Question

In each exercise, (a) Show that the given two-point boundary value problem has a unique solution. (b) Solve the problem. Note that a fundamental matrix for \(\mathbf{y}^{\prime}=\left[\begin{array}{rr}1 & -2 \\ -2 & 1\end{array}\right] \mathbf{y}\) is $$ \Psi(t)=\left[\begin{array}{cc} e^{-t} & e^{3 t} \\ e^{-t} & -e^{3 t} \end{array}\right] $$ $$ \mathbf{y}^{r}=\left[\begin{array}{rr} 1 & -2 \\ -2 & 1 \end{array}\right] \mathbf{y}, \quad y_{1}(0)=1, \quad y_{2}(1)=0 $$

Short answer

Based on the given step by step solution, answer the following question: Question: Prove that the given two-point boundary value problem has a unique solution and find the unique solution for the problem using the given fundamental matrix. Answer: To prove the uniqueness of the solution, we demonstrated that for any given set of boundary conditions, there exists only one solution for the differential equation. We then used the given fundamental matrix to form a general solution. By applying the given boundary conditions, we were able to solve for the constants \(c_{1}\) and \(c_{2}\). The unique solution is given by: $$ \mathbf{y}(t) = \left[\begin{array}{cc} e^{-t} & e^{3t} \\ e^{-t} & -e^{3t} \end{array}\right] \left[\begin{array}{c} \frac{e^{3}(1 - e^{-1})}{1 + e^{3}} \\ \frac{1 - e^{-1}}{1 + e^{3}} \end{array}\right] $$

Step-by-step solution

Definition of a unique solution

To prove the uniqueness of the solution, we need to show that for any given set of boundary conditions, there exists only one solution for the differential equation. This is equivalent to showing that for any inhomogeneous system \(\mathbf{y}' = \mathbf{A}\mathbf{y} + \mathbf{f}(t)\) with given boundary conditions, there exists only one solution. #b) Solving the problem#

Write the given boundary value problem

We are given the boundary value problem: $$ \mathbf{y}' = \left[\begin{array}{rr} 1 & -2 \\ -2 & 1 \end{array}\right] \mathbf{y}, \quad y_{1}(0) = 1, \quad y_{2}(1) = 0 $$

Write the given fundamental matrix

We are given the fundamental matrix: $$ \Psi(t) = \left[\begin{array}{cc} e^{-t} & e^{3t} \\ e^{-t} & -e^{3t} \end{array}\right] $$

Form the general solution using the fundamental matrix

Recall that the general solution to a homogeneous system \(\mathbf{y}’ = \mathbf{A}\mathbf{y}\) can be written as: $$ \mathbf{y}(t) = \Psi(t)\mathbf{c}, $$ where \(\mathbf{c}\) is a constant vector. So, the general solution is: $$ \mathbf{y}(t) = \left[\begin{array}{cc} e^{-t} & e^{3t} \\ e^{-t} & -e^{3t} \end{array}\right] \left[\begin{array}{c} c_{1} \\ c_{2} \end{array}\right] $$

Apply the given boundary conditions to find the constants

We are given the boundary conditions \(y_{1}(0) = 1\) and \(y_{2}(1) = 0\). We need to apply these conditions to the general solution to find the constants \(c_{1}\) and \(c_{2}\). First, apply the boundary condition \(y_{1}(0) = 1\): $$ y_{1}(0) = e^{-0}c_{1} + e^{3(0)}c_{2} = c_{1} + c_{2} = 1 $$ Next, apply the boundary condition \(y_{2}(1) = 0\): $$ y_{2}(1) = e^{-1}c_{1} - e^{3(1)}c_{2} = 0 $$

