UDC 517.946.9 : 534.11

THE PROBLEM OF LONGITUDINAL VIBRATIONS OF A FINITE ELASTIC-VISCOUS ROD

S. I. GAIDUK

A number of works have been devoted to problems on longitudinal vibrations of finite elastic-viscous rods; among them we note [1—3]. In [1] the problem of free vibrations of a rod with fixed ends is considered, with simultaneous allowance for aftereffect and relaxation; moreover, the study is carried out mainly from the standpoint of the physical interpretation of the results obtained. Works [2] and [3] are devoted to the solution of problems connected with the theory of longitudinal impact on a finite elastic-viscous rod without allowance for relaxation.

In the present paper, the problem of free longitudinal vibrations of a finite elastic-viscous rod without allowance for relaxation is also considered; the case most often encountered in practice is examined, when one end of the rod is fixed and the other is free. In contrast to [1—3], where the Fourier method or an operational method is used, in this paper, in order to obtain a rigorous solution of the problem and its mathematical justification, the subtraction method and the contour-integral method developed in [4] are applied. Such a scheme for solving and studying the problem under consideration can easily be extended also to more complicated problems on vibrations of composite and nonhomogeneous rods.

§ 1. FORMULATION OF THE PROBLEM

It is required to find a solution of the equation

\[ \frac{\partial^2 u}{\partial t^2} = \alpha \frac{\partial^2 u}{\partial x^2} + \beta \frac{\partial^3 u}{\partial x^2 \partial t} \qquad (0<x<l,\ 0<t<T) \tag{1.1} \]

under the initial conditions

\[ u\big|_{t=0}=\varphi(x),\qquad \frac{\partial u}{\partial t}\bigg|_{t=0}=\psi(x) \qquad (0<x<l) \tag{1.2} \]

and boundary conditions

\[ u\big|_{x=0}=0,\qquad \alpha \frac{\partial u}{\partial x} + \beta \frac{\partial^2 u}{\partial x \partial t}\bigg|_{x=l} =0 \qquad (0<t<T) \tag{1.3} \]

under the assumption that a) the function \(\varphi(x)\) on the interval \([0,l]\) has three continuous derivatives and a fourth piecewise-continuous one, b) the function \(\psi(x)\) on the same interval has a continuous first derivative and a piecewise-continuous second one, and c) the conditions \(\varphi(0)=\varphi'(l)=\varphi''(0)=0\) and \(\psi(0)=0\) are satisfied.

In equation (1.1), the constant \(\alpha>0\) characterizes the elastic properties of the rod material, the constant \(\beta>0\) characterizes the viscous properties of the rod material, and \(u(x,t)\) is the displacement function of the cross sections of the rod (\(x\) is the spatial coordinate, and \(t\) the time coordinate).

§ 2. RESIDUE REPRESENTATION OF THE SOLUTION OF PROBLEM (1.1)—(1.3)

We put the basic problem (1.1)—(1.3) into correspondence with the following auxiliary, so-called spectral problem:

\[ (\alpha+\beta\lambda^2)\frac{d^2y}{dx^2}-\lambda^4y=-f(x,\lambda)\quad (0<x<l), \tag{2.1} \]

\[ y\big|_{x=0}=0,\qquad \frac{dy}{dx}\bigg|_{x=l}=0, \tag{2.2} \]

where \(f(x,\lambda)=\lambda^2\varphi(x)+\psi(x)-\beta\varphi''(x)\), and where the operation of differentiation with respect to \(t\) in equation (1.1) is put in correspondence with the complex parameter \(\lambda^2\); moreover, in contrast to [4], we take the right-hand side of the equation containing the parameter with a minus sign.

The solution of problem (2.1), (2.2) can be represented by means of the Green function \(G(x,\xi,\lambda)\) in the form [5]

\[ y(x,\lambda)=\int_0^l G(x,\xi,\lambda)F(\xi,\lambda)\,d\xi, \tag{2.3} \]

where

\[ G(x,\xi,\lambda)= \]

\[ = \begin{cases} \displaystyle G_1(x,\xi,\lambda)= \frac{ \sqrt{\alpha+\beta\lambda^2}\, \operatorname{ch}\dfrac{\lambda^2(l-x)}{\sqrt{\alpha+\beta\lambda^2}}\, \operatorname{sh}\dfrac{\lambda^2\xi}{\sqrt{\alpha+\beta\lambda^2}} }{ \lambda^2\operatorname{ch}\dfrac{\lambda^2l}{\sqrt{\alpha+\beta\lambda^2}} } & \text{for } \xi\le x, \\[3ex] \displaystyle G_1^*(x,\xi,\lambda)= \frac{ \sqrt{\alpha+\beta\lambda^2}\, \operatorname{ch}\dfrac{\lambda^2(l-\xi)}{\sqrt{\alpha+\beta\lambda^2}}\, \operatorname{sh}\dfrac{\lambda^2x}{\sqrt{\alpha+\beta\lambda^2}} }{ \lambda^2\operatorname{ch}\dfrac{\lambda^2l}{\sqrt{\alpha+\beta\lambda^2}} } & \text{for } \xi\ge x, \end{cases} \tag{2.4} \]

\[ F(x,\lambda)=\frac{f(x,\lambda)}{\alpha+\beta\lambda^2}. \tag{2.5} \]

