Variational Principles of Elastodynamics

Abstract

In this chapter both the classical Hamilton-Kirchhoff Principle and a convolutional variational principle of Gurtin’s type that describes completely a solution to an initial-boundary value problem of elastodynamics are used to solve a number of typical problems of elastodynamics.

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In this chapter both the classical Hamilton-Kirchhoff Principle and a convolutional variational principle of Gurtin’s type that describes completely a solution to an initial-boundary value problem of elastodynamics are used to solve a number of typical problems of elastodynamics.

1 The Hamilton-Kirchhoff Principle

To formulate H-K principle we introduce a notion of kinematically admissible process, and by this we mean an admissible process that satisfies the strain-displacement relation, the stress-strain relation, and the displacement boundary condition.

(H-K) The Hamilton-Kirchhoff Principle. Let $P$ denote the set of all kinematically admissible processes ${p}=[\mathbf{u,E,S}]$ on ${\overline{\mathrm{{B}}}}\times [0,\infty )$ satisfying the conditions

$$\begin{aligned} \mathbf{{u}}(\mathbf{{x}},\mathrm{{t}}_1 )=\mathbf{{u}}_1 (\mathbf{{x}}),\quad \mathbf{{u}}(\mathbf{{x}},\mathrm{{t}}_2 )=\mathbf{{u}}_2 (\mathbf{{x}})\quad \mathrm{{on}}\quad {\overline{\mathrm{{B}}}} \end{aligned}$$

(5.1)

where $t_1$ and $t_2$ are two arbitrary points on the $t$-axis such that $0\le {t}_1 <{t}_2 <\infty$, and $\mathbf{u}_1 (\mathbf{x})$ and $\mathbf{u}_2 (\mathbf{x})$ are prescribed fields on $\overline{\mathrm{B}}$. Let $\mathsf{K }=\mathsf{K }\{{p}\}$ be the functional on $P$ defined by

$$\begin{aligned} \mathsf{{K}}\{{p}\}=\int \limits _{{t}_1 }^{{t}_2 } {[\mathrm{F}({t})-\mathrm{K}({t})]\,{dt}} \end{aligned}$$

(5.2)

where

$$\begin{aligned} {{F}(t)}=\mathrm{U}_\mathrm{C} \{\mathbf{E}\}-\int \limits _\mathrm{B} {\mathbf{b}\cdot \mathbf{u}} \,{ dv}-\int \limits _{\partial \mathrm{B}_2} {\hat{\mathbf{s}}\cdot \mathbf{u}\,{da}} \end{aligned}$$

(5.3)

and

$$\begin{aligned} {{K}(t)}=\frac{1}{2}\int \limits _\mathrm{B} {\rho \,\dot{\mathbf{u}}^2{dv}} \end{aligned}$$

(5.4)

for every ${p}=[\mathbf{u, E, S}]\in {P}$. Then

$$\begin{aligned} \delta \,\mathsf{K }\{{p}\}=0 \end{aligned}$$

(5.5)

if and only if p satisfies the equation of motion and the traction boundary condition.

Clearly, in the (H-K) principle a displacement vector $\mathbf{u}=\mathbf{u}(\mathbf{x},\mathrm{t})$ needs to be prescribed at two points $t_1$ and $t_2$ of the time axis. If $t_1 =0$, then $\mathbf{u}(\mathbf{x},0)$ may be identified with the initial value of the displacement vector in the formulation of an initial-boundary value problem, however, the value $\mathbf{u}(\mathbf{x},\mathrm{t}_2 )$ is not available in this formulation. This is the reason why the (H-K) principle can not be used to describe the initial-boundary value problem. A full variational characterization of an initial-boundary value problem of elastodynamics is due to Gurtin, and it has the form of a convolutional variational principle. The idea of a convolutional variational principle of elastodynamics is now explained using a traction initial-boundary value problem of incompatible elastodynamics. In such a problem we are to find a symmetric second-order tensor field $\mathbf{S}=\mathbf{S}(\mathbf{x},{t})\;\mathrm{on}\;\overline{\mathrm{B}}\times [0,\infty )$ that satisfies the field equation

$$\begin{aligned} \hat{\nabla }[\rho ^{-1}(\mathrm{div}\,\mathbf{S})]-{\mathbf{\mathsf{{K}} }}[\ddot{\mathbf{S}}]=-\mathbf{B}\quad \mathrm{on}\quad \mathrm{B}\times [0,\infty ) \end{aligned}$$

(5.6)

subject to the initial conditions

$$\begin{aligned} \mathbf{S}(\mathbf{x},0)=\mathbf{S}_0 (\mathbf{x}),\quad \dot{\mathbf{S}}(\mathbf{x},0)=\dot{\mathbf{S}}_0 (\mathbf{x})\quad \mathrm{for}\quad \mathbf{x}\in \mathrm{B} \end{aligned}$$

(5.7)

and the boundary condition

$$\begin{aligned} \mathbf{s}=\mathbf{Sn}=\hat{\mathbf{s}}\quad \mathrm{on}\quad \partial \mathrm{B}\times [0,\infty ) \end{aligned}$$

(5.8)

Here $\mathbf{S}_0$ and $\dot{\mathbf{S}}_0$ are arbitrary symmetric tensor fields on B, and B is a prescribed symmetric second-order tensor field on $\overline{\mathrm{B}}\times [0,\infty )$. Moreover, $\rho ,{\mathbf{\mathsf{{K}} }},$ and $\hat{\mathbf{s}}$ have the same meaning as in classical elastodynamics.

First, we note that the problem is equivalent to the following one. Find a symmetric second-order tensor field on $\overline{\mathrm{B}}\times [0,\infty )$ that satisfies the integro-differential equation

$$\begin{aligned} \hat{\nabla }[\rho ^{-1}{ t}*(\mathrm{div}\,\mathbf{S})]-{\mathbf{\mathsf{{K}} }}[\mathbf{S}]=-\hat{\mathbf{B}}\quad \mathrm{on}\quad \mathrm{B}\times [0,\infty ) \end{aligned}$$

(5.9)

subject to the boundary condition

$$\begin{aligned} \mathbf{s}=\mathbf{Sn}=\hat{\mathbf{s}}\quad \mathrm{on}\quad \partial \mathrm{B}\times [0,\infty ) \end{aligned}$$

(5.10)

where

$$\begin{aligned} \hat{\mathbf{B}}={t}*\mathbf{B}+{\mathbf{\mathsf{{K}} }}[\mathbf{S}_0 +\mathrm{t}\,\dot{\mathbf{S}}_0 ] \end{aligned}$$

(5.11)

and $*$ stands for the convolution product, that is, for any two scalar functions ${a}={a}(\mathbf{x},{t})\;\mathrm{and}\;{b}={b}(\mathbf{x},{t})$

$$\begin{aligned} ({a}*{b})(\mathbf{x},{t})=\int \limits _0^{t} {{a}(\mathbf{x},\tau ){b}(\mathbf{x},{t}-\tau )} \,{d}\tau \end{aligned}$$

(5.12)

Next, the convolutional variational principle is formulated for the problem described by Eqs. (5.9)–(5.10).

Principle of Incompatible Elastodynamics. Let $N$ denote the set of all symmetric second-order tensor fields S on $\overline{\mathrm{B}}\times [0,\infty )$ that satisfy the traction boundary condition (5.8) $\equiv$ (5.10). Let $\mathrm{I}_{t} \{\mathbf{S}\}$ be the functional on $N$ defined by

$$\begin{aligned} \mathrm{I}_{t} \{\mathbf{S}\}=\frac{1}{2}\int \limits _\mathrm{B} {\{\,\rho ^{-1}{t}*(\mathrm{div}\,\mathbf{S})\,*}\, (\mathrm{div}\,\mathbf{S})+\mathbf{S}*{\mathbf{\mathsf{{K}} }}[\mathbf{S}]-2\,\mathbf{S}*\hat{\mathbf{B}}\,\}\,{dv} \end{aligned}$$

(5.13)

Then

$$\begin{aligned} \delta \,\mathrm{I}_\mathrm{t} \{\mathbf{S}\}=0 \end{aligned}$$

(5.14)

at a particular $\mathbf{S}\in \mathrm{N}$ if and only if S is a solution to the traction problem described by Eqs.  (5.6)–(5.8).

Note. When the fields B, $\mathbf{S}_0$, and $\dot{\mathbf{S}}_0$ are arbitrarily prescribed, the principle of incompatible elastodynamics may be useful in a study of elastic waves in bodies with various types of defects.

2 Problems and Solutions Related to Variational Principles of Elastodynamics

Problem 5.1.

