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3.7: Uniqueness and Existence for Second Order Differential Equations

( \newcommand{\kernel}{\mathrm{null}\,}\)

Recall that for a first order linear differential equation

y+p(t)y=g(t)y(t0)=y0

if p(t) and g(t) are continuous on [a,b], then there exists a unique solution on the interval [a,b].

We can ask the same questions of second order linear differential equations. We need to first make a few comments. The first is that for a second order differential equation, it is not enough to state the initial position. We must also have the initial velocity. One way of convincing yourself, is that since we need to reverse two derivatives, two constants of integration will be introduced, hence two pieces of information must be found to determine the constants.

A second comment is that of notation. Let

y

be a second order linear differential equation. Then we call the operator

L(y) = y'' + p(t)y' + q(t)y \nonumber

the corresponding linear operator. Thus we want to find solutions to the equation

L(y) = g(t), \;\;\; y(t_0) = y_0, y' (t_0) = y'_0. \nonumber

We will state the following theorem without proof. The proof is well above the level of this course.

Theorem: Existence and Uniqueness

Let p(t), q(t), and g(t) be continuous on [a,b], then the differential equation

y'' + p(t) y' + q(t) y = g(t), \;\;\; y(t_0) = y_0, \;\;\; y'(t_0) = y'_0 \label{EE}

has a unique solution defined for all t in [a,b].

Example \PageIndex{1}

Find the largest interval where

(t^2 -1 )y'' + 3ty' + \cos t y = e^t, \;\;\; y(0) = 4, \;\;\; y'(0) = 5 \nonumber

is guaranteed to have a unique solution.

Solution

We first put it into standard form

y'' + \frac {3t}{t^2 - 1} y' + \frac {\cos t }{ t^2 -1} y = \frac {e^t}{ t^2 -1}, \;\;\; y(0) = 4, \;\;\; y'(0) = 5. \nonumber

p, q, and g are all continuous except at t = -1 and t = 1 . The Existence and Uniqueness theorem (Equation \red{EE}) tells us that there is a unique solution on [-1,1].

Homogeneous Linear Second Order Differential Equations

Next we will investigate solutions to homogeneous differential equations. Consider the homogeneous linear differential equation

L(y) = 0. \nonumber

We have the following theorem

Theorem

Let L (y) = 0 be a homogeneous linear second order differential equation and let y_1 and y_2 be two solutions. Then c_1y_1 + c_2y_2 is also a solution for any pair or constants c_1 and c_2.

Using the terminology of linear algebra, we know that L is a linear transformation of the vector space of differentiable functions into itself. The theorem reminds us that the kernel of a linear transformation is a vector subspace.

Proof: The Wronskian

\begin{align*} L ( c_1y_1 + C_2y_2) &= (c_1y_1 + c_2y_2)'' + p(t)(c_1y_1 + c_2y_2)' + q(t)(c_1y_1 + c_2y_2) \\[4pt] &= c_1y''_1 + c_2y''_2 + p(t)c_1y'_1 + p(t) c_2y'_2 + q(t) c_1y_1 + q(t) c_2y_2 \\[4pt] &= c_1y''_1 + p(t)c_1y'_1 + q(t)c_1y_1 + q(t) c_2y''_2 + p(t) c_2y'_2 + q(t)c_2y_2 \\[4pt] &= c_1(y''_1 + p(t)y'_1 + q(t)y_1) + c_2(y''_2 + p(t)y'_2 + q(t) y_2) \\[4pt] &= c_1L(y_1) + c_2L(y_2) \\[4pt] &= 0 + 0 = 0. \end{align*}\nonumber

Next, we investigate the initial conditions. If we find a general solution to the homogenous system, can we choose constants such that the solution satisfies the initial conditions? That is can we find c_1 and c_2 such that

c_1y_1(t_0) + c_2y_2(t_0) = y_0 \nonumber

c_1y'_1(t_0) + c_2y'_2(t_0) = y'_0. \nonumber

We can put this into a matrix equation

{ \begin{pmatrix} y_1(t_0) y_2(t_0) \\ y'_1(t_0) y'_2(t_0) \end{pmatrix} \begin{pmatrix} c_1\\ c_2 \end{pmatrix} = \begin{pmatrix} y_0 \\ y'_0 \end{pmatrix} } \label{Wronskian}\nonumber

This has a unique solution if and only if the determinant of the matrix is not zero; this determinant is called the Wronskian.

This proves the following theorem:

Theorem

Let

L(y) = 0 \;\;\; y(t_0) = y_0 \;\;\;y'(t_0) = y'_0 \nonumber

be a homogeneous linear second order differential equation and let y_1 and y_2 be two general solutions (No initial value). Then if the Wronskian

y_1y'_2 - y'_1y_2 \nonumber

is nonzero, there exists a solution to the initial value problem of the form

y = c_1y_1 + c_2y_2. \nonumber

Example \PageIndex{2}

Consider the differential equation

y'' + 2y' - 8y = 0 \nonumber

It is easy to check that the general solution is given by

y = c_1e^{2t} + c_2e^{-4t}. \nonumber

The Wronskian of

y_1 = e^{2t}, \;\;\; y_2 = e^{-4t} \nonumber

is given by

e^{2t}(-4e^{-4t}) - (2e^{2t})e^{-4t} = -4e^{-2t} - 2e^{-2t} = -6e^{-2t}. \nonumber

Which is never zero. We can conclude that any initial value problem will have a unique solution of the form

y = c_1e^{2t} + c_2e^{-4t}. \nonumber

Contributors and Attributions


This page titled 3.7: Uniqueness and Existence for Second Order Differential Equations is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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