THE WAVE EQUATION 1. Longitudinal Vibrations We describe the ...

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$Math 664 Homework #2: Solutions 1. Let Î© = R, F = {A â R : either A ...$

4 R. E. SHOWALTERNote that if X(·) and T (·) are solutions of these respective equations, then it followsdirectly that their product is a solution (3a). We have already found the non-null solutionsof the boundary-value problem{X ′′ (x) + λX(x) = 0, 0 < x < l,(4)X(0) = 0, X(l) = 0.The solutions are the normalized eigenfunctionsX n (x) =√2l sin(nπ l x)corresponding to the eigenvalues λ n = (nπ/l) 2 . If we combine these with the correspondingtime-dependent solutions cos(α √ λ n t) and sin(α √ λ n t) of the first differentialequation and take linear combinations, we obtain a large class of solutions of the partialdifferential equation (3a) and boundary conditions (3b) in the form of a seriesu(x, t) =∞∑ (An cos(α √ λ n t) + B n sin(α √ λ n t) ) X n (x),n=1where the sequences {A n } and {B n } are to be determined. From the initial conditions(3c), it follows that these coefficients must satisfyso we obtain∞∑A n X n (x) = u 0 (x),n=1∞∑B n α √ λ n X n (x) = v 0 (x), 0 < x < l,n=1A m = (u 0 (·), X m (·)), B m = (v 0(·), X m (·))α √ λ m, m ≥ 1 .In summary, the solution of the initial-boundary-value problem (3) is given by the seriesu(x, t) =∞∑ ( √cos(α λn t)(u 0 (·), X n (·)) + sin(α √ λ n t) (v 0(·), X n (·))α √ )Xn (x).λ nn=1Denote the second term in the preceding formula by[S(t)v 0 ](x) =∞∑ ( √sin(α λn t) (v 0(·), X n (·))α √ )Xn (x).λ nn=1This defines the operator S(t) on the space of functions on [0, l]. We can use this operatorto represent the solution by(5) u(·, t) = S ′ (t)u 0 + S(t)v 0 .Example 2. Suppose the rod (0, l) is perfectly elastic and set α 2 = k ρ. Assume that bothends of the rod are fixed, the initial displacement and velocity are both null, and that there
THE WAVE EQUATION 5are distributed forces f(x, t) along its length. The initial-boundary-value problem for thiscase is(6a)(6b)(6c)u tt (x, t) = α 2 u xx (x, t) + f(x, t), 0 < x < l, t > 0,u(0, t) = 0, u(l, t) = 0, t > 0,u(x, 0) = 0, u t (x, 0) = 0, 0 < x < l.For this problem with a non-homogeneous partial differential equation (6a), we lookfor the solution in the form∞∑(7) u(x, t) = u n (t)X n (x).n=1If we assume that f(x, t) has the eigenfunction expansion∞∑f(x, t) = f n (t)X n (x) ,then the coefficients are given byf n (t) ≡∫ l0n=1f(s, t)X n (s) ds ,and substituting this expansion into equation (6a) yields∞∑ [ün (t) + λ n α 2 u n (t) ] ∞∑X n (x) = f n (t)X n (x) .n=1By equating the coefficients of the series given in this last equation, we are led to thesequence of initial-value problems(8a)(8b)The solution to (8) isn=1ü n (t) + λ n α 2 u n (t) = f n (t), t > 0u n (t) =u n (0) = 0, ˙u n (0) = 0 .∫ t0lnπα sin(nπα l(t − τ))f n(τ) dτNow, if we use this in (7), we find that the solution to our non-homogeneous initialboundary-valueproblem (6) is∫ t ∞∑ l(9) u(x, t) =nπα sin(nπα l(t − τ))f n(τ)X n (x) dτ.0n=1Note that we can use the operator S(t) to represent this formula as(10) u(·, t) =∫ t0S(t − τ)f(τ) dτ.
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THE WAVE EQUATION 1. Longitudinal Vibrations We describe the ...

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