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    First-Order Partial Differential Equations, Vol. 1
    First-Order Partial Differential Equations, Vol. 1
    First-Order Partial Differential Equations, Vol. 1
    Ebook863 pages6 hours

    First-Order Partial Differential Equations, Vol. 1

    By Hyun-Ku Rhee, Rutherford Aris and Neal R. Amundson

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    This first volume of a highly regarded two-volume text is fully usable on its own. After going over some of the preliminaries, the authors discuss mathematical models that yield first-order partial differential equations; motivations, classifications, and some methods of solution; linear and semilinear equations; chromatographic equations with finite rate expressions; homogeneous and nonhomogeneous quasilinear equations; formation and propagation of shocks; conservation equations, weak solutions, and shock layers; nonlinear equations; and variational problems. Exercises appear at the end of most sections. This volume is geared to advanced undergraduates or first-year grad students with a sound understanding of calculus and elementary ordinary differential equations. 1986 edition. 189 black-and-white illustrations. Author and subject indices.
    LanguageEnglish
    PublisherDover Publications
    Release dateMay 5, 2014
    ISBN9780486146201
    First-Order Partial Differential Equations, Vol. 1

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      First-Order Partial Differential Equations, Vol. 1 - Hyun-Ku Rhee

      Mathematical Preliminaries

      0

      0.1 Functions and Their Derivatives

      In analytic geometry it is convenient to think of the function y = f(x) as a curve in the (x, y)-plane. The curve is said to be continuous at a point (x0, y0) if given an ε > 0 there exists a δ > 0 such that

      The curve is said to be uniformly continuous in the interval (a, b) if δ does not depend on x. The function f(x) is said to be single valued at x0 if only one value of y0 is defined by y0 = f(x0). The function y = f(x) has a derivative at x0 if the limit

      exists. It is denoted by f’(x0), (df/dx)x=x0, dy/dx, or y’, each evaluated at x0. The geometric interpretation of the derivative is the slope of the curve y = f(x) or the slope of the tangent to the curve at the point x0.

      Figure 0.1

      Similarly, for functions of several variables, say of two in particular, z = f(x, y), the function f(x, y) is said to be single valued at (x0, y0) if only one value of z0 is defined at that point by z0 = f(x0, y0). A function of two variables is said to be continuous at a point (x0, y0) if given an ε > 0 there exists a δ > 0 such that

      We think of z = f(x, y) as defining a surface in the space of x, y, and z. For example, ax + by + c = z is a plane, whereas z = 2x² + 3y² is a paraboloid defined for all x and y and for z > 0. Any plane parallel to the (x, y)-plane cuts the paraboloid in an ellipse; planes passing through the z-axis give parabolas.

      The partial derivatives of f(x, y) = z at (x0, y0) are defined as

      These are the rates of change of f(x, y) in planes parallel to the (x, z)-plane and (y, z)-plane, respectively, and are therefore the slopes of the curves of intersection of these planes with the surface at (x0, y0), as shown in Fig. 0.1.

      In general, if u = f(x1, . . . , xn) is a function of n variables, the partial derivative ∂f/∂xi is defined by

      As before, a change is made in the ith coordinate only, and the limiting value of the quotient, when it exists, defines the partial derivative.

      0.2 Functions of Functions and Their Derivatives

      Given a function u = f(x1 x2 x3, . . . , xn) with the independent variables x1 x2, . . . , xn, the total differential of u is defined as

      If we suppose that each xi is a function of m variab1es (tj, j = 1, ... , m,

      is the total differential of u considered as a function of m variables tj.

      If the partial derivative of u with respect to tj is desired, then, since u can be regarded as a function of the tj, its total derivative is

      This is the familiar chain rule of the differential calculus. If all the xi are functions of a single variable t, then

      As an example, let us consider the differential equation

      where V is a constant, and let us make the transformation to new independent variables

      Hence the equation becomes

      Again, if we consider the equation

      we may ask whether there are linear relations

      We can remove both the pure second derivatives, by setting

      so that the equation becomes

      Now Eq. (0.2.11) can be satisfied by taking

      But if we choose the same sign in each case, the coefficient (a1a2 – α²b1b2) in Eq. (0.2.12) will also vanish. However, if we take opposite signs, giving

      then a1a2 – α²b1b2 = ±2α²b1b2 ≠ 0, and Eq. (0.2.12) reduces to

      which is the form for which we were aiming.