Theorem 1. Under conditions a)—e) of § 1 the formulas hold

\[ \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} y(x,\lambda)\lambda^{2s+1}\,d\lambda = \begin{cases} \varphi(x) & \text{for } s=0,\\ \psi(x) & \text{for } s=1, \end{cases} \tag{2.6} \]

where the convergence of the series is uniform with respect to \(x\in[0,l]\), and \(C_\nu\) is a simple closed contour containing inside it only one singular point \(\lambda_\nu\) of the integrand; the singular points are arranged in the order of increasing moduli.

Proof. Evaluating by parts the integral standing on the right-hand side of equality (2.3), taking into account conditions a)—e) of § 1, we obtain

\[ y(x,\lambda)=\frac{\varphi(x)}{\lambda^2} +\frac{\psi(x)}{\lambda^4} +\frac{\alpha\varphi''(x)}{\lambda^6} - \left[ \frac{\alpha\varphi'''(l)}{\lambda^8} +\frac{\psi'(l)}{\lambda^6} \right]\times \]

\[ \times \frac{ \sqrt{\alpha+\beta\lambda^{2}}\, \operatorname{sh}\dfrac{\lambda^{2}x}{\sqrt{\alpha+\beta\lambda^{2}}} }{ \operatorname{ch}\dfrac{\lambda^{2}l}{\sqrt{\alpha+\beta\lambda^{2}}} } + \frac{\alpha}{\lambda^{6}}\int_{0}^{l}G(x,\xi,\lambda)\varphi^{\mathrm{IV}}(\xi)\,d\xi + \frac{1}{\lambda^{4}}\int_{0}^{l}G(x,\xi,\lambda)\psi'(\xi)\,d\xi . \tag{2.7} \]

Let us find an asymptotic representation of the Green’s function \(G(x,\xi,\lambda)\) for large \(|\lambda|\). Since for large \(|\lambda|\) the equality

\[ \frac{\lambda^{2}}{\sqrt{\alpha+\beta\lambda^{2}}} = \frac{\lambda}{\sqrt{\beta}}+O\!\left(\frac{1}{\lambda}\right) \]

holds, where the notation \(O\!\left(\frac{1}{\lambda}\right)\) denotes a function of the form \(\dfrac{\omega(\lambda)}{\lambda}\), with \(\omega(\lambda)\) bounded in modulus for large \(|\lambda|\), it follows that for these values of \(|\lambda|\) we have

\[ e^{\frac{\lambda^{2}}{\sqrt{\alpha+\beta\lambda^{2}}}} = e^{\frac{\lambda}{\sqrt{\beta}}} \left[ 1+O\!\left(\frac{1}{\lambda}\right) \right]. \tag{2.8} \]

Taking (2.8) into account, the Green’s function \(G(x,\xi,\lambda)\) for large \(|\lambda|\) can be represented in the form

\[ G(x,\xi,\lambda)= \begin{cases} G_1(x,\xi,\lambda)= \dfrac{\Phi(x,\xi,\lambda)} {\lambda\left[ 1+e^{k\frac{2l}{\sqrt{\beta}}\lambda} +O\!\left(\dfrac{1}{\lambda}\right) \right]}, & \text{for } \xi \leq x,\\[2.2ex] G_1^{*}(x,\xi,\lambda)= \dfrac{\Phi^{*}(x,\xi,\lambda)} {\lambda\left[ 1+e^{k\frac{2l}{\sqrt{\beta}}\lambda} +O\!\left(\dfrac{1}{\lambda}\right) \right]}, & \text{for } \xi \geq x, \end{cases} \tag{2.9} \]

where the functions \(\Phi(x,\xi,\lambda)\) and \(\Phi^{*}(x,\xi,\lambda)\) are bounded in modulus for large \(|\lambda|\), and \(k=-1\) in the right \(\lambda\)-half-plane and \(k=1\) in the left \(\lambda\)-half-plane.

Following [6], it can be shown that the roots of the equation

\[ H(\lambda)=1+e^{k\frac{2l}{\sqrt{\beta}}\lambda} +O\!\left(\frac{1}{\lambda}\right)=0 \tag{2.10} \]

have the form

\[ \lambda_{\nu} = \pm\frac{\sqrt{\beta}}{2l}(2\nu+1)\pi i + O\!\left(\frac{1}{2\nu+1}\right) \quad (\nu=0,1,2,\ldots) \]

and that for every \(\delta>0\) one can specify a number \(N_{\delta}>0\) such that, for all \(\lambda\) satisfying the inequality \(|\lambda-\lambda_{\nu}|\geq \delta\), the inequality

\[ |H(\lambda)|\geq N_{\delta} \tag{2.11} \]

will hold.