A symmetrical elastic beam of flexural rigidity $EI$, density $\rho$, and length $L$, is acted upon by: (i) the transverse force ${F}={F}({x_{1}}, {t})$, (ii) the end shear forces $V_0 \;\mathrm{and}\;{V}_{L}$, and (iii) the end bending moments ${M}_0 \;\mathrm{and}\;{M}_{L}$ shown in Fig. 5.1. The strain energy of the beam is

$$\begin{aligned} {F}({t})=\frac{1}{2}\int \limits _0^{L} {{E\,I}({u}^{\prime \prime }_2 )^2\,{dx}_1 } \end{aligned}$$

(5.15)

the kinetic energy of the beam is

$$\begin{aligned} {K(t)}=\frac{1}{2}\int \limits _0^{L} {\rho \,(\dot{u}_2 )^2\,{dx}_1 } \end{aligned}$$

(5.16)

Fig. 5.1

The symmetrical beam

and the energy of external forces is

$$\begin{aligned} V(t)&=-\int \limits _0^L {F(x_1 ,t)\,u_2 (x_1 ,t)\,dx_1 +V_0 } u_2 (0,t)\nonumber \\&\quad \;+\,M_0 {u}^{\prime }_2 (0,t)-V_L u_2 (L,t)-M_L {u}^{\prime }_2 (L,t) \end{aligned}$$

(5.17)

where the prime denotes differentiation with respect to ${x}_1$. Let $U$ be the set of functions ${u}_2 ={u}_2 ({x}_1, {t})$ that satisfies the conditions

$$\begin{aligned} {u_2 (x_1, t_1 )=u(x_1 ),\quad u_2 (x_1, t_2 )=v(x_1 )} \end{aligned}$$

(5.18)

where ${t}_1 \;\mathrm{and}\;{t}_2$ are two arbitrary points on the $t$-axis such that $0\le {t}_1 <{t}_2 <\infty$, and ${u(x}_1 )$ and ${v(x}_1 )$ are prescribed fields on $[0,{L}]$. Define a functional $\hat{K}\{{u}_2 \}$ on $U$ by

$$\begin{aligned} \hat{K}\{{u}_2 \}=\int \limits _{{t}_1 }^{{t}_2 } {[F(t)+V(t)-K(t)]\,dt} \end{aligned}$$

(5.19)

Show that

$$\begin{aligned} \delta \hat{K}\{{u}_2 \}=0 \end{aligned}$$

(5.20)

if and only if ${u}_2$ satisfies the equation of motion

$$\begin{aligned} ({EI{u}}^{\prime \prime }_2 {)}^{\prime \prime }+\rho \,\ddot{u}_2 ={F}\quad \mathrm{on}\quad [0,{L}]\times [0,\infty ) \end{aligned}$$

(5.21)

and the boundary conditions

$$\begin{aligned}{}[({EI\,{u}}^{\prime \prime }_2 {)}^{\prime }]\,(0,{t})&=-{V}_0 \quad \; \mathrm{on}\quad [0,\infty ) \end{aligned}$$

(5.22)

$$\begin{aligned} {[}({EI\,{u}}^{\prime \prime }_2 )]\,(0,{t})&= {M}_0 \qquad \mathrm{on}\quad [0,\infty ) \end{aligned}$$

(5.23)

$$\begin{aligned} {[}({EI\,{u}}^{\prime \prime }_2 {)}^{\prime }]\,({L,t})&= -{V}_{L} \quad \; \mathrm{on}\quad [0,\infty ) \end{aligned}$$

(5.24)

$$\begin{aligned} {[}({EI\,{u}}^{\prime \prime }_2 )]\,({L,t})&= {M}_{L} \qquad \mathrm{on}\quad [0,\infty ) \end{aligned}$$

(5.25)

The field equaton (5.21) and the boundary conditions (5.22) through (5.25) describe flexural waves in the beam.

Solution.

Introduce the notation

$$\begin{aligned} { u}_{2}(x_{1},{ t})\equiv { u}({ x,t}) \end{aligned}$$

(5.26)

Then the functional $\hat{ K}\{{ u}_{2}\}$ takes the form

$$\begin{aligned} \hat{ K}\{{ u}\}&=\frac{1}{2}\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{0}^{ L}{ dx}\ [{ EI}({ u}^{\prime \prime })^{2}-\rho (\dot{ u})^{2}] \nonumber \\&\quad +\int \limits _{{ t}_{1}}^{{ t}_{2}}\left\{ -\int \limits _{0}^{ L}\!{ Fu \, dx }+{ V}_{0}{ u}(0,{ t})+{ M}_{0} { u}^{\prime }(0,{ t})-{ V}_{ L}{ u}({ L,t})-{ M}_{ L} { u}^{\prime }({ L,t})\!\right\} \!{ dt} \end{aligned}$$

(5.27)

Let ${ u}\in { U}$ and ${ u}+\omega \tilde{ u}\in { U}$. Then

$$\begin{aligned} \tilde{ u}({ x},{ t}_{1})=\tilde{ u}({ x},{ t}_{2})=0\qquad { x}\in [0,{ L}] \end{aligned}$$

(5.28)

Computing $\delta \hat{ K}\{{ u}\}$ we obtain

$$\begin{aligned} \delta \hat{ K}\{{ u}\}&= \left. \frac{{ d}}{{ d}\omega }\ \hat{ K}\{{ u}+\omega \tilde{ u}\}\right| _{\omega =0} =\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{0}^{ L}{ dx}\ [EI{u}^{\prime \prime }\,{\tilde{u}}{\,}^{\prime \prime }-\rho \,\dot{u}\,\dot{\tilde{u}}]\nonumber \\&\;\quad +\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\left\{ -\int \limits _{0}^{ L}{ F}\tilde{ u}\,{ dx}+{ V}_0\tilde{ u}(0, { t})+{ M}_0\tilde{ u}^{\prime }(0,{ t})\right. \nonumber \\&\;\quad \left. -{ V}_{ L}\tilde{ u}({ L,t})-{ M}_{ L}\tilde{ u}^\prime ({ L, t})\right\} \end{aligned}$$

(5.29)

Next, note that integrating by parts we obtain

$$\begin{aligned} \left. \int \limits _{0}^{ L}{ dx}({ EIu}^{\prime \prime }\tilde{ u}{\,}^{\prime \prime })\right.&=\left. ({ EIu}^{\prime \prime })\tilde{ u}{\,}^{\prime }\right| _{{ x}\,=\,0}^{{ x}\,=\,{ L}}-\int \limits _{0}^{ L}{ dx}({ EIu}^{\prime \prime })^{\prime }\tilde{ u}{\,}^{\prime }\nonumber \\&=\left. ({ EIu}^{\prime \prime })\tilde{ u}{\,}^{\prime }\right| _{{ x}\,=\,0}^{{ x}\,=\,{ L}}\left. -({ EIu}^{\prime \prime })^{\prime }\tilde{ u}\right| _{{ x}\,=\,0}^{{ x}\,=\,{ L}}+\int \limits _{0}^{ L}{ dx}({ EIu}^{\prime \prime })^{\prime \prime }\tilde{ u} \end{aligned}$$

(5.30)

and

$$\begin{aligned} -\int \limits _{{ t}_{1}}^{{ t}_{2}}\rho \dot{ u}\dot{\tilde{u}}\,{ dt}=-\rho {\dot{ u}\tilde{ u}}\bigg |_{{ t}\,=\,{ t}_{1}}^{{ t}\,=\,{ t}_{2}}+\int \limits _{{ t}_{1}}^{{ t}_{2}}\rho \ddot{ u}\tilde{ u}\,{ dt} \end{aligned}$$

(5.31)

Hence, using the homogeneous conditions (5.28) we reduce (5.29) to the form

$$\begin{aligned} \delta \hat{ K}\{{ u}\}&=\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{0}^{{ L}}{ dx}\ [({ EIu}^{\prime \prime })^{\prime \prime }+\rho \ddot{ u}-{ F}]\tilde{ u}({ x, t})\nonumber \\&\quad +\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\{[{ V}_{0}+({ EIu}^{\prime \prime })^{\prime }(0,{ t})]\tilde{ u}(0,{ t}) -[{ V_{L}}+({ EIu}^{\prime \prime })^{\prime }({ L,t})]\tilde{ u}({ L,t})\nonumber \\&\quad +[{ M}_{0}-({ EIu}^{\prime \prime })(0,{ t})]\tilde{ u}{\,}^{\prime }(0,{ t}) -[{ M}_{ L}-({ EIu}^{\prime \prime })({ L,t})]\tilde{ u}{\,}^{\prime }({ L,t})\} \end{aligned}$$

(5.32)

Now, if ${ u}={ u}({ x,t})$ satisfies (5.21)–(5.25) then $\delta \hat{ K}\{{ u}\}=0$. Conversely, if ${\delta \hat{ K}\{{ u}\}=0}$ then selecting $\tilde{ u}=\tilde{ u}({ x,t})$ in such a way that $\tilde{ u}=\tilde{ u}({ x,t})$ is an arbitrary smooth function on $[{ 0,L}]\times [{ t}_{1},{ t}_{2}]$ and such that $\tilde{ u}({ 0,t})=\tilde{ u}({ L,t})=0$ on $[{ t_{1},t_{2}}]$ and $\tilde{ u}^{\prime }({ 0,t})=\tilde{ u}{\,}^{\prime }({ L,t})=0$ on $[{ t_{1},t_{2}}]$, from Eq. (5.32) we obtain

$$\begin{aligned} \int \limits _{{ t}_{1}}^{{ t}_{2}}\int \limits _{0}^{ L}[({ EIu}^{\prime \prime })^{\prime \prime }+\rho {\ddot{u}}-{ F}]\tilde{ u}\ { dtdx}=0 \end{aligned}$$

