Linear Independence Dependence of a Set of Functions

# Linear Independence Dependence of a Set of Functions

Recall from the Wronskian Determinants and Higher Order Linear Homogenous Differential Equations page that if we have an $n^{\mathrm{th}}$ order linear homogenous differential equation $\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0$ where $p_1$, $p_2$, …, $p_n$ are continuous on an open interval $I$ and if $y = y_1(t)$, $y = y_2(t)$, …, $y = y_n(t)$ are solutions to this differential equation, then provided that $W(y_1, y_2, ..., y_n) \neq 0$ for at least one point $t \in I$, then $y_1$, $y_2$, …, $y_n$ form a fundamental set of solutions to this differential equation - that is, for constants $C_1$, $C_2$, …, $C_n$, then every solution to this differential equation can be written in the form:

(1)
\begin{align} \quad y = C_1y_1(t) + C_2y_2(t) + ... + C_ny_n(t) \end{align}

We will now look at the connection between the solutions $y_1$, $y_2$, …, $y_n$ forming a fundamental set of solutions and the linear independence/dependence of such solutions. We first define linear independence and linear dependence below.

 Definition: The functions $f_1$, $f_2$, …, $f_n$ are said to be Linearly Independent on an interval $I$ if for constants $k_1$, $k_2$, …, $k_n$ we have that $k_1f_1(t) + k_2f_2(t) + ... + k_nf_n(t) = 0$ implies that $k_1 = k_2 = ... = k_n = 0$ for all $t \in I$. This set of functions is said to be Linearly Dependent if $k_1f_1(t) + k_2f_2(t) + ... + k_nf_n(t) = 0$ where $k_1$, $k_2$, …, $k_n$ are not all zero for all $t \in I$.

Perhaps the simplest linearly independent sets of functions is that set that contains $f_1(t) = 1$, $f_2(t) = t$, and $f_3(t) = t^2$. Let $k_1$, $k_2$, and $k_3$ be constants and consider the following equation:

(2)
\begin{align} \quad k_1f_1(t) + k_2f_2(t) + k_3f_3(t) = 0 \\ \quad k_1 + k_2t + k_3t^2 = 0 \end{align}

It's not hard to see that equation above is satisfied if and only if the constants $k_1 = k_2 = k_3 = 0$.

For another example, consider the functions $f_1(t) = \sin t$ and $f_2(t) = \sin (t + \pi)$ defined on all of $\mathbb{R}$. This set of functions is not linearly independent. To show this, let $k_1$ and $k_2$ be constants and consider the following equation:

(3)
\begin{align} \quad k_1f_1(t) + k_2f_2(t) = 0 \\ \quad k_1 \sin t + k_2 \sin (t + \pi) = 0 \end{align}

Now choose $t = \pi$. Then we have that:

(4)
\begin{align} \quad k_1 \sin \pi + k_2 \sin 2\pi = 0 \\ \end{align}

But the above equation is true for any choice of constants $k_1$ and $k_2$ since $\sin \pi = \sin 2\pi = 0$, and thus $f_1$ and $f_2$ do not form a linearly independent set on all of $\mathbb{R}$.

From the concept of linear independence/dependence, we obtain the following theorem on fundamental sets of solutions for $n^{\mathrm{th}}$ order linear homogenous differential equations.

 Theorem 1: Let $\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0$ be an $n^{\mathrm{th}}$ order linear homogenous differential equation. If $y = y_1(t)$, $y = y_2(t)$, …, $y = y_n(t)$ are solutions to this differential equation then $y_1$, $y_2$, …, $y_n$ form a fundamental set of solutions to this differential equation on the open interval $I$ if and only if $y_1$, $y_2$, …, $y_n$ are linearly dependent on $I$.
• Proof: Consider the following $n^{\mathrm{th}}$ order linear homogenous differential equation:
(5)
\begin{align} \quad \frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = 0 \end{align}
• $\Rightarrow$ Suppose that $y = y_1(t)$, $y = y_2(t)$, …, $y = y_n(t)$ form a fundamental set of solutions to this differential equation on the open interval $I$. Then this implies that for all $t_0 \in I$ we have that :
(6)
\begin{align} \quad W(y_1, y_2, ..., y_n) \biggr \rvert_{t_0} \neq 0 \end{align}
• Thus this implies that following system of equations have only trivial solution $k_1 = k_2 = ... = k_n = 0$:
(7)
• Thus the equation $k_1y_1(t) + k_2y_2(t) + ... + k_ny_n(t) = 0$ implies that $k_1 = k_2 = ... = k_n = 0$. Thus $y_1$, $y_2$, …, $y_n$ are linearly independent on $I$.
• $\Leftarrow$ We will prove the converse of Theorem 1 by contradiction. Suppose that $y_1$, $y_2$, …, $y_n$ are linearly independent on $I$, and assume that instead $y_1$, $y_2$, …, $y_n$ do NOT form a fundamental set of solutions on $I$. Then for some $t_0 \in I$, the Wronskian $W(y_1, y_2, ..., y_n) \biggr \rvert_{t_0} = 0$. Thus the system of equations above does not have only the trivial solution. Let the constants $k_1^*$, $k_2^*$, …, $k_n^*$ be a nontrivial solution to this system. Define $\phi(t)$ as:
• Note that $y = \phi(t)$ satisfies the initial conditions $y(t_0) = 0$, $y'(t_0) = 0$, …, $y^{(n-1)} (t_0) = 0$, and $\phi(t)$ satisfies our $n^{\mathrm{th}}$ order linear homogenous differential equation because $\phi(t)$ is a linear combination of the solutions $y_1$, $y_2$, …, $y_n$.
• Now note that the function $y = 0$ also satisfies the differential equation and the initial conditions. By the existence/uniqueness theorem for $n^{\mathrm{th}}$ order linear homogenous differential equations, this implies that $\phi(t) = 0$ for all $t \in I$, so:
• But $y_1$, $y_2$, …, $y_n$ are linearly independent which implies that $k_1^* = k_2^* = ... = k_n^* = 0$. Thus $k_1^*$, $k_2^*$, …, $k_n^*$ is a trivial solution to the system above, which is a contradiction. Therefore our assumption that $y_1$, $y_2$, …, $y_n$ do not form a fundamental set of solutions was false. $\blacksquare$