Question

In Exercises 26-30 find necessary and sufficient conditions on \(\alpha, \beta, \rho,\) and \(\delta\) for the boundary value problem to have a unique solution for every continuous \(F,\) and find the Green's function. $$ y^{\prime \prime}-2 y^{\prime}+2 y=F(x), \quad \alpha y(0)+\beta y^{\prime}(0)=0, \quad \rho y(\pi / 2)+\delta y^{\prime}(\pi / 2)=0 $$

Short answer

2. What is the Green's function for this problem? Answer: 1. The necessary and sufficient conditions on α, β, ρ, and δ for the boundary value problem to have a unique solution for every continuous F are α + β = 0 and ρ + δ = 0. 2. The Green's function for this problem is given by: $$ G(x) = \begin{cases} -e^x \sin(x) & \text{for } 0 \le x \le \pi/2 \\ 0 & \text{for } \pi/2 \le x \le \pi \end{cases} $$

Step-by-step solution

Analyze the Boundary Value Problem

The given boundary value problem is: $$ y^{\prime \prime}-2 y^{\prime}+2 y=F(x), \quad \alpha y(0)+\beta y^{\prime}(0)=0, \quad \rho y(\pi/2)+\delta y^{\prime}(\pi/2)=0 $$ Notice that the given ODE has a unique solution if the associated homogeneous equation, $$ y^{\prime \prime}-2 y^{\prime}+2 y=0 $$ has a unique solution satisfying the boundary conditions for every continuous \(F(x)\).

Find the General Solution of the Homogeneous Equation

To find the general solution of the homogenous equation, solve its characteristic equation: $$ r^2-2r+2=0 $$ The characteristic equation has complex roots \(r_{1,2}=1\pm i\). Hence, the general solution of the homogenous equation is: $$ y_h(x) = e^x (C_1 \cos(x) + C_2 \sin(x)) $$

Apply Boundary Conditions to the Homogeneous Solution

Apply the boundary conditions to the general solution \(y_h(x)\): $$ \alpha y(0)+\beta y^{\prime}(0)=0 \\ \rho y(\pi/2)+\delta y^{\prime}(\pi/2)=0 $$ This gives us: $$ \alpha [e^0 (C_1 \cos(0) + C_2 \sin(0))] + \beta [e^0 (C_1 \cos(0) - C_2 \sin(0))] = 0 \\ \rho [ e^{(\pi/2)}(C_1 \cos(\pi/2)+C_2 \sin(\pi/2))]+\delta [ e^{(\pi/2)}(-C_1 \sin(\pi/2) + C_2 \cos(\pi/2))]=0 $$ Simplify and solve for \(C_1\) and \(C_2\): $$ (\alpha + \beta)C_1=0 \\ \rho C_2 + (\rho+\delta)C_1=0 $$ For a unique solution to exist for every continuous \(F(x)\), we need \(C_1=C_2=0\). This implies that \(\alpha+\beta = 0\) and \(\rho+\delta = 0\).

Find the Green's Function

We've found that \(\alpha+\beta = 0\) and \(\rho+\delta = 0\). To find the Green's function, we first consider the associated homogeneous equation with a delta function source term: $$ G^{\prime \prime}(x)-2G^{\prime}(x)+2 G(x)=\delta(x) $$ Assuming the boundary conditions \(G(0) = G(\pi/2) = 0\), we can find the Green's function for this problem by applying the same steps as before (Steps 2-3). The Green's function is given by: $$ G(x) = \begin{cases} A_1 e^x \sin(x) & \text{for } 0 \le x \le \pi/2 \\ A_2 e^x \sin(x) & \text{for } \pi/2 \le x \le \pi \end{cases} $$ Applying boundary conditions and continuity conditions, we find \(A_1=-1\) and \(A_2=0\). Thus, the Green's function is: $$ G(x) = \begin{cases} -e^x \sin(x) & \text{for } 0 \le x \le \pi/2 \\ 0 & \text{for } \pi/2 \le x \le \pi \end{cases} $$ In summary, the necessary and sufficient conditions on \(\alpha, \beta, \rho,\) and \(\delta\) for the boundary value problem to have a unique solution for every continuous \(F\) are \(\alpha+\beta = 0\) and \(\rho+\delta = 0\). The Green's function for this problem is given by: $$ G(x) = \begin{cases} -e^x \sin(x) & \text{for } 0 \le x \le \pi/2 \\ 0 & \text{for } \pi/2 \le x \le \pi \end{cases} $$

Key concepts

Homogeneous Differential Equation

In the realm of differential equations, a homogeneous differential equation is an equation in which every term is a function of the unknown variable and its derivatives.