From (2.9) and (2.11) it follows that in the \(\lambda\)-plane outside the circles \(|\lambda-\lambda_{\nu}|\leq \delta\) the estimate

\[ |G(x,\xi,\lambda)|\leq \frac{A}{|\lambda|} \tag{2.12} \]

holds.

Therefore equality (2.7) can be represented in the form

\[ y(x,\lambda)=\frac{\varphi(x)}{\lambda^{2}}+\frac{\psi(x)}{\lambda^{4}}+\frac{1}{\lambda^{5}}W(x,\lambda), \tag{2.13} \]

where the function \(W(x,\lambda)\) is bounded in modulus for large \(|\lambda|\) outside the circles \(|\lambda-\lambda_\nu|=\delta\).

Let us now draw in the \(\lambda\)-plane a sequence of expanding concentric circles \(O_\nu\) \((\nu=1,2,3,\ldots)\) with common center at the origin and with radii \(r_\nu\to\infty\) as \(\nu\to\infty\), lying from the circles \(|\lambda-\lambda_\nu|=\delta\) at a distance greater than \(\delta\), and consider the integrals

\[ \lim_{\nu\to\infty}\frac{1}{2\pi i}\int_{O_\nu} y(x,\lambda)\lambda^{2s+1}\,d\lambda \quad (s=0,1). \tag{2.14} \]

Substituting (2.13) into (2.14), we obtain

\[ \begin{aligned} \lim_{\nu\to\infty}\frac{1}{2\pi i}\int_{O_\nu} y(x,\lambda)\lambda^{2s+1}\,d\lambda &= \frac{\varphi(x)}{2\pi i}\lim_{\nu\to\infty}\int_{O_\nu}\lambda^{2s-1}\,d\lambda \\ &\quad+ \frac{\psi(x)}{2\pi i}\lim_{\nu\to\infty}\int_{O_\nu}\lambda^{2s-3}\,d\lambda + \frac{1}{2\pi i}\lim_{\nu\to\infty}\int_{O_\nu}\lambda^{2s-4}W(x,\lambda)\,d\lambda . \end{aligned} \tag{2.15} \]

In view of the fact that the integrand in the last term on the right-hand side of equality (2.15) decreases as \(\lambda^{-\alpha}\) \((\alpha>1)\) for \(s=0\) and \(s=1\), this term is equal to zero, and we obtain

\[ \lim_{\nu\to\infty}\frac{1}{2\pi i}\int_{O_\nu}y(x,\lambda)\lambda^{2s+1}\,d\lambda = \begin{cases} \varphi(x), & \text{for } s=0,\\ \psi(x), & \text{for } s=1. \end{cases} \]

Since, according to the fundamental theorem on residues,

\[ \lim_{\nu\to\infty}\frac{1}{2\pi i}\int_{O_\nu}y(x,\lambda)\lambda^{2s+1}\,d\lambda = \sum_\nu \int_{C_\nu} y(x,\lambda)\lambda^{2s+1}\,d\lambda, \tag{2.16} \]

where the sum over \(\nu\) on the right-hand side of (2.16) is extended over all singular points of the integrand, the theorem is proved.

Theorem 2. Under the conditions a)—c) of § 1, if problem (1.1)—(1.3) has a solution \(u(x,t)\), continuous for \(x\in[0,l]\), \(t\in[0,T]\) and having the continuous derivatives entering the problem: \(\dfrac{\partial u}{\partial t}\) for \(0<x<l\), \(0\leq t<T\), \(\dfrac{\partial u}{\partial x}\), \(\dfrac{\partial^2 u}{\partial x\partial t}\) for \(0<x\leq l\), \(0<t<T\), and the remaining derivatives for \(0<x<l\), \(0<t<T\), then this solution is representable in the form of the complete integral residue

\[ u(x,t)=\frac{1}{2\pi i}\sum_\nu \int_{C_\nu} y(x,\lambda)e^{\lambda^2 t}\,d\lambda . \tag{2.17} \]

