Question

In this problem we show how to generalize Theorem 3.3 .2 (Abel's theorem) to higher order equations. We first outline the procedure for the third order equation $$ y^{\prime \prime \prime}+p_{1}(t) y^{\prime \prime}+p_{2}(t) y^{\prime}+p_{3}(t) y=0 $$ Let \(y_{1}, y_{2},\) and \(y_{3}\) be solutions of this equation on an interval \(I\) (a) If \(W=W\left(y_{1}, y_{2}, y_{3}\right),\) show that $$ W^{\prime}=\left|\begin{array}{ccc}{y_{1}} & {y_{2}} & {y_{3}} \\\ {y_{1}^{\prime}} & {y_{2}^{\prime}} & {y_{3}^{\prime}} \\ {y_{1}^{\prime \prime \prime}} & {y_{2}^{\prime \prime \prime}} & {y_{3}^{\prime \prime \prime}}\end{array}\right| $$ Hint: The derivative of a 3 -by-3 determinant is the sum of three 3 -by-3 determinants obtained by differentiating the first, second, and third rows, respectively. (b) Substitute for \(y_{1}^{\prime \prime \prime}, y_{2}^{\prime \prime \prime},\) and \(y_{3}^{\prime \prime \prime}\) from the differential equation; multiply the first row by \(p_{3},\) the second row by \(p_{2},\) and add these to the last row to obtain $$ W^{\prime}=-p_{1}(t) W $$ (c) Show that $$ W\left(y_{1}, y_{2}, y_{3}\right)(t)=c \exp \left[-\int p_{1}(t) d t\right] $$ It follows that \(W\) is either always zero or nowhere zero on \(I .\) (d) Generalize this argument to the \(n\) th order equation $$ y^{(n)}+p_{1}(t) y^{(n-1)}+\cdots+p_{n}(t) y=0 $$ with solutions \(y_{1}, \ldots, y_{n} .\) That is, establish Abel's formula, $$ W\left(y_{1}, \ldots, y_{n}\right)(l)=c \exp \left[-\int p_{1}(t) d t\right] $$ for this case.

Short answer

Question: Generalize Abel's theorem for higher-order differential equations and establish Abel's formula for the nth-order equation. Answer: To generalize Abel's theorem for higher-order differential equations, we just need to replace the third-order equation with the nth-order equation and follow the same process as before, considering we have \(n\) functions \(y_1, \ldots, y_n\) that are the solutions of this equation. Abel's formula for the nth-order equation is: $$ W\left(y_{1}, \ldots, y_{n}\right)(t) = c \exp \left[-\int p_{1}(t) dt\right] $$

Step-by-step solution

Find the derivative of Wronskian for the third-order equation

To start, we first find Wronskian, W, for the functions y1, y2, and y3. Using the given hint that the determinant can be found by differentiating each row, we compute the derivative of the Wronskian, W': $$ W^{\prime}=\left|\begin{array}{ccc}{y_{1}} & {y_{2}} & {y_{3}}\\\ {y_{1}^{\prime}} & {y_{2}^{\prime}} & {y_{3}^{\prime}}\\\ {y_{1}^{\prime\prime\prime}} & {y_{2}^{\prime\prime\prime}} & {y_{3}^{\prime\prime\prime}}\end{array}\right| $$

Substitute third-order derivatives from the given differential equation

Rewrite the given differential equation for \(y_1^{\prime\prime\prime}, y_2^{\prime\prime\prime}\), and \(y_3^{\prime\prime\prime}\): $$ y_1^{\prime\prime\prime} = -p_1(t)y_1^{\prime\prime} - p_2(t)y_1^{\prime} - p_3(t)y_1 \\ y_2^{\prime\prime\prime} = -p_1(t)y_2^{\prime\prime} - p_2(t)y_2^{\prime} - p_3(t)y_2 \\ y_3^{\prime\prime\prime} = -p_1(t)y_3^{\prime\prime} - p_2(t)y_3^{\prime} - p_3(t)y_3 $$ Substitute these values into the expression for \(W'\): $$ W^{\prime}=\left|\begin{array}{ccc}{y_{1}} & {y_{2}} & {y_{3}}\\\ {y_{1}^{\prime}} & {y_{2}^{\prime}} & {y_{3}^{\prime}}\\\ {-p_1(t)y_1^{\prime\prime} - p_2(t)y_1^{\prime} - p_3(t)y_1} & {-p_1(t)y_2^{\prime\prime} - p_2(t)y_2^{\prime} - p_3(t)y_2} & {-p_1(t)y_3^{\prime\prime} - p_2(t)y_3^{\prime} - p_3(t)y_3}\end{array}\right| $$

