Transformation to an Equivalent First Order System. Consider the nth order scalar differential equation in explicit form

Supplement II: Differential Equations in the Sense of Carat héodory

I. Transformation to an Equivalent First Order System. Consider the nth order scalar differential equation in explicit form

§ 11. Initial Value Problems for Equations of Higher Order 125

The estimate (17) holds in an interual J_ = — a, to the left of e

if

p' < —w(x,p)

a.e. in J_,

Proof. Let = y(x)I. From proposition (a) and (16), it follows that

S _<w(x,q5).

The conclusion now follows from Theorem XXII with f replacedby w.

I

Exercise. Show that, with the Eudidean norm, the estimate (17) also holds

under the weaker assumption (y,(f(x,y))

Here (.,.) ^is the scalar product in

XXIV. Exercise.

Separated Variables.

Show that the Theorems 1 .VII—VIII hold for the differential equation with separated variables

= f(x)g(y) under the assumption

f

g E and that the representation formula (1.8) remains valid.

XXV. Exercise. Solve the initial value problem (n =3)

I

^Y2Y3\

⁽⁰

= ^YiY3 ^with. ^y(O) = ¹

)

by the method of successive approximation. What is the kth approximation if yo(x) =y(O)?

11. Initial Value Problems for Equations of Higher Order

Transformation to an Equivalent First Order System.

Consider

Equation (1) and the system of equations (2) are equivalent in the following sense: If y(x) is a solution of (1), then the vector function y = _{(yi, Y2,••}.,yn) (y,

y',..

^.^,^y(fl_l)) is a solution of (2). Conversely, if y is a (differentiable) solution of (2) and one sets Yl (x) := y(x), then y(x) is n-times differentiable, y2(x) y'(x),.^.

=

and equation (1) holds.

In a similar manner, systems of rith order equations can be transformed into systems of first order equations by introducing new functions for the lower order derivatives. For example, the equation of motion of a point mass of mass 1 in three-dimensional space in the presence of a force field k(t, x) is given by

x=k(t,x)

for

x=x(t),

or, written in expanded form with x= ^(x,^{y, z), k =} ^{(f,g, h),}

th=f(t,x,y,z)

ü =g(t,x,y,z) 2 =h(t,x,y,z).

This system is equivalent to the following system of six first order differential equations for six unknown functions x, y, z, u, v, w.

x=u, ü=f(t,x,y,z) y=v,

w, th = h(t,x,y,z).

We return to equation (1). An initial value problem for the equivalent system (2) prescribes the values of the functions yi,. ..^, ^at a point Corresponding initial conditions for (1) are given by

v(e) ^{= ?7o,} y'(e)^{= ?7j,}

...

⁼ ⁽³⁾

For example, the initial value problem for a second order differential equation reads

y"_—f(x,y,y'),

Because of the equivalence of the two initial value problems, aM of the the- orerns derived earlier for the system (2) can be applied to (1), (3). Here, the right-hand side of (2), f = (fr,.^.^. takes the special form

(i=1,...,n—1)

Thus f is continuous if I is continuous, holomorphic if f is holomorphic, and satisfies a Lipschitz condition if f satisfies a Lipschtiz condition

If(x,y) — (4)

In particular, if f(x,yi,.. is continuous in a domain DC and if the partial derivatives arecontinuous in D, then both f and f satisfy a local Lipschitz condition in y. We summarize.

§ 11. Initial Value Problems for Equations of Higher Order 127

II. Peano Existence Theorem.

Let f(x, y) be continuous in a domain

D c

and let . E D. Then the initial value problem (1), (3) has at least one solution. Every solution can be extended up to the boundary of D.

III. Uniqueness Theorem.

Let f satisfy the hypotheses of Theorem II.

If in addition, f satisfies a local Lipschitz condition (4), then there exists exactly one solution. This is the case, in particular, if Df/3y E C(D).

IV.

Complex Existence Theorem.

