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Chapter 11: Problem 2

Find the displacement \(u(r, t)\) in a vibrating circular elastic membrane of radius 1 that satisfies the boundary condition $$ u(1, t)=0, \quad t \geq 0 $$ and the initial conditions $$ u(r, 0)=0, \quad u_{t}(r, 0)=g(r), \quad 0 \leq r \leq 1 $$ where \(g(1)=0\) Hint: The differential equation to be satisfied is Eq. ( 3 ) of the text.

Short Answer

Expert verified
Answer: The final expression for the displacement function \(u(r,t)\) is: $$ u(r,t)=\sum_{n=1}^{\infty}c_nJ_0(k_nr)\left(\frac{1}{2}b_n\sin(k_nt)\right) $$ where $$ b_n=\frac{\int_0^1 r g(r) J_0(k_nr)dr}{c_nk_n \int_0^1 rJ_0(k_nr)^2 dr} $$

Step by step solution

01

Write down the wave equation for the circular membrane

The wave equation for a circular membrane is given by $$ \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{\partial^2 u}{\partial t^2}=0 $$
02

Apply the boundary and initial conditions

Using the given boundary condition \(u(1,t)=0\) and the initial conditions \(u(r,0)=0\) and \(u_{t}(r,0)=g(r)\), we have Boundary condition: $$ u(1,t)=0 $$ Initial conditions: $$ u(r,0)=0, \quad u_{t}(r,0)=g(r) $$
03

Apply separation of variables

We will use separation of variables, with the assumption that \(u(r,t)=R(r)T(t)\). Substituting this into the wave equation and dividing by \(u\) yields: $$ \frac{1}{R}\frac{d}{dr} \left(r\frac{dR}{dr}\right)+\frac{1}{T}\frac{d^2T}{dt^2}=0 $$ Since the left side depends on \(r\) and the right side depends on \(t\), each side must be equal to a constant. Let this constant be \(-k^2\). Then, we have two equations: $$ \frac{1}{R}\frac{d}{dr}\left(r\frac{dR}{dr}\right)=-k^2 $$ $$ \frac{1}{T}\frac{d^2T}{dt^2} = k^2 $$
04

Solve the spatial and temporal equations

Solving the spatial equation, we have the following Bessel differential equation: $$ \frac{d}{dr}\left(r\frac{dR}{dr}\right)=-k^2rR $$ Taking the Bessel function as the solution, we have $$ R(r)=cJ_0(kr) $$ Solving the temporal equation, we have the following harmonic oscillator equation: $$ \frac{d^2T}{dt^2}=k^2T $$ Taking the sinusoidal function as the solution, we have $$ T(t)=\frac{1}{2}a\cos(kt)+\frac{1}{2}b\sin(kt) $$ Now, we have $$ u(r,t)=R(r)T(t)=cJ_0(kr)\left(\frac{1}{2}a\cos(kt)+\frac{1}{2}b\sin(kt)\right) $$
05

Apply boundary and initial conditions to find coefficients

Applying the boundary condition \(u(1,t)=0\), we have $$ cJ_0(k)=0 $$ Making use of the fact that \(J_0(t)\) has an infinite number of zeros, we have the summation: $$ u(r,t)=\sum_{n=1}^{\infty}c_nJ_0(k_nr)\left(\frac{1}{2}a_n\cos(k_nt)+\frac{1}{2}b_n\sin(k_nt)\right) $$ Now, applying the initial conditions \(u(r,0)=0\), we have $$ 0=\sum_{n=1}^{\infty}c_n J_0(k_nr)a_n $$ Multiplying both sides by \(rJ_0(k_mr)\) and integrating from \(0\leq r \leq 1\), we get $$ \int_0^1rJ_0(k_mr)J_0(k_nr)dr = 0 $$ From the orthogonality of Bessel functions, we find that $$ a_n=0 $$ Finally, applying the initial condition \(u_{t}(r,0)=g(r)\), we have $$ g(r)=\sum_{n=1}^\infty c_nk_nJ_0(k_nr)b_n $$ Now, we need to find the coefficients \(b_n\). Multiplying both sides by \(rJ_0(k_mr)\) and integrating from \(0\leq r \leq 1\), we get $$ \int_0^1 r g(r) J_0(k_mr)dr=\sum_{n=1}^\infty c_n k_n b_n \int_0^1 rJ_0(k_mr)J_0(k_nr)dr $$ Making use of the orthogonality of Bessel functions, we find that $$ c_nk_n b_n \int_0^1 rJ_0(k_mr)J_0(k_nr)dr=\int_0^1 r g(r) J_0(k_nr)dr $$ Therefore, we have $$ b_n=\frac{\int_0^1 r g(r) J_0(k_nr)dr}{c_nk_n \int_0^1 rJ_0(k_nr)^2 dr} $$
06