Proof. The validity of the theorem is verified by substituting formula (2.17) into equation (1.1) and into conditions (1.2), (1.3). As a result of substituting (2.17) into equation (1.1), we have

\[ \alpha \frac{\partial^2 u}{\partial x^2} +\beta \frac{\partial^3 u}{\partial x^2 \partial t} -\frac{\partial^2 u}{\partial t^2} = \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} \left[ (\alpha+\beta\lambda^2)\frac{d^2 y}{d x^2} -\lambda^4 y \right]\lambda e^{\lambda^2 t}\,d\lambda = \]

\[ = \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} f(x,\lambda)\lambda e^{\lambda^2 t}\,d\lambda =0 \]

by virtue of the analyticity of the integrand. Substituting (2.17) into the boundary conditions (1.3), we obtain

\[ u\big|_{x=0} = \lim_{x\to 0} u(x,t) = \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} y(x,\lambda)\big|_{x=0}\lambda e^{\lambda^2 t}\,d\lambda =0, \]

\[ \left. \alpha \frac{\partial u}{\partial x} +\beta \frac{\partial^2 u}{\partial x\partial t} \right|_{x=l} = \lim_{x\to l} \left( \alpha \frac{\partial u}{\partial x} +\beta \frac{\partial^2 u}{\partial x\partial t} \right) = \]

\[ = \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} (\alpha+\beta\lambda^2) \left.\frac{\partial y}{\partial x}\right|_{x=l} \lambda e^{\lambda^2 t}\,d\lambda =0. \]

Checking the fulfillment of the initial conditions, by virtue of Theorem 1 we have

\[ \left.u\right|_{t=0} = \lim_{t\to 0} u(x,t) = \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} y(x,\lambda)\lambda\,d\lambda = \varphi(x), \]

\[ \left.\frac{\partial u}{\partial t}\right|_{t=0} = \lim_{t\to 0} \frac{\partial u}{\partial t} = \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} y(x,\lambda)\lambda^3\,d\lambda = \psi(x). \]

§ 3. PROOF OF EXISTENCE OF A SOLUTION

Theorem 3. If conditions a)–e) of § 1 are satisfied, problem (1.1)—(1.3) has a solution possessing the properties indicated in Theorem 2 and representable by the formula

\[ u(x,t)=\frac{1}{\pi i}\int_L y(x,\lambda)e^{\lambda^2 t}\lambda\,d\lambda, \tag{3.1} \]

where \(L\) is an open contour, the choice of which is described below.

Proof. Let \(D_h\) be a strip of finite width \(h\), with its center at the origin of coordinates of the \(\lambda\)-plane, whose boundaries are parallel to the imaginary axis. Denote by \(R(h,\beta)\) the set of values of \(\lambda\) lying outside the strip \(D_h\) and satisfying the conditions

\[ -\frac{\pi}{2}+\beta < \arg\lambda < \frac{\pi}{2}+\beta, \]

where the constant number \(\beta\) is chosen so that the inequality

\[ -\frac{\pi}{4}<\beta<\frac{\pi}{4}. \tag{3.2} \]

is satisfied.

Next, let \(L\) and \(L^*\) be open contours symmetric with respect to the imaginary axis, the first of which is located in the region \(R(h,\beta)\), and let the straight lines defined by the equations

\[ \arg \lambda=-\frac{\pi}{2}+\beta,\qquad \arg \lambda=\frac{\pi}{2}-\beta, \tag{3.3} \]

be their asymptotes; and let, in addition, \(O_\nu\) \((\nu=1,2,3,\ldots)\) be the sequence of circles described in § 2. Denote by \(L_1^{(\nu)}\) and \(L_2^{(\nu)}\) the parts of the contours \(L\) and \(L^*\), respectively, located inside \(O_\nu\). Suppose that \(L-L_1^{(\nu)}\) and \(L^*-L_2^{(\nu)}\) coincide with sufficiently distant parts of the straight lines defined by equations (3.3). The arcs of the circles \(O_\nu\) located between \(L_1^{(\nu)}\) and \(L_2^{(\nu)}\) and lying in the upper and lower \(\lambda\)-half-planes shall be denoted by \(M_1^{(\nu)}\) and \(M_2^{(\nu)}\), respectively. Let \(\Gamma_\nu\) \((\nu=1,2,3,\ldots)\) be the closed contour consisting of the arcs \(L_1^{(\nu)}\), \(L_2^{(\nu)}\), \(M_1^{(\nu)}\), \(M_2^{(\nu)}\) (see the figure).

We shall show that the function defined by formula (3.1) satisfies equation (1.1) and the boundary conditions (1.3). For this it is necessary to prove the uniform convergence of the integrals

\[ \int_L \lambda^{2s+1}\frac{d^m y}{dx^m} e^{\lambda^2 t}\,d\lambda \tag{3.4} \]

\[ (s=0,\ m=0,1,2;\ s=1,\ m=1,2;\ s=2,\ m=0) \]

with respect to \(x\in[0,l]\) and \(t\in(0,T]\), which will make it possible to differentiate under the integral sign in formula (1.3) with respect to \(x\) and \(t\), and to pass to the limit as \(x\to0\) and \(x\to l\).