Let f(z, w1,^. .. , be a function of n+1 complex variables z, w1,..., 1ff is defined and holomorphic in D C

then there exists exactly one function w(z), defined and holomorphic in a neighborhood of the point (zo, Co, (i,. .. , E D, that satisfies the differential

equation

= f(z,w,w',..^.

andthe initial conditions

w(zo) w'(zo) = (1, ^..., w(n_1)(zo)

=

The solution can be expanded in a power series

w(z) = ^—^z0)k.

For example, the ansatz y(x) = a0 + a1x + a2x2 in the initial value problem

y"+y,

^y(O)=?7o, ^y'(O)=iji ^(*)

leads to the recursion formulas

k(k—1)ak+ak_2=O (k=2,3,...),

from which it is easy to see that

x3 x5

77i1

and hence the solution of problem (*) is y = m cosx + sin x.

In the next three parts we consider some differential equations of second order that either can be integrated directly or reduced to simpler differential equations.

V. y" f(x, 1/')

This is equivalent to a first order differential equation for z(x) = y'(x),

=

f(x,z).

If the initial condition is = ^y' = ^onefirst looks for the solution z with

=

Theny is found by quadrature:

y(x) =

+ f z(t) dt.

VI.

=

^f(y,^y')

Suppose y(x) is a solution and x(y) its inverse. We introduce the function p(y) defined by

p(y) =y'(x(y)).

Note that dx(y)/dy = l/p(y). The expression for the derivative of p, dp/dy = y"(x(y))/p(y), leads to the first order differential equation

dp 1

for

p=p(y).

Suppose p(y) is a solution of this equation. Then we may calculate

11

The solution y(x) is the inverse of this function.

Example.

=y •smy,

. _y(O)=O,

y(O)=l.

The equation for p(y) is

dp/dy = p.^siny ^with p(O) =^y'(O) =^1.

Therefore

p(y) = x(y)

j

⁰ ^eC088l^ds.

§ 11. Initial Value Problems for Equations of Higher Order 129

vu.

= 1(n)

This is a special case of VI. Suppose f is continuous, y(x) is a solution, and F(y) =

f

f(y) dy is a primitive of f(y).

Iviultiplying the differential equation by 2y', we obtain (y'2)'

=

2y'y" = 2y'f(y) =

which implies

= 2F(y) + C.

Solving for y' gives a differential equation y'

which does not depend explicitly on the independent variable.

If 0, then the value of the constant C and the sign are uniquely determined by the initial condition y(O) = y'(O) = and the differential equation has a unique solution y(x) with y(O) = (Theorem 1.VII). In the case where = 0, the situation is somewhat more complicated, and there may be more than one solution.

In the remainder of this section, we discuss some examples from physics.

VIII. The Catenary. A flexible chain or cable with no

sus- pended from two points, hangs under the influence of gravity in a shape called a catenary (from Latin catena, chain). Let (a,ya), (b,yb), a < b, be the two suspension points in the xy-plane, p the density (mass per unit length) of the chain, which is assumed to be constant, and g the constant of gravitational acceleration. We choose a reference frame such that the gravitational force op- erates in the of the negative y-axis. In order to determine the function y(x) that describes the shape of the curve, we conduct a thought experiment:

We cut the chain at a point (x, y(x)) and remove and replace the right-hand part by a force = (Hr, in such a way that the left-hand part of the chain remains at rest. Then

with

(this relation reflects the assumption that the chain has no stiffness). If, on the other hand, the left-hand part is removed and the right retained, then an opposite force is required.

Now consider the segment of the chain between a and [3. It is kept at rest by forces = Va) on the left and (Hp, Vp) on the right. Since the system is at rest, the sum of the forces is zero. Separating into horizontal and vertical components, one is led to a pair of equations

—Ha = ⁰ const. = H,

Vp — Va 0.

