Question

Deal with the Sturm-Liouville problem $$ y^{\prime \prime}+\lambda y=0, \quad \alpha y(0)+\beta y^{\prime}(0), \quad \rho y(L)+\delta y^{\prime}(L)=0 $$ where \(\alpha^{2}+\beta^{2}>0\) and \(\rho^{2}+\delta^{2}>0\) Show that \(\lambda=0\) is an eigenvalue of (SL) if and only if $$ \alpha(\rho L+\delta)-\beta \rho=0 $$

Short answer

Based on the solution provided above, create a short answer question for the exercise: Question: Prove that λ = 0 is an eigenvalue of the Sturm-Liouville problem below if and only if the given condition holds: Sturm-Liouville problem: $$ y^{\prime\prime}+\lambda y=0, \\ \alpha y(0)+\beta y^{\prime}(0)=0, \\ \rho y(L)+\delta y^{\prime}(L)=0 $$ Condition: $$ \alpha(\rho L+\delta)-\beta\rho=0 $$

Step-by-step solution

Write the given Sturm-Liouville equation and boundary conditions

The Sturm-Liouville problem is given by: $$ y^{\prime\prime}+\lambda y=0, \\ \alpha y(0)+\beta y^{\prime}(0)=0, \\ \rho y(L)+\delta y^{\prime}(L)=0 $$

Solve the equation for the case λ = 0

When λ = 0, the equation becomes: $$ y^{\prime\prime}=0 $$ Integrating both sides twice with respect to x yields: $$ y(x)=Ax + B, $$ where A and B are constants.

Apply the first boundary condition

Substitute x = 0 into the equation and apply the first boundary condition: $$ \alpha y(0)+\beta y^{\prime}(0)=\alpha B + \beta A = 0, $$

Apply the second boundary condition

Substitute x = L into the equation and apply the second boundary condition: $$ \rho y(L)+\delta y^{\prime}(L)=\rho (AL + B) + \delta A = 0. $$

Show that the condition must hold for λ = 0 to be an eigenvalue

From the first boundary condition in Step 3, we can write A in terms of B: $$ A=-\frac{\alpha}{\beta}B. $$ Now substitute this expression for A in the second boundary condition: $$ \rho\left(-\frac{\alpha}{\beta}BL + B\right) + \delta\left(-\frac{\alpha}{\beta}B\right) = 0. $$ Now, we simply need to simplify this equation to reach the desired result: $$ \left(-\alpha\rho L - \alpha\delta +\beta\rho\right)B=0, $$ Since we need a non-trivial solution (B ≠ 0) for λ = 0 to be an eigenvalue, the condition must be: $$ \alpha(\rho L+\delta)-\beta\rho=0. $$ Thus, we have proved that λ = 0 is an eigenvalue of the Sturm-Liouville problem iff the given condition holds true.

Key concepts

Eigenvalues

In the context of the Sturm-Liouville problem, eigenvalues play a crucial role. Eigenvalues are special numbers associated with differential equations that reveal many properties about the solutions.
In this exercise, eigenvalues are represented by the symbol \( \lambda \), which appears in the main differential equation of the problem. When \( \lambda = 0 \), it simplifies the equation considerably.
For \( \lambda = 0 \) to be an eigenvalue, certain conditions must be satisfied. The boundary conditions, combined with solving the differential equation, determine whether \( \lambda = 0 \) qualifies as an eigenvalue.
In this scenario, \( \lambda = 0 \) is indeed an eigenvalue if the specific condition \( \alpha(\rho L+\delta)-\beta \rho=0 \) holds true. Understanding eigenvalues help to solve and analyze the behavior of solutions in differential equations like these.

Boundary Conditions

Boundary conditions are essential in solving differential equations. They define values at the boundaries of the interval where the solution is sought.
In our problem, we encounter two boundary conditions:

• \( \alpha y(0) + \beta y'(0) = 0 \)
• \( \rho y(L) + \delta y'(L) = 0 \)

These conditions are crucial because they help in determining the constants involved in the solution of the differential equation. Each boundary condition is a mathematical expression that the solution must satisfy at specific points, often at the start and end of an interval.
For the exercise, these conditions are applied to simplify and rearrange the differential equation until they match up with known resembling structures, further emphasizing their importance. They are key to establishing whether \( \lambda = 0 \) is an eigenvalue.

Differential Equations

Differential equations form the backbone of the Sturm-Liouville problem. They relate functions with their derivatives and are used to describe physical phenomena.
In this case, the differential equation is \( y'' + \lambda y = 0 \). When \( \lambda = 0 \), the equation simplifies to \( y'' = 0 \), indicating that the second derivative of the solution is zero, suggesting a linear function as a solution.
Integrating this equation twice, we derive the general solution \( y(x) = Ax + B \), where \( A \) and \( B \) are constants. These constants are then determined by applying the boundary conditions.
Differential equations are key to understanding how changes happen within a system, and when complemented with boundary conditions, they help in finding specific and meaningful solutions that satisfy all aspects of the problem at hand.