Calculus of Variations Revisited Plus the Gamma and Bessel Functions

Abstract

A variety of topics that use the Calculus of Variations are explored. After developing general conditions for constrained optimization in a pastoral interlude they are applied to Queen Dido’s isoperimetric problem. Then the Euler-Lagrange equations of motion for conservative dynamical systems are derived in a pastoral interlude and applied to a linear spring, a double plane pendulum, and a vibrating string, respectively. Finally Bessel functions needed in the problems are introduced and examined which requires a similar preliminary examination of the gamma function as well since the former are defined in terms of the latter. In this context the Method of Stationary Phase and Laplace’s Method are developed and employed to derive asymptotic representations for the Bessel functions and Stirling’s formula for n!, respectively. There are eight problems: The first seven deal with a variety of Calculus of Variations applications and asymptotic representations for the special functions as well as a complete examination of the Bessel equation of order one-half. The eighth problem requires a Calculus of Variations approach to derive the wave equation governing the motion of a vibrating circular membrane which when solved by separation of variables gives rise to the eigenvalue example involving Bessel’s equation of order zero handled in the final section of the chapter.

Problems

19.1.

Consider the surface area functional for $W = W(x, y) > 0$ in $\mathcal {D}$ and prescribed to be 0 on $\partial \mathcal {D}$

$$\begin{aligned} I(W) = \iint \limits _{\mathcal {D}}{\underbrace{\sqrt{1 + W_x^2 + W_y^2}}_f\,dx\, dy} \end{aligned}$$

subject to the fixed volume constraint that

$$\begin{aligned} J(W) = \iint \limits _{\mathcal {D}}{\underbrace{W}_g\,dx\, dy} \equiv C, \end{aligned}$$

where C is a given positive constant.

• (a) Using a constrained optimization Calculus of Variations approach in conjunction with the introduction of a Lagrange multiplier $\lambda$, employ the governing partial differential equation

  $$\begin{aligned} \frac{\partial f^*}{\partial w} - \frac{\partial }{\partial x} \left( \frac{\partial f^*}{\partial w_x}\right) - \frac{\partial }{\partial y}\left( \frac{\partial f^*}{\partial w_y}\right) = 0 \text{ where } f^*= f + \lambda g, \end{aligned}$$

  and hence show that (see Problem 10.1)

  $$\begin{aligned} \lambda = \kappa (w) = \frac{w_{xx}(1 + w_y^2) - 2w_xw_yw_{xy} + w_{yy}(1 + w_x^2)}{(1 + w_x^2 + w_y^2)^{3/2}} \end{aligned}$$

  for the minimizing function $w = w(x, y)$.

• (b) The result of part (a) implies that this surface $z = w(x, y)$ of fixed volume which minimizes its area must be one of constant curvature $\lambda$. Excluding the trivial planar solution $w(x, y) \equiv 0$ since $C > 0$, the only function satisfying this condition is a hemisphere. Demonstrate that

  $$\begin{aligned} \lambda = -\frac{2}{a} = -2\left( \frac{2\pi }{3C}\right) ^{1/3} \end{aligned}$$

  for the hemisphere of radius a

  $$\begin{aligned} w(x, y) = (a^2 - x^2 - y^2)^{1/2} \text{ with } \mathcal {D} = \{(x, y)| 0 \le x^2 + y^2 \le a^2\} \end{aligned}$$

  by computing its curvature $\kappa (w) = -2/a$ and using the integral constraint to find a relationship between that radius a and the fixed volume parameter C. In doing so, the volume of this hemisphere may assumed to be $(2/3)\pi a^3$ without derivation.

19.2.

Deduce the explicit equations of motion for the double plane pendulum problem from the Euler–Lagrange equations

$$\begin{aligned} \frac{d}{dt}\frac{\partial \mathcal {L}}{\partial \dot{\theta }_j} - \frac{\partial \mathcal {L}}{\partial \theta _j} = 0 \text{ for } j = 1 \text{ and } 2 \end{aligned}$$

where

$$\begin{aligned} \mathcal {L}&= \frac{m_1 + m_2}{2}\ell _1^2 \dot{\theta }_1^2 + \frac{m_2}{2}\ell _2^2\dot{\theta }_2^2 + m_2\ell _1\ell _2 \cos (\theta _1 - \theta _2)\dot{\theta }_1\dot{\theta }_2 \\&- (m_1 + m_2)g\ell _1[1 - \cos (\theta _1)] - m_2g\ell _2[1 - \cos (\theta _2)]. \end{aligned}$$

19.3.

