Mean-Value Theorem (Several Variables)

Transcription

1 Mean-Value Theorem (Several Variables) 1 Mean-Value Theorem (Several Variables) THEOREM THE MEAN-VALUE THEOREM (SEVERAL VARIABLES) If f is ifferentiable at each point of the line segment ab, then there exists on that line segment a point c between a an b such that f(b) f(a) = f(c) (b a). Proof. As t ranges from 0 to 1, a + t(b a) traces out the line segment ab. The iea of the proof is to apply the one-variable mean-value theorem to the function g(t) = f(a + t[b a]), t [0, 1]. To show that g is ifferentiable on the open interval (0, 1), we take t (0, 1) an form g(t + h) g(t) = f(a + (t + h)[b a]) f(a + t[b a]) = f(a + t[b a] + h[b a]) f(a + t[b a]) = f(a + t[b a]) h[b a] + o(h[b a]). Since f(a + t[b a]) h(b a) = [ f(a + t[b a]) (b a)]h an the o(h(b a)) term is obviously o(h), we can write g(t + h) g(t) = [ f(a + t[b a]) (b a)]h + o(h). Diviing both sies by h, we see that g is ifferentiable an g (t) = f(a + t[b a]) (b a). The function g is clearly continuous at 0 an at 1. Applying the one-variable meanvalue theorem to g, we can conclue that there exists a number t 0 between 0 an 1 such that g(1) g(0) = g (t 0 )(1 0). Since g(1) = f(b), g(0) = f(a), an g (t 0 ) = f(a + t 0 [b a]) (b a), the above gives f(b) f(a) = f(a + t 0 [b a]) (b a).

2 Mean-Value Theorem (Several Variables) 2 Setting c = a + t 0 [b a], we have f(b) f(a) = f(c) (b a). A nonempty open set U (in the plane or in three-space) is sai to be connecte if any two points of U can be joine by a polygonal path that lies entirely in U. THEOREM. Let U be an open connecte set. If f(x) = 0 for all x U, then f is constant in U. Proof. Let a an b be any two points in U. Since U is open an connecte, we can join these points by a polygonal path with vertices a = c 0, c 1,, c n 1, c n = b. By the mean-value theorem there exist points c 1 on c 0 c 1 such that f(c 1 ) f(c 0 ) = f(c1) (c 1 c 0 ), c 2 on c 1 c 2 such that f(c 2 ) f(c 1 ) = f(c2) (c 2 c 1 ), c 3 on c 2 c 3 such that f(c 3 ) f(c 2 ) = f(c3) (c 3 c 2 ), c n on c n 1 c n such that f(c n ) f(c n 1 ) = f(cn) (c n c n 1 ). If f(x) = 0 for all x in U, then f(c 1 ) f(c 0 ) = 0, f(c 2 ) f(c 1 ) = 0,, f(c n ) f(c n 1 ) = 0 This shows that f(a) = f(c 0 ) = f(c 1 ) = f(c 2 ) = = f(c n 1 ) = f(c n ) = f(b). Since a an b are arbitrary points of U, f must be constant on U. THEOREM. Let U be an open connecte set. If f(x) = g(x) for all x in U, then f an g iffer by a constant on U. Proof. If f(x) = g(x) for all x in U, then [f(x) g(x)] = f(x) g(x) = 0 for all x in U. Therefore, f g must be constant on U.

3 Continuity on a Close Region 3 We have seen that a function continuous on an interval skips no values. (The intermeiate-value theorem.) There are analogous results for functions of several variables. Here is one of them. THEOREM AN INTERMEDIATE-VALUE THEOREM (SEVERAL VARIABLES) Suppose that f is continuous on an open connecte set U an A < C < B. If, somewhere on U, f takes on the value A an, somewhere on U, f takes on the value B, then, somewhere on U, f takes on the value C. Proof. Let a an b be points of U for which f(a) = A an f(b) = B. We must show that there exists a point c in U for which f(c) = C. Since U is polygonally connecte, there is a polygonal path in U r = r(t), t [a, b] that joins a to b. Since r is continuous on [a, b], the composition g(t) = f(r(t)) is also continuous on [a, b]. Since g(a) = f(r(a)) = f(a) = A an g(b) = f(r(b)) = f(b) = B, we know from the intermeiate-value theorem of one variable that there is a number c in [a, b] for which g(t) = C. Setting c = r(t) we have f(c) = C. Continuity on a Close Region An open connecte set is calle an open region. If we start with an open region an ajoin to it the bounary, then we have what is calle a close region. (A close region is therefore a close set, the interior of which is an open region.) Continuity on a close region Ω requires continuity at the bounary points of Ω as well as the interior points. If x 0 is an interior point of Ω, then all points x sufficiently close to x 0 are in Ω an, by efinition, f is continuous at x 0 if as x approaches x 0, f(x) approaches f(x 0 ).

