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

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