Simplify W' using row operations

Multiply the first row by \(p_3(t)\), second row by \(p_2(t)\), and add them to the last row. This will give us: $$ W^{\prime} = -p_1(t) \left|\begin{array}{ccc}{y_{1}} & {y_{2}} & {y_{3}}\\\ {y_{1}^{\prime}} & {y_{2}^{\prime}} & {y_{3}^{\prime}}\\\ {y_{1}^{\prime\prime}} & {y_{2}^{\prime\prime}} & {y_{3}^{\prime\prime}}\end{array}\right| = -p_1(t)W $$

Determine the general formula for W

We have the differential equation for the Wronskian, \(W^{\prime} = -p_1(t)W\). Now, separate variables and integrate: $$ \frac{dW}{W} = -p_1(t) dt $$ Integrate both sides: $$ \int \frac{dW}{W} = - \int p_1(t) dt $$ We get: $$ \ln |W| = - \int p_1(t) dt + C \Rightarrow W(y_1, y_2, y_3)(t) = c \exp \left[-\int p_1(t) dt\right] $$

Generalize the argument for nth-order equation

To generalize this for an nth-order equation, we just need to replace the third-order equation with the nth-order equation and follow the same process as before, considering we have \(n\) functions \(y_1, \ldots, y_n\) that are the solutions of this equation. We will establish the Abel's formula for the nth-order equation: $$ W\left(y_{1}, \ldots, y_{n}\right)(t) = c \exp \left[-\int p_{1}(t) dt\right] $$ This generalization follows the same outline we used for the third-order equation and applies to any higher-order equation.

Key concepts

Abel's Theorem

Abel's Theorem is a powerful tool in the analysis of linear differential equations. It allows us to determine the behavior of solutions by examining the Wronskian, which is a determinant constructed from the solutions. The theorem tells us that the Wronskian's rate of change is proportional to a function drawn from the equation's coefficients. In simple terms, Abel's Theorem provides a way to predict solutions' dependency without solving the entire differential equation.
In practice, if you can compute the Wronskian for a set of solutions, Abel's Theorem helps you understand whether these solutions are linearly independent. Linearly independent solutions imply that the general solution is a combination of these solutions.
The remarkable feature of Abel's Theorem is seen in its assertion for the Wronskian, where the formula is: \[ W(y_1, y_2, ..., y_n)(t) = c \cdot e^{\int -p_1(t) \ dt} \] The constant \(c\) stresses the unchanging value of the Wronskian or indicates the absence of linear independence. This makes Abel’s Theorem an indispensable asset in the study of differential equations.

Higher Order Differential Equations

Higher order differential equations extend the concept of ordinary differential equations beyond second-order. An equation of this type involves derivatives higher than the second derivative. For example, a third-order differential equation includes derivatives up to the third degree. The process of solving these equations often involves generalizing techniques used in typical second-order equations.
The standard form of a higher order linear differential equation looks like:\[ y^{(n)} + p_1(t)y^{(n-1)} + \cdots + p_n(t)y = 0 \] Here, \(y^{(n)}\) represents the nth derivative of \(y\).
When solving higher order differential equations, one of the goals is to establish a set of linearly independent solutions. This is done by determining if the solutions span a complete solution space for that order of the differential equation. Take note that, like in lower-order equations, we still need boundary or initial conditions to find specific solutions.

Determinant Calculus

Determinant Calculus is integral in the study and solution of linear differential equations, particularly when using the Wronskian to ascertain linear independence among solutions. The determinant provides a scalar value that can ascertain properties of a matrix, such as invertibility or linear independence of columns.

• The determinant of a 3x3 matrix, for example, is calculated using a specific formula involving the elements of the matrix, allowing us to quickly determine essential properties about the matrix.
• In differential equations, the Wronskian itself is a determinant, measuring the area spanned by solution vectors in their respective function space.

The mathematical operation of computing a determinant and its derivative forms the backbone of establishing relations in Abel's Theorem. Adjustments to different elements, such as row or column operations, can be applied to simplify or compute the determinant, which often appears in problems involving linear independence.

Matrix Operations in Differential Equations

Matrix operations are a key component of handling systems of differential equations. By organizing sets of equations or solutions within matrices, we simplify the operations required to solve them. Matrix techniques leverage familiarity with linear transformations: they provide a systematic method to deal with multiple equations simultaneously.
Here’s how matrix operations help:

• Using matrices allows for compact representations of systems, such as expressing a system of equations in matrix form as \(AX = B\).
• Operations like matrix multiplication, inversion, and transposition streamline the steps needed to arrive at solutions or explore stability, particularly in linear algebra equations.
• When dealing with Wronskians in higher order differential equations, determinantly defined expressions represent matrices whose rows are derivatives of solution sets.

Understanding matrix operations aids in visualizing how transformations and rotations apply not just to spatial data but abstract solutions to differential equations as well. This visualization helps in grasping complex relationships in differential systems.