V'3

Theintegral in the second equation measures the length L of the curve segment (cf. A.I), and the term pgL gives the downward pull due to gravity. The second equation can be rewritten in the form

(iiy"

^— + ^dx =^0,

sinceVp — V =Hy'. Now we use a well-known argument:

Since a, can be chosen arbitrarily, the integrand must vanish, and it follows that

= +y'2 with c

= ^> ⁰ Catenary equation.

This equation_belongs to the type studied in V. From the differential equation

=

cVl

+ z2 for z =

y' one easily obtains arcsinhz = c(x+ A) and hence z =^y' = sinhc(x +A). The general solution is then given by

y=B+?coshc(x+A)

(A, B arbitrary).

Remark. The equation of the catenary (and the above derivation) remains valid if the density p is variable.

The Boundary Value Problem. Suppose the distance b — aand Yb — Ya

are given, together with the length of the cable L; naturaily, we require that

L2>(b—a)2+(yb—ya)2. ⁽⁵⁾

Without loss of generality, let A =0, i.e., a reference frame is chosen such that the minimum of y is at x =^0. Then we have

f

^a ^{+ y'2 =}

^jb

^a ^y"^{dx =} ^—^y'(a)] (6)

—sinhca] = sinli ^{c(b —}^a) cosh c(a + b)

c c 2 2

a x b a

Initial Value Problems for Equations of Higher Order 131 and

1 - 2 . c(b—a) . c(a+b)

Yb —Ya = ^{— [cosh}^{cb —}^cosh^{ca] =} ^{— sinh} ₂ ^smh

From this pair of equations, it follows that

2 c(b—a)

L 2 (8)

The value of c can be determined from this equation. The equation is first written in the form of a fixed point equation. = ^can then be solved by iteration. Setting =c/2 one obtains = (sinh(b — or, in terms of the inverse function,

= sinh'(L'e). (9)

If denotes the right-hand side of (9), then L' > b—a implies that > 1.

Since arcsinh (t) is concave for t 0 and grows like ln(t) for large t, there exists exactly one positive fixed point Thereader should verify using a sketch that the sequence obtained by iteration

=

converges to for every starting value > 0. In the special case Ya = Yb, we have a = —b < 0 and L = L'. In the general case, once c has been determined, a can be determined

from (7), (8), and a+b=(b—a)+2a:

(10)

Historical Remark. In the second discussion in the Discorsi (1638), Galileo expressed the opinion that the catenary was a parabola. The true form was found, independently, in 1691 by Leibniz, Johann Bernoulli, and Huygens. Leib- niz originated the name catenarij.

Example. Let b—a = 200, = 100, L = 240, hence L' 20V144 — 25 = 218.1742. By iteration in (9), one obtains c = = 0.007287 and a = —39.114.

Thus the chain is described by the function y(x) = 137.237. cosh(0.007287x) + B, its sag S = min(Ya,Yb) —miny(x) is given by

S = y(a) _—y(O) = ^— ¹⁾ = 5.612.

IX. Catenary

Problems.

(a) For fixed b—a and the horizontal component of the force H is a function of L. Show: H = H(L) is monotone decreasing and tends to oo as L ^— ^a)2 _{+ (Yb —}ya)2 and toward 0 as

L oo.

(b) Let b —^a = 200, Ya = Yb, L = 240. How large is the sag?

(d) Let Ya _{= Yb} and span =sag= ¹m. How long is the chain?

(e) Prove: If y(x) = (1/c)coshcx describes a catenary and z(x) =a+ fix2 is a parabola with y(O) =^z(O), y(b) =z(b), then y < z in (0, b).

(f) (Galileo Vindicated?) For which density function p =p(x)is the parabola y =^a+ fix2 a solution of the catenary equation? Interpret this result in terms of (e).

X. Nonlinear Oscillations. We consider the differential equation

x+h(x)=O for x=x(t).

(11)

Here h is locally Lipsch.itz continuous in IR and satisfies the conditions h(0) = 0

and x• h(x) > 0 for x 0. This equation is of type VII.

