Question

(a) State a condition on \(a\) and \(b\) such that the boundary value problem $$ y^{\prime \prime}+y=F(x), \quad y(a)=0, \quad y(b)=0 $$ has a unique solution for every continuous \(F\), and find the solution by the method used to prove Theorem 13.1 .3 (b) In the case where \(a\) and \(b\) don't satisfy the condition you gave for (a), state necessary and sufficient on \(F\) such that (A) has a solution, and find all solutions by the method used to prove Theorem 13.1 .4 .

Short answer

Answer: The linearly independent solutions are ψ₁(x) = cos(x) and ψ₂(x) = sin(x). 2. What is the condition on a and b for the given problem to have a unique solution? Answer: The condition is that (b - a) ≠ nπ for any integer n. 3. State the necessary and sufficient conditions on F(x) to have a solution when (b - a) = nπ. Answer: The necessary and sufficient condition is that ∫ₐᵦ F(x) cos(x) dx = 0. 4. How can we find the particular solution y_p(x) using the variation of parameters method? Answer: We can find the particular solution by using the following formula: y_p(x) = -ψ₁(x) ∫ (ψ₂(x)F(x)/w) dx + ψ₂(x) ∫ (ψ₁(x)F(x)/w) dx. 5. How can we find the solution to the problem when (b - a) = nπ using the method of Theorem 13.1.4? Answer: We can find the solution using the following formula: y(x) = C₁ψ₁(x) + C₂ψ₂(x) - ψ₂(x) ∫ (ψ₁(x)F(x)/w) dx., where C₁ and C₂ are arbitrary constants determined from the boundary conditions.

Step-by-step solution

Write down the given boundary value problem

We are given the following boundary value problem: $$ y^{\prime \prime} + y = F(x), \quad y(a)=0, \quad y(b)=0 $$

Check if the given problem satisfies the conditions of Theorem 13.1.3

Theorem 13.1.3 states that a second-order linear homogeneous boundary value problem has a unique solution if the following condition is satisfied: $$ w(a, b)= \begin{vmatrix} \psi_1(a) & \psi_2(a) \\ \psi_1(b) & \psi_2(b) \end{vmatrix} \neq 0 $$ where \(\psi_1(x)\) and \(\psi_2(x)\) are two linearly independent solutions of the homogeneous part of the given differential equation. In this case, the homogeneous part is \(y'' + y = 0\). The corresponding linearly independent solutions are \(\psi_1(x) = \cos(x)\) and \(\psi_2(x) = \sin(x)\). Compute the Wronskian of these solutions:

Compute the Wronskian

We need to compute the determinant of the following matrix: $$ w(a, b) = \begin{vmatrix} \cos(a) & \sin(a) \\ \cos(b) & \sin(b) \end{vmatrix} = \cos(a) \sin(b) - \cos(b) \sin(a) $$ Now, we set this determinant equal to zero and find the condition on a and b:

Find the condition on a and b

Setting the determinant equal to zero, we get: $$ \cos(a) \sin(b) - \cos(b) \sin(a) = 0 $$ This equation holds true if and only if \((b-a) = n\pi\) for some integer n.

Use the method of Theorem 13.1.3 to find the solution

We need to find the particular solution \(y_p(x)\) and the homogeneous solution \(y_h(x)\) for the given problem. For the homogeneous solution, we have \(\psi_1(x) = \cos(x)\) and \(\psi_2(x) = \sin(x)\). Now, we need to find the particular solution using the variation of parameters method, which is as follows: $$ y_p(x) = -\psi_1(x) \int \frac{\psi_2(x)F(x)}{w}dx + \psi_2(x) \int \frac{\psi_1(x)F(x)}{w}dx $$ Substitute the expressions for \(\psi_1(x)\), \(\psi_2(x)\) and w in the expression above.

