The Lipschitz Condition and Its Difference from Continuity

1Introduction

This lecture distinguishes continuity from Lipschitz continuity and defines a standard sufficient condition for uniqueness in an initial value problem.

2Terminology and Definition

Consider $$y^{\prime }=f(x,y)$$y'=f(x,y) on a region $$R$$R. The Lipschitz condition in $$y$$y means that there exists a constant $$L\text{[PARSE ERROR: Undefined("Command("ge")")]}0$$L\ge0, independent of $$x,y_1,y_2$$x,y_1,y_2, such that

whenever $$(x,y_1),(x,y_2)\in R$$(x,y_1),(x,y_2)\in R have the same value of $$x$$x. The essential point is that one constant $$L$$L applies throughout $$R$$R rather than being chosen separately for each $$x$$x.

If this condition holds on a sufficiently small region around every point, then $$f$$f is locally Lipschitz in $$y$$y. If one constant $$L$$L applies on the entire region under consideration, then $$f$$f is globally Lipschitz in $$y$$y.

3Motivation for the Lipschitz Condition

For each fixed $$x$$x, continuity in the $$y$$y direction ensures that sufficiently small changes in $$y$$y produce sufficiently small changes in $$f(x,y)$$f(x,y). The Lipschitz condition in $$y$$y gives the stronger quantitative bound $$L|y_1-y_2|$$L|y_1-y_2|, uniformly on the region under consideration. This condition alone does not imply continuity in the $$x$$x direction. Its linear control in the $$y$$y direction is what prevents two candidate solutions issuing from the same initial value from separating.

4How the Condition Produces Uniqueness

Assume that $$f$$f is continuous and satisfies a Lipschitz condition in $$y$$y with one constant $$L$$L on the region under consideration. Also assume that two solutions $$y_1,y_2$$y_1,y_2 with the same initial value already exist; the purpose is only to prove that they coincide. The existence of a solution does not follow from the Lipschitz condition alone. A subsequent existence theorem will treat existence by using continuity separately from uniqueness.

First suppose that $$x\text{[PARSE ERROR: Undefined("Command("ge")")]}{x}_0$$x\ge x_0. The integral form of each solution is

Taking the difference gives

If $$f$$f is Lipschitz in $$y$$y with constant $$L$$L, then

The following zero-initial-value form of Gronwall's inequality converts this estimate into equality of the two solutions.

4.1Proposition: Zero-Initial-Value Form of Gronwall's Inequality

Let $$u:[a,b]\to [0,\mathrm{\infty })$$u:[a,b]\to[0,\infty) be continuous, and let $$L\text{[PARSE ERROR: Undefined("Command("ge")")]}0$$L\ge0. If

for every $$x\in [a,b]$$x\in[a,b], then $$u\left(x\right)=0$$u(x)=0 throughout $$[a,b]$$[a,b].

4.1.1Proof

Define

Because $$u$$u is continuous and nonnegative, $$H$$H is differentiable and satisfies $$H\left(a\right)=0$$H(a)=0, $$H\left(x\right)\text{[PARSE ERROR: Undefined("Command("ge")")]}0$$H(x)\ge0, and $$H^{\prime }\left(x\right)=u\left(x\right)$$H'(x)=u(x). The hypothesis gives

Consequently,

It follows that $$e^{-Lx}H\left(x\right)\text{[PARSE ERROR: Undefined("Command("le")")]}{e}^{-La}H\left(a\right)=0$$e^{-Lx}H(x)\le e^{-La}H(a)=0. Since $$H\left(x\right)\text{[PARSE ERROR: Undefined("Command("ge")")]}0$$H(x)\ge0, we must have $$H\left(x\right)=0$$H(x)=0, and hence $$u\left(x\right)=H^{\prime }\left(x\right)=0$$u(x)=H'(x)=0.

Fix an arbitrary point $$x\text{[PARSE ERROR: Undefined("Command("ge")")]}{x}_0$$x\ge x_0 in the common domain. On the compact interval joining $$x_0$$x_0 to $$x$$x, the same Lipschitz constant $$L$$L applies along both solution graphs. Apply the proposition to $$u\left(t\right)=\left|y_1\right(t)-y_2(t\left)\right|$$u(t)=|y_1(t)-y_2(t)| with $$a=x_0$$a=x_0 and $$b=x$$b=x. It follows that $$y_1\left(x\right)=y_2\left(x\right)$$y_1(x)=y_2(x). If instead a point $$x<x_0$$x<x_0 is fixed, the corresponding estimate is

Set $$s=x_0-t$$s=x_0-t and $$w\left(s\right)=u(x_0-s)$$w(s)=u(x_0-s) for $$0\text{[PARSE ERROR: Undefined("Command("le")")]}s\text{[PARSE ERROR: Undefined("Command("le")")]}{x}_0-x$$0\le s\le x_0-x. Then

so the same proposition gives $$w=0$$w=0 and therefore $$y_1\left(x\right)=y_2\left(x\right)$$y_1(x)=y_2(x). Since $$x$$x was arbitrary, the two solutions coincide throughout their common domain. This is the mechanism by which the Lipschitz condition yields uniqueness.

Continuity alone does not provide an estimate that controls the difference by a constant multiple. The distinction is not merely whether nearby inputs have nearby outputs, but whether the rate of their separation can be uniformly controlled.

5Examples

5.1A Function That Satisfies a Lipschitz Condition

For $$f\left(y\right)=3y+1$$f(y)=3y+1,

so one may choose $$L=3$$L=3.

5.2A Continuous Function That Is Not Lipschitz

The function $$f\left(y\right)=\sqrt{\left|y\right|}$$f(y)=\sqrt{|y|} is continuous at $$y=0$$y=0. However,

diverges as $$y\to 0$$y\to0. Hence no single constant $$L$$L controls the difference on a neighborhood of $$0$$0.

5.3Relation to Continuity of $$\mathit{\partial }f/\mathit{\partial }y$$\partial f/\partial y

Let $$R$$R be a closed and bounded rectangle whose interior contains the initial point. If $$\mathit{\partial }f/\mathit{\partial }y$$\partial f/\partial y is continuous on $$R$$R, the extreme value theorem gives the finite constant

For each fixed $$x$$x, the mean value theorem then establishes the Lipschitz condition in $$y$$y on $$R$$R. On an open or unbounded region, continuity alone does not imply boundedness. Moreover, continuity of the partial derivative is sufficient for the Lipschitz condition but is not necessary.

6Scope

The Lipschitz condition is a standard sufficient condition for uniqueness. Failure of the condition does not necessarily imply nonuniqueness; failure of a theorem's hypothesis does not imply the negation of its conclusion.

Local Lipschitz continuity alone does not establish existence for all time. Global existence also requires examination of the domain and growth of the right-hand side. The next lecture distinguishes local existence obtained from continuity from uniqueness obtained from Lipschitz continuity, and then studies continuation of a local solution through its maximal interval of existence.

7Previous Lecture

8Next Reference

9Exercises