(5.33)

and by the Fundamental Lemma of the calculus of variations we obtain

$$\begin{aligned} ({ EIu}^{\prime \prime })^{\prime \prime }+\rho {\ddot{ u}}={ F} \end{aligned}$$

(5.34)

Next, by selecting $\tilde{ u}=\tilde{ u}({ x,t})$ in such a way that $\tilde{u}$ is an arbitrary smooth function on $[{ 0,L}]\times [{ t_{1},t_{2}}]$ that complies with the conditions $\tilde{ u}({ 0,t})\ne 0$ on $[{ t_{1},t_{2}}],\ \tilde{ u}({ L,t})=0,\tilde{ u}{\,}^{\prime }({ 0,t})=\tilde{ u}{\,}^{\prime }({ L,t})=0$ on $[{ t_{1},t_{2}}]$, and by using (5.32) and (5.34), we obtain

$$\begin{aligned} \int \limits _{{ t}_{1}}^{{ t}_{2}}[{ V}_{0}+({ EIu}^{\prime \prime })^{\prime }(0,{ t})]\tilde{ u} (0,{ t})=0 \end{aligned}$$

(5.35)

This together with the Fundamental Lemma of calculus of variations yields

$$\begin{aligned} ({ EIu}^{\prime \prime })^{\prime }(0,{ t})=-{ V}_{0} \end{aligned}$$

(5.36)

Next, by selecting $\tilde{ u}$ to be an arbitrary smooth function on $[{ 0,L}]\times [{ t_{1},t_{2}}]$ that satisfies the conditions $\tilde{ u}({ L,t})\ne 0$ on $[{ t_{1},t_{2}}],\ \tilde{ u}{\,}^{\prime }({ 0,t})=0$, and $\tilde{ u}{\,}^{\prime }({ L,t})=0$ on $[{ t_{1},t_{2}}]$, we find from Eqs. (5.34), (5.36), and (5.32) that

$$\begin{aligned} \int \limits _{{ t}_{1}}^{{ t}_{2}}[{ V}_{ L}+({ EIu}^{\prime \prime })^{\prime }({ L,t})]\tilde{ u}({ L,t}){ dt}=0 \end{aligned}$$

(5.37)

Equation (5.37) together with the Fundamental Lemma of calculus of variations imply that

$$\begin{aligned} ({ EIu}^{\prime \prime })^{\prime }({ L,t})=-{ V}_{ L} \end{aligned}$$

(5.38)

Next, by selecting $\tilde{u}$ to be an arbitrary smooth function on $[{ 0,L}]\times [{ t_{1},t_{2}}]$ that meets the conditions $\tilde{ u}{\,}^{\prime }({ 0,t})\ne 0$ on $[{ t_{1},t_{2}}]$, and $\tilde{ u}{\,}^{\prime }({ L,t})=0$ on $[{ t_{1},t_{2}}]$, by virtue of Eqs. (5.34), (5.36), (5.38), and (5.32), we obtain

$$\begin{aligned} \int \limits _{{ t}_{1}}^{{ t}_{2}}[{ M}_{0}-({ EIu}^{\prime \prime })(0,{ t})]\tilde{ u}{\,}^{\prime }(0,{ t}){ dt}=0 \end{aligned}$$

(5.39)

This together with the Fundamental Lemma of calculus of variations yields

$$\begin{aligned} ({ EIu}^{\prime \prime })(0,{ t})={ M}_{0} \end{aligned}$$

(5.40)

Finally, by letting $\tilde{u}$ to be an arbitrary smooth function on $[{ 0,L}]\times [{ t_{1},t_{2}}]$ and such that $\tilde{ u}{\,}^{\prime }({ L,t})\ne 0$, from Eqs. (5.34), (5.36), (5.38), (5.40), and (5.32) we obtain

$$\begin{aligned} \int \limits _{{ t}_{1}}^{{ t}_{2}}[{ M_{L}}-({ EIu}^{\prime \prime })({ L,t})]\tilde{ u}{\,}^{\prime }({ L,t})=0 \end{aligned}$$

(5.41)

Equation (5.41) together with the Fundamental Lemma of calculus of variations yields

$$\begin{aligned} ({ EIu}^{\prime \prime })({ L,t})={ M_{L}} \end{aligned}$$

(5.42)

This completes a solution to Problem 5.1.

Problem 5.2.

A thin elastic membrane of uniform area density $\hat{\rho }$ is stretched to a uniform tension $\hat{T}$ over a region ${C}_0$ of the ${x}_1, {x}_2$ plane. The membrane is subject to a vertical load ${f}={f}(\mathbf{x},{t})\;\mathrm{on}\;{C}_0 \times [0,\infty )$ and the initial conditions

$$\begin{aligned} {u}(\mathbf{x},0)={u}_0 (\mathbf{x}),\quad \dot{u}(\mathbf{x},0)=\dot{u}_0 (\mathbf{x})\quad \mathrm{for}\quad \mathbf{x}\in {C}_0 \end{aligned}$$

where ${u}={u}(\mathbf{x},{t})$ is a vertical deflection of the membrane on $\overline{C}_0 \times [0,\infty )$, and ${u}_0 (\mathbf{x})$ and $\dot{u}_0 (\mathbf{x})$ are prescribed functions on ${C}_0$. Also, ${u}={u}(\mathbf{x},{t})$ on $\partial {C}_0 \times [0,\infty )$ is represented by a given function ${g}={g}(\mathbf{x},{t})$. The strain energy of the membrane is

$$\begin{aligned} {F(t)}=\frac{\hat{T}}{2}\int \limits _{\mathrm{C}_0 } {{u}_{,\alpha }} {u}_{,\alpha } \,{da} \end{aligned}$$

(5.43)

The kinetic energy of the membrane is

$$\begin{aligned} {K(t)}=\frac{\hat{\rho }}{2}\int \limits _{\mathrm{C}_0 } {(\dot{u})^2} \,{da} \end{aligned}$$

(5.44)

The external load energy is

$$\begin{aligned} {V(t)}=-\int \limits _{\mathrm{C}_0} {f\,u\,da} \end{aligned}$$

(5.45)

Let $U$ be the set of functions ${u}={u}(\mathbf{x},{t})$ on ${C}_0 \times [0,\infty )$ that satisfy the conditions

$$\begin{aligned} {u}(\mathbf{x},{t}_1)={a}(\mathbf{x}),\quad {u}(\mathbf{x},{t}_2 )={b}(\mathbf{x})\quad \mathrm{for}\quad \mathbf{x}\in {C}_0 \end{aligned}$$

(5.46)

and

$$\begin{aligned} {u}(\mathbf{x},{t})={g}(\mathbf{x},{t})\quad \mathrm{on}\quad \partial {C}_0 \times [0,\infty ) \end{aligned}$$

(5.47)

where ${t}_1 \;\mathrm{and}\;{t}_2$ have the same meaning as in Problem 5.1, and $a$(x) and $b$(x) are prescribed functions on $C_0$. Define a functional $\hat{K}\{.\}$ on $U$ by

$$\begin{aligned} \hat{K}\{{u}\}=\int \limits _{{t}_1 }^{{t}_2}{[F(t)+V(t)-K(t)]\,dt} \end{aligned}$$

(5.48)

Show that the condition

$$\begin{aligned} \delta \,\hat{K}\{{u}\}=0\quad \mathrm{on}\quad {U} \end{aligned}$$

(5.49)

implies the wave equation

$$\begin{aligned} \left( {\nabla ^2-\frac{1}{{c}^2}\frac{\partial ^2}{\partial {t}^2}} \right) \,{u}=-\frac{{f}}{\hat{T}}\quad \mathrm{on}\quad {C}_0 \times [0,\infty ) \end{aligned}$$

(5.50)

where

$$\begin{aligned} {c}=\sqrt{\frac{\hat{T}}{\hat{\rho }}} \end{aligned}$$

(5.51)

Note that $[\hat{T}]=[\mathrm{Force}\times {L}^{-1}],\;[\hat{\rho }]=[\mathrm{Density}\times {L}],\;[{c}]=[{LT}^{-1}],$ where $L$ and $T$ are the length and time units, respectively.

Solution.

