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Assume that \(p\) and \(q\) are continuous, and that the functions \(y_{1}\) and \(y_{2}\) are solutions of the differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) on an open interval \(I\) Prove that if \(y_{1}\) and \(y_{2}\) have maxima or minima at the same point in \(I,\) then they cannot be a fundamental set of solutions on that interval.

Short Answer

Expert verified
Question: Prove that if \(y_1\) and \(y_2\) have the same maxima or minima point within the open interval \(I\), they cannot be a fundamental set of solutions on that interval for the given second-order differential equation \(y^{\prime\prime} + p(t)y^{\prime} + q(t)y = 0\). Answer: Since the Wronskian, \(W\), evaluates to zero at the point where both \(y_1\) and \(y_2\) have maxima or minima, it shows that \(y_1\) and \(y_2\) cannot form a fundamental set of solutions on that interval.

Step by step solution

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01

Find first and second derivative of \(y_1\) and \(y_2\)

To determine if \(y_1\) and \(y_2\) have maxima or minima at the same point, we need to find their first and second derivatives. Let \(y_i^{\prime}\) and \(y_i^{\prime\prime}\) denote the first and second derivatives of \(y_i\), respectively. According to the given differential equation, we have: $$y_1^{\prime\prime} + p(t)y_1^{\prime} + q(t)y_1 = 0$$ $$y_2^{\prime\prime} + p(t)y_2^{\prime} + q(t)y_2 = 0$$
02

Explore the properties of maxima and minima

At the points of maxima or minima, the first derivative must be equal to zero, i.e., \(y_1^{\prime}(t_0) = y_2^{\prime}(t_0) = 0\), for some \(t_0\) in \(I\). Moreover, the second derivative determines the type of extremum - a positive value means a minimum and a negative value means a maximum. Thus, we should have either \(y_1^{\prime\prime}(t_0) > 0\) and \(y_2^{\prime\prime}(t_0) > 0\), or \(y_1^{\prime\prime}(t_0) < 0\) and \(y_2^{\prime\prime}(t_0) < 0\).
03

Calculate the second derivatives at \(t_0\)

Using the given second order differential equations and properties of maxima and minima, we can rewrite the differential equation for the functions at \(t_0\), having \(y_1^{\prime}(t_0)=y_2^{\prime}(t_0)=0\): $$y_1^{\prime\prime}(t_0) + p(t_0)y_1^{\prime}(t_0) + q(t_0)y_1(t_0) = 0$$ $$y_2^{\prime\prime}(t_0) + p(t_0)y_2^{\prime}(t_0) + q(t_0)y_2(t_0) = 0$$ Since \(y_1^{\prime}(t_0) = y_2^{\prime}(t_0) = 0\), we can simplify the equations to: $$y_1^{\prime\prime}(t_0) + q(t_0)y_1(t_0) = 0$$ $$y_2^{\prime\prime}(t_0) + q(t_0)y_2(t_0) = 0$$
04

Evaluate the Wronskian

In order to claim that \(y_1\) and \(y_2\) are a fundamental set of solutions, their Wronskian \(W\) must be non-zero on the interval \(I\). The Wronskian is defined as: $$W = \begin{vmatrix} y_1(t) & y_2(t) \\ y_1^{\prime}(t) & y_2^{\prime}(t) \end{vmatrix} = y_1(t)y_2^{\prime}(t) - y_2(t)y_1^{\prime}(t)$$ At \(t_0\), where both functions have maxima or minima, we have \(W(t_0) = y_1(t_0)y_2^{\prime}(t_0) - y_2(t_0)y_1^{\prime}(t_0) = y_1(t_0)(0) - y_2(t_0)(0) = 0\). As a result, the Wronskian vanishes, and \(y_1\) and \(y_2\) cannot form a fundamental set of solutions on that interval. Thus, we have shown that if both \(y_1\) and \(y_2\) have the same maxima or minima point within the open interval \(I\), they cannot be a fundamental set of solutions on that interval.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Set of Solutions
In solving second-order linear differential equations like \(y'' + p(t)y' + q(t)y = 0\), finding a fundamental set of solutions is crucial. These solutions provide the general solution to this differential equation.

To qualify as a fundamental set of solutions, two functions, \(y_1\) and \(y_2\), must be linearly independent. This means that one cannot be expressed as a constant multiple of the other. When \(y_1\) and \(y_2\) are linearly independent, any solution to the differential equation can be written as a linear combination of these two functions:
  • \(y(t) = c_1y_1(t) + c_2y_2(t)\), where \(c_1\) and \(c_2\) are constants.
In the context of the exercise, if \(y_1\) and \(y_2\) share a maximum or minimum at the same point, their Wronskian is zero at that point, implying they cannot form a fundamental set.

This information is crucial in differential equations because it determines the completeness and generality of the solution gathered from these functions.
Wronskian
The Wronskian is a determinant used to test whether two functions are linearly independent, making it a key tool in differential equations.

For two functions \(y_1\) and \(y_2\), the Wronskian \(W\) is defined as:
  • \(W(y_1, y_2)(t) = \begin{vmatrix} y_1(t) & y_2(t) \ y_1'(t) & y_2'(t) \end{vmatrix} = y_1(t)y_2'(t) - y_2(t)y_1'(t)\)
If \(W(t) eq 0\) for all \(t\) in an interval, \(y_1\) and \(y_2\) are linearly independent on that interval, thus a fundamental set of solutions.

However, if \(W(t) = 0\) at any point, as in the exercise where both functions have maxima or minima at the same point, they become linearly dependent at that point.

The Wronskian provides insight into the behaviour of solutions and validates the uniqueness of them within differential equations.
Maxima and Minima Properties
Understanding the properties of maxima and minima is important in analyzing solutions of differential equations. A maximum or minimum occurs at a critical point where the first derivative is zero:
  • \(y'(t_0) = 0\)
The second derivative tells us if it's a maximum (\(y''(t_0) < 0\)) or a minimum (\(y''(t_0) > 0\)).

In the exercise, both \(y_1\) and \(y_2\) having maxima or minima at the same point means their first derivatives satisfy \(y_1'(t_0) = y_2'(t_0) = 0\). This causes the Wronskian to vanish since:
  • \(W(t_0) = y_1(t_0)y_2'(t_0) - y_2(t_0)y_1'(t_0) = 0\)
Thus, the point of extremum affects the linear independence, highlighting how maxima and minima properties interact with the concepts of linear dependence and independence in solutions.

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