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

THE WAVE EQUATIONR. E. SHOWALTER1. Longitudinal VibrationsWe describe the longitudinal vibrations in a long narrow cylindrical rod of cross sectionarea S. The rod is located along the x−axis, and we identify it with the interval [a, b] inR. The rod is assumed to stretch or contract in the horizontal direction, and we assumethat the vertical plane cross-sections of the rod move only horizontally. Denote by u(x, t)the displacement in the positive direction from the point x ∈ [a, b] at the time t > 0. Thecorresponding displacement rate or velocity is denoted by v(x, t) ≡ u t (x, t).Let σ(x, t) denote the local stress, the force per unit area with which the part of therod to the right of the point x acts on the part to the left of x. Since force is positiveto the right, the stress is positive in conditions of tension. For a section of the rod,x 1 < x < x 2 , the total (rightward) force acting on that section due to the remainder ofthe rod is given by(σ(x2 , t) − σ(x 1 , t) ) S .If the density of the rod at x is given by ρ > 0, the momentum of this section is just∫ x2x 1ρ u t (x, t) S dx .If we let F (x, t) denote any external applied force per unit of volume in the positivex−direction, then we obtain from Newton’s second law thatddt∫ x2x 1ρ u t (x, t) S dx = ( σ(x 2 , t) − σ(x 1 , t) ) S +∫ x2x 1F (x, t) S dxfor any such x 1 < x 2 . For a sufficiently smooth displacement u(x, t), we obtain theconservation of momentum equation(1) ρ u tt (x, t) − σ x (x, t) = F (x, t) , a < x < b, t > 0 .The stress σ(x, t) is determined by the type of material of which the rod is composedand the amount by which the neighboring region is stretched or compressed, i.e., on theelongation or strain, ε(x, t). In order to define this, first note that a section [x, x+h] of therod is deformed by the displacement to the new position [x+u(x), (x+h)+u(x+h)]. Theelongation is the limiting increment of the change in the length due to the deformationas given by[u(x + h) + (x + h)] − [u(x) + x] − hlimh→0hso the strain is given by ε(x, t) ≡ u x (x, t).1= d u(x)dx ,

2 R. E. SHOWALTERThe relation between the stress and strain is a constitutive law, usually determinedby experiment, and it depends on the type of material. In the simplest case, with smalldisplacements, we find by experiment that σ(x, t) is proportional to ε(x, t), i.e., thatthere is a constant k called Young’s modulus for whichσ(x, t) = k ε(x, t) .The constant k is a property of the material, and in this case we say the material ispurely elastic. A rate-dependent component of the stress-strain relationship arises whenthe force generated by the elongation depends not only on the magnitude of the strainbut also on the speed at which it is changed, i.e., on the strain rate ε t (x, t) = v x (x, t).The simplest such case is that of a visco-elastic material defined by the linear constitutiveequationσ(x, t) = k ε(x, t) + µ ε t (x, t) ,in which the material constant µ is the viscosity or internal friction of the material.Finally, if we include the effect of the transverse motions of the rod that result from theelongations under conditions of constant volume or mass, we will get an additional termto represent the transverse inertia. If the constant P denotes Poisson’s ratio, and r isthe average radius of that cross section, then the corresponding stress-strain relationshipis given as before byσ(x, t) = k ε(x, t) + µ ε t (x, t) + ρ r 2 P ε tt (x, t) .In terms of displacement, the total stress is(2) σ(x, t) = ku x (x, t) + µu xt (x, t) + ρ r 2 P u xtt (x, t) .The partial differential equation for the longitudinal vibrations of the rod is obtained bysubstituting (2) into (1).1.1. Initial and Boundary conditions. Since the momentum equation is second–orderin time, one may expect that in order to have a well–posed problem, two initial conditionsshould be specified. Thus, we shall specify the initial conditionsu(x, 0) = u 0 (x), u t (x, t) = v 0 (x), a < x < b ,where u 0 (·) and v 0 (·) are the initial displacement and the initial velocity, respectively.We list a number of typical possibilities for determining the two boundary conditions.Each of these is illustrated as before with a condition at the right end, x = b, and wenote that another such condition will also be prescribed at the left end, x = a.1. The displacement could be specified at the end point:u(b, t) = d b (t), t > 0 .This is the Dirichlet boundary condition, or boundary condition of first type. It could beobtained from observation of the endpoint position, or it could be imposed directly onthe endpoint. The homogeneous case d b (t) = 0 corresponds to a clamped end.2. The horizontal force on the rod could be specified at the end point:σ(b, t) = f b (t), t > 0 .