The functional $\hat{ K}=\hat{ K}\{{ u}\}$ takes the form

$$\begin{aligned} \hat{ K}\{{ u}\}=\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{{ C}_{0}}\left( \frac{\widehat{ T}}{2}{ u}, _{\alpha }{ u}, _{\alpha }-\frac{\hat{\rho }}{2}\dot{ u}^{2}-{ fu}\right) { da}\nonumber \\ \qquad \qquad \qquad \qquad \qquad \mathrm{for\ every}\qquad u\in {U} \end{aligned}$$

(5.52)

Let ${ u}\in { U}$ and ${ u}+{ \omega }\tilde{ u}\in { U}$. Then

$$\begin{aligned} \tilde{ u}(\mathbf x ,{ t}_{1})=\tilde{ u}(\mathbf x ,{ t}_{2})=0\quad \mathrm{for}\ \mathbf x \in { C}_{0} \end{aligned}$$

(5.53)

and

$$\begin{aligned} \tilde{ u}(\mathbf x ,{ t})=0\quad \mathrm{on}\ \partial { C}_{0}\times [0,\infty ) \end{aligned}$$

(5.54)

Computing $\delta \widehat{ K}\{{ u}\}$ we obtain

$$\begin{aligned} \delta \widehat{ K}\{{ u}\}&=\frac{ d}{{ d}\omega }\widehat{ K}\{{ u}+\omega \tilde{ u}\}|_{\omega \,=\,0} \nonumber \\&=\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{{ C}_{0}}(\widehat{ T}{ u}, _{\alpha }\tilde{ u}, _{\alpha }-\hat{\rho }\dot{ u}\dot{\tilde{ u}}-{ f}\tilde{ u}){ da} \end{aligned}$$

(5.55)

Since

$$\begin{aligned} { u}, _{\alpha }\tilde{ u}, _{\alpha }=({ u}, _{\alpha }\tilde{ u}), _{\alpha }-{ u}, _{\alpha \alpha }\,\tilde{ u} \end{aligned}$$

(5.56)

and

$$\begin{aligned} \dot{ u}\dot{\tilde{ u}}=(\dot{ u}\tilde{ u})^{.}-\ddot{ u}\tilde{ u} \end{aligned}$$

(5.57)

therefore, using the divergence theorem and the homogeneous conditions (5.53) and (5.54), we reduce (5.55) into the form

$$\begin{aligned} \delta \widehat{ K}\{{ u}\}=\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{{ C}_{0}}(-\widehat{ T}{ u}, _{\alpha \alpha }+\hat{\rho }\ddot{ u}-{ f})\tilde{ u}\ { da} \end{aligned}$$

(5.58)

Hence, the condition

$$\begin{aligned} \delta \widehat{ K}\{{ u}\}=0\quad \mathrm{on}\ U \end{aligned}$$

(5.59)

together with the Fundamental Lemma of calculus of variations imply Eq. (5.50). This completes a solution to Problem 5.2.

Problem 5.3.

Transverse waves propagating in a thin elastic membrane are described by the field equation (see Problem 5.2.)

$$\begin{aligned} \left( {\nabla ^2-\frac{1}{{c}^2}\frac{\partial ^2}{\partial {t}^2}} \right) \,{u}=-\frac{{f}}{\hat{T}}\quad \mathrm{on}\quad {C}_0 \times [0,\infty ) \end{aligned}$$

(5.60)

the initial conditions

$$\begin{aligned} {u}(\mathbf{x},0)={u}_0 (\mathbf{x}),\quad \dot{u}(\mathbf{x},0)=\dot{u}_0 (\mathbf{x})\quad \mathrm{for}\quad \mathbf{x}\in {C}_0 \end{aligned}$$

(5.61)

and the boundary condition

$$\begin{aligned} {u}(\mathbf{x},{t})={g}(\mathbf{x},{t})\quad \mathrm{on}\quad \partial {C}_0 \times [0,\infty ) \end{aligned}$$

(5.62)

Let $\hat{{U}}$ be a set of functions ${u}={u}(\mathbf{x},{t})$ on ${C}_0 \times [0,\infty )$ that satisfy the boundary condition (5.62). Define a functional $\mathcal F _{ t} \{.\}\;\mathrm{on}\;\hat{{U}}$ in such a way that

$$\begin{aligned} \delta {F}_{t} \{{u}\}=0 \end{aligned}$$

(5.63)

if and only if ${u}={u}(\mathbf{x},{t})$ is a solution to the initial-boundary value problem (5.60) through (5.62).

Solution.

By transforming the initial-boundary value problem (5.60)–(5.62) to an equivalent integro-differential boundary-value problem in a way similar to that of the Principle of Incompatible Elastodynamics [see Eqs. (5.6)–(5.12)] we find that the functional $\mathcal F _{ t}\{{ u}\}$ on $\hat{ U}$ takes the form

$$\begin{aligned} \mathcal F _{ t}\{{ u}\}=\frac{1}{2}\int \limits _{{ C}_{0}}({ i}*{ u}, _{\alpha }*\,{u}, _{\alpha }+\frac{1}{{ c}^{2}}{ u}*{ u}-2{ g}*{ u}){ da} \end{aligned}$$

(5.64)

where

$$\begin{aligned} { i}={ i(t)}={ t} \end{aligned}$$

(5.65)

and

$$\begin{aligned} { g}={ i}*\frac{ f}{\widehat{ T}}+\frac{1}{{ c}^{2}}({ u}_{0}+{ t}\dot{ u}_{0}) \end{aligned}$$

(5.66)

The associated variational principle reads:

$$\begin{aligned} \delta \mathcal F _{ t}\{{ u}\}=0\quad \mathrm{on}\ \hat{ U} \end{aligned}$$

(5.67)

if and only if $u$ is a solution to the initial-boundary value problem (5.60)–(5.62). This completes a solution to Problem 5.3.

Problem 5.4.

A homogeneous isotropic thin elastic plate defined over a region ${C}_0$ of the ${x}_1, {x}_2$ plane, and clamped on its boundary $\partial C_0$, is subject to a transverse load ${p}={p}(\mathbf{x},{t})\;\mathrm{on}\;{C}_0 \times [0,\infty )$. The strain energy of the plate is

$$\begin{aligned} {F(t)}=\frac{D}{2}\int \limits _{\mathrm{C}_0}{(\nabla ^2{w})^2} \,{da} \end{aligned}$$

(5.68)

The kinetic energy of the plate is

$$\begin{aligned} {K(t)}=\frac{\hat{\rho }}{2}\int \limits _{\mathrm{C}_0} {(\dot{w})^2} \,{da} \end{aligned}$$

(5.69)

The external energy is

$$\begin{aligned} {V(t)}=-\int \limits _{\mathrm{C}_0} {p\,w\,da} \end{aligned}$$

(5.70)

Here, ${w}={w}(\mathbf{x},{t})$ is a transverse deflection of the plate on ${C}_0 \times [0,\infty )$, $D$ is the bending rigidity of the plate (${[D]} = \mathrm{[Force}\times \mathrm{Length}]$), and $\hat{\rho }$ is the area density of the plate ([$\hat{\rho }$] $=$ [Density $\times$ Length]).

Let $W$ be the set of functions ${w}={w}(\mathbf{x},{t})$ on ${C}_0 \times [0,\infty )$ that satisfy the conditions

$$\begin{aligned} {w}(\mathbf{x},{t}_1 )={a}(\mathbf{x}),\quad {w}(\mathbf{x},{t}_2 )={b}(\mathbf{x})\quad \mathrm{for}\quad \mathbf{x}\in {C}_0 \end{aligned}$$

(5.71)

and

$$\begin{aligned} {w}=0,\quad \frac{\partial {w}}{\partial {n}}=0\quad \mathrm{on}\quad \partial {C}_0 \times [0,\infty ) \end{aligned}$$

(5.72)

where ${t}_1, \;{t}_2$, $a$(x) and $b$(x) have the same meaning as in Problem 5.2, and $\partial /\partial {n}$ is the normal derivative on $\partial {C}_0$. Define a functional $\hat{K}\{.\}$ on $W$ by

$$\begin{aligned} \hat{K}\{{w}\}=\int \limits _{{t}_1 }^{{t}_2 } {[F(t)+V(t)-K(t)]\,dt} \end{aligned}$$

(5.73)

Show that

$$\begin{aligned} \delta \hat{K}\{{w}\}=0\quad \mathrm{on}\quad {W} \end{aligned}$$

(5.74)

if and only if ${w}={w}(\mathbf{x},{t})$ satisfies the differential equation

$$\begin{aligned} \nabla ^2\nabla ^2{w}+\frac{\hat{\rho }}{{D}}\frac{\partial ^2{w}}{\partial {t}^2}=\frac{{p}}{{D}}\quad \mathrm{on}\quad {C}_0 \times [0,\infty ) \end{aligned}$$

(5.75)

and the boundary conditions

$$\begin{aligned} {w}=0,\quad \frac{\partial {w}}{\partial {n}}=0\quad \mathrm{on}\quad \partial {C}_0 \times [0,\infty ) \end{aligned}$$

(5.76)

Solution.

