Lemma on the Extension of Solutions. Let D C and f e

Chapter VI: Boundary Value and Eigenvalue Problems

W. Integrating Factors (or Euler Multipliers). The differential equa- tion

VI. Lemma on the Extension of Solutions. Let D C and f e

§ 6. An Existence and Uniqueness Theorem 67 V. Lemma. Let f be defined in D and let = (A be a set of functions, where is a solution to the initial value problem (1) in the interval Ja containing Assume that the following property (U) holds:

for

(a,/3EA).

(U)

Then there exists exactly one solution j defined in the interval J = U _Ja with

czEA

the property cb for every a E A.

This solution can be constructed as follows: For each x e J determine an a E A such that x E and then define = If /3 is another index with x E Jp, then by hypothesis = i.e., q5(x) is well-defined.

If a is an arbitrary point from J, then there exists an a E A such that

a E Ja. Thus a] C Jo, and qS(x) q5o,(x) holds in a]. It follows that is a solution of (1) in J.

If this lemma is applied to the set of all solutions to the initial value problem, then (U) implies the uniqueness of the solution. In summary, one obtains the following

Corollary. If the initial value problem (1) has at least one solution and if the uniqueness statement (U) holds for every pair of solutions, then there exists a solution of (1) which cannot be extended. All other solutions are restrictions

of this solution.

holdsfore<x<b.

Therefore 4) is differentiable (to the left) at b and 41(b) = f(b,4)(b)).

(b) It is

to check that u satisfies the differential equation at b.

The function u is differentiable to the left and the right at this point, and both

derivatives are equal to f(b, 4)(b)).

I

Wecome now to the main theorem of this section.

VII. Existence and Uniqueness Theorem.

Let f E C(D) satisfy a local Lipschitz condition with respect to y in D, where D C is open. Then for every ij) E D the initial value problem

y'—f(x,y),

(7)

has a solution 4) that cannot be eztended and that to the left and to the right comes arbitrarily close to the boundary of D. The solution is uniquely deter- mined in the sense that every solution of (7) is a restriction of 4).

Definition. The statement "4) comes arbitrarily close to the boundary of D to the right" is defined as follows: If G is the closure of graph 4) and if is the set of points (x, y) C

with x

then

(a) G+ is not a compact subset of D.

An equivalent formulation that gives a better understanding reads as follows:

4) exists to the right in an interval e x < ^b (b = oois allowed), and one of the following cases applies:

(b) b = 00; thesolution exists for all x

(c) b < ^{00 and}urn sup j4)(x) = 00; thesolution "becomes infinite."

(d) b <00 andUrn mi p(z, 4)(x)) =0,where p(x, y) denotes the distance from the point (x, y) to the boundary of D; the solution "comes arbitrarily close to the boundary of D."

Indeed, statement (a) says that is either unbounded (case (b) or (c)) or is bounded and contains boundary points of D (case (d)).

We have repeatedly encountered these three types of behavior. In the exam- ple y' = sinx of 1.VIII, (b) or (c) holds to the left and to the right, depending on the value of y(0). For the equation y' =

(2y)'

in the upper half plane y >0,

all solutions are given by y =

(x>

—c). Here, case (b) prevails to the right and case (d) to the left.

Proof. Uniqueness. We prove the statement "If 4) and are two solutions of the initial value problem and if J is a common interval of existence of both solutions with E J, then 4) = in J."

Let us assume on the contrary that there exist, say to the right of points x EJ with 4)(x) Then there also exists a first point ZO E Jto the right of where the two solutions separate. This x0 is the largest number with the property that 4)(x) = for

x

xo (xo = is not excluded).

§ 6. An Existence and Uniqueness Theorem 69

However, we know from IV that there exists a local solution through the point (ZO, and that it is uniquely determined. Tn other words, ç1(x) =

in a right neighborhood of This is a contradiction to our assumption about xO. The uniqueness to the left is proved similarly.

