Bridge to Ordinary Differential Equations

1From Calculus to Determining an Unknown Function

This lecture introduces an ordinary differential equation as a relation containing derivatives from which an unknown function is to be determined. A differential equation with one independent variable is an ordinary differential equation (ODE). Its order is the order of its highest derivative.

For example, the general solution of $y^{\prime }=2x$y'=2x is

The arbitrary constant $C$C represents information lost by differentiation. A condition such as $y\left(x_0\right)=y_0$y(x_0)=y_0, which specifies a value at one point, is an initial condition. A differential equation together with initial conditions is an initial-value problem. Here the condition uniquely determines $C=y_0-x_{0^2}$C=y_0-x_0^2.

2Equation Form and Method Selection

On an interval where $y\ne 0$y\ne0, the equation $y^{\prime }=xy$y'=xy can be separated as

Integration gives $y=C{e}^{x^2/2}$y=Ce^{x^2/2}. Because $C=0$C=0 also satisfies the original equation, the zero solution lost during division must be checked separately.

The equation $y^{\prime }\text{'}+y=0$y''+y=0 is a second-order linear homogeneous equation, with general solution

Determining a particular solution of a second-order equation normally requires two initial values, such as $y\left(x_0\right)$y(x_0) and $y^{\prime }\left(x_0\right)$y'(x_0). A method should be selected only after identifying the order, linearity, homogeneity, and coefficient properties of the equation.

3Existence and Uniqueness

For the initial-value problem

merely writing the equation guarantees neither existence nor uniqueness. By the Peano existence theoremPeano existence theorem, if $f$f is continuous near $(x_0,y_0)$(x_0,y_0), at least one local solution exists. By the Picard--Lindelöf uniqueness theoremPicard--Lindelöf uniqueness theorem, if $f$f is also locally Lipschitz continuous with respect to $y$y, that local solution is unique. These are local conclusions and do not exclude finite-time blow-up.

For example, in $y^{\prime }=\sqrt{\left|y\right|}$y'=\sqrt{|y|} with $y\left(0\right)=0$y(0)=0, the right-hand side is continuous but is not locally Lipschitz in $y$y near zero. Besides the zero solution $y\equiv 0$y\equiv0, every $c\text{[PARSE ERROR: Undefined("Command("ge")")]}0$c\ge0 gives a solution

At $x=c$x=c, the function values and derivatives from both sides are zero, so $y_c$y_c is $C^1$C^1 and satisfies the equation at the joining point. The solution is therefore not unique. Consequently, one must both substitute a candidate into the equation and initial conditions and verify the hypotheses of every theorem invoked.

4Subsequent Study