The functional $\hat{ K}=\hat{ K}\{{ w}\}$ on $W$ takes the form

$$\begin{aligned} \widehat{ K}\{{ w}\}=\frac{1}{2}\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{{ C}_{0}}[{ D}(\nabla ^{2}{ w})^{2}-\hat{\rho }\dot{ w}^{2}-2{ pw}]{ da} \end{aligned}$$

(5.77)

Let ${ w}\in { W,w}+\omega \tilde{ w}\in { W}$. Then

$$\begin{aligned} \tilde{ w}(\mathbf x ,{ t}_{1})=\tilde{ w}(\mathbf x ,{ t}_{2})=0\quad \mathrm{for}\ \mathbf x \in { C}_{0} \end{aligned}$$

(5.78)

and

$$\begin{aligned} \tilde{ w}=0,\quad \frac{\partial \tilde{ w}}{\partial { n}}=0\quad \mathrm{on}\ \partial { C}_{0}\times [0,\infty ) \end{aligned}$$

(5.79)

Hence, we obtain

$$\begin{aligned} \delta \widehat{ K}\{{ w}\}&=\frac{ d}{{ d}\omega }\widehat{ K}\{{ w}+\omega \tilde{ w}\}|_{\omega \,=\,0} \nonumber \\&=\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{{ C}_{0}}[{ D}(\nabla ^{2}{ w})(\nabla ^{2}\tilde{ w})-\hat{\rho }\dot{ w}\dot{\tilde{ w}}-{ p}\tilde{ w}]{ da} \end{aligned}$$

(5.80)

Since

$$\begin{aligned} (\nabla ^{2}{ w})(\nabla ^{2}\tilde{ w})&={ w}, _{\alpha \alpha }\ \tilde{ w}, _{\beta \beta }=({ w}, _{\alpha \alpha }\ \tilde{ w}, _{\beta }), _{\beta } \nonumber \\ -{ w}, _{\alpha \alpha \beta }\ \tilde{ w}, _{\beta }&=({ w}, _{\alpha \alpha }\tilde{ w}, _{\beta }-{ w}, _{\alpha \alpha \beta }\tilde{ w}), _{\beta }+{ w}, _{\alpha \alpha \beta \beta }\tilde{ w} \end{aligned}$$

(5.81)

and

$$\begin{aligned} \dot{ w}\dot{\tilde{ w}}=(\dot{ w}\tilde{ w})^{.}-\ddot{ w}\tilde{ w} \end{aligned}$$

(5.82)

therefore, using the divergence theorem as well as the homogeneous conditions (5.78) and (5.79), we reduce (5.80) to the form

$$\begin{aligned} \delta \widehat{ K}\{{ w}\}=\int \limits _{{ t}_{1}}^{{ t}_{2}}{ dt}\int \limits _{{ C}_{0}}({ D}\nabla ^{4}{ w}+\hat{\rho }\ddot{ w}-{ p})\tilde{ w}{ da} \end{aligned}$$

(5.83)

Hence, by virtue of the Fundamental Lemma of calculus of variations

$$\begin{aligned} \delta \widehat{ K}\{{ w}\}=0\quad \mathrm{on}\ { W} \end{aligned}$$

(5.84)

if and only if $w$ satisfies the differential equation

$$\begin{aligned} \nabla ^{2}\nabla ^{2}{ w}+\frac{\hat{\rho }}{ D}\frac{\partial ^{2}{ w}}{\partial { t}^{2}}=\frac{ p}{ D}\quad \mathrm{on}\ { C}_{0}\times [0,\infty ) \end{aligned}$$

(5.85)

and the boundary conditions

$$\begin{aligned} { w}=\frac{\partial { w}}{\partial { n}}=0\quad \mathrm{on}\ \partial { C}_{0}\times [0,\infty ) \end{aligned}$$

(5.86)

This completes a solution to Problem 5.4.

Problem 5.5.

Transverse waves propagating in a clamped thin elastic plate are described by the equations (see Problem 5.4)

$$\begin{aligned} \nabla ^2\nabla ^2{w}+\frac{\hat{\rho }}{{D}}\frac{\partial ^2{w}}{\partial {t}^2}=\frac{{p}}{{D}}\quad \mathrm{on}\quad {C}_0 \times [0,\infty ) \end{aligned}$$

(5.87)

$$\begin{aligned} {w}(\mathbf{x},0)={w}_0 (\mathbf{x}),\quad \dot{{w}}(\mathbf{x},0)=\dot{{w}}_0 (\mathbf{x})\quad \mathrm{for}\quad \mathbf{x}\in { C}_0 \end{aligned}$$

(5.88)

and

$$\begin{aligned} {w}=0,\quad \frac{\partial {w}}{\partial {n}}=0\quad \mathrm{on}\quad \partial {C}_0 \times [0,\infty ) \end{aligned}$$

(5.89)

where ${w}_0 (\mathbf{x})$ and $\dot{{w}}_0 (\mathbf{x})$ are prescribed functions on ${C}_0$. Let ${W}^*$ denote the set of functions ${w}={w}(\mathbf{x},{t})$ that satisfy the homogeneous boundary conditions (5.89). Find a functional $\hat{\mathcal{F }}_{t} \{.\}\;\mathrm{on}\;{W}^*$ with the property that

$$\begin{aligned} \delta \,\hat{\mathcal{F }}_{t} \{{w}\}=0\quad \mathrm{on}\quad {W}^*\end{aligned}$$

(5.90)

if and only if $w$ is a solution to the initial-boundary value problem (5.87) through (5.89).

Solution.

First, we note that the initial-boundary value problem (5.87)–(5.89) is equivalent to the following boundary-value problem. Find ${ w}={ w}(\mathbf x ,{ t})$ on ${ C}_{0}\times [0,\infty )$ that satisfies the integro-differential equation.

$$\begin{aligned} { i}*\nabla ^{4}{ w}+\frac{1}{{ c}^{2}}{ w}={ h}\quad \mathrm{{on}}\quad { C}_{0}\times [0,\infty ) \end{aligned}$$

(5.91)

subject to the boundary conditions

$$\begin{aligned} {w}=\frac{\partial { w}}{\partial {n}}=0\quad \mathrm{{on}}\quad \partial { C}_{0}\times [0,\infty ) \end{aligned}$$

(5.92)

Here,

$$\begin{aligned}&{ i}={ i(t)}={ t},\quad { h}(\mathbf x ,{ t})={ i}*\frac{ p}{ D}+\frac{1}{{ c}^{2}}({ w}_{0}+{ t}\dot{ w}_{0}),\nonumber \\&\mathrm{{and}}\quad \frac{1}{{ c}^{2}}=\frac{\hat{\rho }}{ D} \end{aligned}$$

(5.93)

Next, we define a functional ${\hat{\mathcal{F }}}_{ t}\ \{{ w}\}$ on ${ W}^{*}$ by

$$\begin{aligned} {\hat{\mathcal{F }}}_{ t}\{{ w}\}=\frac{1}{2}\int \limits _{{ C}_{0}}\left( { i}*\nabla ^{2}{ w}*\nabla ^{2}{ w}+\frac{1}{{ c}^{2}}{ w}*{ w}-2{ h}*{ w}\right) { da} \end{aligned}$$

(5.94)

By computing $\delta \mathcal{\hat{F} }_{ t}\ \{{ w}\}$, we obtain

$$\begin{aligned} \delta {\hat{\mathcal{F }}}_{ t}\{{ w}\}=\int \limits _{{ C}_{0}}\left( { i}*\nabla ^{4}{ w}+\frac{1}{{ c}^{2}}{ w}-{ h}\right) *\tilde{ w}{ da} \end{aligned}$$

(5.95)

where $\tilde{ w}$ is an arbitrary smooth function on ${ C}_{0}$ such that

$$\begin{aligned} \tilde{ w}=\frac{\partial \tilde{ w}}{\partial {n}}=0\quad \mathrm{{on}}\ \partial { C}_{0}\times [0,\infty ) \end{aligned}$$

(5.96)

Therefore, using the Fundamental Lemma of calculus of variations, it follows from Eq. (5.95) that the condition

$$\begin{aligned} \delta {\hat{\mathcal{F }}}_{ t}\{{ w}\}=0\quad \mathrm{{on}}\ W^{*} \end{aligned}$$

(5.97)

holds true if and only if $w$ is a solution to the initial-boundary value problem (5.87)–(5.89). This completes a solution to Problem 5.5.

Problem 5.6.

Free longitudinal vibrations of an elastic bar are defined as solutions of the form

$$\begin{aligned} {u(x,t)}=\phi ({x})\,\sin (\omega \,{t}+\gamma ) \end{aligned}$$

(5.98)

to the homogeneous wave equation

$$\begin{aligned} \frac{\partial }{\partial {x}}\left( {{E}\frac{\partial {u}}{\partial {x}}} \right) -\rho \frac{\partial ^2{u}}{\partial {t}^2}=0\quad \mathrm{{on}}\quad [0,{L}]\times [0,\infty ) \end{aligned}$$

(5.99)

subject to the homogeneous boundary conditions

$$\begin{aligned} {u(0,t)}={u(L,t)}=0\quad \mathrm{{on}}\quad [0,\infty ) \end{aligned}$$

(5.100)

or

$$\begin{aligned} \frac{\partial {u}}{\partial {x}}({0,t})=\frac{\partial {u}}{\partial {x}}({L,t})=0\quad \mathrm{{on}}\quad [0,\infty ) \end{aligned}$$

(5.101)

Here, $\omega$ is a circular frequency of vibrations, $\gamma$ is a dimensionless constant, and $\phi =\phi ({x})$ is an unknown function that complies with Eqs. (5.99) and (5.100), or Eqs. (5.99) and (5.101). Substituting ${u}={u(\mathbf x ,t)}$ from Eq. (5.98) into (5.99) through (5.101) we obtain

$$\begin{aligned} \frac{{d}}{{dx}}\left( {{E}\frac{{d}\phi }{{dx}}} \right) +\lambda \phi =0\quad \mathrm{on}\quad [{0,L}] \end{aligned}$$