      Exercises

      0.2.1. Consider Eq. (0.2.7) under the change of variables σ = z – Vt, τ = t. Give a physical interpretation to this change of variables and to that of Eq. (0.2.8).

      0.2.2. Consider the pair of equations in which x and t are independent variables and V, v, k2, γ, and K are all positive constants,

      and make the transformation

      Could you make a further transformation to remove explicit mention of the constant k? What advantages or disadvantages might this have?

      0.3 Implicit Functions

      We will often want to change the form of an equation defining the relationship between two or more variables. Here, for example, are three of the types of questions we may want to ask.

      1. If y = f(x), can we define the inverse relationship x = g(y) uniquely?

      2. If F(x, y, z) = 0, can we solve the equation and express z in the form z = f(x, y)?

      3. Does the transformation u = u(x, y), v = v(x, y) map a region of the (x, y)-plane onto a region of the (u, v)-plane in a one-to-one fashion? (This question, although it is related to the first two, will be answered in Sec. 0.4.)

      The first question can be illustrated by Figs. 0.2 and 0.3. The relation y = f(x) is plotted in Fig. 0.2 as a curve in the (x, y)-plane, and x0 and y0 = f(x0) are the coordinates of a point on the curve. It seems clear that if the curve is monotonically increasing, as it is to the left of

      x

      0 in the figure, then for any y in the immediate neighborhood of y0 we can find the corresponding x uniquely. We write this inverse relation as x = g(y) and recognize that it is single valued. However, we should be in difficulty in the neighborhood of a point like (

      x

      0,

      y

      0), where f’(

      x

      0) = 0. The figure shows that, for y <

      y

      0, two values of x correspond to y, whereas for y >

      y

      0 there is no value of x at all. Thus in an interval a ≤ x ≤ b for which f’(x) does not become zero (it could be negative just as well as positive), we can find x = g(y), and the derivative g’(y) is defined. In Fig. 0.3 there is an inflection point at x = x0 with f’(x0) = 0. In this case x = g(y) is uniquely defined, but g’(y) does not exist (i.e., is infinite) at y = y0. Evidently, monotonicity is a necessary and sufficient condition for the inversion of the relation y = f(x). What we are asking for in the third question is a generalization of this idea to a pair of functions of two variables.

      In the case of the second question we have raised, the example F(x, y, z) = x² + y² + z² + 1 = 0 shows that it may be impossible to find real solutions of the form z = f(x, y). However, if a solution of this form can be found, then F(x, y, f(x, y)) = O is an identity in x and y. The possibility of the solution in such a case is embodied in the following more general theorem. (See, for example, Courant, Vol. II, p. 117.)

      Figure 0.2

      Figure 0.3

      THEOREM: Let F(x1, x2, . . . , xn, z) be a continuous function of the independent variables x1 x2, . . . , xn, and z with continuous partial derivatives, Fx1, Fx2, . . . , Fz in some region R’. Let be an interior point of this region, and let but Then there is an interval z1 < z < z2 containing z⁰ and a region R containing ( z⁰) within the region R’ such that for every (x1, x2, . . . , xn, z) in R the equation F(x1, x2, . . . , xn, z) = 0 is satisfied by exactly one value of z in the interval zx< z < z2. For this value of z, which is denoted by z = f(x1 x2, . . . , xn), the equation

      holds identically in R and The function f is a continuous function and has continuous partial derivatives in R.

      For a function F(x, y, z), the differential dF is defined as

      where x, y, and z are independent variables and dx, dy, and dz are the differentials of these independent variables. If F = K, a constant, then

      and, since the partial derivatives are computed at some definite point, dx, dy, and dz cannot be independent. However, from the theorem on implicit functions, assuming the hypotheses are satisfied, z = f(x, y), and therefore

      Substituting this in the preceding equation, one obtains

      which must hold for all dx and dy. Thus

      and this gives the differentiation formulas for the function f in terms of the partial derivatives of F.