The integrals (3.4) will converge uniformly if the parts of these integrals over the remote pieces \(a_\nu, a_{\nu+n}\) and \(a'_\nu, a'_{\nu+n}\) tend to zero as \(\nu\to\infty\) for all \(n\), uniformly with respect to \(x\in[0,l]\) and \(t\in(0,T]\). We shall show that this is indeed the case.

Differentiating equality (2.3) twice with respect to \(x\), we have

\[ \frac{d^m y}{dx^m}=\gamma_m(x)+\int_0^l \frac{\partial^m G}{\partial x^m}F(\xi,\lambda)\,d\xi, \tag{3.5} \]

where

\[ \gamma_m(x)= \begin{cases} 0, & \text{for } m=0,1,\\ -F(x,\lambda), & \text{for } m=2. \end{cases} \]

Finding asymptotic representations of the functions \(\dfrac{\partial^m G}{\partial x^m}\) \((m=1,2)\) for large \(|\lambda|\) and estimating them in modulus in the same way as was done for the function \(G(x,\xi,\lambda)\) in § 2, we obtain the estimates

\[ \left|\frac{\partial^m G}{\partial x^m}\right| \leqslant A|\lambda|^{m-1}\quad (m=1,2). \]

From the last estimate and from (2.12) and (3.5) it follows that

\[ \left|\frac{d^m y}{dx^m}\right| \leqslant B|\lambda|^{m-1}\quad (m=0,1,2), \tag{3.6} \]

therefore

\[ J=\left|\int_{a_\nu}^{a_{\nu+n}} \lambda^{2s+1}\frac{d^m y}{dx^m} e^{\lambda^2 t}\,d\lambda\right| \leqslant B\int_{r_\nu}^{r_{\nu+n}} r^{2s+m} e^{r^2 t\cos 2\varphi}\,dr, \tag{3.7} \]

where \(\lambda = r e^{i\varphi}\).

Since on the line on which the segment \(a_\nu a_{\nu+n}\) lies, \(\arg \lambda = \dfrac{\pi}{2}-\beta\), on this line \(\cos 2\varphi = -\cos 2\beta = -\varepsilon\), where \(\varepsilon>0\), since \(\beta\) satisfies condition (3.2).

Thus,

\[ J \leqslant B \int_{r_\nu}^{r_{\nu+n}} r^{2s+m} e^{-r^2\varepsilon t}\,dr \leqslant B e^{-\frac12 r_\nu^2\varepsilon t}\int_0^\infty e^{-\frac12 r^2\varepsilon t} r^{2s+m}\,dr \leqslant \]

\[ \leqslant B\left(\sqrt{\frac{2}{\varepsilon t}}\right)^{2s+m+1} e^{-\frac12 r_\nu^2\varepsilon t} \int_0^\infty e^{-\beta^2}\beta^{2s+m}\,d\beta = \]

\[ = C_{ms}\left(\sqrt{\frac{2}{\varepsilon t}}\right)^{2s+m+1} e^{-\frac12 r_\nu^2\varepsilon t}. \]

The quantity \(J\) is estimated analogously on the segment \(a'_\nu a'_{\nu+n}\). Hence follows the uniform convergence of the integrals (3.4) with respect to \(x\in[0,l]\) and \(t\in(0,T]\). Substituting now formula (3.1) into equation (1.1) and the boundary conditions (1.3), we verify, as in § 2, that the equation and boundary conditions are satisfied.

To prove that the initial conditions are fulfilled, it is necessary to prove the uniform convergence of the integrals

\[ \int_L \lambda^{2s+1} y(x,\lambda)e^{\lambda^2 t}\,d\lambda\quad (s=0,1) \tag{3.8} \]

with respect to \(x\in[0,l]\), \(t\in[0,T]\).

By virtue of the symmetry of the contours \(L_1^{(\nu)}\) and \(L_2^{(\nu)}\), and by virtue of the evenness with respect to \(\lambda\) of the function \(y(x,\lambda)e^{\lambda^2 t}\),

\[ \int_{L_1^{(\nu)}} \lambda^{2s+1}y(x,\lambda)e^{\lambda^2 t}\,d\lambda = \int_{L_2^{(\nu)}} \lambda^{2s+1}y(x,\lambda)e^{\lambda^2 t}\,d\lambda. \tag{3.9} \]

Moreover, by virtue of the choice of the contours \(M_1^{(\nu)}\) and \(M_2^{(\nu)}\), for \(t>0\)

\[ \lim_{\nu\to\infty}\int_{M_1^{(\nu)}} \lambda^{2s+1} y(x,\lambda)e^{\lambda^2 t}\,d\lambda = \lim_{\nu\to\infty}\int_{M_2^{(\nu)}} \lambda^{2s+1} y(x,\lambda)e^{\lambda^2 t}\,d\lambda =0. \tag{3.10} \]