Consider a Laplace integral of the form

$$\begin{aligned} I(v) = \int _{a}^{b}{\varphi (\theta )e^{vh(\theta )}\, d\theta } \end{aligned}$$

where $\varphi \in C[a, b]$ and $h\in C^\infty [a, b]$. In this problem any results for Laplace’s Method of Section 19.6 and any arguments appearing in Section 19.8 on the Method of Stationary Phase may be employed.

• (a) For an h such that $h^\prime (\alpha ) = 0$ and $h^{\prime \prime }(\theta ) < 0$, as in Section 19.6, conclude

  $$\begin{aligned} I(v) \sim \varphi (\alpha )e^{vh(\alpha )}\left[ \frac{-\beta (\alpha )\pi }{vh^{\prime \prime }(\alpha )}\right] ^{1/2} \text{ as } v \rightarrow \infty \end{aligned}$$

  where

  $$\begin{aligned} \beta (\alpha ) = {\left\{ \begin{array}{ll} 2, &{} \text{ if } a< \alpha < b \\ 1/2, &{} \text{ if } \alpha = a \text{ or } b \end{array}\right. }. \end{aligned}$$

  Explain why this asymptotic formula is different when $\alpha$ occurs at either end point of the interval as opposed to at an interior point.

• (b) We now consider

  

  where h is such that $h^\prime (0) = h^\prime (\pi ) = 0$, $h^\prime (\theta ) < 0$ for $0< \theta < \pi$; and $h^{\prime \prime }(\theta _c) = 0$, $h^{\prime \prime }(\theta ) < 0$ for $0\le \theta < \theta _c$, $h^{\prime \prime }(\theta ) > 0$ for $\theta _c < \theta \le \pi$, as depicted in Fig. 19.9. Deduce from part (a) and the results of Section 19.6 that:

  
  
  

  and thus, employing the reasoning of Section 19.6 and the fact that $h(0) > h(\theta _c)$, conclude that the contribution from dominates those from and , respectively. Hence again obtain that

  $$\begin{aligned} I(v) \sim \varphi (0)e^{vh(0)}\left[ \frac{-\pi }{2vh^{\prime \prime }(0)}\right] ^{1/2} \text{ as } v \rightarrow \infty . \end{aligned}$$

• (c) Defining $f(\theta ) = e^{h(\theta )}$, show that under the conditions of part (b),

  $$\begin{aligned} \int _{a = 0 = \alpha }^{b = \pi }{\varphi (\theta )[f(\theta )]^v\, d\theta } \sim \varphi (\alpha )[f(\alpha )]^{v + 1/2}\left[ \frac{-\pi }{2vf^{\prime \prime }(\alpha )}\right] ^{1/2} \text{ as } v \rightarrow \infty ; \end{aligned}$$

  where $f^\prime (\alpha ) = 0$ and $f^{\prime \prime }(\alpha ) < 0$.

• (d) Recalling from Section 12.7 the integral relationship for the Legendre polynomial

  $$\begin{aligned} P_v(\mu ) = \frac{1}{\pi }\int _{0}^{\pi }{\left[ \underbrace{[\mu + (\mu ^2 - 1)^{1/2}\cos (\theta )]^n}_{f(\theta )}\right] ^v\, d\theta } \end{aligned}$$

  for $\mu > 1$, employ the one term asymptotic representation of part (c) to deduce that

  $$\begin{aligned} P_v(\mu ) \sim \frac{1}{\sqrt{2\pi v}}\frac{[\mu + (\mu ^2 - 1)^{1/2}]^{v + 1/2}}{(\mu ^2 - 1)^{1/4}} \text{ as } v \rightarrow \infty . \end{aligned}$$

Fig. 19.9

Plot of $y = h(\theta )$ relevant to Problem 19.3(b).

19.4.

In the Saddle Point Method (see Copson, [22]) the locus in the complex z-plane where

$$\begin{aligned} \frac{1}{2}\psi ^{\prime \prime }(\zeta )(z - \zeta )^2 \end{aligned}$$

is real and negative plays a pivotal role. Here $\zeta$ is the saddle point and $\psi$ an analytic function. Demonstrate that this locus is a straight line through the pole by representing both $z -\zeta$ and $\psi ^{\prime \prime }(\zeta )/2$ in polar form. That is let

$$\begin{aligned} z - \zeta = re^{i\theta } \text{ and } \frac{\psi ^{\prime \prime }(\zeta )}{2} = R_0e^{i\varphi _0} \text{ where } r, R_0 > 0. \end{aligned}$$

19.5.