      A word should be inserted here about notation. Given a function f(x, y, z) with x, y, and z as independent variables, the meaning of ∂f/∂x, ∂f/∂y, and ∂f/∂z is clear. In applications, and particularly in thermodynamics, it is customary to write

      in which the variables under consideration are explicitly specified. This is necessary, for a quantity like the free energy may be considered as being a function of different sets of variables, some being more convenient in one situation than another. For example, (∂F/∂T)p is not the same as (∂F/∂T)v, and this is obvious when written in this form. In purely mathematical analyses, the variables that are taken to be independent are usually apparent, whereas in physics there is frequently a choice.

      The function F(x, y, z) = K represents a one-parameter family of surfaces since, in principle, provided that ∂F/∂z is not zero, one can solve for z as z

      0.4 Sets of Functions

      In mathematical applications we are frequently led to the problem of inverting a set of functions, such as

      That is, we are asking whether it is possible to find an inverse pair of functions

      The answer to this question is embodied in the following theorem. (See Courant, Vol. II, p. 152.)

      THEOREM: If in the neighborhood of a point (x0, y0) the functions u(x, y) and v(x, y) are continuously differentiable with u0 = u(x0, y0), v0 = v(x0, y0), and if in addition the Jacobian determinant

      is not zero at (x0, y0), then in the neighborhood of the point (x0, y0) the system of equations u = u(x, y), v = v(x, y) has a unique inverse

      in some neighborhood of (u0, v0). Also in that neighborhood

      These last formulas may be easily obtained by differentiation of Eqs. (0.4.4) and (0.4.5), since

      where the first and fourth are obtained by differentiating with respect to u and the second and third by differentiating with respect to v; Eqs. (0.4.6) and (0.4.7) are then obtained by solving the first and fourth and then the second and third as simultaneous algebraic equations. These formulas have an obvious generalization to a higher number of dimensions.

      If the Jacobian J vanishes at a point (x0, y0), the possibility exists that the two functions u = u(x, y), v = v(x, y) cannot be inverted in the neighborhood of the point, and, even if the inverse does exist, the derivatives of the inverse do not, as may be seen from the preceding formulas for the partial derivatives. For example,

      u = a cos x cos y, v = a cos x sin y

      has a Jacobian

      This Jacobian will vanish if x = 0, π/2, and so on. Now if x = 0, u = a cos y, v = a sin y, and (u, v) traverses the boundary of the circle of radius a as y varies from 0 to 2π. If 0 < x < π/2, the point (u, v) goes around a circle of smaller radius a cos x. But if x = π/2, that radius has shrunk to zero and the whole line segment x = π/2, 0 ≤ y ≤ 2π, is mapped onto a point. Since the locus of x = c coincides with that of x = –c, it is evident that the line x = 0, on which J = 0, is the boundary of a unique relationship between (x, y) and (u, v). This is also true for x = π/2, since x = π/2 – c and x = π/2 + c would have the same image in the (u, v)-plane. In addition, we see a geometric degeneracy associated with x = π/2, for the line x = π/2, 0 ≤ y ≤ 2π, which is a one-dimensional structure, is mapped into a point, u = v = 0, that is of zero dimension.

      The general theorem (see, for example, Goursat, Vol. I, p. 45) on implicit functions is as follows.

      THEOREM: Consider a system of n equations

      Suppose that these equations are satisfied for a set of values

      and that the functions Fj are continuous along with all their first partial derivatives in the neighborhood of this point. Suppose the Jacobian

      does not vanish at the point. Then there exists one and only one system of continuous functions

      An interesting situation, however, arises when the Jacobian vanishes identically in a region. We will show in this case that the two functions u and v are dependent; that is, there exists a function F(u, v) = 0 in that region. For example, if u = x + y, v = x² + 2xy + y², J ≡ 0, and obviously u² – v = 0.

      Suppose first that ux – 0 and uy = 0 (then J ≡ 0); it follows that u is a constant. As x and y range over a region, then the image u and v will range over u = constant, and hence a region of the (x, y)-plane is mapped into a single line u = constant in the (u, v)-plane, and there is no possibility of a one-to-one mapping from one plane to the other. On the other hand, if ux = 0 and vx = 0, then u and v ard each only functions of y, and therefore u is some function of v. If ux = 0 and the Jacobian vanishes, then either vx = 0 or uy = 0. Thus we need consider further only the case where J = 0, but the derivatives themselves do not vanish.

