Existence, Uniqueness, and the Maximal Interval for an Initial Value Problem

1Introduction

This lecture distinguishes the existence of a solution to an initial value problem from the uniqueness of that solution.

2Initial Value Problem under Consideration

The standard form is

The principal questions are not whether an explicit formula is available, but whether a solution exists near the initial point $$(x_0,y_0)$$(x_0,y_0) and whether that solution is uniquely determined.

3Why Existence and Uniqueness Must Be Distinguished

Existence and uniqueness can often be decided even when no analytic formula is available. Continuity of the right-hand side $$f$$f is the basic hypothesis for existence. Local Lipschitz continuity in the $$y$$y-variable is a standard sufficient condition for uniqueness, but it is not necessary.

4Local Existence and Uniqueness Theorems

4.1Peano Existence Theorem

Let $$D\subset R^2$$D\subset\mathbb{R}^2 be open and let $$(x_0,y_0)\in D$$(x_0,y_0)\in D. If $$f:D\to \mathbb{R}$$f:D\to\mathbb{R} is continuous, then there exists $$h>0$$h>0 such that the initial value problem

has at least one solution on $$|x-x_0|<h$$|x-x_0|<h. This theorem does not assert uniqueness.

4.2Picard--Lindelof Theorem

Let $$D\subset R^2$$D\subset\mathbb{R}^2 be open and let $$(x_0,y_0)\in D$$(x_0,y_0)\in D. If $$f:D\to \mathbb{R}$$f:D\to\mathbb{R} is continuous and locally Lipschitz in $$y$$y, then there exists $$h>0$$h>0 such that the initial value problem has a unique solution on $$|x-x_0|<h$$|x-x_0|<h. Uniqueness means that any two solutions with the same initial condition coincide wherever both are defined.

The two levels of conclusion can be summarized as

and the useful sufficient test

5Why the Lipschitz Condition Produces Uniqueness

The preceding lecture bounded the difference between two solutions with the same initial value by a Lipschitz constant and then applied the zero-initial-value form of Gronwall's inequality. Here that mechanism is used through the hypotheses and conclusion of the Picard--Lindelof theorem.

6Decision Table

7Maximal Interval of Existence and Continuation

Assume that $$f$$f is continuous on an open region $$D$$D and locally Lipschitz in $$y$$y. Local unique solutions furnished by the Picard--Lindelof theorem agree on overlaps and therefore combine into a unique solution on a maximal open interval $$(\alpha ,\beta )$$(\alpha,\beta) containing the initial time. This interval is the maximal interval of existence.

In the plane, a closed and bounded subset is called a compact set.

If continuation stops at a finite endpoint, for example if $$\beta <\mathrm{\infty }$$\beta<\infty, then as $$x\to {\beta }^{-}$$x\to\beta^- the graph $$(x,y(x\left)\right)$$(x,y(x)) cannot remain in any compact subset of $$D$$D. If it eventually remained in a compact set $$K\subset D$$K\subset D, continuity would bound $$f$$f on $$K$$K, so $$\left|y^{\prime }\right|\text{[PARSE ERROR: Undefined("Command("le")")]}M$$|y'|\le M. Hence $$y\left(x\right)$$y(x) would be Cauchy as $$x\to {\beta }^{-}$$x\to\beta^- and would have a finite limit $$y_{\beta }$$y_\beta. Since $$K$$K is closed, $$(\beta ,y_{\beta })\in K\subset D$$(\beta,y_\beta)\in K\subset D. Reapplying the local theorem at this point would continue the solution beyond $$\beta$$\beta, contradicting maximality. In particular, when $$D=R^2$$D=\mathbb{R}^2, termination at a finite endpoint requires $$\left|y\right(x\left)\right|$$|y(x)| to become unbounded. Local uniqueness alone does not imply $$\alpha =-\mathrm{\infty }$$\alpha=-\infty and $$\beta =\mathrm{\infty }$$\beta=\infty.

Now suppose that $$f$$f is continuous on all of $$R^2$$\mathbb{R}^2 and satisfies a global Lipschitz condition in $$y$$y, uniform in $$x$$x:

For any $$T>0$$T>0, continuity gives a bound $$\left|f\right(x,0\left)\right|\text{[PARSE ERROR: Undefined("Command("le")")]}{M}_T$$|f(x,0)|\le M_T on the finite interval $$|x-x_0|\text{[PARSE ERROR: Undefined("Command("le")")]}T$$|x-x_0|\le T. Hence

For $$x\text{[PARSE ERROR: Undefined("Command("ge")")]}{x}_0$$x\ge x_0, set $$A=\left|y_0\right|+M_{T}T$$A=|y_0|+M_TT. The integral equation gives $$u\left(x\right)=\left|y\right(x\left)\right|\text{[PARSE ERROR: Undefined("Command("le")")]}A+L{\int }_{x_0}^{x}u\left(t\right)dt$$u(x)=|y(x)|\le A+L\int_{x_0}^x u(t)\,dt. Defining $$z\left(x\right)=A+L{\int }_{x_0}^{x}u\left(t\right)dt$$z(x)=A+L\int_{x_0}^x u(t)\,dt gives $$z^{\prime }\text{[PARSE ERROR: Undefined("Command("le")")]}Lz$$z'\le Lz and $$z\left(x_0\right)=A$$z(x_0)=A, so monotonicity of $$e^{-Lx}z\left(x\right)$$e^{-Lx}z(x) yields $$u\left(x\right)\text{[PARSE ERROR: Undefined("Command("le")")]}A{e}^{L(x-x_0)}$$u(x)\le Ae^{L(x-x_0)}. Reversing the integration interval gives the analogous estimate for $$x\text{[PARSE ERROR: Undefined("Command("le")")]}{x}_0$$x\le x_0. Thus $$\left|y\right(x\left)\right|$$|y(x)| remains finite on every finite time interval. Finite-time blow-up is impossible, and the continuation criterion implies that the maximal interval of existence is all of $$\mathbb{R}$$\mathbb{R}.

8Examples

8.1An Initial Value Problem with a Unique Solution

For

we have $$f(x,y)=y$$f(x,y)=y and $$\mathit{\partial }f/\mathit{\partial }y=1$$\partial f/\partial y=1, which is continuous. A unique local solution therefore exists. In fact, the solution is $$y=e^x$$y=e^x.

8.2Failure of Uniqueness

For

the function $$f\left(y\right)=\sqrt{\left|y\right|}$$f(y)=\sqrt{|y|} is continuous, so the Peano theorem guarantees a local solution. It is not Lipschitz near $$y=0$$y=0. The function $$y\left(x\right)=0$$y(x)=0 is a solution, and for every $$a\text{[PARSE ERROR: Undefined("Command("ge")")]}0$$a\ge0,

is also a solution. Thus the same initial condition produces multiple solutions, and uniqueness fails.

8.3A Finite Maximal Interval of Existence

For

separation of variables gives

The solution diverges at $$x=1$$x=1, so the maximal interval containing the initial time $$0$$0 is $$(-\mathrm{\infty },1)$$(-\infty,1). Although the right-hand side is smooth on the entire plane, local Lipschitz continuity alone does not prevent finite-time blow-up.

9Cautions in Applying the Theorems

• Continuity of $$\mathit{\partial }f/\mathit{\partial }y$$\partial f/\partial y must not be treated as a necessary condition for uniqueness.
• Existence of a solution must not be confused with representability by elementary functions.
• Local existence must not be confused with global existence.

10Scope

The theorems treated here constitute a local theory for first-order initial value problems. Boundary value problems and partial differential equations require different hypotheses and theories.

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12Exercises