Consider the Bessel equation of order $v = 1/2$ for $y = y(x)$:

$$\begin{aligned} L[y] = x^2 y^{\prime \prime } + xy^\prime + \left( x^2 - \frac{1}{4}\right) y = 0, \, x > 0. \end{aligned}$$

• (a) Show directly by the method of Frobenius that this equation has a general solution given by

  $$\begin{aligned} y(x) = c_1\frac{\cos (x)}{x^{1/2}} + c_2 \frac{\sin (x)}{x^{1/2}}. \end{aligned}$$

  That is let

  $$\begin{aligned} y(x) = \sum _{n=0}^{\infty }a_n(r)x^{n+r}, \end{aligned}$$

  find the two values of $r = r_{1,2}$ satisfying the indicial equation, and determine the $a_n(r_{1,2})$ from the recursion relation. Do not automatically take the coefficients with odd indices to be zero as is traditionally done for deriving the Bessel function formulae and deduce the $a_n(r_{1,2})$ from a bottom-up approach by examining $n = 2,4,\dots$, which depend upon $a_0$ and $n = 3,5,\dots$, which depend on $a_1$ separately and in that order until its general form can be ascertained and identified with the desired Maclaurin series for $\cos (x)$ and $\sin (x)$ given by

  $$\begin{aligned} \cos (x) = \sum _{n=0}^{\infty }(-1)^n\frac{x^{2n}}{(2n)!} \text{ and } \sin (x) = \sum _{n=0}^{\infty }(-1)^n\frac{x^{2n+1}}{(2n+1)!}. \end{aligned}$$

• (b) Now introduce the transformation

  $$\begin{aligned} y(x) = x^{-1/2}v{(x)} \end{aligned}$$

  and by obtaining and solving the differential equation resulting from $L[x^{-1/2}v] = 0$ for v(x), demonstrate in this alternate manner that the general solution in part (a) is correct.

• (c) Finally show, upon substitution of $v = 1/2$ into the series representations for $J_{\pm v}(x)$ of Section 19.7, that

  $$\begin{aligned} J_{1/2}(x) = \sqrt{\frac{2}{\pi x}}\sin (x) \text{ and } J_{-1/2}(x) = \sqrt{\frac{2}{\pi x}}\cos (x). \end{aligned}$$

  Here the result from Section 19.5

  $$\begin{aligned} \Gamma \left( m + \frac{1}{2}\right) = \frac{(2m)!}{2^{2m}m!}\sqrt{\pi } \end{aligned}$$

  may be employed directly and used in conjunction with $\Gamma (z+1) = z\Gamma (z)$ to deduce that

  $$\begin{aligned} \Gamma \left( m + \frac{3}{2}\right) = \Gamma \left( m + \frac{1}{2} + 1\right) = \left( m + \frac{1}{2}\right) \Gamma \left( m + \frac{1}{2}\right) = \frac{(2m+1)!}{2^{2m+1}m!} \sqrt{\pi }. \end{aligned}$$

19.6.

Consider the integral representation for $J_0(x)$

$$\begin{aligned} J_0(x) = \frac{1}{\pi }\int _{0}^{\pi }{\cos [x\sin (\theta )]\, d\theta } \end{aligned}$$

which follows from that for $J_n(x)$ of Section 19.8 with $n = 0$.

• (a) The purpose of this part is to show that

  $$\begin{aligned} J_0(x) = \frac{1}{\pi }\int _{0}^{\pi }{e^{ix\cos (\varphi )}\, d\varphi } \end{aligned}$$

  by justifying each step in the following transformation procedure:

  $$\begin{aligned} \underbrace{\int _{0}^{\pi }{\cos [x\sin (\theta )]\, d\theta }}_{\begin{array}{c} \theta = \pi /2 - \varphi \\ d\theta = -d\varphi \end{array}}&= \int _{-\pi /2}^{\pi /2}{\cos [x\cos (\varphi )]\, d\varphi } = 2\int _{0}^{\pi /2}{\cos [x\cos (\varphi )]\, d\varphi } \\&= \int _{0}^{\pi }{\cos [x\cos (\theta )]\, d\theta } \\&= \int _{0}^{\pi }{e^{ix\cos (\varphi )}\, d\varphi } \text{ since } \int _{0}^{\pi }{\sin [x\cos (\theta )]\, d\theta } = 0. \end{aligned}$$