      Since ∂v/∂y ≠ 0, we can solve ν = v(x, y) for y to obtain y = y(x, v), and therefore u = u(x, y(x, v)) = F(x, v). We will now show that if J = 0 then F cannot depend on x. Since

      it follows by subtraction that

      Because x and y are independent variables, dx and dy are independent, and hence

      But since J = 0, the function F does not depend on x, and thus we have obtained a sufficient condition for dependence. The condition J = 0 is obviously also a necessary one. For, if there exists a functional relation π(u, v) = 0 between u = u(x, y) and v = v(x, y), then

      which, considered as a set of linear simultaneous algebraic equations in πu, πν, demands that J = 0 if a nontrivial solution is to exist.

      The general theorem may be stated. (See, for example, Goursat, Vol. I, p. 52.)

      THEOREM: Let u1, u2, . . . , un be n functions of the n independent variables x1, x2, . . . , xn. In order that there exists a function π(u1, u2, . . . , un) = 0 that does not involve the xi explicitly, it is necessary

      and sufficient that the Jacobian

      should vanish identically.

      Remark: If the functions u1, u2, . . . , un involve certain other variables y1, y2, . . . , ym besides the xi, then the vanishing of the same Jacobian is necessary and sufficient for the existence of a functional relation among the ui independent of the xi but perhaps still containing the yi.

      A simple computation will show that if ui, = ui(x1, x2, . . . , xn) and xj = xj(y1, y2, . . . , yn), i = 1 to n, j = 1 to n, then the Jacobian of the transformation from the yj’s directly to the ui’s is the product of the two individual Jacobians; that is,

      From this formula it is apparent that ∂(u1, u2, . . . , un)/∂(u1, u2, . . . , un) = 1 or that ∂(u1, u2, . . . , un)/∂(u1, u2, . . . , un) and ∂(u1, u2, . . . , un)/∂(u1, u2, . . . , un) are reciprocals.

      0.5 Differentiation of Implicit Functions

      Given a function F(x, y, z) = 0, the partial derivatives may be computed from the formula

      For example, if this is written as

      and y is regarded as a function of x and z, so that

      With a pair of functions, say

      The second of these may be solved for

      and substituted into the first to give

      with similar formulas for other partial derivatives. They may be easily generalized to other situations with a larger number of functions and variables. For example, suppose we are given n functions

      where we have assumed that the independent variables are x1, x2, . . . , xp (Note that we must always decide which variables are to be taken as independent.) Then

      where the Jacobian in the denominator is assumed to be nonzero.

      0.6 Surfaces

      Previously, we mentioned that the relation F(x, y, z) = K represents a surface for a fixed value of K and a one-parameter family of surfaces if the parameter K is allowed to vary. Clearly, G(x, y, z, K) = 0 is a one-parameter family of surfaces where the parameter is implicit in the function G. Consider now the three functions

      where the functions ϕ, ψ, and Χ are continuous with continuous partial derivatives. We will assume also that the three Jacobians

      do not vanish; then the functions ϕ, ψ, and Χ are independent, and the first two may be solved for u and v and substituted into the third to give

      to give the equations of a surface. (Note: Actually only one of the Jacobians need be nonzero for the substitution to be valid.) Thus the representation given by Eq. (0.6.1) is a two-parameter representation of a surface. As the point (u, v) ranges over a plane, the point (x, y, z) sweeps out a surface.

      Consider next the parametric representation

      If t is eliminated between the first two, one obtains a function f(x, y) = 0, and if t is eliminated between the second and third, a function g(y, z) = 0. These two functions then define cylinders, the first with generators parallel to the z-axis and the second parallel to the z-axis. The intersection of these cylinders will generally define a curve in space. Thus Eqs. (0.6.3) are the parametric representation of a space curve.

      In Fig. 0.4 we have a surface defined by the two-parameter representation given by Eq. (0.6.1) and a set of curves defined on the surface given by the same set of equations, but with v fixed in one direction and u fixed in the other. Thus the numbers (u, v) may be treated as coordinates on the surface.