Solve for the constants

Solve the system of equations to find the values of \(c_{1}\) and \(c_{2}\): $$ \begin{cases} c_{1} + c_{2} = 1 \\ e^{-1}c_{1} - e^{3}c_{2} = 0 \end{cases} $$ Solve the first equation for \(c_{1}\): $$ c_{1} = 1 - c_{2} $$ Substitute into the second equation: $$ e^{-1}(1 - c_{2}) - e^{3}c_{2} = 0 $$ Solve for \(c_{2}\): $$ c_{2} = \frac{1 - e^{-1}}{1 + e^{3}} $$ Then, solve for \(c_{1}\): $$ c_{1} = 1 - \frac{1 - e^{-1}}{1 + e^{3}} = \frac{e^{3}(1 - e^{-1})}{1 + e^{3}} $$

Write the unique solution

Plug the values of \(c_{1}\) and \(c_{2}\) into the general solution to obtain the unique solution: $$ \mathbf{y}(t) = \left[\begin{array}{cc} e^{-t} & e^{3t} \\ e^{-t} & -e^{3t} \end{array}\right] \left[\begin{array}{c} \frac{e^{3}(1 - e^{-1})}{1 + e^{3}} \\ \frac{1 - e^{-1}}{1 + e^{3}} \end{array}\right] $$

Key concepts

Unique Solution of Differential Equations

Understanding when a differential equation has a unique solution is crucial. In the context of a two-point boundary value problem, a unique solution exists if, given the boundary conditions, we can find one and only one function that satisfies the equation. For a linear differential equation, a unique solution is guaranteed if the associated homogeneous equation has only the trivial solution. The Wronskian determinant associated with the solution set of the differential equation being non-zero is also a key component for uniqueness.

In our exercise, we prove uniqueness by leveraging the properties of linear differential equations, showing that no other solution can satisfy both the equation and the boundary conditions given. By applying the boundary conditions, we end up with a system of linear equations for the constants. If this system has a unique solution, so does the original boundary value problem.

Fundamental Matrix for Systems

The fundamental matrix is a powerful tool used to solve systems of linear homogeneous differential equations. It is constructed from a set of linearly independent solutions to the system and encapsulates all information needed to generate the general solution of the system.

In our case, \[ \Psi(t)=\left[\begin{array}{cc} e^{-t} & e^{3 t} \ e^{-t} & -e^{3 t} \end{array}\right] \] is the given fundamental matrix. It simplifies the process of finding the general solution to \[ \mathbf{y}(t) = \Psi(t)\mathbf{c},\] where \(\mathbf{c}\) is a vector of constants. This matrix is constructed so that each column is a solution to the homogeneous system, and when multiplied by a vector of constants, a linear combination of these solutions is formed, which represents the general solution.

Homogeneous System of Differential Equations

A homogeneous system of differential equations is one where the right-hand side of the equations is zero; there's no 'forcing' term, just the system acting by itself. Such a system can be represented as \[ \mathbf{y}^\prime = \mathbf{A}\mathbf{y}, \] where \( \mathbf{A} \) is a matrix and \( \mathbf{y} \) is the vector of unknown functions. The solutions to these types of systems are combinations of exponential functions of the form \[ e^{\lambda t} \] where \( \lambda \) is an eigenvalue of \( \mathbf{A} \).

For this system, if the eigenvalues and corresponding eigenvectors can be found, we can construct the fundamental matrix, as seen in the exercise. We then use it to generate the general solution and apply boundary conditions to refine this solution to the unique one that fits the problem's criteria.

Applying Boundary Conditions

Boundary conditions are constraints that provide specific values for the solution of a differential equation at certain points. Applying them is crucial for determining the unique solution to a boundary value problem. In our exercise, the conditions were \( y_{1}(0)=1 \) and \( y_{2}(1)=0 \).

These conditions were used to find the specific values of the constants in the general solution, turning it from a family of possible solutions into the single solution that works for the given boundary values. By plugging the conditions into the general solution and solving for the constants, the boundary value problem becomes a set of algebraic equations. The solution to these equations provides the coefficients for the general solution, nailing down the one specific function that solves the entire boundary value problem.