Existence. By Theorem IV there exists a local solution to (7), and as we have just proved, the uniqueness statement (U) of V holds. Thus Corollary V guarantees the existence of a nonextendable solution ^and we have only to show that it comes arbitrarily close to the boundary of D (we consider only the case "to the right" in the direction of increasing x, x C).

Assume that (a) is false. Then G+ is a compact subset of D, and exists in a finite interval x or 5 b. In the first case, Lemma VL(a) can be applied, i.e., can be extended to [e,b]. In the second case, e D, and there exists a local solution that passes through this point. Applying VI. (b), one again obtains an extension of cb.

In either case, we have a contradiction to the assumption that cannot be extended. This completes the proof of the theorem.

I

VIII.

Exercise. Let k(x, t, z) be continuous for 0 <t <x < a, —00 <

z < oo and satisfy a Lipschitz condition in z,

ik(x,t,z)^— —

andlet g(x) be continuous for 0 5 x 5 a. Show, by the fixed point theorem 5.IX, that the Volterra integral equation"

u(x) = g(x)

+ f k(x, t, u(t)) dt

has exactly one continuous solution in 0 x 5 a.

IX. Exercise. Prove: If f(x, y) satisfies a local condition with respect to y in the set D C 1R2 and if A C D is compact and f bounded on A, then f satisfies a Lipschitz condition with respect to y in A. In particular, if v,w E C([a,bl) and graphv, graphw CD, then there exists L >0 such that

jf(x,v(x)) — f(x,w(x))j 5 Llv(x) —w(x)j ^{in [a,b].}

X. Exercise. Prove: If f is continuous in the open set D and qS is a solution of (7) in the interval b) with b < oo that comes arbitrarily close to the boundary of D to the right, then at least one of the following two cases applies (both can happen at the same time):

(c') or —00 as x —+^b—;

(d') p(x, q5(x)) 0as x b—.

This sharpens the statement in VII.

Hint: Show: If Gb is the intersection of graph q5 with the line x = b, then C,, C t9D (the boundary of D).

Xi. Exercise. Rosenblatt's Condition.

Let the function f(x, y) be

continuousin the strip S =J

x R, J =

^[0,a] and satisfy the condition

f(x,y)—f(x,z)I

^{for 0< x a and y,z}

ER

with q < 1. Show that the initial value problem

y'=f(x,y) inJ,

y(O)=?)

has exactly one solution and that this solution can be obtained by the method of successive approximations. The above condition was introduced by Rosenblatt (1909).

Hint. In the Banach space B of all functions u E C(J) with finite norm

lull := sup {lu(x)l/x : 0

<x

a}, the operator T,

(Tu)(x)

I

satisfies the Lipschitz condition (5.3). If u is a fixed point of T, then y = u+ is a solution of the initial value problem.

Supplement: Singular Initial Value Problems

Here we consider a singular initial value problem for a differential equation of second order,

in

J0=(0,b],

y(O)=?7, y'(O)=O. (8) This problem is closely connected to the problem of finding rotationally sym- metric solutions of the nonlinear effiptic equation

= f(r,u), where x E R'2 and r = lxi.

XII. The Operators La and Ia.

In what follows, J = ^[0,b], Jo (0, b],

> 0, and La is the differential operator Lay = y"+ y' =

Lemma. ^Lety E C(J) fl C2(J0), y' bounded, and f(x) E C(J). If

Lay =1(x) in Jo, y(0) = ⁽⁹⁾

then y E C2(J), y'(O) = 0,

Em y'(x)/x =

y"(O) = f(0)/(a+ 1), and

y(x) = +

With

laf

=

f f

taf(t) dt ds. (10) Conversely, if y is defined by (10), then y is a solution of (9) with the above properties; in particular, y E C2 (J), y' (0) = 0.

§ 6. An Existence and Uniqueness Theorem 71

Proof. Because y' is bounded, we have xay!(x) 0 as x —p By inte-

grating = one obtains

=

f

^{taf(t) dt. (11)}

The substitution t

xr, dt =

xdT in (11) leads to y'(x)

(11')

Since f(xr) f(0) as x 0+ uniformly in T E [0,1], one derives from (11')

that y'(x)/x f(0)/(a

+ 1) and, in particular, y'(x) —k 0 as x —* 0+. By a well-known theorem from analysis (see C.VI.(b)), y e C'(J) and y'(O) 0.

