Question

In each exercise, (a) Can you use Theorem \(11.2\) or Theorem \(11.3\) to decide whether the given boundary value problem has a unique solution? (b) If your answer to part (a) is yes, find the unique solution. (c) If your answer to part (a) is no, use the Fredholm alternative theorem to decide whether the given boundary value problem has a unique solution. (d) If the Fredholm alternative theorem indicates there is a unique solution of the given boundary value problem, find that solution. (e) If the Fredholm alternative theorem indicates the given boundary value problem has either infinitely many solutions or no solution, find all the solutions or state that there are none. $$ \begin{aligned} &y^{\prime \prime}+y=2 \\ &y(0)=8 \\ &y(\pi)+y^{\prime}(\pi)=5 \end{aligned} $$

Short answer

Answer: The unique solution to the given boundary value problem is $$y(x) = 6 \cos x - \sin x + 2(\cos x - 1)$$

Step-by-step solution

Check Applicability of Theorem 11.2 and 11.3

For this boundary value problem: $$ \begin{aligned} y^{\prime \prime}+y=2 \\ y(0)=8 \\ y(\pi)+y^{\prime}(\pi)=5 \end{aligned} $$ Let's check if we can apply Theorem 11.2 or Theorem 11.3.\ Note that neither Theorem 11.2 nor 11.3 can be applied here directly. Therefore, we have to apply the Fredholm Alternative Theorem.

Application of Fredholm Alternative Theorem

First, let's rewrite the given differential equation as a homogeneous equation and the related adjoint equation. For the homogeneous equation: $$ y^{\prime \prime} + y = 0 $$ For the adjoint equation: $$ y^{\prime \prime}* + y* = 0 $$ where \(y*\) is the adjoint function.

Solve the Homogeneous Equation and Adjoint Equation

Solving the homogeneous equation: $$ y^{\prime \prime} + y = 0 $$ Let the characteristic equation be $$ r^2 + 1 = 0 $$ The roots are \(r = \pm i\), so the general solution for the homogeneous equation is $$ y_h = C_1 \cos x + C_2 \sin x $$ Now, solve the adjoint equation: $$ y^{\prime \prime}* + y* = 0 $$ The characteristic equation and its roots are the same as the homogeneous equation, so the solution for the adjoint equation is $$ y* = C_{1*} \cos x + C_{2*} \sin x $$

Apply the Fredholm Alternative Theorem

We will now apply the Fredholm Alternative Theorem to the given boundary value problem. In order to do so, we need to find the Green's function as $$ G(x,t) = \frac{ y_{1}(t) y_{2}(x) - y_{1}(x) y_{2}(t)}{W(t)} $$ Here, \(y_{1}(t) = \cos t\), \(y_{1}(x) = \cos x\), \(y_{2}(t) = \sin t\), \(y_{2}(x) = \sin x\) and the Wronskian \(W(t) = y_{1}(t) y_{2}^{\prime}(t) - y_{1}^{\prime}(t) y_{2}(t) = \sin^{2}t + \cos^{2}t = 1\). So, we have $$ G(x,t) = \sin t \cos x - \cos t \sin x = \sin (t - x) $$ According to the Fredholm Alternative Theorem, if $$ \int_{0}^{\pi} \sin(t-x) [y(\pi) + y^{\prime}(\pi) - 5] dt = 0 $$ for all functions \(y(x)\) satisfying the boundary value problem, the problem has at least one solution.