• (b) The purpose of this part is to use part (a) and the Method of Stationary Phase to deduce the one-term asymptotic representation for $J_0(x)$ of Section 19.8 in an alternate manner. Justify

  $$\begin{aligned} J_0(x) = \int _{0 = \alpha _1}^{\pi = \alpha _2}{\underbrace{\frac{1}{\pi }}_{f(\varphi )}\exp [ix\underbrace{\cos (\varphi )}_{g(\varphi )}]\, d\varphi } \sim \frac{1}{2}\sum _{j=1}^{2}F(\alpha _j, x) \text{ as } x \rightarrow \infty \end{aligned}$$

  where

  $$\begin{aligned} F(\alpha , x) = \sqrt{\frac{2\pi }{|g^{\prime \prime }(\alpha )|}}\frac{f(\alpha )}{x^{1/2}}e^{i\{xg(\alpha ) + \frac{\pi }{4} sgm[g^{\prime \prime }(\alpha )]\}} \end{aligned}$$

  or

  $$\begin{aligned} J_0(x) \sim \sqrt{\frac{2}{\pi x}} \frac{e^{i(x - \pi /4)} + e^{-i(x - \pi /4)}}{2} = \sqrt{\frac{2}{\pi x}}\cos \left( x -\frac{\pi }{4}\right) \text{ as } x \rightarrow \infty ; \end{aligned}$$

  by filling in the details.

• (c) The purpose of this part is to use the integral relations for the Legendre polynomial and the Bessel function of the first kind of order zero to deduce that

  $$\begin{aligned} \lim _{v \rightarrow \infty }P_v\left[ \cos \left( \frac{\theta }{v}\right) \right] = J_0(\theta ). \end{aligned}$$

  Recall from Section 12.7 that the integral representation for $P_v(\mu )$ can be extended by analytic continuation to those $|\mu | < 1$ and taking $\mu = \cos (\theta /v)$ show that

  $$\begin{aligned} P_v\left[ \cos \left( \frac{\theta }{v}\right) \right] = \frac{1}{\pi }\int _{0}^{\pi }{\left[ \cos \left( \frac{\theta }{v}\right) + i\sin \left( \frac{\theta }{v}\right) \cos (\varphi )\right] ^v \, d\varphi }. \end{aligned}$$

  Rewriting

  $$\begin{aligned} \left[ \cos \left( \frac{\theta }{v}\right) + i \sin \left( \frac{\theta }{v}\right) \cos (\varphi )\right] ^v = \exp \left( \frac{\ln [\cos (\theta \alpha ) + i\sin (\theta \alpha )\cos (\varphi )]}{\alpha }\right) , \end{aligned}$$

  where $\alpha = \frac{1}{v}$, and using L’Hospital’s Rule to calculate

  $$\begin{aligned} \lim _{\alpha \rightarrow 0}\frac{\ln [\cos (\theta \alpha ) + i\sin (\theta \alpha )\cos (\varphi )]}{\alpha }&= \lim _{\alpha \rightarrow 0}\frac{-\theta \sin (\theta \alpha ) + i\theta \cos (\theta \alpha )\cos (\varphi )}{\cos (\theta \alpha ) + i\sin (\theta \alpha )\cos (\varphi )} \\&= i\theta \cos (\varphi ), \end{aligned}$$

  conclude that

  $$\begin{aligned} \lim _{v\rightarrow \infty }P_v\left[ \cos \left( \frac{\theta }{v}\right) \right]&= \frac{1}{\pi }\int _{0}^{\pi }{\lim _{v \rightarrow \infty }\left[ \cos \left( \frac{\theta }{v}\right) + i \sin \left( \frac{\theta }{v}\right) \cos (\varphi )\right] ^v\, d\varphi } \\&= \frac{1}{\pi }\int _{0}^{\pi }{e^{i\theta \cos (\varphi )}\, d\varphi } = J_0(\theta ). \end{aligned}$$

19.7.