(a) If x(t) is a solution of (11), then the functions x(t + c) and x(—t) are also solutions.

By VII, the function

E(x, ±) = + H(x) with H(x)

f

^{h(s) ds} ⁽¹²⁾

is constant for every solution of (11), and for the initial conditions x(0) =

th(0)

=

v0,

=

E(xo,vo) =: a. For vo > 0, we have ± = ^—H(x)), and using 1.V, we obtain the solution in the form

px(t) ds

I with

a=E(xo,vo)>0.

⁽¹³⁾

JXO

In applications to mechanics, x(t) describes the motion of a point mass

of mass 1, x =

⁰ corresponds to the equilibrium state, and —h(x) gives the magnitude of a "restoring force" (its sign is opposite to the displacement x, whence its name). The function E in (12) is the total energy, the sum of the kinetic energy and the potential energy H(x) (the work done in moving the mass from x to x+dx is h(x)dx). The equation E(x(t),th(t)) = const has the following interpretation: In the oscillations of a frictionless system, like the one under consideration here, there is a continual exchange between kinetic and potential energy, while the total energy remains constant.

Equation (11) can be written as an autonomous system

± = ^y, = —h(x). (11')

In 3.V, we described how to construct a phase portrait, from which the qualita- tive behavior of the solutions of such systems can be read. The trajectories in the xy-plane are level sets of E(x, y). Clearly, H(x) > 0 for x 0. Therefore, o = E(O,O)

<E(x,y) for (x,y)

Theorem. Suppose H(x) —i oo as x —p Then every solution to the differential equation (11) is periodic. A solution x(t) takes on its extreme values when ±(t) 0; the extreme values of ±(t) occur when x(t) = 0.

§11. Initial Value Problems for Equations of Higher Order 133

Proof. By (11'), x(t) is strongly monotone increasing for y> 0 and decreasing for y < 0; a similar statement holds for y(t). This implies the statement about the extreme values. The remaining conclusions follow from A.VIII—IX.

We indicate a direct proof.

The equation E(x, y) = a > ⁰ describes a closed Jordan curve that surrounds the origin and is symmetric to the x-axis and that is represented by

for

r1 < 0 < r2

and H(ri) =

H(r2) = a. These values are uniquely determined because of the strong monotonicity of H in the intervals (—oo, 0] and

[0,oo).

If (x(t), y(t)) is a solution of (11') with the initial values x(0) xO > 0,

y(O) = = 0, then since the curve Ka is bounded, the solution exists in IR;

cf. 10.XI.(c). Using an argument similar to the one used for the predator—prey model in 3.VI, one shows that the solution runs over the entire curve Ka and is periodic. (Initially < 0; y(t) is strongly monotone decreasing, as long as x(t) > 0; from y —e it follows that th —e, hence

there is a t1 > 0 with

x(t1) = 0; etc.)

I

We discuss some examples.

(b) The Harmonic Oscillator. In the linear case h(s) = w2xwith w> 0, the equation is

+ w2x =0 harmonic oscillator equation.

The general solution is x(t) = (r 0), and the total energy is given by 2E(x, y) = ^y2

+

^Thus the trajectories are ellipses y2 + w2x2 = 2a.

The (minimal) period T = is the same for all solutions. The number v = 1/T =w/(2ir) is called the frequency (the number of oscillations per second), w the circular frequency, and r the amplitude of the oscillation.

In mathematical models of elastic objects, a linear relationship between displacement and tension is known as Hooke's law. The classical example is a mass hanging on a spiral spring; here k =^w2 is the spring constant.

