Question

deal with column buckling problems. In some buckling problems the eigenvalue parameter appears in the boundary conditions as well as in the differential equation. One such case occurs when one end of the column is clamped and the other end is free. In this case the differential equation \(y^{i v}+\lambda y^{\prime \prime}=0\) must be solved subject to the boundary conditions $$ y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(L)=0, \quad y^{\prime \prime \prime}(L)+\lambda y^{\prime}(L)=0 $$ Find the smallest eigenvalue and the corresponding eigenfunction.

Short answer

Answer: The smallest eigenvalue is 1, and the corresponding eigenfunction is \(y(x) = x(Ae^{ix}+Be^{-ix})\).

Step-by-step solution

Write down the given differential equation and boundary conditions

We are given the following problem: Differential equation: \(y^{i v}+\lambda y^{\prime\prime}=0\) Boundary conditions: $$ y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(L)=0, \quad y^{\prime\prime\prime}(L)+\lambda y^{\prime}(L)=0 $$

Try a trial solution

Let's assume a trial solution of the form: $$ y(x)=e^{rx} $$ where r is an unknown constant. We will differentiate the trial solution to find \(y^{\prime}\), \(y^{\prime\prime}\), \(y^{\prime\prime\prime}\), and \(y^{iv}\), and substitute them into the differential equation.

Find the characteristic equation

Differentiating the trial solution, we have: $$ y^{\prime}(x)=re^{rx}, \quad y^{\prime\prime}(x)=r^2e^{rx}, \quad y^{\prime\prime\prime}(x)=r^3e^{rx}, \quad y^{iv}(x)=r^4e^{rx} $$ Substituting these into the differential equation, we get: $$ r^4e^{rx}+\lambda r^2e^{rx}=0 $$ Divide by \(r^2e^{rx}\), assuming \(r \neq 0\) and \(e^{rx}\neq 0\), we obtain the characteristic equation: $$ r^2+\lambda=0 $$

Solve the characteristic equation

Solving the characteristic equation, we get: $$ r=\pm\sqrt{-\lambda} $$ An important thing to notice here is that the eigenvalue \(\lambda\) must be positive in order to get real roots for r. Thus, we have two real roots for r, and the general solution is: $$ y(x)=Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x} $$

Apply the boundary conditions

First, let's apply the boundary conditions \(y(0)=0\) and \(y^{\prime}(0)=0\): $$ y(0)=Ae^{0}+Be^{0}=A+B=0\quad\Longrightarrow\quad B = -A \\ y^{\prime}(0)=A\sqrt{-\lambda}e^{0}-B\sqrt{-\lambda}e^{0}=\sqrt{-\lambda}(A-B)=0\quad\Longrightarrow\quad A = B = 0 $$ We have a contradiction. The only solution is the trivial one, \(A = B = 0\). We need to change our trial solution. Since we are looking for non-trivial solutions, we try a solution of the form: $$ y(x)=x(Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x}) $$

Reapply the boundary conditions

We apply the boundary conditions again: $$ y(0)=0( Ae^{0}+Be^{0})=0\\ y^{\prime}(x)=Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x}+\sqrt{-\lambda}x(Ae^{\sqrt{-\lambda}x}-Be^{-\sqrt{-\lambda}x}) \\ y^{\prime}(0)=A+B=0\quad\Longrightarrow\quad B = -A \\ y^{\prime \prime}(x)=-\lambda(Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x})+\sqrt{-\lambda}(Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x}) \\ y^{\prime\prime}(L)=0 \\ y^{\prime\prime\prime}(x)=\lambda\sqrt{-\lambda}(Ae^{\sqrt{-\lambda}x}-Be^{-\sqrt{-\lambda}x})+\sqrt{-\lambda}(Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x}) \\ y^{\prime\prime\prime}(L)+\lambda y^{\prime}(L)=0 $$

Solve the boundary conditions for eigenvalue and eigenfunction

To find the smallest eigenvalue, we need to solve the system of equations for \(y^{\prime\prime}(L)=0\) and \(y^{\prime\prime\prime}(L)+\lambda y^{\prime}(L)=0\). We have: $$ (-\lambda+1)(Ae^{\sqrt{-\lambda}L}-Be^{-\sqrt{-\lambda}L})=0 \\ (1-\lambda\sqrt{-\lambda})(Ae^{\sqrt{-\lambda}L}+Be^{-\sqrt{-\lambda}L})=0 $$ Since B = -A, we can rewrite the equations as: $$ (-\lambda+1)(2Ae^{\sqrt{-\lambda}L})=0 \\ (1-\lambda\sqrt{-\lambda})(0)=0 $$ Since the second equation is an identity, we need to find the smallest value of \(\lambda\) that satisfies the first equation. Thus, we need to find the smallest positive value of \(\lambda\) for which \(-\lambda+1 = 0\). This value is \(\lambda = 1\). The corresponding eigenfunction is: $$ y(x)=x(Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x})=x(Ae^{\sqrt{-1}x}+Be^{-\sqrt{-1}x}) $$ So, the smallest eigenvalue is 1, and the corresponding eigenfunction is \(y(x)=x(Ae^{ix}+Be^{-ix})\).

Key concepts

Eigenvalue Problems

Understanding eigenvalue problems is fundamental when dealing with physical scenarios such as column buckling, where stability is a key concern. An eigenvalue problem is a type of mathematical situation in an equation where a differential operator is applied to a function (called an eigenfunction), resulting in the original function scaled by a factor (the eigenvalue). In the context of column buckling, eigenvalues represent the load factors that induce a state of buckling, signifying a critical transition from stability to instability.

In the given exercise, we are asked to solve an eigenvalue problem where the differential equation includes an eigenvalue parameter, and this parameter is also present in the boundary conditions. The goal is to find the smallest eigenvalue that satisfies both the differential equation and the boundary conditions, as this represents the lowest load at which the column will start to buckle. Solving such problems typically involves proposing a trial solution, substituting it into the given conditions, and then fine-tuning the trial solution until all conditions are met.

Boundary Conditions

Boundary conditions are essential constraints that are required for the unique solution of differential equations that describe physical situations like column buckling. They represent the physical constraints or behaviors at the boundaries of the domain over which a problem is defined. There are several types of boundary conditions, such as Dirichlet, Neumann, or mixed, each specifying fixed values, derivatives, or a combination of both.

In the column buckling example, we are provided with four boundary conditions, two at the point where the column is clamped, representing zero displacement and zero slope, and two at the free end, corresponding to zero bending moment and a condition that combines deflection and rotation at the end. Correct application of these boundary conditions leads to a system of equations that define the possible eigenvalues, specifying the critical loads under which the column will buckle.

Differential Equations

Differential equations are a type of equation that involves derivatives, representing how a particular quantity changes with respect to another. Within the context of engineering and physics, these equations are pivotal as they are used to model a wide array of phenomena, including the behavior of columns under load, the motion of celestial bodies, or the flow of fluids. They can be simple or highly complex, involving multiple variables and derivatives, as well as nonlinear terms.

In the context of the column buckling problem, we are presented with a fourth-order differential equation, indicating that the physical system is influenced by factors that necessitate consideration of the rate of change up to the third derivative of displacement. This high-order differential equation arises from the relationship between bending moments and deflection in beams and columns. To tackle such problems, engineers and mathematicians must utilize trial solutions that might initially satisfy the differential equation. Integrating these solutions with the appropriate boundary conditions leads to a better understanding of the critical points at which structures like columns will begin to fail.