From the last two equalities it follows that

\[ u(x,t)=\frac{1}{\pi i}\int_L \lambda^{2s+1}y(x,\lambda)e^{\lambda^2 t}\,d\lambda = \frac{1}{2\pi i}\lim_{\nu\to\infty}\int_{\Gamma_\nu} \lambda^{2s+1}y(x,\lambda)e^{\lambda^2 t}\,d\lambda. \tag{3.11} \]

Substituting into the right-hand side of the last equality, for \(s=0\), (2.13), we obtain

\[ u(x,t)=\frac{1}{2\pi i}\lim_{\nu\to\infty} \int_{\Gamma_\nu} y(x,\lambda)\lambda e^{\lambda^2 t}\,d\lambda = \varphi(x)+\frac{1}{2\pi i}\lim_{\nu\to\infty} \int_{\Gamma_\nu} \left[ \frac{\psi(x)}{\lambda^3} + \frac{1}{\lambda^4}W(x,\lambda) \right]e^{\lambda^2 t}\,d\lambda . \]

It is clear that for \(t>0\) the integral on the right-hand side of the last equality converges uniformly and is equal to zero for \(t=0\), since the contour of integration \(\Gamma_\nu\) may be replaced by the circle \(O_\nu\), on which the integrand decreases as \(\lambda^{-\alpha}\) (\(\alpha>1\)). Hence it follows that in the integral (3.8), for \(s=0\), one may pass to the limit as \(t\to0\). The legitimacy of passing to the limit as \(t\to0\) in the same integral for \(s=1\) is justified analogously.

Taking the above into account, we have

\[ u(x,t)\big|_{t=0} = \lim_{t\to0}u(x,t) = \frac{1}{2\pi i}\lim_{\nu\to\infty} \int_{\Gamma_\nu} y(x,\lambda)\lambda\,d\lambda = \]

\[ = \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} y(x,\lambda)\lambda\,d\lambda = \varphi(x), \]

\[ \left.\frac{\partial u}{\partial t}\right|_{t=0} = \lim_{t\to0}\frac{\partial u}{\partial t} = \frac{1}{2\pi i}\lim_{\nu\to\infty} \int_{\Gamma_\nu} y(x,\lambda)\lambda^3\,d\lambda = \]

\[ = \frac{1}{2\pi i}\sum_\nu \int_{C_\nu} y(x,\lambda)\lambda^3\,d\lambda = \psi(x). \]

§ 4. UNIQUENESS OF THE SOLUTION AND ITS CONTINUOUS DEPENDENCE ON THE INITIAL CONDITIONS

Theorem 4. Under conditions a)—c) of § 1, the solution of problem (1.1)—(1.3) is unique and depends continuously on the initial conditions.

Proof. Suppose that problem (1.1)—(1.3) has two sufficiently smooth solutions \(u_1(x,t)\) and \(u_2(x,t)\). Then their difference is a sufficiently smooth solution of problem (1.1)—(1.3) with zero initial conditions. This solution, according to Theorem 2, is representable in the form (2.17), where \(\varphi(x)=0\) and \(\psi(x)=0\). Consequently, it is identically equal to zero. Thus, \(u_1=u_2\).

Substituting into (2.3), instead of the function \(F(\xi,\lambda)\), its value and twice integrating by parts the integral whose integrand contains the function \(\varphi(x)\), after simple transformations we obtain

\[ y(x,\lambda)=\frac{\varphi(x)}{\lambda^2} +\frac{1}{\lambda^3}\int_0^l \omega_1(x,\xi,\lambda)\psi(\xi)\,d\xi+ \]

\[ +\frac{1}{\lambda^5}\int_0^l \omega_2(x,\xi,\lambda)\varphi''(\xi)\,d\xi, \tag{4.1} \]

where \(\omega_j(x,\xi,\lambda)\) \((j=1,2)\) are bounded for large \(|\lambda|\) outside the circles \(|\lambda-\lambda_\nu|=\delta\). We now substitute (4.1) into the right-hand side of equality (3.11) and change the order of integration by virtue of the uniform convergence of the integrals

\[ \Phi_j(x,\xi,t)=-\frac{1}{2\pi i}\lim_{\nu\to\infty} \int_{\Gamma_\nu}\frac{\omega_j(x,\xi,\lambda)}{\lambda^2}e^{\lambda^2 t}\,d\lambda \quad (j=1,2) \]

with respect to \(x,\xi\in[0,l]\), \(t\in[0,T]\); then we obtain

\[ u(x,t)=\varphi(x)+\int_0^l \psi(\xi)\Phi_1(x,\xi,t)\,d\xi +\int_0^l \varphi''(\xi)\Phi_2(x,\xi,t)\,d\xi, \]

whence follows the inequality

\[ |u(x,t)|\leq |\varphi(x)|+C\int_0^l|\psi(\xi)|\,d\xi +D\int_0^l|\varphi''(\xi)|\,d\xi, \]

from which the continuous dependence of the solution on the initial conditions follows.

