Linear Independence of Functions

If we have an $n^{\mathrm {th} }$ order linear homogenous differential equation ${\frac {d^{n}y}{dt^{n}}}+p_{1}(t){\frac {d^{n-1}y}{dt^{n-1}}}+...+p_{n-1}(t){\frac {dy}{dt}}+p_{n}(t)y=0$ 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:

\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 $k_{1}f_{1}(t)+k_{2}f_{2}(t)+...+k_{n}f_{n}(t)=0$ 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 $k_{1}f_{1}(t)+k_{2}f_{2}(t)+...+k_{n}f_{n}(t)=0$ 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:

\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:

\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:

\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 ${\frac {d^{n}y}{dt^{n}}}+p_{1}(t){\frac {d^{n-1}y}{dt^{n-1}}}+...+p_{n-1}(t){\frac {dy}{dt}}+p_{n}(t)y=0$ 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:

\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 :

\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 $k_{1}=k_{2}=...=k_{n}=0$:

\begin{align} \quad k_1y_1(t) + k_2y_2(t) + ... + k_ny_n(t) = 0 \\ \quad k_1y_1'(t) + k_2y_2'(t) + ... + k_ny_n'(t) = 0 \\ \quad \quad \quad \quad \quad \quad \vdots \quad \quad \quad \quad \quad \quad \\ \quad k_1y_1^{(n-2)}(t) + k_2y_2^{(n-2)}(t) + ... + k_ny_n^{(n-2)}(t) = 0 \\ \quad k_1y_1^{(n-1)}(t) + k_2y_2^{(n-1)}(t) + ... + k_ny_n^{(n-1)}(t) = 0 \end{align}

• Thus the equation $k_{1}y_{1}(t)+k_{2}y_{2}(t)+...+k_{n}y_{n}(t)=0$ 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 Failed to parse (syntax error): {\displaystyle 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:

\begin{align} \quad \phi(t) = k_1^* y_1(t) + k_2^* y_2(t) + ... + k_n^* y_n(t) \end{align}

• 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:

\begin{align} \quad 0 = k_1^* y_1(t) + k_2^* y_2(t) + ... + k_n^* y_n(t) \end{align}

• 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. $\blacksquare$