• (a) Consider

  $$\begin{aligned} I(v) = \int _{0}^{\pi }{f(\theta )e^{ivg(\theta )}\, d\theta } \end{aligned}$$

  where $f(\theta )$ is an analytic function of the real variable $\theta$ and $g(\theta )$ is a real-valued function possessing only one point $\theta = \alpha$ such that $g^\prime (\alpha ) = 0$ for $\theta \in [0,\pi ]$. Given that, $g^{\prime \prime }(\alpha ) = 0$ and $g^{\prime \prime \prime }(\alpha ) \not = 0$, show

  $$\begin{aligned} I(v) \sim \frac{\Gamma (1/3)}{\sqrt{3}}\beta (\alpha )f(\alpha )&\left[ \frac{6}{|g^{\prime \prime \prime }(\alpha )|v}\right] ^{1/3}e^{ivg(\alpha )} \text{ as } v \rightarrow \infty \\ \text{ where } \beta (\alpha ) =&{\left\{ \begin{array}{ll} 1, &{} \text{ for } 0< \alpha < \pi \\ 1/2, &{} \text{ for } \alpha = 0 \text{ or } \pi \end{array}\right. }. \end{aligned}$$

• (b) Use the result of part (a) to deduce that

  $$\begin{aligned} J_v(v) = \text {Re}\left\{ \int _{0}^{\pi }{\frac{1}{\pi }e^{iv[\sin (\theta ) - \theta ]}\, d\theta }\right\} \sim \frac{\Gamma (1/3)}{2^{2/3}3^{1/6}\pi v^{1/3}} \text{ as } v \rightarrow \infty . \end{aligned}$$

Hints:

• For (a): For $0< \alpha < \pi$,

  $$\begin{aligned} I(v) \sim \int _{\alpha - \delta }^{\alpha + \delta }{f(\theta )e^{ivg(\theta )}\, d\theta } \text{ as } v \rightarrow \infty . \end{aligned}$$

  Let $g(\theta ) \cong g(\alpha ) + g^{\prime \prime \prime }(\alpha )(\theta -\alpha )^3/6$ and $z {=} (vA)^{1/3}(\theta - \alpha )$ where $A = |g^{\prime \prime \prime }(\alpha )|/6$. Thus

  $$\begin{aligned} I(v) \sim f(\alpha )\left[ \frac{6}{|g^{\prime \prime \prime }(\alpha )|v}\right] ^{1/3}e^{ivg(\alpha )}\int _{-\infty }^{\infty }{e^{isgm[g^{\prime \prime \prime }(\alpha )]z^3}\, dz} \text{ as } v \rightarrow \infty . \end{aligned}$$

  Note:

  $$\begin{aligned} \int _{-\infty }^{\infty }{e^{iz^3}\, dz}&= \int _{-\infty }^{0}{e^{iz^3}\, dz} + \int _{0}^{\infty }{e^{iz^3}\, dz} \\&= \int _{0}^{\infty }{(e^{iz^3} + e^{-iz^3})\, dz} = 2\text {Re}\left\{ \int _{0}^{\infty }{e^{iz^3}\, dz}\right\} \end{aligned}$$

  and

  $$\begin{aligned} \int _{-\infty }^{\infty }{e^{-iz^3}\, dz} = 2\text {Re}\left\{ \int _{0}^{\infty }{e^{-iz^3}\, dz}\right\} = 2\text {Re}\left\{ \int _{0}^{\infty }{e^{iz^3}\, dz}\right\} \text{ as } \text{ well. } \end{aligned}$$

  Consider the contour $\Gamma _R$ depicted in Fig. 19.7a with $n = 3$. Then

  $$\begin{aligned} \oint \limits _{\Gamma _R}{e^{iz^3}\, dz} = 0 \end{aligned}$$

  and, in the limit as $R\rightarrow \infty$, this implies

  $$\begin{aligned} \int _{0}^{\infty }{e^{iz^3}\, dz} = e^{i\pi /6}\int _{0}^{\infty }{e^{-r^3}\, dr} = e^{i\pi /6}\frac{\Gamma (1/3)}{3} = (\sqrt{3} + i)\frac{\Gamma (1/3)}{6}. \end{aligned}$$

  For $\alpha = 0$ or $\pi$, one obtains 1 / 2 the result for $0< \alpha < \pi$.

• For (b): Considering

  $$\begin{aligned} I(v) = \int _{0}^{\pi }{\frac{1}{\pi }e^{iv[\sin (\theta ) - \theta ]}\, d\theta }; \end{aligned}$$

  make the identifications $f(\theta )\equiv 1/\pi$ and $g(\theta ) = \sin (\theta ) - \theta$. Hence $\alpha = 0$.

19.8.