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.

6 R. E. SHOWALTERAs before, each S(t) is an integral operator of the form∞∑∫ll[S(t)v 0 ](x) =nπα sin(nπα lt) (v 0 (s) sin( nπ l s)) ds2 l sin(nπ l x)=n=1∫ l02( ∑∞ln=10lnπα sin(nπα lt) sin(nπ l s) sin(nπ l x)) v 0 (s) dsn=1=∫ l0H(x, s, t)v 0 (s) dsfor which the kernelH(x, s, t) = 2 ( ∑∞ ll nπα sin(nπα lt) sin(nπ l s) sin(nπ l x))is the Green’s function for the problem.Example 3. Suppose the left end of the elastic rod is free while the right end has anelastic constraint given by u(l, t) + u x (l, t) = 0. The initial-boundary-value problem forthis situation is(11a)(11b)(11c)u tt (x, t) = α 2 u xx (x, t), 0 < x < l, t > 0,u x (0, t) = 0, u(l, t) + u x (l, t) = 0, t > 0,u(x, 0) = u 0 (x), u t (x, 0) = v 0 (x), 0 < x < l.We seek a solution in the formu(x, t) = X(x)T (t),where the boundary conditions imply that X x (0) = 0 and X(l)+X x (l) = 0. The methodof separation of variables leads us to the boundary-value problemX ′′ (x) + λX(x) = 0, 0 < x < l,X ′ (0) = 0, X(l) + X ′ (l) = 0.We obtain a sequence of eigenvalues λ n and corresponding eigenfunctions given byX n (x) = ( 2) 12cos (λ 1 2(12)n x).lNote that for large λ n we haveλ n ≈ ( nπ l )2 ,so the eigenvalues are asymptotically close to those of the preceding example. Combiningthese results with the time-dependent solutions and using the orthogonality of theeigenfunctions, we find solutions of the initial-boundary-value problem (11) in the form∞∑ ( √u(x, t) = cos(α λn t)(u 0 (·), X n (·)) + sin(α √ λ n t) (v 0(·), X n (·))α √ )Xn (x),λ nn=1

THE WAVE EQUATION 7with the eigenfunctions given by (12). Once again, this can be represented in the form(5) for an appropriate family of operators {S(t) : t ≥ 0}.Exercise 1. Suppose the rod (0, l) is elastic and that we account for the inertia of lateralextension. Set α 2 = k and ρβ2 = r 2 P , where P is Poisson’s ratio and r is the averageradius of a cross section as above. Assume there are no internal forces, i.e., f(x, t) = 0,and that both ends of the rod are fixed. The initial displacement and velocity are givenby u 0 (x) and v 0 (x), respectively. The initial-boundary-value problem is(13a)(13b)(13c)u tt (x, t) = α 2 u xx (x, t) + β 2 u xxtt (x, t), 0 < x < l, t > 0,u(0, t) = 0, u(l, t) = 0, t > 0,u(x, 0) = u 0 (x), u t (x, 0) = v 0 (x), 0 < x < l.Find the solution by separation of variables. Find the family of operators {S(t) : t ≥ 0}for which this can be represented in the form (5).Department of Mathematics, Oregon State University, Corvallis, OR 97331 - 4605E-mail address: show@math.oregonstate.edu