Using equation (9) to determine y", we get y"(x) f(0)/(a+ 1) as x 0+.

A second application of the previously mentioned lemma shows that y E C2(J) and that y"(O) has the specified value.

Solving equation (11) for y' and then integrating gives (10). The integrand taf(t)_{dt in} is a continuous function in J vanishing at 0. This follows from the boundeciness of f together with the inequality < 1.

Conversely, one obtains (11) by differentiating equation (10). As we have seen, the specified properties of y follow from (11). Finally, if equation (11) for xayl is differentiated, (9) follows.

XIII. Existence and Uniqueness Theorem.

Let the function f(x, y) be continuous in J x R and satisfy a Lipschitz condition (2) in y. Then for given

a >

^0, the initial value problem (8) has exactly one solution y E C2(J).

Proof. By the previous lemma, (8) is equivalent -to the Volterra integral equation y = + Iaf(•, y), which can be written in the form

y(x) = + f k(x, t)f(t, y(t)) dt

(12)

with

k(x,t) =

^ta

Since

1 implies that 0 k(x, t) x —

t, it follows that the "kernel"

k(x, t) is continuous in the triangle D :

0 t x b. The assertion now

follows from the theorem in Exercise VIII. U

XIV.

Rotationally Symmetric Solutions of Elliptic Differential Equations.

In the following, x E ^RTh,

n 2

^{2, r} = (Eucidean norm), and

is the Laplace operator

+'UX2X2+ +

The Laplace operator for rotationally symmetric functions u(x) = y(jxl) (they are also called radial functions) is given by

for proof, use the formulas for u2, given below. The results established here lead at once to the following

Existence and Uniqueness Theorem. Let the function f(r, z) be con-

tinuous in J x R and Lipschitz continuous in z. Denote by Bb the closed ball

Bb: xI

b. Then the differential equation

in Bb

has exactly one rotationally symmetric solution u e C2(Bb) satisfying the initial condition u(0) =u0.

This result follows immediately from the previous theorem. The solution u is obtained in the form u(x) = where y(r) is the solution of (12) with

=

n— 1, u0, and r in the place of x. One question needs clarification.

While the C2-property of y carries over at once to u as long as x 0, the same is not immediately obvious for x =0. The next lemma gives information about this.

Lemma. The function u(x) = y(IxI) (x E is twice continuously differ- entiable in the ball Bb : _xI b if and only if y E C2(J) and y'(O) = 0.

Proof. Since u(t, 0,.^{.. ,}0) = is an even function of t, it follows easily that u E C2(Bb) implies y C2(J) and y'(O) =0. For the proof of the converse proposition, the partial derivatives of u will be denoted by u2 and For x 0, we have

'uj = y', = _y'+ (n"^—

ii).

Because 1x2/rI < 1, 0 as x — 0. Setting := 0, one obtains a continuous function in Bb for which = 0; cf. B.VI.(a) and (b).

Thus u C'(Bb). ^One proceeds in exactly the same manner with the second derivatives. From y'(r)/r = (y'(r)—y'(O))/r y"(O) it follows that y"—y'/r

0

as r

0+, thus u13(x) S23y"(O) as x 0. Taking this value to define (0) and using Theorem VI from Appendix B again, we see that this defines a continuous function in Bb and that =

at x =

^0. This completes the proof of this lemma and also the proof of the previous theorem.

I

Radial solutions of effiptic equations have been studied extensively. Such solutions are important in differential geometry and in many areas of applied mathematics. The question of the existence of entire radial solutions (i.e., those that exist in RTh) has been completely solved in the case of the equation

+

=0. The first comprehensive results for more general equations of the form + K(r)uP =⁰were given by Ni (1982).

§ 7. The Peano Existence Theorem 73

XV. Exercise.

Comparison Theorem.