Consider a thin elastic membrane extended over a closed non-self-intersecting curve C. The planar domain $\mathcal {D}$ enclosed by C coincides with the equilibrium configuration of the membrane and is of area D. Assume the membrane is permitted to vibrate so that each of its points undergoes a motion in a direction perpendicular to this x-y plane and its stretch-resistant restorative force per unit length or surface tension has a constant value $\tau _0$. Denote that vertical displacement from equilibrium by $w = w(x,y, t)$, and assume a surface mass density $\rho = \rho (w)$ for the membrane. Then its kinetic and potential energies at any time t are given by

$$\begin{aligned} T = \frac{1}{2}\iint \limits _{\mathcal {D}}{\rho (w)\dot{w}^2\,dx\, dy} \text{ and } V = \tau _0\iint \limits _{\mathcal {D}}{[(1 + w_x^2 + w_y^2)^{1/2} - 1]\,dx\, dy} \end{aligned}$$

where $\dot{w} \equiv \partial w/\partial t$, $w_x \equiv \partial w/\partial x$, and $w_y \equiv \partial w /\partial y$ while $w(x,y, t) = 0$ for $(x, y)\in C$.

• (a) Noting that D is equal to

  $$\begin{aligned} \iint \limits _{\mathcal {D}}{dx\, dy}, \end{aligned}$$

  explain the physical significance of the integral appearing in V.

• (b) Construct the Lagrangian for this system

  $$\begin{aligned} L = T - V = \iint \limits _{\mathcal {D}}{f(w,\dot{w},w_x,w_y)\,dx\, dy} \end{aligned}$$

  where

  $$\begin{aligned} f(w,\dot{w},w_x, w_y) = \rho (w)\frac{\dot{w}^2}{2} - \tau _0[(1 + w_x^2 + w_y^2)^{1/2} - 1]. \end{aligned}$$

  Then according to Hamilton’s Principle, the action integral

  $$\begin{aligned} I = \int _{t_1}^{t_2}{\iint \limits _{\mathcal {D}}f(w,\dot{w},w_x,w_y)\,dx\,dy\, dt} \end{aligned}$$

  is minimized by the function w(x, y, t) that describes the actual motion of the membrane over all such functions which have continuous second partial derivatives, vanish on C, and coincide with the actual membrane configuration at $t = t_1$ and $t_2$, where the latter times are completely arbitrary. Show that the Euler–Lagrange equation for this integral

  $$\begin{aligned} \frac{\partial f}{\partial w} - \frac{\partial }{\partial x}\left( \frac{\partial f}{\partial w_x}\right) - \frac{\partial }{\partial y}\left( \frac{\partial f}{\partial w_y}\right) - \frac{\partial }{\partial t}\left( \frac{\partial f}{\partial \dot{w}}\right) = 0 \end{aligned}$$

  yields the equation of motion (see Problem 19.1 and Section 19.4)

  $$\begin{aligned} \rho (w)\ddot{w} + \rho ^\prime (w)\frac{\dot{w}^2}{2} = \tau _0\kappa (w) \end{aligned}$$

  where

  $$\begin{aligned} \kappa (w) = \frac{w_{xx}(1 + w_y^2) - 2w_xw_yw_{xy} + w_{yy}(1 + w_x^2)}{(1 + w_x^2 + w_y^2)^{3/2}} \end{aligned}$$

• (c) Assuming that w, the deviation of the membrane from its equilibrium configuration, is small enough to allow $\rho (w)\cong \rho (0) = \rho _0$, $|w_x|<< 1$, and $|w_y|<< 1$, deduce that this equation of motion reduces to the two-dimensional wave equation

  $$\begin{aligned} \ddot{w} = c_0^2 \nabla _2^2 w \text{ where } c_0^2 \equiv \frac{\tau _0}{\rho _0} \text{ and } \nabla _2^2w \equiv w_{xx} + w_{yy}. \end{aligned}$$

  Further, since $[\tau _0] = (ML/\tau ^2)/L = M/\tau ^2$ and $[\rho _0] = M/L^2$, conclude that $[c_0] = L/\tau$.

• (d) We now examine the problem of a circular membrane of radius R vibrating under the conditions already stipulated with the additional stipulation that its initial vibration at time $t = 0$ is axisymmetric in that it only depends on the radial distance from the center of this circle. Under these conditions, upon introduction of polar coordinates $(r,\theta )$, it can be deduced that

  $$\begin{aligned} w(x,y,t) = W(r, t); \end{aligned}$$

  i.e., the displacement is then independent of $\theta$ for all later times $t > 0$ as well. If F(r) and K(r) represent the initial displacement and velocity functions of this membrane, respectively, explain why the governing partial differential equation (PDE), boundary conditions (BC’s), and initial conditions (I.C.’s) for this circular membrane undergoing axisymmetric vibrations are given by (the form of $\nabla _2^2$ in polar coordinates developed in Section 12.2 may be assumed)

