c) arc length α (t) = 1 + u2 dx ≃ l.0 x2. String is p erfectly flexible (it bends). This implies the tension is in the tan-gent direction and the horizontal tension is constant, or else there would be a preferred direction of motion for the string. Consider an element of the string between x and x + Δx. Let T (x, t) be tension and θ (x, t) be the angle wrt the horizontal x-axis. Note that ∂u tan θ (x, t) = slope of tangent at (x, t) in ux-plane = (x, t) . (1) ∂x 1 Newton’s Second Law (F = ma) states that ∂2u F = (ρΔx) (2) ∂t2 where ρ is the linear density of the string (ML−1) and Δx is the length of the segment. The force comes from the tension in the string only - we ignore any external forces such as gravity. The horizontal tension is constant, and hence it is the vertical tension that moves the string vertically (obvious). Balancing the forces in the horizontal direction gives T (x + Δx, t) cos θ (x + Δx, t) = T (x, t) cos θ (x, t) = τ = const (3) where τ is the constant horizontal tension. Balancing the forces in the vertical direc-tion yields F = T (x + Δx, t) sin θ (x + Δx, t) −T (x, t) sin θ (x, t) = T (x + Δx, t) cos θ (x + Δx, t) tan θ (x + Δx, t) − T (x, t) cos θ (x, t) tan θ (x, t) Substituting (3) and (1) yields F = τ (tan θ (x + Δx, t) − tan θ (x, t)) ∂u ∂u = τ (x + Δx, t) − (x, t) . (4) ∂x ∂x Substituting F from (2) into Eq. (4) and dividing by Δx gives ∂2∂u ∂u (ξ, t) = τ ∂x (x + Δx, t) −∂x (x, t)u ρ ∂t2 Δx for ξ ∈ [x, x + Δx]. Letting Δx 0 gives the 1-D Wave Equation →∂2u 2 ∂2u 2 τ = c , c = > 0. (5) ∂t2 ∂x2 ρ hi1/2 Force Note that c has units [c] = Density = LT −1 of speed. 1.1 Boundary conditions Ref: Guenther & Lee §4.2 (p. 94), Myint-U & Debnath §4.4 In order to guarantee that Eq. (5) has a unique solution, we need initial and boundary conditions on the displacement u (x, t). There are now 2 initial conditions and 2 boundary conditions. 2E.g. The string is fixed at both ends, u (0, t) = u (l, t) = 0, t > 0 (homogeneous type I BCs) E.g. The string is connected to frictionless cylinders of mass m that move vertically on tracks at x = 0, l. Performing a force balance at either x = 0 or x = 1 gives T sin θ = mg (6) In other words, the vertical tension in the string balances the mass of the cylinder. But τ = T cos θ = const and tan θ = ux, so that (6) becomes τux = T cos θ tan θ = mg Rearranging yields mg , x = 0, 1ux = τ These are Type II BCs. If the string is really tight and the cylinders are very light, then mg/τ ≪ 1 and we approximate ux ≈ 0 at x = 0, 1, and the BCs become Type II homogeneous BCs. 1.2 Initial conditions Ref: Guenther & Lee §4.2 (p. 94) We need to specify both the initial position of the string and the initial velocity: u (x, 0) = f (x) and ut (x, 0) = g (x), 0 < x < l. The idea is the same as finding the trajectory of a falling body; we need to know both the initial position and initial velocity of the particle to compute its trajectory. We can also see this mathematically. The Taylor series of u (x, t) about t = 0 is t2 ∂3u t3 u (x, t) = u (x, 0) + ut (x, 0) t + utt (x, 0) + (x, 0) + 2 ∂t3 3! ··· From the initial conditions, u (x, 0) = f (x), ut (x, 0) = g (x) and the PDE gives utt (x, 0) = c 2 uxx (x, 0) = c 2f ′′ (x) , ∂3u 2 2 ′′ (x, 0) = c utxx (x, 0) = c g (x) . ∂t3 Higher order terms can be found similarly. Therefore, the two initial conditions for u (x, 0) and ut (x, 0) are sufficient to determine u (x, t) near t = 0. To summarize, the dimensional basic 1-D Wave Problem with Type I BCs (fixed ends) is PDE : utt = c 2 uxx, 0 < x < l (7) BC : u (0, t) = 0 = u (l, t) , t > 0, (8) IC : u (x, 0) = f (x) , ut (x, 0) = g (x) , 0 < x < l (9) 3 1.3 Non-dimensionalization We now scale the basic 1-D Wave Problem. The characteristic quantities are length L∗ and time T∗. Common sense suggests choosing L∗ = l, the length of the string. We introducing the non-dimensional variables x t u (x, t) f (x) T∗g (x) xˆ = , tˆ= , ˆ ˆ t = , x) = , g (ˆ =u x, ˆfˆ(ˆ ˆ x) . L∗ T∗ L∗ L∗ L∗ From the chain rule, ∂u ∂uˆ ∂xˆ ∂uˆ ∂u ∂uˆ∂tˆL∗ ∂uˆ= L∗ = = L∗ = ∂x ∂xˆ ∂x ∂xˆ, ∂t ∂tˆ∂t T∗ ∂tˆand similarly for higher derivatives. Substituting the dimensionless variables into 1-D Wave Equation (7) gives T 2c2 ∗ uˆtˆtˆ= uˆxˆxˆL2 ∗ This suggests choosing T∗ = L∗/c = l/c, so that uˆˆt = ˆxˆ0 x < 1, t > 0. (10) tˆuˆx, < ˆˆThe BCs (8) become uˆ 0, ˆ= 0 = u 1, ˆ, t > 0.t ˆ t ˆ(11) The ICs (9) become uˆ (ˆ = x) , utˆx, 0) = ˆ x) , 0 < ˆ (12) x, 0) fˆ(ˆ ˆ (ˆ g (ˆ x < 1. 1.3.1 Dimensionless 1-D Wave Problem with fixed ends Dropping hats, the dimensionless 1-D Wave Problem is, from (10) – (12), PDE : utt …
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