Fundamental Solutions to Linear Homogenous Differential Equations

# Fundamental Solutions to Linear Homogenous Differential Equations

 Theorem 1: Let $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = 0$ be a second order linear homogenous differential equation where $p$ and $q$ are continuous on an open interval $I$ such that $t_0 \in I$, and let $y = y_1(t)$ and $y = y_2(t)$ be two solutions to this differential equation. The set of all linear combinations of these two solutions, $y = Cy_1(t) + Dy_2(t)$ where $C$ and $D$ are constants contains all solutions to this differential equation if and only if there exists a point $t_0$ for which the Wronksian of $y_1$ and $y_2$ at $t_0$ is nonzero, that is $W(y_1, y_2) \biggr \rvert_{t_0} \neq 0$.
• Proof: Let $y = y_1(t)$ and $y = y_2(t)$ both be solutions to the differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = 0$ and suppose that $y = \phi (t)$ is any arbitrary solution as well. We want to show that $\phi (t)$ is a linear combination of $y_1$ and $y_2$ for some constants $C$ and $D$.
• $\Leftarrow$ Let $t_0$ be such that the Wronskian of $y_1$ and $y_2$ evaluated at $t_0$ is nonzero, that is:
(1)
\begin{align} \quad W(y_1, y_2) \biggr \rvert_{t_0} \neq 0 \end{align}
• Take this value $t_0$ and evaluate both $\phi$ and $\phi'$ at this point. Then $\phi (t_0) = y_0$ and $\phi'(t_0) = y'_0$ (since $y = \phi(t)$ is a solution to our differential equation). Now consider the initial value problem $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ with the initial conditions $y(t_0) = y_0$ and $y'(t_0) = y'_0$. The function $\phi$ satisfies this differential equation. Since $W(y_1, y_2) \biggr \rvert_{t_0} \neq 0$ then we have that there exists constants $C$ and $D$ such that $y = Cy_1(t) + Dy_2(t)$ satisfies this initial value problem. But since $p$ and $q$ are continuous on the open interval $I$ containing $t_0$ then this implies that a unique solution exists, and so:
(2)
\begin{align} \phi(t) = Cy_1(t) + Dy_2(t) \end{align}
• So all solutions for this differential equation are a linear combination of the solutions $y = y_1(t)$ and $y = y_2(t)$.
• $\Rightarrow$ Suppose that every point $t_0 \in I$ is such that $W(y_1, y_2) \biggr \rvert_{t_0} = 0$, that is, there exists no point $t_0$ on $I$ where the Wronskian of $y_1$ and $y_2$ evaluated at $t_0$ is nonzero. Let $y_0$ and $y'_0$ be values for which the system $\left\{\begin{matrix} Cy_1(t_0) + Dy_2(t_0) = y_0 \\ Cy_1'(t_0) + Dy_2'(t_0) = y'_0 \end{matrix}\right.$ has no solutions for a set of constants $C$ and $D$.
• Now since $p$ and $q$ are continuous on an open interval $I$ containing $t_0$, such a solution $\phi(t)$ satisfies the initial conditions $y(t_0) = y_0$ and $y'(t_0) = y'_0$. Note though this solution is not a linear combination of $y_1$ and $y_2$ though which completes our proof. $\blacksquare$

Theorem 1 above implies that if we can find two solutions $y = y_1(t)$ and $y = y_2(t)$ for which the Wronskian $W(y_1, y_2) \neq 0$, then for constants $C$ and $D$, all solutions of the second order linear homogenous differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ are given by:

(3)
\begin{align} \quad y = Cy_1(t) + Dy_2(t) \end{align}

Also note that thus far we have not said that $y = y_1(t)$ and $y = y_2(t)$ need to be distinct. However, with the Theorem above, we see that if $y_1(t) = y_2(t)$ then the Wronskian $W(y_1, y_2) = W(y_1, y_1) = W(y_2, y_2)$ is zero (as you should verify) and so not all solutions to a second order linear homogenous differential are given by the linear combination of just $y_1$.

 Definition: Let $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = 0$ where $p$ and $q$ are continuous on an open interval $I$ such that $t_0 \in I$ and let $y = y_1(t)$ and $y = y_2(t)$ be solutions to this differential equation. If the Wronskian $W(y_1, y_2) \neq 0$ then the set of linear combinations of $y_1$ and $y_2$ is known as the Fundamental Set of Solutions to this differential equation.

From the definition above, we see that if we can find two solutions $y = y_1(t)$ and $y = y_2(t)$ for which the Wronskian $W(y_1, y_2)$ is nonzero, then $y_1$ and $y_2$ form a fundamental set of solutions. The next question that we might pose is whether or not a second order linear homogenous differential equation always has a fundamental set of solutions.

 Theorem 2: Let $\frac{d^2 y}{dt} + p(t) \frac{dy}{dt} + q(t) y = 0$ be a second order linear homogenous differential equation where $p$ and $q$ are continuous on an open interval $I$ such that $t_0 \in I$. If $y = y_1(t)$ is a solution to this differential equation that satisfies the initial conditions $y_1(t_0) = 1$ and $y_1'(t_0) = 0$, and if $y = y_2(t)$ be a solution to this differential equation that satisfies the initial conditions $y_2(t_0) = 0$ and $y_2'(t_0) = 1$. Then $y_1$ and $y_2$ form a fundamental set of solutions for this differential equation.
• Proof: We note that:
(4)
\begin{align} \quad W(y_1, y_2) \biggr \rvert_{t_0} = \begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0)\end{vmatrix} = \begin{vmatrix} 1 & 0\\ 0 & 1 \end{vmatrix} = 1 \neq 0 \end{align}
• Thus Theorem 1 implies that ALL solutions to this differential equation are given by $y = Cy_1(t) + Dy_2(t)$ where $C$ and $D$ are constants. Thus $y_1$ and $y_2$ form a fundamental set of solutions for this differential equation. $\blacksquare$