Write the final solution

The final solution of the boundary value problem is given by the sum of the particular solution \(y_p(x)\) and the homogeneous solution \(y_h(x)\): $$ y(x) = y_p(x) + y_h(x) $$ Part (b)

State necessary and sufficient conditions on F(x) for the case when (b - a) = nπ

When \((b-a) = n\pi\), Theorem 13.1.4 states that the boundary value problem has a solution if and only if the following condition is satisfied: $$ \int_a^b F(x) \psi_1(x) dx = 0 $$ where \(\psi_1(x)\) is a linearly independent solution of the homogeneous part of the differential equation (in this case, \(\psi_1(x) = \cos(x)\)).

Find all solutions using the Theorem 13.1.4 method

The method used in Theorem 13.1.4 to find the solution is as follows: $$ y(x) = C_1\psi_1(x) + C_2\psi_2(x) - \psi_2(x) \int \frac{\psi_1(x)F(x)}{w}dx $$ where \(C_1\) and \(C_2\) are arbitrary constants that can be determined from the boundary conditions. Substitute the expressions for \(\psi_1(x)\), \(\psi_2(x)\), and w in the expression above and find the solution.

Key concepts

Differential Equations

Differential equations are a central concept in mathematics and physics because they describe how things change over time or space. Specifically, they relate a function to its derivatives, expressing a rate of change or a slope of the function at any point. A boundary value problem is a specific type of differential equation where the solution is determined not just by the equation itself, but also by the conditions (or 'boundaries') it must meet at specific points.

For instance, if you're examining the temperature distribution along a metal rod, you can't fully determine it just by knowing how temperature tends to change; you must also know the temperature at the rod’s ends. This transforms the situation into a boundary value problem, with the differential equation describing the general tendency of temperature change and the boundary values anchoring your solution in physical reality.

Wronskian

The Wronskian is a useful tool in the analysis of differential equations, and it's named after the mathematician Hoene Wronski. It’s constructed from the solutions of a differential equation and their derivatives, arranged in a determinant. The value of the Wronskian can tell you a lot about the solutions of the differential equation; if it’s never zero on an interval, the solutions used to construct it are linearly independent on that interval.

Linear independence is crucial because, in the realm of differential equations, it allows us to construct the general solution for a homogeneous equation as a linear combination of these independent solutions. In the context of our exercise, we use the Wronskian to check the uniqueness of the solution to our boundary value problem. If the Wronskian is non-zero, it implies a unique solution exists under certain conditions.

Variation of Parameters

Variation of Parameters is a method to find particular solutions to non-homogeneous differential equations. This technique is based on the idea that a particular solution can be expressed as a variable combination of the homogeneous equation's solutions. Essentially, you're using the form of the homogeneous solutions, but allowing the parameters (constants) to vary to satisfy the non-homogeneous part of the equation.

Once the form of the particular solution is assumed, the variable parameters are found by solving a system of equations derived from the original differential equation. This method requires the use of integrals and is quite powerful because it can handle many different types of source functions, not just simple polynomials or sines and cosines.

Homogeneous Solution

A homogeneous solution to a differential equation is one in which the right-hand side of the equation is zero. These solutions are incredibly significant because they form the backbone of the general solution to the related non-homogeneous problem. When we solve a homogeneous equation, we're essentially finding the 'natural' modes of the system -- how it behaves without any external influences or forcing functions.

In practice, finding the homogeneous solution often involves identifying a set of basis solutions that are linearly independent. Once these solutions are found, any linear combination of them will also be a solution to the homogeneous equation. Thus, the homogeneous solution often involves constants that will be specified by boundary or initial conditions.

Particular Solution

In contrast to the homogeneous solution, a particular solution to a differential equation deals with the actual non-zero right-hand side of the equation, which represents a specific external influence or forcing function. The purpose of finding a particular solution is to see how the system behaves under this specific influence.

The particular solution is only one piece of the puzzle and doesn't include the general behavior described by the homogeneous solution. However, the overall or general solution to a non-homogeneous differential equation is acquired by adding together the particular solution and the homogeneous solution. This combined solution will satisfy both the differential equation and the given boundary conditions.