Question

The purpose of this problem is to show that if \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right) \neq 0\) for some \(t_{0}\) in an interval \(I,\) then \(y_{1}, \ldots, y_{n}\) are linearly independent on \(I,\) and if they are linearly independent and solutions of $$ L(y]=y^{(n)}+p_{1}(t) y^{(n-1)}+\cdots+p_{n}(t) y=0 $$ on \(I,\) then \(W\left(y_{1}, \ldots, y_{n}\right)\) is nowhere zero in \(I .\) (a) Suppose that \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right) \neq 0,\) and suppose that $$ c_{1} y_{1}(t)+\cdots+c_{n} y_{n}(t)=0 $$ for all \(t\) in \(I\). By writing the equations corresponding to the first \(n-1\) derivatives of Fa. (ii) at \(t_{0}\), show that \(c_{1}=\cdots=c_{n}=0 .\) Therefore, \(y_{1}, \ldots, y_{n}\) are linearly independent. (b) Suppose that \(y_{1}, \ldots, y_{n}\) are linearly independent solutions of Eq. (i). If \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right)=0\) for some \(t_{0},\) show that there is a nonzero solution of Eq. (i) satisfying the initial conditions $$ y\left(t_{0}\right)=y^{\prime}\left(t_{0}\right)=\cdots=y^{(n-1)}\left(t_{0}\right)=0 $$ since \(y=0\) is a solution of this initial value problem, the uniqueness part of Theorem 4. 1. I yields a contradiction. Thus \(W\) is never zero.

Short answer

In summary, we have proven that: (a) If the Wronskian of \(n\) functions is nonzero for some \(t_0\) in the interval \(I\), then the functions are linearly independent on \(I\). This was done by showing that the constants in the linear combination are all equal to zero. (b) If the functions are linearly independent and solutions of a given linear differential equation, then the Wronskian is nowhere zero in \(I\). This was shown by demonstrating that if the Wronskian was zero, then there would be a contradiction due to the uniqueness of solutions in the given ODE. This helps us understand the relationship between the Wronskian and linear independence of functions in the context of linear differential equations.

Step-by-step solution

(a) Proving linear independence

Suppose the Wronskian of \(n\) functions \(y_1(t), \ldots, y_n(t)\) is nonzero for some \(t_0\) in an interval \(I\), i.e., \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right) \neq 0\). Also, suppose there exist constants \(c_1, \ldots, c_n\) such that the following equation holds for all \(t\) in \(I\): $$ c_{1} y_{1}(t)+\cdots+c_{n} y_{n}(t)=0 $$ To prove linear independence, we need to show that \(c_1=\cdots=c_n=0\). Applying one derivative to the above equation, we have $$ c_{1} y_{1}'(t)+\cdots+c_{n} y_{n}'(t)=0 $$ Continue applying derivatives \(n-1\) times to get $$ c_{1} y_{1}^{(n-1)}(t)+\cdots+c_{n} y_{n}^{(n-1)}(t)=0 $$ Now, we have a system of \(n\) linear equations involving \(c_1, \ldots, c_n\), they are: $$ c_{1} y_{1}(t_0)+\cdots+c_{n} y_{n}(t_0)=0 $$ $$ c_{1} y_{1}'(t_0)+\cdots+c_{n} y_{n}'(t_0)=0 $$ $$ \vdots $$ $$ c_{1} y_{1}^{(n-1)}(t_0)+\cdots+c_{n} y_{n}^{(n-1)}(t_0)=0 $$ The determinant of this system is the Wronskian \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right) \neq 0\). Since the determinant is nonzero, the system has a unique solution, which in this case, is the trivial solution: \(c_1=\cdots=c_n=0\). This implies that the functions \(y_1, \ldots, y_n\) are linearly independent.

(b) Proving the Wronskian is never zero

Suppose \(y_1, \ldots, y_n\) are linearly independent solutions of the linear differential equation: $$ L(y)=y^{(n)}+p_{1}(t) y^{(n-1)}+\cdots+p_{n}(t) y=0 $$ If the Wronskian \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right)=0\) for some \(t_0\), we need to show that there exists a nonzero solution of the linear ODE satisfying the initial conditions: $$ y\left(t_{0}\right)=y^{\prime}\left(t_{0}\right)=\cdots=y^{(n-1)}\left(t_{0}\right)=0 $$ The zero function, \(y(t)=0\), is a solution to this initial value problem and the linear ODE. But the uniqueness part of Theorem 4.1.1 implies that there cannot be another nonzero solution with the same initial conditions. Thus, we arrive at a contradiction, meaning the Wronskian \(W\left(y_{1}, \ldots, y_{n}\right)\) can never be zero in the interval \(I\).

Key concepts

Linear Independence

Linear independence is a fundamental concept when dealing with differential equations and their solutions. Essentially, a set of functions \(y_1(t), y_2(t), \ldots, y_n(t)\) are linearly independent over an interval \(I\) if the only solution to the equation \(c_1 y_1(t) + c_2 y_2(t) + \cdots + c_n y_n(t) = 0\) for all \(t\) in \(I\) is \(c_1 = c_2 = \ldots = c_n = 0\).

In practical terms, linear independence implies that no function in the set can be written as a linear combination of the others. This concept is vital for ensuring that the solutions to differential equations provide a complete description of the system's behavior without redundancy. The Wronskian, a determinant involving the functions and their derivatives, is often used to test for linear independence. If it is non-zero at a particular point, the functions are linearly independent on that interval.

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They describe how a particular quantity changes over time and are widely used to model real-world phenomena such as physics, engineering, and economics.

A common form is the linear differential equation given by \[L(y) = y^{(n)} + p_1(t)y^{(n-1)} + \cdots + p_n(t)y = 0,\] where each \(p_i(t)\) is a known function of \(t\), and \(y^{(n)}\) denotes the \(n\)-th derivative of \(y\). Solving these equations typically involves finding functions that satisfy the equation for specific initial conditions or constraints. The solutions can then help to predict the system's future behavior or understand its intrinsic properties.

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific values for the function and its derivatives at a particular point, typically called the initial conditions.

For example, if we have a differential equation like \[L(y) = y^{(n)} + p_1(t)y^{(n-1)} + \ldots + p_n(t)y = 0,\] the initial value problem would specify that \(y(t_0) = y_0, y'(t_0) = y_1, \ldots, y^{(n-1)}(t_0) = y_{n-1}\). These initial conditions are crucial because they ensure the differential equation has a unique solution from the multiple possibilities that might satisfy the differential equation alone.

Uniqueness Theorem

The uniqueness theorem in the context of differential equations affirms that under certain conditions, a differential equation has exactly one solution that satisfies a given set of initial conditions. This is especially important when dealing with initial value problems, as it ensures predictability and consistency of the solutions.

For instance, consider the linear differential equation \[L(y) = y^{(n)} + p_1(t)y^{(n-1)} + \ldots + p_n(t)y = 0\]. The uniqueness theorem, sometimes pointed out in textbooks as Theorem 4.1.1, implies that if the Wronskian \(W(y_1, \ldots, y_n)(t_0) eq 0\) at some point \(t_0\), then the solutions of the differential equation are unique in that interval when initial conditions are given. This prevents ambiguity in scientific modeling and analysis.