Finding the Solution

Since the condition from the Fredholm Alternative Theorem is satisfied, we can proceed to finding the particular solution using the variation of parameters method: $$ y_p(x) = - \int_{0}^{\pi} G(x,t) f(t) dt $$ where \(f(t) = 2\). So, we have $$ y_p(x) = -2 \int_{0}^{\pi} \sin(t-x) dt $$ We then integrate by parts to get $$ y_p(x) = 2 (\cos x - 1) $$ Finally, the general solution of the given boundary value problem is $$ y(x) = y_h(x) + y_p(x) = C_1 \cos x + C_2 \sin x + 2(\cos x - 1) $$ Using the boundary conditions \(y(0) = 8\) and \(y(\pi) + y^{\prime}(\pi) = 5\), we can find the constants \(C_1\) and \(C_2\): $$ C_1 = 6 \\ C_2 = -1 $$ Thus, the unique solution to the given boundary value problem is $$ y(x) = 6 \cos x - \sin x + 2(\cos x - 1) $$

Key concepts

Unique Solution of Differential Equations

Understanding whether a differential equation has a unique solution is fundamental in mathematical analysis, especially when dealing with boundary value problems. In these cases, the existence of a unique solution helps ensure that the problem is well-posed, meaning that it can be solved with certainty and the solution behaves predictably with changes to the input.

For a differential equation to have a unique solution, it must satisfy specific criteria known as the existence and uniqueness theorems. These theorems are based on the form of the differential equation and the nature of the boundary conditions. The boundary conditions offer the additional information necessary to pinpoint the exact solution out of a potentially infinite set of possibilities.

When assessing a boundary value problem, as shown in the textbook exercise, we first look to see if there's an applicable theorem like Theorem 11.2 or Theorem 11.3 that confirms the uniqueness of a solution under the given conditions. If these theorems are not applicable, we can turn to more complex methods such as the Fredholm alternative theorem to further investigate the existence and uniqueness of the solution.

In the event that there is a unique solution, this solution can generally be obtained by superimposing the homogeneous part (related to the associated homogeneous equation) and a particular part that incorporates the non-homogeneous nature of the equation. This methodology is highlighted in the step-by-step solution provided above, where the unique solution is found by solving the homogeneous equation and then adjusting for the non-homogeneous term.

Fredholm Alternative Theorem

The Fredholm Alternative Theorem is a profound concept within the context of boundary value problems involving differential equations. It essentially offers two outcomes for a non-homogeneous linear problem: either a unique solution or infinitely many solutions. This theorem is closely associated with the concept of an adjoint operator and self-adjoint operators in relation to the given differential equation.

This theorem gets its name from Ivar Fredholm, a mathematician who contributed significantly to the field of integral equations. In simple terms, the theorem stipulates that for the non-homogeneous linear equation \(Ax = f\), where \(A\) is a linear operator, there will either be a unique solution or none, depending on whether the corresponding homogeneous equation \(Ax = 0\) has only the trivial solution.

In the exercise, the Fredholm Alternative Theorem is invoked after finding that Theorems 11.2 and 11.3 do not apply. By rewriting the differential equation in its homogeneous form and working with its adjoint, the theorem guides whether a solution is present. As it often involves calculations and conditions on the Green's function, the theorem is both a high-level conceptual and practical tool for solving complex boundary value problems, as demonstrated in the solution steps.

Homogeneous Equation

A homogeneous equation, within the scope of differential equations, refers to an equation where the function and its derivatives are set to zero. This is represented succinctly as \(y'' + p(x)y' + q(x)y = 0\), where the absence of a non-zero function on the right-hand side signifies its homogeneous nature.

The importance of the homogeneous equation lies in its simplicity and the information it provides about the structure of the solution space. In the context of boundary value problems, solving the homogeneous equation tied to the original non-homogeneous equation allows us to understand the complementary function, which is the part of the general solution that deals with the associated homogeneous equation.

As illustrated in the textbook solution, by solving the homogeneous version of the equation \(y^{\text{prime prime}} + y = 0\), we gain a general solution involving constants that can later be determined by the boundary conditions. Only by combining the solutions to the homogeneous and the particular equation can we fulfill the conditions given at the boundary points, thus solving the boundary value problem in full.

The homogeneous equation is at the heart of the method for finding the unique solution to differential equations, as the complementary function it provides is essential when constructing the general solution.