(5.102)

$$\begin{aligned} \phi (0)=\phi ({L})=0 \end{aligned}$$

(5.103)

or

$$\begin{aligned} {\phi }^{\prime }(0)={\phi }^{\prime }({L})=0 \end{aligned}$$

(5.104)

where the prime stands for derivative with respect to $x$, and

$$\begin{aligned} \lambda =\rho \omega ^2 \end{aligned}$$

(5.105)

Therefore, introduction of (5.98) into (5.99) through (5.101) results in an eigenproblem in which an eigenfunction $\phi =\phi ({x})$ corresponding to an eigenvalue $\lambda$ is to be found. An eigenproblem that covers both boundary conditions (5.100) and (5.101) can be written as

$$\begin{aligned} \frac{{d}}{{dx}}\left( {{E}\frac{{d}\phi }{{dx}}} \right) +\lambda \phi =0\quad \mathrm{on}\quad [0,{L}] \end{aligned}$$

(5.106)

$$\begin{aligned} {\phi }^{\prime }(0)-\alpha \,\phi (0)=0,\quad {\phi }^{\prime }({L})+\beta \,\phi ({L})=0 \end{aligned}$$

(5.107)

where $\left| \alpha \right| +\left| \beta \right| >0.$ Let $U$ be the set of functions $\phi =\phi ({x})$ on [0, $L$] that satisfy the boundary conditions (5.107). Define a functional $\pi \{.\}\;\mathrm{on}\;{U}$ by

$$\begin{aligned} \pi \{\phi \}=\frac{1}{2}\int \limits _0^{L} {\left[ {{E}\left( {\frac{{d}\phi }{{dx}}} \right) ^2-\lambda \phi ^2} \right] \,{dx}} +\frac{1}{2}\alpha \,{E}(0)\,\,[\phi (0)]^2+\frac{1}{2}\beta \,{E}({L})\,\,[\phi ({L})]^2 \end{aligned}$$

(5.108)

Show that

$$\begin{aligned} \delta \,\pi \{\phi \}=0\quad \mathrm{over}\quad {U} \end{aligned}$$

(5.109)

if and only if $\phi =\phi ({x})$ is an eigenfunction corresponding to an eigenvalue $\lambda$ in the eigenproblem (5.106) and (5.107).

Solution.

Let $\phi \in { U}$ and $\phi +\omega \ \tilde{\phi }\in { U}$. Then

$$\begin{aligned} \tilde{\phi }^{\prime }(0)-\alpha \tilde{\phi }(0)=0,\quad \tilde{\phi }^{\prime }({ L})+\beta \tilde{\phi }({ L})=0 \end{aligned}$$

(5.110)

and

$$\begin{aligned} \pi \{\phi +\omega \tilde{\phi }\}&=\frac{1}{2}\int \limits _{0}^{ L}[{ E}(\phi ^{\prime }+\omega \tilde{\phi }^{\prime })^{2}-\lambda (\phi +\omega \tilde{\phi })^{2}]{ dx}\nonumber \\&\quad +\frac{1}{2}\alpha { E}(0)[\phi (0)+\omega \tilde{\phi }(0)]^{2}+\frac{1}{2}\beta { E(L)}[\phi ({ L})+\omega \tilde{\phi }({ L})]^{2}\quad \end{aligned}$$

(5.111)

Hence, we obtain

$$\begin{aligned} \delta \pi \{\phi \}&= \left. \frac{ d}{{ d}\omega }\pi \{\phi +\omega \tilde{\phi }\}\right| _{\omega \,=\,0} \nonumber \\&=\int \limits _{0}^{ L}[{ E}\phi ^{\prime }\tilde{\phi }^{\prime }-\lambda \phi \tilde{\phi }]\ { dx} +\alpha { E}(0)\,\phi (0)\,\tilde{\phi }(0)+\beta { E}({ L})\,\phi ({ L})\,\tilde{\phi }({ L}) \end{aligned}$$

(5.112)

Since

$$\begin{aligned} \int \limits _{0}^{ L}{ E}\phi ^{\prime }\tilde{\phi }^{\prime }{ dx}=\left. { E}\phi ^{\prime }\tilde{\phi }\right| _{{x}\,=\,0}^{{x}\,=\,{ L}}-\int \limits _{0}^{ L}({ E}\phi ^{\prime })^{\prime }\tilde{\phi }{ dx} \end{aligned}$$

(5.113)

therefore, Eq. (5.112) takes the form

$$\begin{aligned} \delta \pi \{\phi \}&=-\int \limits _{0}^{ L}[({ E}\phi ^{\prime })^{\prime }+\lambda \phi ]\tilde{\phi }{ dx} -{ E}(0)[\phi ^{\prime }(0)-\alpha \phi (0)]\ \tilde{\phi }(0)\nonumber \\&\quad +{ E(L)}[\phi ^{\prime }({ L})+\beta \phi ({ L})]\tilde{\phi }({ L}) \end{aligned}$$

(5.114)

Now, if $\phi =\phi ({ x})$ is an eigenfunction corresponding to an eigenvalue $\lambda$ in the problem (5.106)–(5.107), then by virtue of (5.114) $\delta \pi \{\phi \}=0$ over $U$. Conversely, if $\delta \pi \{\phi \}=0$ then selecting $\tilde{\phi }=\tilde{\phi }({ x})$ to be a smooth function on [0, $L$] such that $\tilde{\phi }(0)=\tilde{\phi }({ L})=0$, and using the Fundamental Lemma of calculus of variations, we obtain

$$\begin{aligned} ({ E}\phi ^{\prime })^{\prime }+\lambda \phi =0\quad \mathrm{{on}}\quad [0,{ L}] \end{aligned}$$

(5.115)

Next, if $\delta \pi \{\phi \}=0$ then selecting $\tilde{\phi }=\tilde{\phi }({ x})$ to be a smooth function on [0, $L$] and such that $\tilde{\phi }({ L})=0$, and $\tilde{\phi }(0)\ne 0$, by virtue of (5.115), we obtain

$$\begin{aligned} {E}(0)[\phi ^{\prime }(0)-\alpha \phi (0)]\tilde{\phi }(0)=0 \end{aligned}$$

(5.116)

Since

$$\begin{aligned} { E}(0)>0 \end{aligned}$$

(5.117)

Equation (5.116) implies that $\phi =\phi ({ x})$ satisfies the boundary condition

$$\begin{aligned} \phi ^{\prime }(0)-\alpha \phi (0)=0 \end{aligned}$$

(5.118)

Finally, if $\delta \pi \{\phi \}=0$ then selecting $\tilde{\phi }$ to be a smooth function on [0, $L$] and such that $\tilde{\phi }({ L})\ne 0$, by virtue of (5.115) and (5.118), we obtain

$$\begin{aligned} {E(L)} [\phi ^{\prime }({ L})+\beta \phi ({ L})]\tilde{\phi }({ L})=0 \end{aligned}$$

(5.119)

Since ${ E(L)}>0$, Eq. (5.119) implies that

$$\begin{aligned} \phi ^{\prime }({ L})+\beta \phi ({ L})=0 \end{aligned}$$

(5.120)

This shows that if Eq. (5.110) holds true then $(\phi ,\lambda )$ is an eigenpair for the problem (5.106)–(5.107). This completes a solution to Problem 5.6.

Problem 5.7.

Free lateral vibrations of an elastic bar clamped at the end ${x}=0$ and supported by a spring of stiffness $k$ at the end ${x}={L}$ are defined as solutions of the form

$$\begin{aligned} {u(x,t)}=\phi ({x})\,\sin (\omega {t}+\gamma ) \end{aligned}$$

(5.121)

to the equation [see Problem 5.1, Eq. (5.127) in which ${u}_2 ={u}$, and ${F}=0$]

$$\begin{aligned} \frac{\partial ^2}{\partial {x}^2}\left( {{EI}\frac{\partial ^2{u}}{\partial {x}^2}} \right) + \ \rho \frac{\partial ^2{u}}{\partial {t}^2}=0\quad \mathrm{on}\quad [0,{L}]\times [0,\infty ) \end{aligned}$$

(5.122)

subject to the boundary conditions

$$\begin{aligned} {u(0,t)={u}^{\prime }(0,t)=0}\quad \mathrm{on}\quad [0,\infty ) \end{aligned}$$

(5.123)

$$\begin{aligned} {u}^{\prime \prime }({L,t})=0,\quad ({EI}\,{u}^{\prime \prime }{)}^{\prime }({L,t})-{k\,u(L,t)}=0\quad \mathrm{on}\quad [0,\infty ) \end{aligned}$$

(5.124)

Let $\rho =\mathrm{const}$, and $\lambda =\rho \,\omega ^2$. Then the associated eigenproblem reads

$$\begin{aligned} ({EI}\,{\phi }^{\prime \prime }{)}^{\prime \prime }-\lambda \phi =0\quad \mathrm{on}\quad [0,{L}] \end{aligned}$$

(5.125)

$$\begin{aligned} \phi (0)={\phi }^{\prime }(0)=0 \end{aligned}$$

(5.126)

$$\begin{aligned} {\phi }^{\prime \prime }({L})=0,\quad ({EI}{\phi }^{\prime \prime }{)}^{\prime }({L})-{k}\phi ({L})=0 \end{aligned}$$

(5.127)

Let $V$ denote the set of functions $\phi =\phi ({x})$ on [0, $L$] that satisfy the boundary conditions (5.126) and (5.127). Define a functional $\pi \{.\}\;\mathrm{on}\;{V}$ by

$$\begin{aligned} \pi \{\phi \}=\frac{1}{2}\int \limits _0^{L} {{EI}\,({\phi }^{\prime \prime })^2{dx}+\frac{1}{2}{k}[\phi ({ L})]^2-\frac{\lambda }{2}\int \limits _0^{L} {\phi ^2{dx}} } \end{aligned}$$

(5.128)

Show that

$$\begin{aligned} \delta \,\pi \{\phi \}=0\quad \mathrm{over}\quad {V} \end{aligned}$$

(5.129)

if and only if $(\lambda , \phi )$ is a solution to the eigenproblem (5.125) through (5.127).