      Now, if we have a curve u = u(t), v = v(t) in the (u, v)-plane, it will be mapped by x = ϕ(u, v), y = ψ(u, v), z = Χ(u, v) into a curve on the surface. In the (x, y, z)-space,

      where s is the length of arc along the curve on the surface. But

      Figure 0.4

      We note that ds, the element of arc length on the surface, is independent of the parameterization of the curve in the (u, v)-plane and depends only on the surface itself. In fact, we may write

      0.7 Tangents and Normals

      If we are given the equation of a surface in explicit form z = f (x, y), the equation of the tangent plane at the point (x0, y0, z0), z0 = f(x0, y0) is

      while the equation of the normal to the plane at that point is

      If the equation of the surface is given in implicit form F(x, y, z) = 0, then the tangent plane is

      and the normal line is

      The direction cosines of the normal are therefore proportional to

      If the equation of the surface is given in parametric form,

      Therefore, the equation of the plane is

      and hence the equation of the normal to the plane is

      where the direction cosines are proportional to the appropriate Jacobians in the denominator.

      Exercise

      0.7.1. Two surfaces are given by f(x, y, z) = a and g(x, y, z) = b. They have a common point at (x0, y0, z0). Show that they also have a common tangent plane if fxgy = fygx = 0 and fxgz – fzgx = 0. What is the value of fygz – fzgy at this point?

      0.8 Direction Cosines and Space Curves

      If we suppose that s is arc length along a curve, the direction cosines of the tangent line to a space curve are given by

      If the curve is given parametrically as x = ξ(t), y = η(t), z = ζ(t), then the cosines are given by

      If the space curve is given as the intersection of two cylinders

      the direction cosines are given by

      If the space curve is given by the intersection of two surfaces

      then the direction cosines are proportional to

      Exercise

      0.8.1. Prove the assertions of this section.

      0.9 Directional Derivatives

      Suppose we are given a function of three variables F(x, y, z) and a space curve given parametrically as x = ξ(t), y = η(t), z = ζ(t). Then, as we move along the curve, the function values of F will in general change continuously. If s is arc length, then it is clear that dF/ds is the rate of change of F along the curve in the direction of the tangent. The arc length s could be chosen as a parameter along the curve, and although the actual parametric representation of a curve in terms of arc length is often not easy to obtain, in general, its use in theoretical considerations greatly facilitates the discussion. Now

      The left side is dF/ds, and it is clear that the directional derivative is a generalization of the partial derivatives ∂F/∂x, ∂F/∂y, and ∂F/∂z, since by appropriate choice of the space curve these can be obtained as special cases. The directional derivative at a point is not dependent on a particular space curve, as is apparent, but its value at a given point does depend on the direction at that point. We can then reasonably ask the question: Of all possible directions at a fixed point, in what direction is the function increasing the most rapidly? To see this most clearly, we note that dF/ds is the scalar product of a vector

      with a unit vector [cos α, cos β, cos γ]. However, the first vector is a vector normal to the surface F(x, y, z) = K, and dF/ds may be written

      where dn/ds is the rate of change of the distance along the normal with respect to distance in the direction given by the vector [cos α, cos β, cos γ]. Now (Fig. 0.5 shows the two-dimensional version)

      Figure 0.5

      where θ is the angle between the normal direction and the vector [cos α, cos β, cos γ]. dF/ds is thus a maximum when θ = 0, and the value of the directional derivative normal to the surface F(x, y, z) = K is

      and is called the gradient.

      In the sequel, much use will be made of the directional derivative, for the theory of first-order partial differential equations leads to the idea of characteristic curves, and solution of the equations is obtained by solving ordinary differential equations in the direction of these characteristic curves.

      0.10 Envelopes

      In this section we will discuss envelopes of curves and surfaces since these will play an important role in finding solutions of first-order partial differential equations.

      Suppose we are given a one-parameter family of curves in the (x, y)-plane expressed analytically as

      Then as α is varied, different members of the family will be generated. Now it may happen when one considers the totality of such curves that each is tangent to a single curve. If this is the case, then the tangent curve is called an envelope. For example, consider the family of straight lines

      As shown in Fig. 0.6, as α varies through positive and negative values, it appears that all straight lines will be tangent to a curve having the appearance of a parabola.

