The Existence/Uniqueness of Solutions to Higher Order Linear Diff. Eqs.

The Existence/Uniqueness of Solutions to Higher Order Linear Differential Equations

We will now begin to look at methods to solving higher order differential equations.

Definition: An $n^{\mathrm{th}}$ Order Differential Equation is a differential equation containing an $n^{\mathrm{th}}$ derivative and can be written in the form $\frac{d^n y}{dt^n} = f \left (t, y, \frac{dy}{dt}, ..., \frac{d^{(n-1)}y}{dt^{(n-1)}} \right )$.

Expanded out, we have that an $n^{\mathrm{th}}$ order linear differential equation has the form:

\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 = g(t) \end{align}

Of course we can also represent an $n^{\mathrm{th}}$ order linear differential equation using Lagrange notation:

\begin{align} \quad y^{(n)} + p_1(t)y^{(n-1)} + ... + p_{n-1}(t)y' + p_n(t)y = g(t) \end{align}

Here it is important to note that the notation $y^{(k)}$ does NOT mean $y$ raised to the $k^{\mathrm{th}}$ power, but instead, the $k^{\mathrm{th}}$ derivative of $y$.

Now recall from The Existence/Uniqueness of Solutions to First Order Linear Differential Equations and The Existence/Uniqueness of Solutions to Second Order Linear Differential Equations that provided that the coefficient functions were continuous on an interval $I$ containing the point $t_0$, then there existed a unique solution $y = \phi (t)$ throughout the interval $I$ that satisfied the initial conditions $y(t_0) = y_0$ (and in the case of second order linear differential equations, also satisfied $y'(t_0) = y_0'$). The following theorem is a higher order analogue to those last two theorems mentioned.

Theorem 1: Let $p_1, p_2, ..., p_n, g$ all be continuous functions on an open interval $I$ such that $t_0 \in I$. Then the $n^{\mathrm{th}}$ order linear differential equation $\frac{d^ny}{dt^n} + p_1(t) \frac{d^{n-1}y}{dt^{n-1}} + ... + p_{n-1}(t) \frac{dy}{dt} + p_n(t)y = g(t)$ with the initial conditions $y(t_0) = y_0$, $y'(t_0) = y_0'$ has a unique solution $y = \phi (t)$ throughout $I$.

Once again, it is important to stress that Theorem 1 above is simply an extension to the Theorems on the existence and uniqueness of solutions to first order and second order linear differential equations. Furthermore, for this theorem to apply, we must have that coefficient in front of the $\frac{d^n}{dt^n}$ term is $1$.

Let's look at an example of verifying that a unique solution to a higher order linear differential equation exists.

Example 1

Show that there exists a unique solution to the third order linear differential equation $t\frac{dy^3}{dt^3}+ 3 \frac{d^2y}{dt^2} + \sin t \frac{dy}{dt} + e^t y = 0$ with the initial conditions $y(1) = 1$, $y'(1) = 1$, $y''(1) = 2$.

We first divide each term by $t$ ($t \neq 0$) to get our third order linear differential equation in the appropriate form:

\begin{align} \quad \frac{dy^3}{dt^3}+ \frac{3}{t} \frac{d^2y}{dt^2} + \frac{\sin t}{t} \frac{dy}{dt} + \frac{e^t}{t} y = 0 \end{align}

Note that the functions $p_1(t) = \frac{3}{t}$, $p_2(t) = \frac{\sin t}{t}$, and $p_3(t) = \frac{e^t}{t}$ are all continuous for $t \neq 0$. However, we have that $t_0 = 1 \neq 0$, and so in fact we have that a unique solution $y = \phi(t)$ exists to this third order linear differential equation through the interval $(0, \infty)$.

Example 2

For what intervals do solutions to the differential equation $(t - 1) y^{(4)} + \frac{1}{t - 3} y^{(3)} - \ln t y = 4$ exist?

We divide both sides of the equation by $t - 1$ to get:

\begin{align} \quad y^{(4)} + \frac{1}{(t - 1)(t - 3)} y^{(3)} - \frac{\ln t}{(t - 1)} y = \frac{4}{(t - 1)} \end{align}

Note that $\frac{1}{(t - 1)(t - 3)}$ is continuous for $t \neq 1$, $t \neq 3$, $\frac{\ln t}{(t - 1)}$ is continuous for $t > 0$ and $t \neq 1$ and $\frac{4}{(t - 1)}$ is continuous for $t \neq 1$. Therefore, solutions exist on the interval $(0, 1)$, $(1, 3)$, and $(3, \infty)$.

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