Soution.

Let $\phi \in { V}$ and $\phi +\omega \tilde{\phi }\in { V}$. Then

$$\begin{aligned} \tilde{\phi }(0)=0,\quad \tilde{\phi }^{\prime }(0)=0 \end{aligned}$$

(5.130)

Computing the first variation of the functional $\pi \{\phi \}$ given by (5.128), we obtain

$$\begin{aligned} \delta \pi \{\phi \}&=\frac{ d}{{ d}\omega }\pi \left. \{\phi +\omega \tilde{\phi }\}\right| _{\omega \,=\,0} \nonumber \\&=\int \limits _{0}^{ L}({ EI}\ \phi ^{\prime \prime }\tilde{\phi }^{\prime \prime }-\lambda \phi \tilde{\phi }){ dx} +{ k}\phi ({ L})\ \tilde{\phi }({ L}) \end{aligned}$$

(5.131)

Since

$$\begin{aligned} \int \limits _{0}^{ L}{ EI}\ \phi ^{\prime \prime }\tilde{\phi }^{\prime \prime }{ dx}=({ EI}\ \phi ^{\prime \prime })\tilde{\phi }^{\prime }\bigg |_{{ x}\,=\,0}^{{ x}\,=\,{ L}} -({ EI}\ \phi ^{\prime \prime })^{\prime }\tilde{\phi }\bigg |_{{ x}\,=\,0}^{{ x}\,=\,{ L}}+\int \limits _{0}^{ L}({ EI}\ \phi ^{\prime \prime })^{\prime \prime }\ \tilde{\phi }{ dx} \end{aligned}$$

(5.132)

therefore, using (5.130) we reduce (5.131) into the form

$$\begin{aligned} \delta \pi \{\phi \}=\int \limits _{0}^{ L}[({ EI}\ \phi ^{\prime \prime })^{\prime \prime }-\lambda \phi ]\tilde{\phi }{ dx} +({ EI}\ \phi ^{\prime \prime })({ L})\tilde{\phi }^{\prime }({ L}) -[({ EI}\ \phi ^{\prime \prime })^{\prime }({ L})-{ k}\phi ({ L})]\tilde{\phi }({ L}) \end{aligned}$$

(5.133)

Now, if $(\lambda ,\phi )$ is a solution to the eigenproblem (5.125)–(5.127), then $\delta \pi \{\phi \}=0$. Conversely, if $\delta \pi \{\phi \}=0$ over $V$, then selecting $\tilde{\phi }$ to be an arbitrary smooth function on [0, $L$] such that $\tilde{\phi }({ x})\not \equiv 0$ for ${ x}\in (0,{ L}),\tilde{\phi }^{\prime }({ L})=0,\tilde{\phi }({ L})=0$, we obtain

$$\begin{aligned} \int \limits _{0}^{ L}[({ EI}\ \phi ^{\prime \prime })^{\prime \prime }-\lambda \phi ]\ \tilde{\phi }\,{ dx}=0 \end{aligned}$$

(5.134)

Equation (5.134) together with the Fundamental Lemma of calculus of variations implies

$$\begin{aligned} ({ EI}\ \phi ^{\prime \prime })^{\prime \prime }-\lambda \phi =0\quad \mathrm{on}\quad [0,{ L}] \end{aligned}$$

(5.135)

Next, by selecting $\tilde{\phi }$ on [0, $L$] in such a way that

$$\begin{aligned} \tilde{\phi }^{\prime }({ L})\ne 0,\quad \tilde{\phi }({ L})=0 \end{aligned}$$

(5.136)

we find that the condition $\delta \pi \{\phi \}=0$ and Eq. (5.135) imply that

$$\begin{aligned} ({ EI}\ \phi ^{\prime \prime })({ L})=0 \end{aligned}$$

(5.137)

Since

$$\begin{aligned} { E(L)}>0,\quad { I(L)}>0 \end{aligned}$$

(5.138)

we obtain

$$\begin{aligned} \phi ^{\prime \prime }({ L})=0 \end{aligned}$$

(5.139)

Finally, by selecting $\tilde{\phi }$ on [0, $L$] in such a way that

$$\begin{aligned} \tilde{\phi }({ L})\ne 0 \end{aligned}$$

(5.140)

we conclude that the condition $\delta \pi \{\phi \}=0$ together with Eqs. (5.135), and (5.139) lead to the boundary condition

$$\begin{aligned} ({ EI}\ \phi ^{\prime \prime })^{\prime }({ L})-{ k}\,\phi ({ L})=0 \end{aligned}$$

(5.141)

This completes a solution to Problem 5.7.

Problem 5.8.

Show that the eigenvalues $\lambda _\mathrm{i}$ and the eigenfunctions $\phi _{i} =\phi _{i} ({x})$ for the longitudinal vibrations of a uniform elastic bar having one end clamped and the other end free are given by the relations

$$\begin{aligned} \omega _\mathrm{i} =\sqrt{\frac{\lambda _\mathrm{i} }{\rho }} =\frac{(2{i}-1)}{2{L}}\sqrt{\frac{{E}}{\rho }} \end{aligned}$$

$$\begin{aligned} \phi _\mathrm{i} ({x})=\sin \frac{(2{i}-1)\pi \,{x}}{2{L}},\quad {i}=1,2,3,\ldots , 0\le {x}\le {L} \end{aligned}$$

(see Problem 5.6).

Solution.

For an elastic bar that is clamped at ${ x}=0$ and free at ${ x}={ L}$ the eigenproblem reads

$$\begin{aligned} { E}\phi ^{\prime \prime }({ x})+\lambda \phi ({ x})=0\qquad x\in [0,{ L}] \end{aligned}$$

(5.142)

$$\begin{aligned} \phi (0)=0,\quad \phi ^{\prime }({ L})=0 \end{aligned}$$

(5.143)

where

$$\begin{aligned} \lambda =\omega ^{2}\rho \end{aligned}$$

(5.144)

There is an infinite sequence of eigensolutions $(\lambda _{ i},\phi _{ i})$ to the problem (5.142)–(5.143) of the form

$$\begin{aligned} \lambda _{ i}&=\frac{(2{ i}-1)^{2}\pi ^{2}}{4{ L}^{2}}{ E} \end{aligned}$$

(5.145)

$$\begin{aligned} \phi _{ i}({ x})&=\sin \frac{(2{ i}-1)\pi { x}}{2{ L}},\quad { i}=1,2,3,\ldots \end{aligned}$$

(5.146)

This can be shown by substituting (5.145) and (5.146) into (5.142), and by showing that $\phi _{ i}({ x})$ satisfies (5.143). By combining (5.144) and (5.145) we obtain

$$\begin{aligned} \omega _{ i}\equiv \sqrt{\frac{\lambda _{ i}}{\rho }}=\frac{(2{ i}-1)\pi }{2{ L}}\sqrt{\frac{ E}{\rho }} \end{aligned}$$

(5.147)

This completes a solution to Problem 5.8.

Problem 5.9.

Show that the eigenvalues $\lambda _{i}$ and the eigenfunctions $\phi _{i} =\phi _{i} ({x})$ for the lateral vibrations of a uniform, simply supported elastic beam are given by the relations

$$\begin{aligned} \omega _{i}&=\sqrt{\frac{\lambda _{i} }{\rho }} =\frac{\pi ^2{i}^2}{{L}^2}\sqrt{\frac{{EI}}{\rho }}\\ \phi _{i} ({x})&=\sin \frac{{i}\,\pi \,{x}}{{L}},\quad {i}=1,2,3\ldots ,0\le {x}\le {L} \end{aligned}$$

(see Problem 5.1).

Solution.