      Figure 0.6

      The method of computation of the envelope is based on the idea of finding the point of intersection of adjacent members of the family. Let α be fixed at α’ ; then

      is also a member of the family. If these two curves intersect at a point P(x, y), then any linear combination of the equations gives a curve that passes through the point of intersection P, and therefore the curve

      Thus we have two equations, each of which is a curve passing through the point P(x, y) at the point of tangency of two adjacent members of the family, and therefore the result of eliminating a’ between these should produce the locus of the tangencies and therefore the envelope. In the case of our example, these equations are

      or the envelope is y² = 2x, a parabola.

      There are examples in which the preceding method leads to loci that are not true envelopes. For instance, (y – α)² = x³ is a family of curves lying in the right half-plane of which the member α has a cusp on the y-axis at y = α. Differentiating with respect to α gives y – α = 0, and substituting back gives x = 0. This is indeed a singular locus, but it is a locus of cusps rather than an envelope, since the tangent to the member of the family does not coincide with the tangent to the locus. A sufficient condition for the locus defined by Eqs. (0.10.2) and (0.10.3) to be an envelope that is valid for most situations is given by the following:

      THEOREM: Let f(x, y, α) be continuous and have continuous partial derivatives (fx, fy, fα, fxα, fyα, fαα) in the neighborhood of a point (x0, y0, α0) so that

      define a curve that is the envelope of the family f(x, y, α) = 0 in the neighborhood of(x0, y0, α0).

      Let us consider now the corresponding situation for surfaces. Let S be a one-parameter family of surfaces

      If there exists a surface T that is tangent to each of the members of S along a curve C, then T is said to be the envelope of S. The tangency curve C is called a characteristic curve. If an envelope to S is to exist, then there must be a curve on each member of S, and the locus of these curves must be tangent to each surface of S along the corresponding curves C on S. As before, then,

      f(x, y, z, α + Δα) – f(x, y, z, α) = 0

      is a surface passing through the intersection of the two surfaces and hence through the point of tangency in the limit of the envelope and the surface. In the limit

      and hence the elimination of α between this equation and f(x, y, z, α) = 0 should give the equation for the envelope.

      We can also consider a two-parameter family of surfaces

      and ask whether there is an envelope. If again we consider a nearby surface

      f(x, y, z, α + Δα, β + Δβ) = 0

      then

      f (x, y, z, α + Δα, β + Δβ) – f(x, y, z, α, β) = 0

      is a surface passing through the intersection of the two surfaces, and in the limit

      Since α and β are independent, so also are dα and dβ, and hence

      and the locus of the point P(x, y, z), the result of the elimination of α and β from these two equations and f(x, y, z, α, β) = 0, is the envelope (if it exists).

      Now the envelope in this case is somewhat different than in the two-dimensional case or the single-parameter case. For if we consider the equations in pairs,

      it is apparent that in general no curve of contact with the envelope can exist here, since for fixed β the first pair will define one curve, while for fixed α the second pair will define a second curve. Since these two lines must lie on a surface, they must intersect in general, and hence the envelope is the locus of points of contact (tangencies) rather than being a locus of characteristic curves.

      If one supposes that β = ϕ (α), then

      f(x, y, z, α, ϕ (α)) = 0

      is a one-parameter family of surfaces, and the envelope is given by the simultaneous elimination of α from

      The envelope then contacts the surface in a characteristic curve, but the characteristic depends on the particular function ϕ (α). The totality of all characteristics for different ϕ (α), in general, will not sweep out a surface envelope.

      As an example, consider the family of unit spheres with centers in z = 0:

      f = (x – α)² + (y – β)² + z² – 1 = 0

      and the envelope is given by z = ±1, that is, the pair of horizontal planes with point contacts with each of the spheres with centers on the (x, y)-plane and unit radius.

      Suppose now that we pick out a one-parameter family whose centers lie on y = x², z = 0 by setting β = α²; then

      The characteristic curves are circles, of which a typical one is the intersection of the plane fα = 0 with the sphere f = 0. The equation of the envelope may be obtained by elimination of α between the two equations, giving a rather complicated expression representing the parabolic tube shown in Fig. 0.7.