Assume that v, w e C2(J) satisfy

LQv ^f(x,v), LaW f(x,w)

^{in J0} (0,b],

v(O) < w(0), ^{v'(O) =}w'(O) =0,

where f(x, y) is increasing in y. Then v' w' and v <w in J.

Hint: Show that v' as long as v <w.

XVI. Exercise. Let the functions p and P' be continuous in J = ^[0,b]

and positive in Jo = (0,b] and let pi(t)/p(x) for 0 < t x b with 0 < < 1. We consider the initial value problem

Ly=f(x,y)

^{in Jo,} y'(O)=O, where

Ly =

Prove: If the function f(x, y) is continuous in J x JR and satisfies a Lipschitz condition in y, then the initial value problem has exactly one solution. For a solution we require y E C1(J) and py' E C'(J0).

Hint: Reduce the problem to a Volterra integral equation and use the theo- rem from Exercise VIII.

§

7. The Peano Existence Theorem

In Chapter I we dealt with instances where the right-hand side of the differ-

ential equation -

= f(x,y) ⁽¹⁾

does not satisfy a Lipschitz condition. An example is the equation y' = Thefundamentally important question whether continuity of f(x, y) is sufficient for existence of a solution was first answered in the affirmative by the Italian mathematician and logician Giuseppe Peano (1858—1932). Peano's paper (1890) is written in logical symbols and was later "translated" by G. Mie (1893) into German.

I.

The Peano Existence Theorem.

If f(x, y) is continuous in a do- main D and ij) is any point in D, then at least one solution of the differential

equation (1) goes through ij). Everysolution can be extended to the right and to the left up to the boundary of D.

The ^{last part} of the statement of this theorem means that every solution has an extension that comes arbitrarily close to the boundary of D both to the right and the left as explained in 6.VII.

The proof of this theorem requires some additional concepts and lemmas.

II. Equicontinuity. A set M =

{f, g,...} of continuous functions on the interval J: a x b is called equi continuous if for every e > 0 there exists a number S =^(5(e) such that for any fE M,

for

(x,±EJ).

⁽²⁾

It is important to note in this definition that for a given e > 0, the same 5 works for every function in the family M.

Example. Let M be a set of functions f(x) that satisfy a Lipschitz condition with a common Lipschitz constant L; i.e.,

for

x,±EJ and IEM.

The set M is equicontinuous. Clearly, one can set 6(e) = ^e/L here.

III.

Lemma.

Let J =

[a,b] and let A c J be a dense set of points in

J.

If the sequence of functions f1(x), f2(x),

...

is equicontinuous in J and converges for every x E A, then it converges uniformly in J. Hence the limit 1(x) is continuous in J.

A point set A is said to be dense in J if every subinterval of J contains at least one point of A (example: A =the set of all rational numbers in J).

Proof. Given e > ^0, let 6 6(e) be determined such that (2) holds for all functions (n 1). Now partition the interval J into p closed subintervals

.11, .. . , such that each J2 is less that S in length. For each choose an

E ^J2 fl A (there exists at least one such point for each i). By hypothesis, there exists an n0 = no(e) such that

Ifm(Xj)fn(Xi)I<E

^for

m,nno and i=1,...,p.

Now let x be an arbitrary point of J and q be such that x E

Jq. It follows from the inequality Ix — Xq_I <5, property (2), and the above inequality that for

m,n n0,

Ifm(x)— _Ifm@) — fm(xq)I+ fm(xq) —fn(Xq)I

±Ifn(xq)_fn(x)l <3e.

This shows that the sequence (x) converges uniformly in J.

I

As a further tool we require the

IV. Ascoli—Arzelà Theorem. Every bounded and equi continuous se- quence of functions in C(J) contains a subsequence that converges uni- formly in J. (Boundedness means that there exists a constant M such that

andxEJ.)

