Question

Find the Green's function for the boundary value problem $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-1 / 4\right) y=F(x), \quad y(\pi / 2)=0, \quad y(\pi)=0 $$ given that $$ y_{1}(x)=\frac{\cos x}{\sqrt{x}} \text { and } y_{2}(x)=\frac{\sin x}{\sqrt{x}} $$ are solutions of the complementary equation. Then use the Green's function to solve (A) with (a) \(F(x)=x^{3 / 2}\) and (b) \(F(x)=x^{5 / 2}\)

Short answer

#Answer# To find the Green's function and use it to solve the given boundary value problem for F(x) = x^(3/2) and F(x) = x^(5/2), we need to follow these steps: Step 1: Verify that the given functions y₁(x) = cos(x)/√(x) and y₂(x) = sin(x)/√(x) are solutions of the complementary equation. Step 2: Construct the Wronskian using the given solutions. W(x) = (cos(x)/√x)(sin(x)/√x)' - (sin(x)/√x)(cos(x)/√x)' Step 3: Find the Green's function using the Wronskian. G(x,ξ) = { y₁(x)y₂(ξ)/W(ξ) for a≤x≤ξ, y₂(x)y₁(ξ)/W(ξ) for ξ≤x≤b, } Step 4: Use the Green's function to solve the boundary value problem for both given F(x). For F(x) = x^(3/2): y(x) = ∫(from π/2 to π) G(x,ξ)F(ξ) dξ For F(x) = x^(5/2): y(x) = ∫(from π/2 to π) G(x,ξ)F(ξ) dξ Calculating the integrals in Step 4 will provide the final solutions y(x) for both cases.

Step-by-step solution

Verify that the given functions are solutions of the complementary equation

First, we need to verify that the given functions, y₁(x) = cos(x)/√(x) and y₂(x) = sin(x)/√(x), are solutions of the complementary equation. To do this, substitute each function into the given differential equation and confirm if the resulting equation is an identity. L(y₁) = x²y₁'' + xy₁' + (x² - ¼)y₁ = L(y₂) = x²y₂'' + xy₂' + (x² - ¼)y₂ = 0 Once we verify that both functions are solutions of the complementary equation, we can proceed to find the Wronskian.

Construct the Wronskian using the given solutions

The Wronskian of two functions y₁(x) and y₂(x) is defined as: W(x) = y₁(x)y₂'(x) - y₂(x)y₁'(x) Computing the derivatives and substituting them into the Wronskian, we get: W(x) = y₁(x) y₂'(x) - y₂(x)y₁'(x) = (cos(x)/√x)(sin(x)/√x)' - (sin(x)/√x)(cos(x)/√x)' Calculate W(x) in order to find the Green's function.

Find the Green's function using the Wronskian

The Green's function for a boundary value problem involving two solutions y₁(x) and y₂(x) and their Wronskian W(x) can be expressed as: G(x,ξ) = { y₁(x)y₂(ξ)/W(ξ) for a≤x≤ξ, y₂(x)y₁(ξ)/W(ξ) for ξ≤x≤b, } In our case, a = π/2 and b = π. Substitute the functions y₁(x), y₂(x), and W(ξ) into the Green's function formula.

Use the Green's function to solve the boundary value problem for both given F(x)

Now that we have the Green's function G(x,ξ), we can solve the boundary value problem for both F(x) = x^(3/2) and F(x) = x^(5/2). To solve the problem for F(x) = x^(3/2): y(x) = ∫(from π/2 to π) G(x,ξ)F(ξ) dξ To solve the problem for F(x) = x^(5/2): y(x) = ∫(from π/2 to π) G(x,ξ)F(ξ) dξ Calculate these integrals to obtain the final solutions y(x) for both cases.

Key concepts

Boundary Value Problem

A boundary value problem (BVP) in mathematics is a differential equation or system of equations that must satisfy certain conditions at the boundaries of the domain. Typically, these boundaries are represented by specific values for the independent variable (like x) where the solutions must meet predefined criteria.

• Consider a domain from \(a\) to \(b\).
• The solution \(y(x)\) must satisfy the differential equation.
• It also needs to meet conditions such as \(y(a)=c_1\) and \(y(b)=c_2\).

In the original problem, the boundary value problem is defined by a second-order differential equation with conditions \((y(\pi/2)=0)\) and \((y(\pi)=0)\). These conditions ensure the solution is not just locally applicable but confirms the behavior of the solution across the whole domain from \(\pi/2\) to \(\pi\). This ensures the problem is well-posed, meaning that a solution certainly exists, is unique, and exhibits continuous dependence on the boundary values.

Wronskian

The Wronskian is a determinant used in the analysis of differential equations named after the Polish mathematician Józef Hoene-Wroński. It provides information about the linear independence of a set of functions. For two functions, \(y_1(x)\) and \(y_2(x)\), used in our problem, it is defined as:\[W(x) = \begin{vmatrix}y_1(x) & y_2(x) \y_1'(x) & y_2'(x)\end{vmatrix} = y_1(x)y_2'(x) - y_2(x)y_1'(x)\]

• If \(W(x) \eq 0\) for all \(x\), then the functions are linearly independent.
• This condition of linear independence is crucial for constructing solutions to differential equations, such as Green's functions.

In the context of Green's functions, the Wronskian helps ensure that the solutions \(y_1(x)\) and \(y_2(x)\) form a basis for the solution space, which is essential for finding the particular solution for the boundary value problem.

Complementary Equation

A complementary equation is a homogeneous differential equation associated with a non-homogeneous equation. When faced with a differential equation of form \((Ly = F(x))\), the complementary equation is simply \((Ly = 0)\), with \(L\) being a linear differential operator. Solving this gives the homogeneous solutions, \(y_1(x)\) and \(y_2(x)\), essential for forming the general solution.

• Solutions to the complementary equation, \(y_1(x) = \frac{\cos(x)}{\sqrt{x}}\) and \(y_2(x) = \frac{\sin(x)}{\sqrt{x}}\), are utilized to construct the Green’s function in the original problem.
• The complementary solutions, though solutions to the simpler form, still capture essential characteristics like oscillation or growth of the solution.

Using these solutions with the non-homogeneous part allows us to construct a specific or particular solution which satisfies the entire equation over the defined domain.

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function, providing a way to model systems that change continuously. They can be first order, second order, and so on, depending on the order of the highest derivative. Key types of differential equations include:

• Ordinary Differential Equations (ODEs): Involves functions of a single variable and their derivatives.
• Partial Differential Equations (PDEs): Involves functions of multiple variables and their partial derivatives.
• Linear vs Non-linear: Linear equations' solutions can be superimposed, while non-linear equations are more complex and can exhibit a broader range of behaviors.

The differential equation in the original problem is a second-order linear ODE with variable coefficients and reflects the dynamics requiring careful analysis using Green's functions. The role of the Green's function here is fundamental because it turns the differential equation solution problem into an integral one, where the function acts as a responding impulse influencing every point in the domain.