For a uniform, simply supported beam with the lateral vibrations, the eigenproblem takes the form

$$\begin{aligned} {EI}\phi ^{(4)}-\lambda \phi&=0\quad \mathrm{on}\quad [0, { L}] \end{aligned}$$

(5.148)

$$\begin{aligned} \phi (0)=\phi ^{\prime \prime }(0)&=0,\quad \phi ({ L})=\phi ^{\prime \prime }({ L})=0 \end{aligned}$$

(5.149)

where

$$\begin{aligned} \lambda =\omega ^{2}{ \rho } \end{aligned}$$

(5.150)

There is an infinite sequence of eigensolutions $(\lambda _{ i},\phi _{ i})$ to the problem (5.148)–(5.149) of the form

$$\begin{aligned} \lambda _{ i}&={ EI}\left( \frac{{ i}\pi }{{ L}}\right) ^{4} \end{aligned}$$

(5.151)

$$\begin{aligned} \phi _{ i}({ x})&=\sin \frac{{ i}\pi { x}}{{ L}}\quad { i}=1,2,3,\ldots \end{aligned}$$

(5.152)

To prove that $(\lambda _{ i},\phi _{ i})$ given by (5.151)–(5.152) satisfies Eqs. (5.148)–(5.149), we note that

$$\begin{aligned} \phi _{ i}^{\prime \prime }({ x})=-\left( \frac{{ i}\pi }{{ L}}\right) ^{2}\phi _{ i}({ x}) \end{aligned}$$

(5.153)

and

$$\begin{aligned} \phi _{ i}^{(4)}({ x})=\left( \frac{{ i}\pi }{{ L}}\right) ^{4}\phi _{ i}({ x}) \end{aligned}$$

(5.154)

Substituting (5.151) and (5.152) into (5.148) and using (5.154) we find that $\phi _{ i}=\phi _{ i}({ x})$ satisfies Eq. (5.148) on [0, $L$]. Also, it follows from Eqs. (5.152) and (5.153) that the boundary conditions (5.149) are satisfied; and Eqs. (5.150) and (5.151) imply that

$$\begin{aligned} \omega _{ i}=\sqrt{\frac{\lambda { i}}{\rho }}=\frac{\pi ^{2}{ i}^{2}}{{ L}^{2}}\sqrt{\frac{{ EI}}{\rho }} \end{aligned}$$

(5.155)

These steps complete a solution to Problem 5.9.

Problem 5.10.

Show that the eigenvalues $\lambda _{mn}$ and the eigenfunctions $\phi _{mn} =\phi _{mn} ({x})$ for the transversal vibrations of a rectangular elastic membrane: $0\le {x}_1 \le {a}_1, \;0\le {x}_2 \le {a}_2$, that is clamped on its boundary, are given by

$$\begin{aligned} \omega _{mn} =\sqrt{\frac{\lambda _{mn}}{\hat{\rho }}} =\pi \sqrt{\frac{\hat{T}}{\hat{\rho }}\left( {\frac{{m}^2}{{a}_1^2 }+\frac{{n}^2}{{a}_2^2 }} \right) } \end{aligned}$$

$$\begin{aligned} \begin{array}{c} \phi _{mn} ({x}_1, {x}_2 )=\sin \,\dfrac{{m}\,\pi \,{x}_1 }{{a}_1 }\,\sin \dfrac{{n}\,\pi \,{x}_2 }{{a}_2 }, \\ {m,n}=1,2,3,\ldots ,0\le {x}_1 \le {a}_1, \,\,0\le {x}_2 \le {a}_2 \\ \end{array} \end{aligned}$$

(See Problem 5.2).

Solution.

Let ${ C}_{0}$ denote the rectangular region

$$\begin{aligned} 0<{ x}_{1}<{ a}_{1},\quad 0<{ x}_{2}<{ a}_{2} \end{aligned}$$

(5.156)

and let $\partial { C}_{0}$ be its boundary. Then the associated eigenproblem reads. Find an eigenpair $(\lambda , \phi )$ such that

$$\begin{aligned} \hat{{ T}}\ \nabla ^{2}\phi +\lambda \phi =0\quad \mathrm{on} \ { C}_{0} \end{aligned}$$

(5.157)

and

$$\begin{aligned} \phi =0\quad \mathrm{on}\ \partial { C}_{0} \end{aligned}$$

(5.158)

where

$$\begin{aligned} \lambda =\omega ^{2}\hat{\rho } \end{aligned}$$

(5.159)

There is an infinite number of eigenpairs $(\lambda _{ mn}, \phi _{ mn}),\ { m, n}=1, 2, 3, \ldots$ that satisfy Eqs. (5.157) and (5.158), and they are given by Equation

$$\begin{aligned} \lambda _{ mn}&=\pi ^{2}\hat{{ T}}\left( \frac{{ m}^{2}}{{ a}_{1}^{2}}+\frac{{ n}^{2}}{{ a}_{2}^{2}}\right) \end{aligned}$$

(5.160)

$$\begin{aligned} \phi _{ mn}({ x_{1},x}_{2})&=\sin \left( \frac{{ m}\pi { x}_{1}}{{ a}_{1}}\right) \sin \left( \frac{{ n}\pi { x}_{2}}{{ a}_{2}}\right) \end{aligned}$$

(5.161)

This can be proved by substituting (5.152) and (5.153) into (5.149) and (5.150).

Also, the eigenvalues $\lambda _{ mn}$ generate the eigenfrequencies $\omega _{ mn}$ by the formulas

$$\begin{aligned} \omega _{ mn}=\sqrt{\frac{\lambda _{ mn}}{\hat{\rho }}}=\pi \sqrt{\frac{\hat{{ T}}}{\hat{\rho }}}\sqrt{\left( \frac{{ m}^{2}}{{ a}_{1}^{2}}\right) +\left( \frac{{ n}^{2}}{{ a}_{2}^{2}}\right) } \end{aligned}$$

(5.162)

This completes a solution to Problem 5.10.

Problem 5.11.

Show that the eigenvalues $\lambda _{mn}$ and the eigenfunctions $\phi _{mn} =\phi _{mn} ({x}_1, {x}_2 )$ for the transversal vibrations of a thin elastic rectangular plate: ${0\le {x}_1 \le {a}_1, \;0\le {x}_2 \le {a}_2}$, that is simply supported on its boundary are given by the relations

$$\begin{aligned} \omega _{mn} =\sqrt{\frac{\lambda _{mn} }{\hat{\rho }}} =\pi ^2\left( {\frac{{m}^2}{{a}_1^2 }+\frac{{n}^2}{{a}_2^2 }} \right) \sqrt{\frac{{D}}{\hat{\rho }}} \end{aligned}$$

$$\begin{aligned} \begin{array}{c} \phi _{{mn}} ({x}_1, {x}_2 )=\sin \,\dfrac{{m}\,\pi \,{x}_1}{{a}_1 }\,\sin \dfrac{{n}\,\pi \,{x}_2 }{{a}_2 }, \\ {m,n}=1,2,3,\ldots ,0\le {x}_1 \le {a}_1, \,\,0\le {x}_2 \le {a}_2 \\ \end{array} \end{aligned}$$

(See Problem 5.4).

Solution.

The eigenproblem associated with the transversal vibrations of a thin elastic rectangular plate that is simply supported on its boundary, reads [see Eq. (5.85) of Problem 5.4]

$$\begin{aligned} { D}\ \nabla ^{2}\nabla ^{2}\ \phi -\lambda \phi =0\quad \mathrm{on}\ { C}_{0} \end{aligned}$$

(5.163)

$$\begin{aligned} \phi =\nabla ^{2}\phi =0\quad \mathrm{on}\ \partial { C}_{0} \end{aligned}$$

(5.164)

where

$$\begin{aligned} \lambda =\omega ^{2}\hat{\rho } \end{aligned}$$

(5.165)

and ${ C}_{0}$ and $\partial { C}_{0}$ are the same as in Problem 5.10.

There are an infinite number of eigenpairs $(\lambda _{ mn},\phi _{ mn})$ that satisfy Eqs. (5.163) and (5.164), and the eigenpairs are given by

$$\begin{aligned} \lambda _{ mn}&=\pi ^{4}{ D}\left( \frac{{ m}^{2}}{{ a}_{1}^{2}}+\frac{{ n}^{2}}{{ a}_{2}^{2}}\right) ^{2} \end{aligned}$$

(5.166)

$$\begin{aligned} \phi _{ mn}({ x_{1},x}_{2})&=\sin \left( \frac{{ m}\pi { x}_{1}}{{ a}_{1}}\right) \sin \left( \frac{{ n}\pi { x}_{2}}{{ a}_{2}}\right) \quad { m,n}=1,2,3,\ldots \end{aligned}$$

(5.167)

This is proved by substituting (5.166) and (5.167) into (5.163) and (5.164).

Also, by using (5.165) the eigenfrequencies $\omega _{mn}$ are obtained

$$\begin{aligned} \omega _{mn}=\sqrt{\frac{\lambda _{mn}}{\hat{\rho }}}=\pi ^{2}\left( \frac{{ m}^{2}}{{ a}_{1}^{2}}+\frac{{ n}^{2}}{{ a}_{2}^{2}}\right) \sqrt{\frac{{ D}}{\hat{\rho }}} \end{aligned}$$

(5.168)

This completes a solution to Problem 5.11.