Download original notes
PAGE 1

2.4 Linear vs. Nonlinear Diff. Eqs.

Given \( y' = f(t, y) \), \( y(t_0) = y_0 \).

  • Is there a solution? existence
  • Is the solution unique? uniqueness

For linear eqs., the answers are straightforward:

\[ y' + p(t)y = g(t) \quad y(t_0) = y_0 \]

Solution: find integrating factor

\[ \mu(t) = e^{\int p(t) dt} \]

For solution to exist, \( \int p(t) dt \) must exist.

\( \rightarrow p(t) \) must be continuous on some interval of \( t \).

\[ \mu(t) (y' + p(t)y) = \mu(t)g(t) \]
PAGE 2
\[ \frac{d}{dt} (\mu(t)y(t)) = \mu(t)g(t) \]

Integrate with respect to \( t \):

\[ \mu(t)y(t) = \int \mu(t)g(t) + C \]

For integral to exist, \( g(t) \) must be continuous on some interval.

\( \mu(t) \) exists on the interval that \( p(t) \) is continuous.

So, for the solution to exist, we must restrict ourselves to an interval on which BOTH \( p(t) \) and \( g(t) \) are continuous (potentially multiple) AND containing \( t_0 \).

And the solution is unique.

PAGE 3

Example: Existence and Uniqueness of Solutions

\[ (4 - t^2)y' + 2ty = 3t^2, \quad y(-3) = 1 \]

First, we rewrite the differential equation in standard form by dividing by \( (4 - t^2) \):

\[ y' + \left( \frac{2t}{4 - t^2} \right) y = \frac{3t^2}{4 - t^2} \]

\( p(t) = \frac{2t}{4 - t^2} \)

\( g(t) = \frac{3t^2}{4 - t^2} \)

Continuity Analysis

\( p(t) \) is continuous on: \( (-\infty, -2), (-2, 2), (2, \infty) \)

\( g(t) \) is continuous on: \( (-\infty, -2), (-2, 2), (2, \infty) \)

To choose the correct interval, look at the initial \( t \): here, it's \( -3 \).

Choose the interval containing that \( t \).

So, the diff. eq. has a unique solution on \( (-\infty, -2) \).

PAGE 4

Example: Existence and Uniqueness with Transcendental Functions

\[ \ln(t) y' + y = \cot(t), \quad y(2) = 3 \]

Rewrite in standard form by dividing by \( \ln(t) \):

\[ y' + \left( \frac{1}{\ln(t)} \right) y = \frac{\cot(t)}{\ln(t)} \]

\( p(t) = \frac{1}{\ln(t)} \)

\( g(t) = \frac{\cot(t)}{\ln(t)} \)

Continuity Analysis

  • \( p(t) \) continuous on: \( (0, 1), (1, \infty) \)
  • \( g(t) \) continuous on: \( (0, 1), (1, \pi), (\pi, 2\pi), (2\pi, 3\pi), \dots \)

The interval where BOTH are continuous AND containing \( t = 2 \) is \( (1, \pi) \).

So, on \( 1 < t < \pi \), there is one and only one solution.

PAGE 5

Uniqueness of Solutions

Uniqueness means only one solution (one solution curve).

Two graphs on t-y axes; one shows a single curve through (t0, y0), the other shows multiple curves.

This also implies that solution curves cannot intersect.

Graph on t-y axes showing two green solution curves intersecting at a single point.

Because if we choose the intersection point as initial condition, then uniqueness is violated.

PAGE 6

Also, for linear eqs, the general solution contains all possible solutions.

Example

For example, \( y' = y \). \( y = 0 \) is obviously a solution.

But if solved as separable:

\[ \frac{1}{y} dy = dt \quad (y \neq 0) \]

\[ \ln|y| = t + C \]

\[ y = Ce^t \]

Notice \( y = 0 \) is included (\( C = 0 \)).

ALWAYS true for linear eq.

(NOT always true for nonlinear)

PAGE 7

Nonlinear Differential Equations

For nonlinear, situation is complicated.

\[ y' = f(t, y) \quad y(t_0) = y_0 \]

has a unique solution on some interval within the rectangle \( \alpha < t < \beta, \quad \gamma < y < \delta \) on which \( f(t, y) \) and \( \frac{\partial f}{\partial y}(t, y) \) are BOTH continuous AND containing \( y(t_0) = y_0 \).

Comparison with Linear Equations

If \( y' = f(t, y) \) is linear, then:

\[ y' = -p(t)y + g(t) \]
\[ \frac{\partial f}{\partial y} = -p(t) \]

Same continuity requirement as earlier.

\[ y = -p(t)y + g(t) \]

Also needs to be continuous (same as before).

PAGE 8

Example Analysis

\[ y' = y^3 \quad y(0) = 0 \]
\[ f = y^3 \] \[ \frac{\partial f}{\partial y} = 3y^2 \]

Continuous on \( -\infty < y < \infty \)

No restrictions on \( t \), so \( -\infty < t < \infty \)

So, the rectangle is:

  • \( -\infty < t < \infty \)
  • \( -\infty < y < \infty \)
Coordinate graph showing a cubic curve passing through the origin (t0, y0) within a dashed rectangular region.

Unique Solution

Unique solution on \( t_0 - h < t < t_0 + h \)

But what is \( h \)?

(Don't know, at least easily)

PAGE 9

Solving a Separable Differential Equation

\[ y' = y^3 \quad y(0) = 0 \]

Note that \( y = 0 \) is a solution.

Separation of Variables

As the equation is separable, we can rewrite it as:

\[ \frac{1}{y^3} dy = dt \quad (y \neq 0) \]
\(\vdots\)
\[ y = \frac{1}{\pm \sqrt{-2t + C}} \]

General solution

\( y \neq 0 \) is NOT here!