Question

Solve the eigenvalue problem. $$ y^{\prime \prime}+\lambda y=0, \quad y^{\prime}(0)=0, \quad \int_{0}^{L} y(x) d x=0 $$

Short answer

Answer: The eigenvalue for this problem is λ = 0, and the corresponding eigenfunction is \(y(x) = A(x-L)x\). This eigenfunction satisfies the orthogonality condition because its integral from 0 to L is equal to zero.

Step-by-step solution

Identify the type of differential equation

The given equation is: $$ y^{\prime\prime} + \lambda y = 0 $$ This is a second-order linear homogeneous ordinary differential equation with constant coefficients.

Solve the differential equation

We assume a solution of the form: $$ y(x) = e^{rx} $$ where r is a constant to be determined. Differentiating this assumed solution twice with respect to x, we obtain: $$ y^{\prime}(x) = re^{rx}, \quad y^{\prime\prime}(x) = r^2e^{rx} $$ Substituting these expressions into the differential equation: $$ (r^2e^{rx}) + \lambda (e^{rx}) = 0 $$ Factor out the e^{rx}: $$ e^{rx}(r^2 + \lambda) = 0 $$ Since the exponential function is never zero, we have: $$ r^2 + \lambda = 0 $$

Apply the boundary conditions

We first apply the boundary condition y'(0) = 0: $$ y^{\prime}(0) = re^{r \cdot 0} = r = 0 $$ Now we substitute r back into the equation \(r^2 + \lambda = 0\): $$ \lambda = -r^2 = 0 $$ Thus, λ can only be zero in this case.

Solve for the eigenfunctions

With λ = 0, the differential equation becomes: $$ y^{\prime\prime} = 0 $$ Integrating twice w.r.t x yields: $$ y(x) = Ax + B $$ We need to satisfy the orthogonality condition: $$ \int_{0}^{L}(Ax + B)dx = 0 $$ Integrating term by term, we have: $$ \int_{0}^{L}(Ax + B)dx = \left[A\frac{x^2}{2} + Bx\right]_{0}^{L} = 0 $$ Substituting the limits, we have: $$ A\frac{L^2}{2} + BL = 0 $$ Since this orthogonality condition must hold for any choice of A and B, we must choose A and/or B such that the eigenfunction satisfies the eigenvalue problem and the orthogonality condition. We can set A to be any non-zero constant and B to be equal to -AL and we will maintain the zero integral condition.

State the solution

With A as a non-zero constant and B = -AL, we have determined that the eigenvalue λ = 0. The corresponding eigenfunction satisfying the given problem is: $$ y(x) = Ax - ALx = A(x-L)x $$

Key concepts

Understanding Second-Order Differential Equations

When we talk about second-order differential equations, we're referring to a relationship that involves an unknown function, its derivatives, and a given function, in which the highest order derivative is the second derivative. In the exercise, the differential equation in question is homogeneous, meaning it's set to zero, and linear, meaning the equation involves linear combinations of the function and its derivatives. The coefficient of \(y\) and \(y''\) are constants, makingsolutution procedures standardized.
To tackle such equations, assumptions are made about the form of the solution; a common assumption is that the solution is exponential in nature, \(y(x) = e^{rx}\). This form is incredibly useful because its derivatives are proportional to the function itself, which simplifies the substitution back into the original equation. Throughout the solution process, various algebraic techniques are used to deduce information about the constant \(r\), with the aim of finding a general form for \(y(x)\).

Why Exponential Assumptions Work Well

Exponential functions, \(e^{rx}\), have a unique property: their derivatives are also exponential functions, just scaled by a factor of \(r\), thus preserving the form of the equation during differentiations. This self-similarity simplifies the process of finding solutions.

Homogeneous vs Non-Homogeneous

It's important to note that if the equation were non-homogeneous, with a term that isn't zero on one side, the solution process would involve finding a particular solution in addition to the general solution derived from the homogeneous part.

The Role of Boundary Conditions

To find a specific solution to a second-order differential equation, we need more information—this comes in the form of boundary conditions. These conditions provide constraints that our solution must satisfy, usually dictated by the physical context or geometric situation the problem represents. In our exercise, the boundary condition given is \(y'(0) = 0\), which puts a restriction on the value of the derivative of our function at a particular point.
Applying boundary conditions often involves substituting initial or specific values into the assumed solution or its derivatives and solving for the constants involved. This tailors the general solution to a particular context, effectively 'pinning' the curve of the solution to specified points or directions, resulting in a solution that fits the given scenario perfectly.

Different Types of Boundary Conditions

Boundary conditions can be of several types, such as Dirichlet, where the values of the function are specified at the boundaries, or Neumann, where derivative values are given. The type used significantly influences the characteristics of the solution.

Physical Significance

Boundary conditions are not arbitrary; they usually represent real-world constraints like fixed ends of a vibrating string, temperature at the surface of an object, or velocity of a fluid at a wall. Understanding their physical meaning is crucial for correctly applying them to the equation.

Orthogonality Condition in Eigenvalue Problems

The orthogonality condition often appears in the context of eigenvalue problems, like the one presented in our exercise. It is a mathematical expression of the requirement that two functions be perpendicular to each other in a certain sense. When discussing functions, being 'perpendicular' is analogous to their inner product being zero—just as two perpendicular vectors in space have a dot product of zero.
In this case, the inner product of a function is defined by the integral over a range. The condition \(\int_{0}^{L} y(x) dx = 0\) essentially says that the area under the curve described by \(y(x)\) over the interval from \(0\) to \(L\) sums to zero. This could represent a balance of mass or charge distribution, or other physical contexts in engineering and physics.

Implications of the Orthogonality Condition

This condition can restrict the forms that the solution \(y(x)\) can take, potentially leading to specific values of constants in the solution. In eigenvalue problems, it is a key concept that allows us to find sets of functions (eigenfunctions) that are mutually orthogonal. This is incredibly useful, particularly in applications like Fourier series, quantum mechanics, and vibrations analysis, which rely on decomposing complex phenomena into simpler, orthogonal components.

Physical Interpretation

In physical terms, orthogonality has interpretations depending on the scenario—acoustic waves in a cavity, for instance, may have modes that are orthogonal so they don't interfere with each other. When engineers or physicists look for such modes, they are seeking functions that satisfy not just the governing differential equation, but also the orthogonality condition relative to the specific problem's geometry and boundary conditions.