§ 5. COMPUTATION OF THE TOTAL INTEGRAL RESIDUE

From §§ 2 and 3 it follows that the solution of problem (1.1)—(1.3) is represented in the form of the total integral residue

\[ u(x,t)=\frac{1}{2\pi i}\sum_\nu \int_{C_\nu} y(x,\lambda)\lambda e^{\lambda^2 t}\,d\lambda. \tag{5.1} \]

It is easy to see from (2.3)—(2.5) that the singular points of the integrand in (5.1) are two essentially singular points
\(\lambda_{1,2}=\pm i\sqrt{\frac{\alpha}{\beta}}\) and an infinite set of poles
\(\lambda_{\nu m}=\pm\sqrt{-p_\nu\pm q_\nu}\)
\((m=1,2,3,4;\ \nu=0,1,2,\ldots)\), where
\[ p_\nu=\frac{\beta\chi_\nu^2}{2},\qquad q_\nu=\frac{1}{2}\sqrt{\beta^2\chi_\nu^4-4\alpha\chi_\nu^2{}^{*}}, \qquad \chi_\nu=\frac{(2\nu+1)\pi}{2l}. \]

\[ \text{*) We shall assume that } \beta^2\chi_\nu^2-4\alpha<0 \text{ for a finite number of values of } \nu. \]

Let us now represent equality (5.1) in the form

\[ u(x,t):=\frac{1}{2\pi i}\sum_{\nu=0}^{\infty}\int_{C_\nu} y(x,\lambda)\lambda e^{\lambda^2 t}\,d\lambda +\frac{1}{2\pi i}\sum_{\nu=1}^{2}\int_{\dot C_\nu} y(x,\lambda)\lambda e^{\lambda^2 t}\,d\lambda, \tag{5.2} \]

where the first sum on the right-hand side of the last equality is extended over all the poles of the integrand, and the second over the essentially singular points. Substituting into the first term on the right-hand side of (5.2), in place of the function \(y(x,\lambda)\), its value (2.3), we obtain

\[ \begin{aligned} \frac{1}{2\pi i}\sum_{\nu=0}^{\infty}\int_{C_\nu} y(x,\lambda)e^{\lambda^2 t}\lambda\,d\lambda &=\sum_{\nu=0}^{\infty}\sum_{m=1}^{4}\Biggl\{ \int_{0}^{l}\left[ \operatorname*{Res}_{\lambda=\lambda_{\nu m}}G(x,\xi,\lambda) \frac{\lambda^3 e^{\lambda^2 t}}{\alpha+\beta\lambda^2} \right]\varphi(\xi)\,d\xi \\ &\quad+\int_{0}^{l}\left[ \operatorname*{Res}_{\lambda=\lambda_{\nu m}}G(x,\xi,\lambda) \frac{e^{\lambda^2 t}\lambda}{\alpha+\beta\lambda^2} \right]\psi(\xi)\,d\xi \\ &\quad-\beta\int_{0}^{l}\left[ \operatorname*{Res}_{\lambda=\lambda_{\nu m}}G(x,\xi,\lambda) \frac{e^{\lambda^2 t}\lambda}{\alpha+\beta\lambda^2} \right]\varphi''(\xi)\,d\xi \Biggr\}, \end{aligned} \tag{5.3} \]

where

\[ \lambda_{\nu1}=\sqrt{-p_\nu-q_\nu},\qquad \lambda_{\nu2}=\sqrt{-p_\nu-q_\nu},\qquad \lambda_{\nu3}=-\sqrt{-p_\nu+q_\nu},\qquad \lambda_{\nu4}=-\sqrt{-p_\nu-q_\nu}. \]

Computing the residues on the right-hand side of equality (5.3), we have

\[ \operatorname*{Res}_{\lambda=\lambda_{\nu m}}G(x,\xi,\lambda) \frac{\lambda^3 e^{\lambda^2 t}}{\alpha+\beta\lambda^2} = -\frac{1}{l}\, \frac{\alpha+\beta\lambda_{\nu m}^{2}}{2\alpha+\beta\lambda_{\nu m}^{2}}\, e^{\lambda_{\nu m}^{2}t}\sin \varkappa_\nu x\sin \varkappa_\nu \xi, \tag{5.4} \]

\[ \operatorname*{Res}_{\lambda=\lambda_{\nu m}}G(x,\xi,\lambda) \frac{e^{\lambda^2 t}\lambda}{\alpha+\beta\lambda^2} = -\frac{1}{l}\, \frac{\alpha+\beta\lambda_{\nu m}^{2}}{\lambda_{\nu m}^{2}(2\alpha+\lambda_{\nu m}^{2})}\, e^{\lambda_{\nu m}^{2}t}\sin \varkappa_\nu x\sin \varkappa_\nu \xi. \tag{5.5} \]