Consider the equation given in the exercise: \[ y^{\text{\textquotesingle \textquotesingle }} - 2y^{\text{\textquotesingle }} + 2y = F(x) \] The corresponding homogeneous equation is obtained by setting the non-homogeneous part, here represented by \( F(x) \), to zero: \[ y^{\text{\textquotesingle \textquotesingle }} - 2y^{\text{\textquotesingle }} + 2y = 0 \] This equation states that any linear combination of derivatives up to the second order, when the function itself is weighted continuously, sums to zero. A key property of the solutions to the homogeneous equation is that they can be freely scaled or added to form new solutions.

In solving boundary value problems, one typically solves the homogeneous equation first to ascertain certain properties of the solutions. These properties include the behavior at initial or boundary conditions and help set up the stage for tackling the non-homogeneous equation.

Characteristic Equation

The characteristic equation is a vital tool for finding solutions to linear homogeneous differential equations with constant coefficients. It converts the problem of solving a differential equation into an algebraic one.

In the provided exercise, the characteristic equation is derived from the homogeneous differential equation by replacing \( y^{\text{\textquotesingle }} \) with \( r \) and \( y^{\text{\textquotesingle \textquotesingle }} \) with \( r^2 \), resulting in: \[ r^2 - 2r + 2 = 0 \] Solving this equation yields the roots, which in this case are complex numbers \( r_{1,2} = 1 \text{\textpm} i \). These roots are crucial in determining the general solution to the original homogeneous equation, as they suggest a combination of exponential, sine, and cosine functions to form the solution set.

Boundary Conditions

The concept of boundary conditions is central to ensuring a unique solution in a boundary value problem. These conditions are additional requirements that a solution to a differential equation must satisfy at specific points or boundaries.

The exercise provided includes two boundary conditions specified at \( x = 0 \) and \( x = \frac{\text{\textpi}}{2} \). They impose constraints on the values of the solution and its derivative at these points. When we introduce these constraints into the general solution of the homogeneous equation, we can potentially determine the arbitrary constants, leading to a unique solution that satisfies the entire set of posed conditions.

Uniqueness of Solution

The uniqueness of solution ensures that, under certain conditions, a boundary value problem has exactly one solution. This property is particularly important because it guarantees that the system described by the problem behaves deterministically.

For the problem in the exercise, a unique solution exists if it is the case for the corresponding homogeneous equation with the given boundary conditions for every continuous function \( F(x) \). The uniqueness can be dictated by parameters such as \( \text{\textalpha}, \text{\textbeta}, \text{\textrho}, \text{ and } \text{\textdelta} \), which must satisfy certain relationships so that the constants \( C_1 \) and \( C_2 \) that appear in the homogeneous solution are uniquely defined. These relationships ensure that the solution space collapses to a single function that meets the specified requirements.

Continuous Function

A continuous function in mathematical analysis is one in which small changes in the input produce small changes in the output. In the context of differential equations and specifically in the exercise at hand, this concept relates to the non-homogeneous term \( F(x) \) of the differential equation.

Importance in Boundary Value Problems

The stipulation that \( F(x) \) be continuous over the interval of interest is important for the existence and uniqueness of solutions. This is because discontinuities can lead to abrupt changes in the behavior of the solution, potentially causing multiple solutions or none at all.

The Green's function method, as applied in the exercise, requires the continuous function \( F(x) \) to construct the particular solution to the non-homogeneous equation. Therefore, the continuity of \( F(x) \) guarantees that the Green's function properly accounts for the inhomogeneities throughout the domain.