[CALCULUS] Existence, uniqueness and linear operators

What guarantees a local IVP solution and why linear solutions form an affine translate of a kernel.

This page answers two questions that the solution methods leave open. Does an initial value problem have a solution near the initial point and do the data determine it uniquely? For linear equations, it also explains why every solution is one particular solution plus a homogeneous one.

The existence theorem accounts for the different IVP outcomes in ODE fundamentals and its Lipschitz condition underlies the error estimates for numerical solutions. The operator structure is the one used for second-order constant-coefficient equations.

Existence and uniqueness

[DEFINITION]

Lipschitz condition in the dependent variable

A function $f(x,y)$ satisfies a Lipschitz condition in $y$ on a region when there is a constant $M$ such that, for each fixed $x$ and all $y_1,y_2$ with $(x,y_1)$ and $(x,y_2)$ in the region,

|f(x,y_1)-f(x,y_2)| \leq M|y_1-y_2|

The condition controls how quickly the slope field can change as $y$ changes. It is stronger than continuity in $y$ and prevents two solution curves through the same initial point from separating while they remain in the region.

[1]:Local existence and uniqueness

[THEOREM]

Local existence and uniqueness

For the IVP

y'=f(x,y),\qquad y(x_0)=y_0,

if $f$ is continuous and satisfies a Lipschitz condition in $y$ on a rectangle contained in its domain and containing $(x_0,y_0)$ , then the IVP has a unique solution on some interval around $x_0$ . The conclusion is local; a global solution needs the hypotheses to remain valid along the solution.

The Lipschitz condition is sufficient, not necessary: if it fails, uniqueness may still hold.

[EXAMPLE]

Reading the hypotheses

The theorem explains the four outcomes from ODE fundamentals:

For $f(x,y)=x$ , the function is continuous and independent of $y$ , so the IVP $y'=x$ , $y(0)=1$ is uniquely determined.
For $f(y)=3y^{2/3}$ (interpreting the real power as $|y|^{2/3}$ ), the Lipschitz condition fails near $y=0$ . This failure alone does not imply nonuniqueness, but here the IVP $y'=3y^{2/3}$ , $y(0)=0$ has the distinct solutions $y=0$ and $y=x^3$ , since $(x^3)'=3x^2=3|x^3|^{2/3}$ .
In $xy'=y-1$ , the coefficient of $y'$ vanishes at $x=0$ , so the equation cannot be put into the usual normal form $y'=f(x,y)$ at the initial point.
The equation $|y'|+|y|=0$ forces $y=0$ , which is incompatible with $y(0)=1$ ; no solution satisfies that data.

[challenge]

[1]:Local existence and uniqueness

Linear differential operators

[DEFINITION]

Linear differential operator

On an interval $I$ where its leading coefficient is nonzero, a linear differential operator of order $n$ can be divided by that coefficient and written in the normalised form

T[y] =y^{(n)}+a_{n-1}(x)y^{(n-1)} +\cdots+a_1(x)y'+a_0(x)y.

It satisfies

T[y_1+y_2]=T[y_1]+T[y_2], \qquad T[\lambda y_1]=\lambda T[y_1].

Linearity follows by distributing derivatives and collecting terms. The constant-coefficient operator $T[y]=y''+ay'+by$ is the operator used in second-order constant-coefficient equations and its characteristic equation.

[1]:All forced solutions are an affine translate

[DEFINITION]

Kernel of an operator

The kernel of a linear differential operator $T$ is

\ker(T)=\{y:T[y]=0\}.

The kernel is exactly the set of homogeneous solutions. Adding an element of the kernel to a particular solution leaves the forcing unchanged. This is the differential-equation version of the kernel of a linear map: the kernel contains the inputs the operator sends to zero.

[1]:Kernel
[2]:All forced solutions are an affine translate

[THEOREM]

All forced solutions are an affine translate

If $T[y]=f$ and $y_p$ is one particular solution, then

\{\text{all solutions of }T[y]=f\} =y_p+\ker(T).

Equivalently, every solution is $y_p+y_h$ for some homogeneous solution $y_h$ .

The solution set is generally not a vector space when $f\neq0$ , because it need not contain the zero function. It is a translate of the homogeneous solution space, as for solution classes of linear systems.

[1]:Linear differential operator
[2]:Kernel of an operator
[3]:Classes of solutions of $Ax=b$
[4]:Proof
[5]:A particular solution translated by the kernel

[PROOF]

If $y_h\in\ker(T)$ , then

T[y_p+y_h]=T[y_p]+T[y_h]=f+0=f,

so every element of $y_p+\ker(T)$ is a solution. Conversely, if $y$ is any solution of $T[y]=f$ , then

T[y-y_p]=T[y]-T[y_p]=f-f=0.

Thus $y-y_p\in\ker(T)$ , so $y=y_p+y_h$ for some $y_h\in\ker(T)$ .

[1]:All forced solutions are an affine translate

[EXAMPLE]

A particular solution translated by the kernel

For

y'+\frac{y}{x}=x,

on an interval with $x\neq0$ , take

T[y]=y'+\frac{y}{x}.

The homogeneous equation $T[y]=0$ has solution

y_h=\frac{C}{x}.

A particular solution is $y_p=x^2/3$ , since

T\left[\frac{x^2}{3}\right] =\frac{2x}{3}+\frac{x}{3} =x.

Therefore the general solution is

y=\frac{C}{x}+\frac{x^2}{3}.

[routine]

[1]:All forced solutions are an affine translate

Existence, uniqueness and linear operators

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