Question

This problem illustrates that the eigenvalue parameter sometimes appears in the boundary conditions as well as in the differential equation. Consider the longitudinal vibrations of a uniform straight elastic bar of length \(L .\) It can be shown that the axial displacement \(u(x, t)\) satisfies the partial differential equation $$ (E / \rho) u_{x x}=u_{u^{i}} \quad 00 $$ $$ \begin{array}{l}{\text { where } E \text { is Young's modulus and } \rho \text { is the mass per unit volume. If the end } x=0 \text { is fixed, }} \\\ {\text { then the boundary condition there is }}\end{array} $$ $$ u(0, t)=0, \quad t>0 $$ $$ \begin{array}{l}{\text { Suppose that the end } x=L \text { is rigidly attached to a mass } m \text { but is otherwise unrestrained. }} \\ {\text { We can obtain the boundary condition here by writing Newton's law for the mass. From }} \\ {\text { the theory of elasticity it can be shown that the force exerted by the bar on the mass is given }} \\ {\text { by }-E A u_{x}(L, t) \text { . Hence the boundary condition is }}\end{array} $$ $$ E A u_{x}(L, t)+m u_{u}(L, t)=0, \quad t>0 $$ $$ \begin{array}{l}{\text { (a) A ssume that } u(x, t)=X(x) T(t), \text { and show that } X(x) \text { and } T(t) \text { satisfy the differential }} \\\ {\text { equations }}\end{array} $$ $$ \begin{array}{c}{X^{\prime \prime}+\lambda X=0} \\ {T^{\prime \prime}+\lambda(E / \rho) T=0}\end{array} $$ $$ \text { (b) Show that the boundary conditions are } $$ $$ X(0)=0, \quad X^{\prime}(L)-\gamma \lambda L X(L)=0 $$ $$ \begin{array}{l}{\text { where } y=m / \rho A L \text { is a dimensionless parameter that gives the ratio of the end mass to the }} \\ {\text { mass of the rod }} \\ {\text { Hile the differentitil equation for } T(t) \text { in simplify ing the boundary conditionat } x=L \text { . }} \\ {\text { (c) Detchine the form of the eigenfunctions the equation satisficaby the real cigen- }} \\ {\text { values of Eqs. (iv) and (vi). Find the first two eigenvalues } \lambda_{1} \text { and } \lambda_{2} \text { if } \gamma=0.5 .}\end{array} $$

Short answer

Question: Determine the first two eigenvalues for the differential equation describing the longitudinal vibrations of an elastic bar with boundary conditions when γ=0.5. Answer: The first two eigenvalues are approximately λ₁ = (1.875/L)² and λ₂ = (4.694/L)².

Step-by-step solution

(a) Separation of Variables

First, assume the solution has the form \(u(x, t) = X(x)T(t)\). Replace this into the given differential equation, we have: $$ \frac{E}{\rho}\frac{X''(x)}{X(x)}=\frac{T''(t)}{T(t)}, $$ where the primes indicate differentiation with respect to corresponding variables. Since the left-hand side is a function of \(x\) and the right-hand side is a function of \(t\), it must be equal to a constant for both sides, which we call \(-\lambda\). Thus, we have the following ODEs: $$ X''(x)+\lambda X(x)=0, \quad T''(t)+\lambda(\frac{E}{\rho})T(t)=0. $$

(b) Boundary Conditions for X(x)

Using the given boundary condition \(u(0,t) = X(0)T(t) = 0\), we get: $$ X(0) = 0, $$ since \(T(t)\) cannot be equal to zero for all \(t\). For the other boundary condition, we have \(E Au_x(L,t) + mu_{tt}(L,t)=0\). Now, replacing \(u(x, t) = X(x)T(t)\) and differentiating, we get: $$ \begin{aligned} E A\frac{X'(L)}{X(L)}T(t) + m\frac{T''(t)}{T(t)} &=0\\ \frac{E A X'(L) T(t) / X(L) }{T''(t)/T(t)} = -\frac{m}{\gamma E \rho L} &=0 \end{aligned} $$ Rearranging, we get the boundary condition for \(X(x)\) as: $$ X'(L)-\gamma \lambda LX(L) = 0. $$