7. The Peano Existence Theorem 75 Proof. LetA= {x1,^x2,...} be a countable dense point set in J(forinstance, the set of all rational numbers in J). The sequence of numbers = f,-,(xi)

(n= ^1,2,

...) is

bounded and hence contains a convergent subsequence, say

The sequence of numbers = ^islikewise bounded and has, accordingly, aconvergent subsequence, say

fqj(x2),fq2(X2), fq3(X2)

Note that is a subsequence of (pa). The sequence = ^(x3) is again a bounded sequence andtherefore ^hasa convergent subsequence

fr2(X3),fr3(X3)

By ^continuing this process one obtains a series of sequences of the form

f11' whichconverges for x

fq,,fq2,fq3,fq4,..., which converges for x

fri,fr2,fra,fr4,..., which convergesfor x =x1,x2,x3,

For each k > 1 the sequence that appears in the kth line is a subsequence of the sequence in the previous, (k — 1)st, line and converges for x =x1,..^. ,

It follows that the diagonalsequence

fp,(x),fq2(x),fra(x),.

converges for every x = xk; i.e., for all x E A, because it is a subsequence of the sequence ^{in the kth}line, at least from the kthterm onward (k = 1,2,...). The

uniform convergence of this diagonalsequence now follows from Lemma III. I We first give a proof of a weaker version of the Peano existence theorem.

V.

Theorem.

Let the functionf(x, y) be continuous and bounded in the strip S = JxlE& with J = [e,e+aI, a> 0. ^Thenthere exists at least one function y(x) defined and differentiable in J for which

y'=f(x,y)

^J, (3)

(and, as a consequence, y(x) is continuously differentiable in J).

Proof. The theorem is proved by finding a function y(x) E C(J) that^satisfies

the integral equation

y(x) = +

f f(t, y(t)) dt

in J; ⁽⁴⁾

76 IL Theory of First Order Differential Equations

—

^The approximate solutions za (x)

e

see 6.1. With this objective in mind we construct, for every a> 0, an "approx- imate solution" E C(J) using the formula

for

xe,

= ⁽⁵⁾

f(t,za(t—a)dt

^for

XEJ.

This formula defines Za unambiguously for a: + a. Indeed, if

a:

+ a, then t —a and Za(t — a) in the integrand; i.e., the integral is well- defined. If e+

a a:

:c e + 2a, then t — a a: — a

+ a.

^Therefore,

(t —a) is determined from the previous step; i.e., the integral is well-defined and so on. After a finite number of such steps we have constructed a continuous function that satisfies the integral equation (5). It follows from fl C that (a:)I C; i.e., the functions z,, (a:) satisfy the Lipschitz condition

— Cia: _—

inJ. If we denote the set of functions Za in C(J) (we consider only their restric- tions to J) by M, then M is equicontinuous. Therefore, by the Ascoli—Arzelà theorem 1V, the sequence z1 (a:), z112 (a:), (a:),... has a uniformly convergent subsequence (a:)) (n = 1,2,3,.^. .; —+ 0), which we relabel as to simplify the notation. Denote the continuous limit of this sequence by y(a:).

Then by (5),

= + f f(t,

^{— an))dt. (6)}

It follows now from the inequality

— — y(t)l lzn(t — — + — y(t)l

+ — y(t)l

that ^— converges uniformly in J to y(t) and hence that f(t, ^—an)) converges uniformly in J to f(t, y(t)) (here we have used the fact that f(a:, y) is uniformly continuous on bounded sets). Therefore, passage to the limit under the integral sign in equation (6) is allowed, and equation (4) follows as a result.

I

§ 7. The Peano Existence Theorem 77

This special case of the Peano theorem can now be used in a manner that parallels the role of Theorem 6.1 in the development of the Lipschitz case. But in other respects, the proof of theorem I differs essentially from that in the corresponding theorem 6.VII.

One first proves, exactly as in §6, that the existence theorem holds for a strip to the left of and also for a rectangle. We formulate the latter case, which corresponds to Theorem 6.111.

VI.

Theorem.

If f is continuous in the rectangle R e

x

+ a,

Jy — b, then the initial value problem y' = f(x,y), = has a solution y(x) existing (at least) in the interval e

x + a, where a

mm (a, b/A), A = maXR A corresponding result holds for a rectangle to the left off.

This establishes the first part of the Peano existence theorem, that an integral curve passes through every point of D.

