Question

Specialize Exercise 28 by considering the system $$ \begin{array}{l} y_{j}^{\prime}=y_{j+1} \quad(j=1, \ldots, k-1), \\ y_{i}^{\prime}=f(x)-\sum_{j=1}^{k} g_{j}(x) y_{j} \end{array} $$ where \(f, g_{1}, \ldots, g_{k}\) are continuous real functions on \([a, b]\), and derive a uniqueness theorem for solutions of the equation $$ y^{(n)}+g_{k}(x) y^{(k-1)}+\cdots+g_{2}(x) y^{\prime}+g_{1}(x) y=f(x) $$ subject to initial conditions $$ y(a)=c_{1}, \quad y^{\prime}(a)=c_{2}, \quad \ldots, \quad y^{(\alpha-1)}(a)=c_{k} . $$

Short answer

Question: Prove the uniqueness of the solution to the given system of equations, using the Existence and Uniqueness theorem, and with given initial conditions. Solution: 1. Rewrite the system of equations in its specialized form. 2. Differentiate the equations and substitute higher-order derivatives. 3. Apply the given initial conditions. 4. Use the Existence and Uniqueness Theorem to claim a unique solution exists with given initial conditions.

Step-by-step solution

Equating Both Sides of the Equations

We are given that the system: $$ \begin{array}{l} y_{j}^{\prime}=y_{j+1} \quad(j=1, \ldots, k-1), \\ y_{i}^{\prime}=f(x)-\sum_{j=1}^{k} g_{j}(x) y_{j} \end{array} $$ Using these equations, we can find the relationship between the variables \(y_j\) and their derivatives. Step 2: Differentiate the Equations

Substituting the Higher-order Derivative Equations

We can differentiate the equation \(y_j'=y_{j+1}\) with respect to \(x\) repeatedly until we reach the highest derivative \(y^{(n)}\). This will allow us to substitute each higher-order derivative in the given equation. For example, differentiate the equation \(y_1'=y_2\) to get \(y_1''=y_2'\). This process can be continued until The derivatives stop at \((j=k-1)\). Next, follow the same method for \(y_{i}^{\prime}=f(x)-\sum_{j=1}^{k} g_{j}(x) y_{j}\): $$y^{(k)}+g_{k}(x) y^{(k-1)}+\cdots+g_{2}(x) y^{\prime}+g_{1}(x) y=f(x).$$ Step 3: Apply Initial Conditions

Applying Initial Conditions

Now, apply the initial conditions required for the uniqueness theorem: $$y(a)=c_{1}, \quad y^{\prime}(a)=c_{2}, \quad \ldots, \quad y^{(\alpha-1)}(a)=c_{k}.$$ Step 4: Existence and Uniqueness Theorem

Referencing the Existence and Uniqueness Theorem

Based on the given information, we can utilize the standard Existence and Uniqueness Theorem to claim that there exists a unique solution to the given equation with initial conditions. In summary, the system of equations leads to the higher-order equation needed, and the problem can be solved by applying the Existence and Uniqueness Theorem together with the initial conditions provided.

Key concepts

Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem is a fundamental concept in differential equations. It assures us that under certain circumstances, a differential equation has a unique solution. This is crucial for disciplines like physics and engineering, where precise solutions are needed.

For a given differential equation, the theorem states that if the function involved is continuous, and the partial derivatives are also continuous in a neighborhood around the initial condition, then there exists a unique solution that passes through the given initial condition. This solution will be defined over some interval.

This theorem plays a pivotal role in understanding whether mathematical models based on differential equations are valid or not, as without uniqueness we cannot be certain of the predictive validity of the equations.

Initial Conditions

Initial conditions are specific values assigned to a differential equation at the beginning of the problem-solving process. These values are crucial because they help determine a unique solution to the differential equation.

For example, if we have an equation such as \( y'(t) = ay(t) \), an initial condition could be \( y(0) = 1 \). By setting these conditions, we can use them along with the differential equation to solve for specific solution curves.

In practical terms, initial conditions are often real-world measurements. For instance, in a physics problem, these might represent initial velocity or initial temperature, providing the specific starting values needed to solve a problem uniquely.

Higher-Order Derivatives

Higher-order derivatives refer to derivatives of a function that are higher than the first derivative. These derivatives are represented as \( y'' \), \( y''' \), etc., and they provide information about the rate of change of the rates of change.

In differential equations, these higher-order derivatives are crucial for describing systems involving changes that depend on not just the current state, but also previous states. For instance, acceleration in physics is a second-order derivative of position with respect to time.

Working with higher-order derivatives means applying differentiation repeatedly to capture more complex relationships within the function described by the equation. This process is essential not only for understanding the behavior of a solution deeply but also for generating the solution itself by satisfying the requirements of the differential equation.

Continuous Functions

Continuous functions are mathematical expressions that do not have any breaks, jumps, or points of discontinuity in their domain. In the context of differential equations, continuity is vital because it ensures that certain theorems, like the Existence and Uniqueness Theorem, can be applied.

When a function is continuous, it can be visualized as a smooth, unbroken curve on a graph. This property guarantees that small changes in input values result in small changes in output values, making calculations and predictions more reliable.

In practical applications, ensuring continuity implies that the physical systems modeled by these equations are stable and predictable, as real-world phenomena often need to behave in a predictable and smooth manner for purposes like engineering design or scientific analysis.