(c) Mass on a Rubber Band. Suppose a rubber band is fixed at the upper end A and hangs vertically downward. Coordinates are chosen such that the s-axis is oriented downward and the position of the lower end B, when the system is at rest, is at the origin. If B is pulled downward, then the restoring force is given by Hooke's law h(s) = kx with k > 0. On the other hand, if B is pushed upward, then, in contrast to the case of the spiral spring, there is no restoring force. A mass m is now attached at B. The vertical motion, described by satisfies the equation

with

(g is the constant of gravitational force). The equilibrium position r0 is given by r0 =mg/k. Writing =ro+ x(t) and setting b =

k/rn>

0, we obtain the differential equation

rñ+h(x)=0 with

Mass on a rubber band

(the reader should make a sketch of h). In this example, the potential is

I ^for x —r0,

H(x) =

for

x<—r0.

For small perturbations, the orbits

Ka: with a>0

are effipses, and the system behaves like a harmonic oscifiator. However, if a > H(ro), then e(t) r0 + x(t) also takes on negative values, and the orbit is a combination of an effipse and a parabola.

The above example is typical of mechanical systems that are not symmetric with respect to their rest state. An interesting system of this type is a suspension bridge, where a linear theory is appropriate for small (vertical, torsional, ...) oscillationsbut fails for large amplitudes. A nonlinear theory of the dynamics of suspension bridges is currently under investigation; cf. in this regard McKenna and Walter (1990) and Laser and McKenna (1990).

(d) The Mathematical Pendulum. One end A of a (weightless) rod of length I is attached to a pivot, and a mass m is attached to the other end B. The system moves in a plane under the influence of the gravitational force of magnitude my, which acts vertically downward. If 4)^denotes the angle between the vertical and the rod, then the tangential component of the downward force —my sin 4)^acts on the mass point at B. If the motion is described in terms of the angle 4)(t),

then s(t) =

^I4)(t) gives the distance traveled by the mass (measured from the lowest point), and from the equation of motion

=

—my sin 4),we obtain

4 +^{a sin 4)}^{= 0} ^with a = g/l mathematical pendulum.

For small values of one may replace sin 4) by⁴⁾ sin_4),which leads to + açb =0 with a = g/l linearized pendulum

as a usable approximation. Thus, for small displacements, the pendulum behaves like a harmonic oscillator with circular frequency w

Initial Value Problems for Equations of Higher Order 135 y

Mathematical pendulum

The potential energy for the mathematical pendulum is given by =

a(1—cos 0 2a, and the total energy is ₀₎ = ₊

H(0) 0.

The assumption > 0, mentioned in earlier results, is violated here. It holds only for <ir, and accordingly, the level sets

Ka : = 2(a ^— ^a(1^—

in the a closed

curves a

a

= ^0, one obtains stationary solutions 0(t) 2kir.

If a >

^a0,

the orbits Kc. are unbounded wavy lines, corresponding to continual rotations about the pivot point A. Such are obtained from the initial conditions g5(O) =0,

0(0) = ^v0 ^with > (the upper wavy lines correspond to v0 >0, the lower wavy lines to v0 <0.) Solutions in the two cases 0 < a < a0

and a> a0 have

completely different behavior and are divided by the set which is called the separatrix. The solution qS(t) that corresponds to the separatrix, for instance the one with initial values 0(0) = 0, is monotone increasing and tends toward ir as t ^{00 (the}reader should give a description of the pendulum motion for this case).

Historical Remark. Galileo writes in the first discussion of the Discorsi (1638) that the frequency of a pendulum does not depend on the maximal angular difiection and is proportional to \/171. ^He is correct in the second assertion; cf. XL(j) below.

XI.

Exercises on Nonlinear Oscillations.

Suppose h is continuous in R and satisfies h(x) > 0for x 0. Let H and E be defined as in X.

(a) Prove: If h is continuous, then every initial value problem for (11) is uniquely solvable (cf. Exercise XIll). Theorem X remains true.

(b) Let H(x) tend toward A as x —' ^{—00 and} toward B as x ^{00, with} 0 < A

B <00.