The second part, the proof of the assertion that every solution can be ex- tended to the boundary, is more in this setting in comparison to the proof in §6. The difficulties come from the fact that we do not have a uniqueness statement at our disposal.

Only the extendibility to the right will be discussed.

We first prove the following intermediate result.

(Z) If is a solution in the interval x < b and A is a compact subset of D, then can_be extended beyond A, i.e., there exists an extension to the right with graph A.

The distance from the set A to the boundary of D is positive, say 3p> 0 (if D = onecan take p =^1). If is the set of points whose distance from A is

<2p, then A2,, is likewise a compact subset of D, and hence is bounded in A2,,, say jfj C. Denote by R(xo,yo) rectangle xo x xo + p, — yol p.

Then R(xo,yo) C A2,, for any (x0,y0) EA.

If graph q5 C A, we begin by extending to bJ in the manner described in 6.VL(a). Then we continue further to the right, by applying Theorem VI in the rectangle R(b, q5(b)). This results in a solution in b x b + a =: ^b1 with a :=min(p,p/C). If this extension still lies entirely in A, then the process is repeated with R(b1, 1)(bl)), etc. Since at each step the interval of existence increases by a fixed number a > 0 (as long as graph C A), one obtains in this manner a solution that extends beyond A after a finite number of steps. This proves (Z).

The remainder of the proof is straightforward. We consider a sequence (An) of compact sets with C C D for all n and such that every compact B C D is contained in one of the (for example, let be the set of points from D with distance — fromthe boundary of D and distance n from the origin). Let with cb(e) =n ^be a solution in an mterval to the right of that

does not approach the boundary of D. Then is a compact subset of D;

i.e., graph q5 C for a suitable p.

We continue beyond using (Z) and call the extension it exists in an interval

=

^br,]. We now construct If does not belong entirely to then we set = ^if does lie in then we continue the function to the right until it leaves and call the continuation

It

exists in an interval with

-

^br,. Continuing inductively in this manner, one obtains a sequence of functions such that is defined in and (ba) is a monotone increasing sequence of numbers. If p n < ^m, then (I'm ill

Thus, by Lemma 6.V, there exists exactly one solution y defined in b),

where b = (b = 00 allowed), with the property that y for every n p. This function y is the extension to the right of the original solution qS mentioned in the conclusion of the theorem. Clearly, graph y is not contained entirely in any and hence is not contained in any compact subset of D. ^U Remark on constractive proofs. The proof of Theorem V is different from that of 6.1 in one important respect. In 6.1 it was possible to calculate explicitly a sequence of successive approximations that converges to the solution (this is the essence of the contraction principle). Here, on the other hand, the initial value problem has several solutions in general, and it cannot be expected that a sequence of approximations like the one we constructed even has a limit, let alone tends toward any particular one of these solutions. Now one applies the Ascoli—Arzelà theorem to a sequence of such approximations. It says that there exists at least one convergent subsequence. However, no procedure is given that would allow the particular subsequence to be identified. An existence proof like the earlier one in §6 is called a constructive proof. By contrast, the proof of the Peano existence theorem given in this section is nonconstructive.

In the proof of the Peano existence theorem that we have just given the approximate solutions Zawerecomputed from the integral relation (5). Another frequently used method of obtaining approximations is

Viii.

The Euler—Cauchy Polygon Method.

In this method, polygo- nal approximations Ua (CE > 0) to the solution to the initial value problem (3) are constructed in the following manner. Let x2 = + ai (i = 0,1,2,.. ^{.). For}

= x0

x xi we set

= ii+ (x — i.e., Ua is the straight line through the point ij) = (xo,yo) with slope f(xo, yo). In the interval

X1 X X2, Ua is the straight line through the point (xi,yi) :=

with slope f(xi, y'). ^In general, one arrives after p steps at a point (xv, with _{Yp =}ua(xp) and defines then for x that Ua is the straight line through (xv, y,,) with slope y,). The advantage to this construction

is It is based on a relatively simple idea that is easily carried out numerically. However, the final step in the proof (passing to the limit to get the solution) is more difficult.