Final expression for the displacement function \(u(r,t)\)

The final expression for the displacement \(u(r,t)\) can be written as: $$ u(r,t)=\sum_{n=1}^{\infty}c_nJ_0(k_nr)\left(\frac{1}{2}b_n\sin(k_nt)\right) $$ where $$ b_n=\frac{\int_0^1 r g(r) J_0(k_nr)dr}{c_nk_n \int_0^1 rJ_0(k_nr)^2 dr} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Equation
The wave equation describes how vibrations propagate through a medium. In the case of a vibrating circular membrane, this medium is the elastic material of the membrane. The wave equation for a circular membrane is given by:
\[\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{\partial^2 u}{\partial t^2}=0\]This equation states that the change in the radial derivative of the displacement is equivalent to the change in time derivatives of the displacement.
The parameters \(u\), \(r\), and \(t\) represent the displacement from equilibrium, radial distance, and time, respectively. For a circular membrane, vibrations spread radially outward, hence the inclusion of the radial term \(r\). The wave equation helps to predict how the membrane will move over time, based on initial conditions.
Boundary Conditions
Boundary conditions specify the behavior of a wave at the edge of its domain. In our exercise, the membrane is fixed at the boundary, meaning it does not displace at the edge. This is represented by the boundary condition:
\[u(1,t)=0\]for all time \(t\geq0\).
This condition physically implies that the edge of the membrane remains at rest, which is a common scenario in physical problems concerning vibrations. Boundary conditions are crucial because they affect the possible solutions to the wave equation, determining the modes in which the membrane can oscillate.
Bessel Functions
Bessel functions appear in problems with cylindrical symmetry, such as a circular membrane. When separating variables in polar coordinates, the radial part of the differential equation often results in a Bessel equation.Bessel functions, particularly the zeroth-order Bessel function of the first kind, \(J_0(x)\), are used to describe the radial component of the membrane's vibration:
\[R(r)=cJ_0(kr)\]These functions have specific properties, including zeros, which are essential for satisfying boundary conditions.
The properties of Bessel functions, such as orthogonality, allow for the expansion of complex functions, like the vibration modes, into simpler terms that can be better analyzed and understood. This is a powerful tool in solving wave-related problems.
Separation of Variables
Separation of variables is an effective method to solve partial differential equations, such as our wave equation,The concept involves assuming a solution that can be written as a product of separate functions, each dependent on a single coordinate:
\[u(r,t)=R(r)T(t)\]Substituting this into the wave equation divides it into two independent ordinary differential equations, each dependent on only their respective variable, \(r\) or \(t\).
By equating each side to a constant, typically denoted as \(-k^2\), separation of variables simplifies the problem into two simpler ODEs that can be solved individually. This technique is so powerful because it allows you to tackle complex, multidimensional situations by breaking them down into more manageable one-dimensional problems.

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Most popular questions from this chapter

The equation $$ v_{x x}+v_{y y}+k^{2} v=0 $$ is a generalization of Laplace's equation, and is sometimes called the Helmholtz \((1821-1894)\) equation. (a) In polar coordinates the Helmholtz equation is $$v_{r r}+(1 / r) v_{r}+\left(1 / r^{2}\right) v_{\theta \theta}+k^{2} v=0$$ If \(v(r, \theta)=R(r) \Theta(\theta),\) show that \(R\) and \(\Theta\) satisfy the ordinary differential equations $$ r^{2} R^{\prime \prime}+r R^{\prime}+\left(k^{2} r^{2}-\lambda^{2}\right) R=0, \quad \Theta^{\prime \prime}+\lambda^{2} \Theta=0 $$ (b) Consider the Helmholtz equation in the disk \(r