  $$\begin{aligned} {\text {PDE:}}\;\frac{\partial ^2 W}{\partial t^2} = c_0^2\left( \frac{\partial ^2 W}{\partial r^2} + \frac{1}{r}\frac{\partial W}{\partial r}\right) , \, 0< r < R, \, t > 0; \end{aligned}$$
  $$\begin{aligned} {\text {BC's:}}\; W(r, t) \;{\text {remain bounded as}}\; r \rightarrow 0 \;{\text {and}}\; W(R, t) = 0 \;{\text {for}}\; t > 0; \end{aligned}$$
  $$\begin{aligned} {\text {I.C.'s:}}\; W(r, 0) = F(r)\ {\text {and}}\; \frac{\partial W}{\partial t}(r, 0) = K(r) \;{\text {for}}\; 0< r < R. \end{aligned}$$

• (e) Solve this mixed initial-boundary value problem by first seeking a separation of variables solution to the PDE and BC’s of the form

  $$\begin{aligned} W(r, t) = H(r)G(t) \end{aligned}$$

  obtaining

  $$\begin{aligned} \begin{array}{cc} \displaystyle \frac{\ddot{G}(t)}{c_0^2G(t)} = \frac{1}{H(r)}\left( \frac{d^2H}{dr^2}(r) + \frac{1}{r}\frac{dH}{dr}(r)\right) \equiv -\lambda , \;{\text {the separation constant;}}\\ \displaystyle {\text {for}}\; 0< r < R \;{\text {and}}\; t > 0;\end{array} \end{aligned}$$
  $$\begin{aligned} H(r) \;{\text {remain bounded as}}\; r \rightarrow 0 \;{\text {and}}\;H(R) = 0. \end{aligned}$$

• (f) Introducing the change of variables $s = \sqrt{\lambda }r$ and employing the results of Section 19.9, transform the boundary value problem for $H(r) = u(s)$ into

  $$\begin{aligned} s^2\frac{d^2u}{ds^2}(s) + s\frac{du}{ds}(s) + s^2 u(s) = 0; \;{\text {for}}\; 0< s < \sqrt{\lambda }R; \end{aligned}$$
  $$\begin{aligned} u(s) \;{\text {remain bounded as}}\; s\rightarrow 0 \;{\text {and}}\; u(\sqrt{\lambda }R) = 0; \end{aligned}$$

  and hence conclude that

  $$\begin{aligned} \lambda _n = \frac{\alpha _n^2}{R^2} \;{\text {and}}\; H_n(r) = J_0\left( \frac{\alpha _nr}{R}\right) \;{\text {for}}\; n = 1,2,3,\dots ; \end{aligned}$$

  while $G_n(t)$ satisfies

  $$\begin{aligned} \ddot{G}_n(t) = \omega _n^2 G_n(t) = 0 \;{\text {where}}\;\omega _n^2 = c_0^2\lambda _n = \frac{c_0^2\alpha _n^2}{R^2}\;{\text {for}}\; t > 0 \end{aligned}$$

  and thus

  $$\begin{aligned} G_n(t) = A_n\cos (\omega _nt) + B_n\sin (\omega _n t)\;{\text {where}}\;\omega _n = \frac{c_0\alpha _n}{R} \;{\text {for}}\; n = 1,2,3,\dots . \end{aligned}$$

• (g) Employing the superposition principle show that the most general solution emerging from the separation of variables approach of the previous two parts is given by

  $$\begin{aligned} W(r, t) = \sum _{n=1}^{\infty }W_n(r, t) = \sum _{n=1}^{\infty }H_n(r)G_n(t) = \sum _{n=1}^{\infty }J_0\left( \frac{\alpha _n r}{R}\right) [A_n\cos (\omega _nt) + B_n\sin (\omega _nt)]. \end{aligned}$$

• (h) Finally making use of the orthonormality property of the $J_0(\alpha _n r/R)$ as demonstrated in Section 19.9 determine that

  $$\begin{aligned} A_n = \frac{2}{J_1^2(\alpha _n)R^2}\int _{0}^{R}{rF(r)J_0\left( \frac{\alpha _n r}{R}\right) \, dr} \end{aligned}$$

  and

  $$\begin{aligned} B_n = \frac{2}{J_1^2(\alpha _n)c_0\alpha _nR}\int _{0}^{R}{rK(r)J_0\left( \frac{\alpha _n r}{R}\right) \, dr}; \end{aligned}$$

  in order to satisfy the I.C.’s.