Since

\[ \frac{\lambda^2}{\sqrt{\alpha+\beta\lambda^2}}=-i\varkappa_\nu, \]

we have

\[ \alpha+\beta\lambda^2=-\frac{\lambda^4}{\varkappa_\nu^2}, \]

whence

\[ \frac{\alpha+\beta\lambda^2}{2\alpha+\beta\lambda^2} = \frac{\lambda^4}{\lambda^4-\alpha\varkappa_\nu^2}. \tag{5.6} \]

Taking (5.6) into account, we obtain

\[ \sum_{m=1}^{4} \frac{\alpha+\beta\lambda_{\nu m}^{2}}{2\alpha+\beta\lambda_{\nu m}^{2}}\, e^{\lambda_{\nu m}^{2}t} = 2e^{-p_\nu t}\left[ \operatorname{ch} q_\nu t-\frac{p_\nu}{q_\nu}\operatorname{sh} q_\nu t \right], \tag{5.7} \]

\[ \sum_{m=1}^{4} \frac{\alpha+\beta\lambda_{\nu m}^{2}}{\lambda_{\nu m}^{2}(2\alpha+\beta\lambda_{\nu m}^{2})}\, e^{\lambda_{\nu m}^{2}t} = 2\frac{1}{q_\nu}\operatorname{sh} q_\nu t\cdot e^{-p_\nu t}. \tag{5.8} \]

After this, the right-hand side of equality (5.3), taking into account (5.4), (5.5) and (5.7), (5.8), can be represented in the form

\[ \frac{1}{2\pi i}\sum_{\nu=0}^{\infty}\int_{C_\nu} y(x,\lambda)e^{\lambda^2 t}\lambda\,d\lambda = \sum_{\nu=0}^{\infty}\left[ A_\nu \operatorname{ch} q_\nu t +\frac{1}{q_\nu}\left(p_\nu A_\nu \operatorname{sh} q_\nu t+\right.\right. \]

\[ + B_\nu)\operatorname{sh} q_\nu t\biggr] e^{-p_\nu t}\sin \chi_\nu x, \tag{5.9} \]

where

\[ A_\nu=\frac{2}{l}\int_0^l \varphi(\xi)\sin \chi_\nu \xi\,d\xi,\qquad B_\nu=\frac{2}{l}\int_0^l \psi(\xi)\sin \chi_\nu \xi\,d\xi. \]

It is difficult to compute the residues at the essential singular points, but in our case their sum is equal to zero, since only in this case, as is easily verified by substituting the right-hand side of equality (5.9), does the function (5.2) satisfy equation (1.1) and the conditions (1.2), (1.3).

Thus,

\[ u(x,t)=\sum_{\nu=0}^{\infty} e^{-p_\nu t}\sin \chi_\nu x \left[A_\nu \operatorname{ch} q_\nu t+\frac{1}{q_\nu}(p_\nu A_\nu+B_\nu)\operatorname{sh} q_\nu t\right]. \]

Putting \(\beta=0\) in \(p_\nu\) and \(q_\nu\) in the last equality, we obtain the known solution of the problem of longitudinal vibrations of an elastic rod [7]:

\[ u(x,t)=\sum_{\nu=0}^{\infty}\left[ A_\nu\cos\frac{(2\nu+1)a\pi t}{2l} +B_\nu\sin\frac{(2\nu+1)a\pi t}{2l} \right]\times \]

\[ \times\sin\frac{(2\nu+1)\pi x}{2l}, \]

where

\[ A_\nu=\frac{2}{l}\int_0^l \varphi(\xi)\sin\frac{(2\nu+1)\pi \xi}{2l}\,d\xi, \]

\[ B_\nu=\frac{4}{(2\nu+1)\pi a}\int_0^l \psi(\xi)\sin\frac{(2\nu+1)\pi \xi}{2l}\,d\xi. \]

References

1. A. Yu. Ishlinskii, PMM, 4, no. 1, 79–92, 1940.
2. V. A. Lazaryan, Proceedings of the Dnepropetrovsk Institute of Railway Transport Engineers, issue XX, 3–32, 1950.
3. A. I. Tveritin, Proceedings of the Dnepropetrovsk Institute of Railway Transport Engineers, issue XXIII, 1953.
4. M. L. Rasulov, The Method of the Contour Integral. Nauka Publishing House, 1964.
5. A. V. Ivanov, Collection “Thermophysics in Foundry Production.” Publishing House of the Academy of Sciences of the BSSR, 11–52, 1963.
6. M. A. Naimark, Linear Differential Operators. Moscow, Gostekhizdat, 1954.
7. N. S. Koshlyakov et al., Basic Differential Equations of Mathematical Physics. Moscow, Fizmatgiz, 1962.

Received by the editors
October 5, 1965

Institute of Mathematics, Academy of Sciences of the BSSR