      A one-parameter family of straight lines generates a ruled surface. If the family is given by the two planes

      where α is a parameter, then a surface will be generated that depends on the four functions f1(α), f2(α), g1(α), and g2(α). Consider now a one-parameter family of planes

      z = αx + f(α)y + ϕ (α)

      The envelope of this family is obtained by eliminating a between this equation and

      x + yf’(α) + ϕ’(α) = 0

      These two equations represent a straight line, and therefore the characteristic curves are straight lines and the envelope is a ruled surface. This envelope is called a developable surface.

      Figure 0.7

      Exercises

      0.10.1. Draw the family of lines

      for various values of α, and observe that an envelope is formed touching the x- and y-axes at (1, 0) and (0, -1), respectively.

      0.10.2. Show that the family of straight lines

      where κ is a constant 0 < κ < 1, envelopes a parabola lying in the fourth quadrant and touching the x- and y-axes at ((1 – κ)/κ, 0) and (0, – (1 -κ)), respectively.

      0.10.3. For the family of lines

      where k: is a positive constant, determine the envelope in terms of the parameter α; that is, find x(α) and y(α) for the envelope. When 0 ≤ α ≤ 1, sketch the lines and envelope for (a) k: = 3, (b) k = 6.75, and (c) k = 12.

      0.10.4. Find the envelope of the planes

      0.11 Differential Equations

      For the system of differential equations

      with the initial conditions

      x = x0, y = y0, z = z0 at t = 0

      we have the fundamental existence and uniqueness theorem for solutions.

      THEOREM: Let the functions f(x, y, z, 0, g(x, y, z, t), and h(x, y, z, t) be defined and continuous in a domain D and satisfy a Lipschitz

      condition in the same domain,

      with similar conditions for g and h. Then for each point

      t

      ,

      x

      ,

      y

      , and z in D there exists a unique solution

      A solution defined in an interval a < t < b is said to be complete if there is no solution defined in a larger interval and agreeing with the given solution in α < t < b. If b < ∞ and the solution is continuous from the left at b but remains in D, then it is not complete and the solution can be extended beyond b. If b < ∞ and the solution is complete, then as t → b from the left a boundary point of D must be approached. A similar argument holds for t = a. Under the hypothesis of the theorem it may be shown that a unique complete solution exists.

      Now suppose we are given a curve in (x, y, z)-space in parametric form:

      And suppose the initial conditions for the system of differential equations are given at a particular value of ξ, say ξ0. Then the solution of Eqs. (0.11.1) will be of the form

      But ξ0 can be any point on the initial curve; so, dropping the subscript zero, the family of solutions that pass through the curve (0.11.2) is given by

      This is a two-parameter family and therefore defines a surface that is generated by the solution curves of the differential equations. It may be shown that the solution depends continuously on the initial data; that is, two solution curves that are close together initially stay close together. Therefore, if the initial curve is continuous, the surface defined by the solution curves will be continuous. (See Fig. 0.8.)

      0.12 Strips

      We have seen that the three differential equations

      subject to x = x0, y = y0, z = z0 at t = 0, define a space curve through the point. We have also seen that the direction normal to the surface z = z(x, y) is defined by the ratio p:q: -1, where p = zx and q = zy are the partial derivatives of z with respect to x and y. The set of five quantities (x, y, z, p, q) may be thought of as defining a surface element located at (x, y, z) with normal in the direction given by the ratio p:q: -1, as shown in Fig. 0.9.

      Now suppose that we have a set of five equations with prescribed va1ues at t = 0:

      These can be integrated (subject to the usual conditions) and will give the quintuple x(t), y(t), z(t), p(t), q(t). Thus we will have a space curve x(t), y(t), z(t) together with a normal direction p(t):q(t): –1 defined at each point. We may think of this as a ribbon or strip, as shown in Fig. 0.9.

      However, the differential equations and initial conditions cannot be quite arbitrary if the strip is to be continuous and not have a break of the sort shown in the insert to Fig. 0.9. In the first place, the functions f, g, h, k, and l must be continuous and satisfy a Lipschitz condition as in the previous

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