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. $$ \begin{aligned} &y^{\prime \prime}+y=2 \\ &y(0)=7 \\ &y(\pi)=3 \end{aligned} $$

Short answer

Based on the given step by step solution, it can be seen that the unique solution for the boundary value problem is \(y(x)=5\cos(x) + 2\). The Fredholm alternative theorem was applied, and the homogeneous and particular solutions were combined, resulting in a unique solution for the given problem.

Step-by-step solution

Determine if Theorem 11.2 or 11.3 can be applied

First, let's analyze if either Theorem 11.2 or 11.3 can be applied to decide whether a unique solution exists for the given boundary value problem. The boundary value problem can be summarized as: $$ \begin{aligned} y''+y=2 \\ y(0)=7 \\ y(\pi)=3 \end{aligned} $$ Unfortunately, we do not know the properties or constraints of these theorems. It would be good if you could provide more information about them. But don't worry, we will continue the solution using the Fredholm alternative theorem.

Apply the Fredholm Alternative Theorem

Now, let's rewrite our given differential equation into the form: \(y''+y=g(x)\), where \(g(x)=2\). Our boundary conditions remain the same: $$ y(0)=7 \\ y(\pi)=3 $$ Now, we can apply the Fredholm alternative theorem to determine the existence of a unique solution.

Check for homogeneous solutions

To check for the existence of homogeneous solutions, let's solve the homogeneous equation \(y''+y=0\). This is a standard linear constant coefficient second order homogeneous differential equation. To find the characteristic equation, we do so as follows: $$ r^2+1=0 \\ r=\pm i $$ Now we can find linearly independent solutions: $$ y_1(x)=\cos(x) \\ y_2(x)=\sin(x) $$ These are our homogeneous solutions, and the general homogeneous solution is given by: $$ y_h(x)=C_1 \cos(x) + C_2 \sin(x) $$ Our particular solution will not have the form of any of our homogeneous solutions, so we can proceed to the next step.

Find a particular solution

Next, we need to find a particular solution to the given nonhomogeneous equation. To do this, we can use the method of undetermined coefficients. Given that our \(g(x)=2\), we'll guess a particular solution of the form: $$ y_p(x)=c $$ Plugging it into the equation, we have: $$ y_p''(x)+y_p(x)=2 $$ Since \(y_p''(x)=0\), we have: $$ c=2 $$ Thus, our particular solution is \(y_p(x)=2\).

Combine the homogeneous and particular solutions

Now, let's combine our homogeneous and particular solutions to find the general solution for the given nonhomogeneous equation: $$ y(x)=y_h(x) + y_p(x)=(C_1 \cos(x) + C_2 \sin(x)) + 2 $$

Determine the constants C1 and C2

Finally, let's use the given boundary conditions to determine the values of constants \(C_1\) and \(C_2\). We have: $$ y(0) = 7 = (C_1 \cos(0) + C_2 \sin(0)) + 2 \\ y(\pi) = 3 = (C_1 \cos(\pi) + C_2 \sin(\pi)) + 2 $$ Given that \(\cos(0)=1\), \(\sin(0)=0\), \(\cos(\pi)=-1\) and \(\sin(\pi)=0\), we have: $$ 7 = C_1 + 2 \\ 3 = -C_1 + 2 $$ Solving the system of equations, we get \(C_1 = 5\) and \(C_2 = 0\).

Finalize the solution

With our determined constants, we can now write the unique solution to the given boundary value problem: $$ y(x) = (5\cos(x) + 0\sin(x)) + 2 \\ y(x) = 5\cos(x) + 2 $$ Thus, our unique solution is \(y(x)=5\cos(x) + 2\).

Key concepts

Boundary Value Problems

Boundary Value Problems (BVPs) are a central concept when working with differential equations. They involve finding a solution to a differential equation that meets specific conditions at the boundaries of the domain. In simple terms, you not only need to solve the equation itself, but the solution also needs to fit certain values at certain points, typically at the start and end points of the interval you're considering.
These are not like initial value problems, which only consider the state of the solution at a single initial point. Instead, BVPs allow for more complex scenarios where the solution needs to pass through predetermined "gate points." This makes them very useful for real-world scenarios, such as modeling temperature distribution or the deflection of beams.

• BVPs are defined by a differential equation and boundary conditions.
• Solutions must satisfy the equation and meet conditions at specific boundary points.
• Examples from real-world applications include physical systems with fixed edges or ends.

Understanding BVPs requires comprehending both the nature of the differential equation and the boundary conditions applied.

Fredholm Alternative Theorem

The Fredholm Alternative Theorem is a powerful tool in the context of linear differential equations and boundary value problems. This theorem provides a framework to decide if a given boundary value problem has a solution, and if it does, whether that solution is unique.
Essentially, the theorem states that either a homogeneous equation has only the trivial solution or that the associated non-homogeneous problem has a particular solution. It's like a set of rules suggesting whether you should expect no solutions, one solution, or infinite solutions. This is especially handy when direct methods provide no clear answers.

• Helps determine existence and uniqueness of solutions.
• Applies both to homogeneous and non-homogeneous systems.
• Reduces complex systems to simpler scenarios by splitting the problem characteristics.

Using this theorem requires understanding of linear algebra concepts such as kernels and orthogonal complements, which also play an essential role in modern applied mathematics.

Homogeneous Solutions

Homogeneous solutions arise from solving a differential equation where the right-hand side is zero. These solutions are crucial because they form the basis for the solution of non-homogeneous differential equations.
For example, if you have \( y'' + y = 0 \), you're looking for solutions where the output is smoothly connected without any external driving forces (which would be your "inhomogeneous" part if it weren't zero).
Finding these solutions involves solving a characteristic equation, which in this case gives complex conjugate roots \( r = ± i \). This leads to solutions involving sine and cosine functions:

• General form of homogeneous solutions: \[ y_h(x) = C_1 \cos(x) + C_2 \sin(x) \]
• These solutions act as a "backbone" for more complex solutions.
• Important in building particular solutions by combining with inhomogeneous solutions.

Understanding homogeneous solutions is like learning a melody on which the symphony builds its complex themes.

Method of Undetermined Coefficients

The Method of Undetermined Coefficients is a technique used to find particular solutions to non-homogeneous differential equations. Once you've solved the homogeneous part, this method allows you to tackle the additional part that doesn't equal zero.
The basic idea is you guess the form of the solution based on the type of non-homogeneous term you have (like constants, polynomials, exponentials, etc.), then determine coefficients to make it work.
If your non-homogeneous term is \( g(x) = 2 \), as in our case, you'd guess a constant solution \( y_p(x) = c \), and simply plug it into the differential equation to solve for \( c = 2 \).

• Formulate a trial solution based on the form of \( g(x) \).
• Plug the trial solution into the differential equation to solve for coefficients.
• Combine with homogeneous solutions to form the complete solution.

This method might require you to adjust your guess, especially if your guess shares characteristics with the homogeneous solution, but overall, it provides a systematic path to finding solutions.