Question

If the matrix \(A\) is invertible, show that \(\mathbf{y}^{\prime}=A \mathbf{y}+\mathbf{b}\) has a unique equilibrium solution. If the matrix \(A\) is not invertible, must the differential equation \(\mathbf{y}^{\prime}=A \mathbf{y}+\mathbf{b}\) possess an equilibrium solution? If an equilibrium solution does exist in this case, is it unique?

Short answer

Question: For a given differential equation \(\mathbf{y}^{\prime}=A\mathbf{y}+\mathbf{b}\), explain the existence and uniqueness of the equilibrium solution based on whether matrix \(A\) is invertible or not. Answer: If matrix \(A\) is invertible, the differential equation has a unique equilibrium solution, given by \(\mathbf{y}_e = -A^{-1}\mathbf{b}\). If matrix \(A\) is non-invertible, the existence and uniqueness of the equilibrium solution cannot be determined in general and must be analyzed for each specific case.

Step-by-step solution

Definition of the equilibrium solution

An equilibrium solution is a constant solution of the differential equation. In other words, we are looking for a constant vector \(\mathbf{y}_e\) such that \(\mathbf{y}^{\prime} = A\mathbf{y}_e + \mathbf{b} = 0\). To find the equilibrium solution, we have to solve the following linear system of equations: $$A\mathbf{y}_e + \mathbf{b} = 0.$$

Invertible matrix \(A\)

If the matrix \(A\) is invertible, we can apply the inverse of \(A\) to both sides of the linear system to obtain: $$A^{-1}(A\mathbf{y}_e + \mathbf{b}) = A^{-1}0 \Rightarrow (A^{-1}A) \mathbf{y}_e + A^{-1}\mathbf{b} = 0 \Rightarrow \mathbf{y}_e = -A^{-1}\mathbf{b}.$$ Since matrix \(A\) has an inverse and we can find a unique solution for \(\mathbf{y}_e\), it shows that the differential equation has a unique equilibrium solution when \(A\) is invertible.

Non-invertible matrix \(A\)

If the matrix \(A\) is non-invertible, we cannot directly find the inverse as before. However, we can still analyze the linear system of equations \(A\mathbf{y}_e + \mathbf{b} = 0.\) - If the system has no solution, it means that the differential equation does not have an equilibrium solution. - If the system has one or more solutions, it means that an equilibrium solution exists. It could be unique (one solution) or non-unique (multiple solutions). In the case of a non-invertible matrix \(A\), we cannot make a general conclusion about the existence and uniqueness of the equilibrium solution. It needs to be analyzed for each specific case.

Key concepts

Equilibrium Solution

An equilibrium solution in the context of a differential equation is a type of solution where the system doesn't change over time. It remains constant. When you have a differential equation like \( \mathbf{y}^{\prime} = A\mathbf{y} + \mathbf{b} \), an equilibrium solution happens when the derivative \( \mathbf{y}^{\prime} \) equals zero. That means the rate of change is zero, so the system remains stable. To find this equilibrium solution, you need a vector \( \mathbf{y}_e \) such that \( A\mathbf{y}_e + \mathbf{b} = 0 \). This is essentially finding a point where the push and pull of the system are perfectly balanced, leading to no movement. Here, you're searching for a fixed point of the system. In practical terms, when you solve the equation \( A\mathbf{y}_e + \mathbf{b} = 0 \), you're finding the state at which the system would rest indefinitely, assuming it starts at that state.

Invertible Matrix

An invertible matrix is a matrix for which there exists another matrix that, when multiplied with the original, yields the identity matrix. In terms of notation, a matrix \( A \) is invertible if there exists a matrix \( A^{-1} \) such that \( A A^{-1} = I \), where \( I \) is the identity matrix. Why is invertibility important? For a differential equation like \( \mathbf{y}^{\prime} = A\mathbf{y} + \mathbf{b} \), if \( A \) is invertible, you can solve \( A\mathbf{y}_e + \mathbf{b} = 0 \) easily. Simply multiply through by \( A^{-1} \) to isolate \( \mathbf{y}_e \) and get \( \mathbf{y}_e = -A^{-1}\mathbf{b} \). This means that not only does the equilibrium solution exist, but it's also unique. Invertible matrices have no eigenvalues equal to zero, and their determinant is non-zero. This property ensures that the matrix can be "reversed," allowing us to find a single, unique equilibrium solution for the system.

Linear System of Equations

A linear system of equations consists of two or more linear equations involving the same set of variables. Solving these systems is critical in finding solutions like equilibrium solutions for differential equations. For the given differential equation \( \mathbf{y}^{\prime} = A\mathbf{y} + \mathbf{b} \), you end up with the linear system \( A\mathbf{y}_e + \mathbf{b} = 0 \). Solving this system involves finding the values of \( \mathbf{y}_e \) that satisfy all the given equations. Depending on the properties of the matrix \( A \), especially whether it's invertible or not, the approach and the number of solutions can vary. Linear systems can have:

• One unique solution, if the system is consistent and the matrix is invertible.
• No solution, if it's inconsistent and the equations contradict each other.
• Infinite solutions, particularly when the matrix is non-invertible and dependent equations lead to free variables.

This makes understanding the structure of these systems essential for determining the behavior of solutions in a differential equation context.

Unique Solution

A unique solution refers to a single set of parameters that satisfy all the conditions of a given problem or equation. In the context of equilibrium solutions for differential equations, uniqueness means that the system settles into one specific state without ambiguity. When dealing with the equation \( \mathbf{y}^{\prime} = A\mathbf{y} + \mathbf{b} \), uniqueness is guaranteed when the matrix \( A \) is invertible. The inverse allows the isolation of \( \mathbf{y}_e \), resulting in the unique solution \( \mathbf{y}_e = -A^{-1}\mathbf{b} \). However, if \( A \) is not invertible, the situation changes. The system may not have a solution or could have multiple solutions, depending on the specific values of \( A \) and \( \mathbf{b} \). In such cases, the possibility of multiple or no unique solutions makes it imperative to analyze each scenario in detail, often by looking at the rank of the matrices involved or employing numerical methods to approximate solutions if necessary.