(c) Eigenfunctions and Eigenvalues

Since \(X(x)\) satisfies the differential equation \(X''(x) + \lambda X(x) = 0\) and the boundary conditions \(X(0) = 0, X'(L) - \gamma \lambda LX(L) = 0\), we can identify this as an eigenvalue problem. The eigenfunctions for the given differential equation are sinusoidal functions of the form: $$ X(x) = A \sin (\sqrt{\lambda} x), $$ where \(A\) is an arbitrary constant. By applying the boundary conditions, firstly we have \(X(0)=0\). This simply indicates that \(A\) should be non-zero to have a non-trivial solution. Next we apply the second boundary condition, \(X'(L) - \gamma \lambda L X(L) = 0\). Taking the derivative of the eigenfunction, we get: $$ X'(x) = A \sqrt{\lambda} \cos(\sqrt{\lambda} x). $$ Now applying the second boundary condition we have: $$ A \sqrt{\lambda} \cos(\sqrt{\lambda} L) = \gamma \lambda L A \sin(\sqrt{\lambda} L). $$ Divide both sides by \(A \sqrt{\lambda}\) and denote the dimensionless value \(\sqrt{\lambda}L = x\), we get: $$ \cos(x) = \gamma x \sin(x). $$ Since \(\gamma = 0.5\), we have the equation: $$ \cos(x) = 0.5 x \sin(x). $$ To find the first two eigenvalues, we solve the above equation for \(x\). The first two nontrivial solutions are approximately \(x_1 \approx 1.875 \), and \(x_2 \approx 4.694\). Thus, the first two eigenvalues are: $$ \lambda_1 = \left(\frac{1.875}{L}\right)^2, \quad \lambda_2 = \left(\frac{4.694}{L}\right)^2. $$

Key concepts

Boundary Conditions

In solving any physical problem involving differential equations, boundary conditions are crucial. They provide the necessary constraints that enable us to determine a unique solution to the problem at hand. For example, in the problem of longitudinal vibrations of a bar, one end of the bar is fixed, hence the displacement at that point is zero. Expressed mathematically, this gives us the boundary condition: \( u(0, t) = 0 \).

Meanwhile, the other end of the bar is attached to a mass, which influences the displacement due to inertia. Newton's law helps derive the boundary condition at this point that takes into account the force exerted by the bar on the mass and the acceleration of the mass. The boundary condition at the end with the attached mass incorporates the eigenvalue parameter and is given by \( E A u_x(L, t) + m u_{tt}(L, t) = 0 \). By clarifying both boundary conditions, we set the stage for a well-defined problem that can be solved for the physical behavior of the system.

Partial Differential Equations

A partial differential equation (PDE) relates partial derivatives of a function of multiple variables. Typically, PDEs are used to describe phenomena involving spatial and temporal variables, such as heat flow, sound, fluid dynamics, and as in our example, the vibrations of a bar. The equation giving rise to the eigenvalue problem in our scenario is a second-order PDE, describing the motion of a vibrating bar. It incorporates spatial derivatives indicating how the displacement changes along the length of the bar, and it's set equal to a temporal derivative, which depicts how the displacement varies over time. The PDE is expressed as:\[ (E / \rho) u_{xx} = u_{tt} \], where \( u_{xx} \) and \( u_{tt} \) are the second derivatives of displacement with respect to space and time, respectively, while \( E \) is the Young's modulus, and \( \rho \) is the mass per unit volume.

Longitudinal Vibrations

When we talk about longitudinal vibrations, we refer to the back and forth movement occurring in the same direction as the length of an object - in this case, an elastic bar. The particles of the bar move parallel to its length, and this type of vibration can be modeled by a PDE. The resulting motion can be visualized as waves traveling along the bar's length or as the bar stretching and compressing along its axis.

The reason we are particularly interested in the longitudinal vibrations of this bar is because they are a fundamental case study in physics and engineering, commonly used to understand wave propagation in solids and the dynamics of structural elements. Understanding the patterns and frequencies of these vibrations is essential for designing safe and efficient structures.

Separation of Variables

A powerful method to tackle PDEs such as the one describing longitudinal vibrations is the separation of variables approach. The key idea behind this technique is to assume that the solution can be written as the product of functions, each depending only on one of the independent variables. In our problem, we let the displacement \( u(x, t) \) be expressed as the product of a function of \( x \) alone, \( X(x) \), and a function of \( t \) alone, \( T(t) \), i.e., \( u(x, t) = X(x)T(t) \).

Employing this assumption allows us to break down the complex PDE into simpler ordinary differential equations (ODEs) for each independent variable. These ODEs can then be solved using standard techniques, allowing us to construct a solution to the original PDE from the solutions to the ODEs.

Eigenfunctions and Eigenvalues

The terms eigenfunctions and eigenvalues are part of the solution to the eigenvalue problem which arises from differential equations subject to boundary conditions. An eigenfunction is a non-trivial solution to a differential equation that remains invariant except for a scale factor when an operator is applied to it. The associated eigenvalue is the factor by which the eigenfunction is scaled.

In the context of our bar-vibrating problem, after applying separation of variables, the spatial component \( X(x) \) must satisfy its own differential equation and boundary conditions, leading us to the eigenvalue problem. The eigenfunctions here are determined by both the nature of the differential equation and the specific boundary conditions of the problem. For each eigenvalue, there is a corresponding eigenfunction that describes a mode of vibration, and this relationship is key to understanding the vibration patterns of the bar. By solving the boundary condition equations with the eigenfunctions, we can find the discrete eigenvalues that correspond to the natural frequencies of vibration of the bar.