19.9.

The purpose of this problem is to use a Calculus of Variations approach to deduce the so-called elastic beam equations for a particular prototype situation related to the behavior of two-ply laminated architectural glass plates used in the automotive industry. Consider a laminated composite beam of length L and width b composed of two layers of glass of thicknesses $h_{1,2}$ separated by a thin plastic interlayer of thickness t. The total potential energy of such a beam subjected to a distributed load q and a point load P at its center can be written as [5]

$$\begin{aligned} \Pi = \int _{0}^{L}{f(x,w, u_1,u_2,w^\prime , u_1^\prime , u_2^\prime ,w^{\prime \prime })\, dx} \end{aligned}$$

where

$$\begin{aligned} f(x,w, u_1,u_2,w^\prime , u_1^\prime , u_2^\prime , w^{\prime \prime }) = \sum _{i=1}^{2}\frac{E}{2}&\left\{ A_i\left[ \frac{du_i}{dx} + \frac{1}{2}\left( \frac{dw}{dx}\right) ^2\right] ^2 + I_i\left( \frac{d^2w}{dx^2}\right) ^2\right\} \\&\quad + \frac{Gb}{2t}\left( u_1 - u_2 - h\frac{dw}{dx}\right) ^2 - qw - \frac{Pw_L}{2L^2}x. \end{aligned}$$

Here x is the axial coordinate of the beam; w, its common lateral displacement with $w_L \equiv w|_{x=L}$; E, the elastic modulus of the glass; G, the shear modulus of the plastic; and $h = \frac{h_1}{2} +\frac{h_2}{2} + t$ while $u_{1,2}$, $A_{1,2}$, and $I_{1,2}$ are the axial displacements, cross-sectional areas, and second moments of area of the top and bottom glass layers, respectively. Assume that the actual displacements are the ones that render the total potential energy a minimum [5].

• (a) Employing a Calculus of Variations approach conclude from the appropriate Euler–Lagrange equation for w

  $$\begin{aligned} \frac{\partial f}{\partial w} - \frac{d}{dx}\frac{\partial f}{\partial w^\prime } + \frac{d^2}{dx^2}\frac{\partial f}{\partial w^{\prime \prime }} = 0, \end{aligned}$$

  that these displacements satisfy the equation

  $$\begin{aligned} \left[ \frac{d^2}{dx^2}\left( EI\frac{d^2}{dx^2}\right) - \frac{Gbh^2}{t}\frac{d^2}{dx^2}\right] w&\\ = q + \left( \frac{dN_1}{dx} + \frac{dN_2}{dx}\right)&\frac{dw}{dx} + (N_1 + N_2)\frac{d^2w}{dx^2} - \frac{Gbh}{t}\left( \frac{du_1}{dx} - \frac{du_2}{dx}\right) , \end{aligned}$$

  where $I = I_1 + I_2$ and $N_{1,2}$, the axial forces in the top and bottom glass layers, are given by

  $$\begin{aligned} N_{1,2} = EA_{1,2}\left[ \frac{du_{1,2}}{dx} + \frac{1}{2}\left( \frac{dw}{dx}\right) ^2\right] . \end{aligned}$$

• (b) Continuing this approach conclude from the appropriate Euler–Lagrange equations for $u_{1,2}$

  $$\begin{aligned} \frac{d}{dx}\frac{\partial f}{\partial u_{1,2}^\prime } = \frac{\partial f}{\partial u_{1,2}}, \end{aligned}$$

  that these displacements, in addition, satisfy

  $$\begin{aligned} \frac{dN_1}{dx} = \frac{Gb}{t}\left( u_1 - u_2 - h\frac{dw}{dx}\right) = -\frac{dN_2}{dx}. \end{aligned}$$

• (c) Deduce from (b) that

  $$\begin{aligned} N_1 + N_2 \equiv N_0, \;{\text {a constant}}, \end{aligned}$$

  and hence the differential equation of (a) reduces to

  $$\begin{aligned} \left[ \frac{d^2}{dx^2}\left( EI\frac{d^2}{dx^2}\right) - \frac{Gbh^2}{t}\frac{d^2}{dx^2}\right] w = q + N_0\frac{d^2w}{dx^2} - \frac{Gbh}{t}\left( \frac{du_1}{dx} - \frac{du_2}{dx}\right) . \end{aligned}$$