^Describe the global behavior (periodicity, behavior for large ti) ofthe solution x(t) with initial values x(0) = b(O)

=

in each of the following cases: (i) 0 <

<A. (ii) A <B. (iii) B.

Make a sketch of the phase portrait.

(c) For which values of A, B (cf. (b)) is the following statement true? Every solution of the differential equation that has a zero and whose first derivative has a zero is periodic.

(d) Suppose, additionally, that h is odd. Prove: If x(t) is a solution to the differential equation, then v(t) :=x(c—t) and w(t) := —x(t) are also solutions.

Further, the relations

x(c)=O

= —x(c—t),

=x(d—t)

hold for every solution x. Thus the solutions have the same symmetry properties as the sine function.

(e) The Period of an Oscillation. Let V denote the duration of the positive quarter oscillation starting at the point (0, vo) and ending at (r, 0) (v0 > 0,

r> 0

is the largest swing). Using (13) with a = H(r) = we get

1 çr

V=V(r)=—j

Jo ^— H(s)

(f) Prove: If h is a symmetric forcing term (h odd), then T = 4V is the period of an oscillation.

(g) Prove: If h(x) h(x) for x > 0, then V(r) V(r).

in addition, h(c) <h*(c) for some c> 0, then V(r) > V*(r) holds for r c.

(h) Prove: If h(x)/x is weakly or strongly increasing [decreasing], then V(r) is weakly or strongly decreasing [increasing]. As an illustration, calculate V(r) for the case h(x) = Za (a> ^0).

(i) Prove: If h'÷(O) = w2 exists, then limr_,o V(r) = ir/(2w), i.e., for small displacements the system is approximately a harmonic oscifiator.

(j) The Mathematical Pendulum. Show that for the mathematical pendulum (cf. X.(d)) (using 1 —cosa =²sin2

— 1 r _ds ₁ _du

V(r)

— ^s — a _— _k2

where k = sinr/2 (substitution sin s/2 = ksin u). This function is an elliptic integral of the first kind. From the binomial expansion for (1— x)'/2, it follows that

2 4

V(r)= (1+aik +a2k

in particular, V(0) = in agreement with (i) and X.(b) (a =w2 here).

Calculate a1, a2. By what percentage is the period of an oscillation of the mathematical pendulum larger than that of the linearized pendulum equation

if the maximum displacement is 5° (10°; 15°; 200)?

Hints: (g) Consider the difference H(r) —H(s).

(b) Consider V(r) and V(qr) with q > ¹ and write V(qr) as an integral

from 0 to r.

The quotient in the corresponding H-differences has the form

§11. Initial Value Problems for Equations of Higher Order 137

Free fall

[1(r) — f(s)]/[g(r) ^—^g(s)]

with f(x) =

H(qx), g(x) = q2H(x). The generalized mean value theorem of differential calculus can be used.

(i) Use (g).

Remark. The dependence of the period of the oscillation on the function h was investigated thoroughly in 1961 by Z. Opial (Ann. Polon. Math. 10,

42—72); (g) and (h) can be found there. The book by Reissig, Sansone, and Conti (1963) contains these results and a number of others.

XII.

Free Fall. En part II of the Introduction, we presented the example of free fall from a great height. The initial value problem

r(0)=R, r(0)=vo

describes the vertical motion of a body with mass m starting at a distance R from the center of the earth with initial velocity v0, where r(t) is the distance from the center of the earth at time t. From the equation

= itfollows that

E(r, r) = — y— total energy functzon

is constant. Here is the kinetic energy term and —yM/r the potential energy. The latter is normalized in such a way that it vanishes at infinity and hence is negative. The trajectories satisfy the equation E(r, = a.

If a >

the trajectories run to infinity; for a < 0, they are return curves (describing a body that falls back to the earth after being thrown vertically upward).

The smallest total energy for a motion without return is a = 0. The corresponding differential equation reads

=

(14)

Dalam dokumen Ordinary Differential Equations (Halaman 138-155)