The method of eigenfunction expansions is often useful for nonhomogeneous problems related to the wave equation or its generalizations. Consider the problem $$ r(x) u_{u}=\left[p(x) u_{x}\right]_{x}-q(x) u+F(x, t) $$ $$ \begin{aligned} u_{x}(0, t)-h_{1} u(0, t)=0, & u_{x}(1, t)+h_{2} u(1, t)=0 \\\ u(x, 0)=f(x), & u_{t}(x, 0)=g(x) \end{aligned} $$ This problem can arise in connection with generalizations of the telegraph equation (Problem 16 in Section 11.1 ) or the longitudinal vibrations of an elastic bar (Problem 25 in Section \(11.1) .\) (a) Let \(u(x, t)=X(x) T(t)\) in the homogeneous equation corresponding to Eq. (i) and show that \(X(x)\) satisfies Eqs. ( 28) and ( 29) of the text. Let \(\lambda_{n}\) and \(\phi_{n}(x)\) denote the eigenvalues and normalized eigenfunctions of this problem. (b) Assume that \(u(x, t)=\sum_{n=1}^{\infty} b_{n}(t) \phi_{n}(x),\) and show that \(b_{n}(t)\) must satisfy the initial value problem $$ b_{n}^{\prime \prime}(t)+\lambda_{n} b_{n}(t)=\gamma_{n}(t), \quad b_{n}(0)=\alpha_{n}, \quad b_{n}^{\prime}(0)=\beta_{n} $$ where \(\alpha_{n}, \beta_{n},\) and \(\gamma_{n}(t)\) are the expansion coefficients for \(f(x), g(x),\) and \(F(x, t) / r(x)\) in terms of the eigenfunctions \(\phi_{1}(x), \ldots, \phi_{n}(x), \ldots\)

Use eigenfunction expansions to find the solution of the given boundary value problem. $$ \begin{array}{l}{u_{t}=u_{x x}+1-|1-2 x|, \quad u(0, t)=0, \quad u(1, t)=0, \quad u(x, 0)=0} \\ {\text { see Problem } 5 .}\end{array} $$

Let \(\phi_{1}, \phi_{2}, \ldots, \phi_{n}, \ldots\) be the normalized eigenfunctions of the Sturm-Liouville problem \((11),(12) .\) Show that if \(a_{n}\) is the \(n\) th Fourier coefficient of a square integrable function \(f,\) then \(\lim _{n \rightarrow \infty} a_{n}=0\) Hint: Use Bessel's inequality, Problem \(9(b)\).

Consider the boundary value problem $$ -\left(x y^{\prime}\right)^{\prime}=\lambda x y $$ \(y, y^{\prime}\) bounded as \(x \rightarrow 0, \quad y^{\prime}(1)=0\) (a) Show that \(\lambda_{0}=0\) is an eigenvalue of this problem corresponding to the eigenfunction \(\phi_{0}(x)=1 .\) If \(\lambda>0,\) show formally that the eigenfunctions are given by \(\phi_{n}(x)=\) \(J_{0}(\sqrt{\lambda_{n}} x),\) where \(\sqrt{\lambda_{n}}\) is the \(n\) th positive root (in increasing order) of the equation \(J_{0}^{\prime}(\sqrt{\lambda})=0 .\) It is possible to show that there is an infinite sequence of such roots. (b) Show that if \(m, n=0,1,2, \ldots,\) then $$ \int_{0}^{1} x \phi_{m}(x) \phi_{n}(x) d x=0, \quad m \neq n $$ (c) Find a formal solution to the nonhomogeneous problem $$ \begin{aligned}-\left(x y^{\prime}\right)^{\prime} &=\mu x y+f(x) \\ y, y^{\prime} \text { bounded as } x \rightarrow 0, & y^{\prime}(1)=0 \end{aligned} $$ where \(f\) is a given continuous function on \(0 \leq x \leq 1,\) and \(\mu\) is not an eigenvalue of the corresponding homogeneous problem.
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