4 Continuity on a Close Region 4 If x 0 is a bounary point of Ω, then we have to moify the efinition an say: f is continuous at x 0 if as x approaches x 0 within Ω, f(x) approaches f(x 0 ). In terms of ɛ δ, f is continuous at a bounary point x 0 if for each ɛ > 0 there exists δ> 0 such that if x x 0 < δ an x Ω, then f(x) f(x 0 ) < ɛ. (This is completely analogous to one-sie continuity at an enpoint of a close interval [a, b].) The intermeiate-value result that we just prove for open connecte sets can be extene to close regions. THEOREM A SECOND INTERMEDIATE-VALUE THEOREM [SEVERAL VARI- ABLES] Suppose that f is continuous on a close region Ω an A < C < B. If, somewhere on Ω, f takes on the value A an, somewhere on Ω, f takes on the value B, then, somewhere on Ω, f takes on the value C. Proof. Let a an b be points of Ω for which f(a) = A an f(b) = B. If a an b are both in the interior of Ω, then the result follows from the previous theorem. But one or both of these points coul lie on the bounary of Ω. To take care of that possibility, we can procee as follows. Take ɛ > 0 small enough that A + ɛ < C < B ɛ. By continuity there exist points x 1, x 2 in the interior of Ω for which f(x 1 ) < A + ɛ an B ɛ < f(x 2 ). Then f(x 1 ) < C < f(x 2 ) an the result follows from the previous theorem.

5 Chain Rules 5 Chain Rules For functions of a single variable there is basically only one chain rule. For functions of several variables there are many chain rules. A vector-value function is sai to be continuous provie that its components are continuous. If f = f(x, y, z) is a scalar-value function (a real-value function), then its graient f is a vector-value function. We say that f is continuously ifferentiable on an open set U if f is ifferentiable on U an f is continuous on U. If a curve r lies in the omain of f then we can form the composition (f r)(t) = f(r(t)). The composition f r is a real-value function of a real variable t. The numbers f(r(t)) are the values taken on by f along the curve r. THEOREM CHAIN RULE (ALONG A CURVE) If f is continuously ifferentiable on an open set U an r = r(t) is a ifferentiable curve that lies entirely in U, then the composition f r is ifferentiable an Proof. We will show that t [(f r)(t)] = f(r(t)) r (t). f(r(t + h)) f(r(t)) lim h 0 h = f(r(t)) r (t). For h 0 an sufficiently small, the line segment that joins r(t) to r(t + h) lies entirely in U. This we know because U is open an r is continuous. For such h, the meanvalue theorem we just prove assures us that there exists a point c(h) between r(t) an r(t + h) such that Diviing both sies by h, we have f(r(t + h)) f(r(t)) = f(c(h)) [r(t + h) r(t)]. f(r(t + h)) f(r(t)) h = f(c(h)) [ r(t + h) r(t) ]. h As h tens to zero, c(h) tens to r(t) an by the continuity of f, f(c(h)) f(r(t)).

6 Chain Rules 6 Since the result follows. r(t + h) r(t) h r (t), Problem 1. Use the chain rule to fin the rate of change of f(x, y) = 1 3 (x3 + y 3 ) with respect to t along the curve r(t) = a cos t i + b sin t j. Solution. The rate of change of f with respect to t along the curve r is the erivative t [f(r(t)]. By the chain rule Here t [f(r(t)] = f(r(t)) r (t). f = x 2 i + y 2 j. With x(t) = a cos t an y(t) = b sin t, we have f(r(t)) = a 2 cos 2 t i + b 2 sin 2 t j. Since r (t) = a sin t i + b cos t j, we see that t [f(r(t)] = f(r(t)) r (t) = (a 2 cos 2 t i + b 2 sin 2 t j) ( a sin t i + b cos t j) = a 3 sin t cos 2 t + b 3 sin 2 t cos t = sin t cos t (b 3 sin t a 3 cos t). Remark. Note that we coul have obtaine the same result without invoking the chain rule by first forming f(r(t)) an then ifferentiating: f(r(t)) = f(x(t), y(t)) = 1 3 ([x(t)]3 + [y(y)] 3 ) = 1 3 (a3 cos 3 t + b 3 sin 3 t) so that t [f(r(t))] = 1 3 [3a3 cos 2 t( sin t) + 3b 3 sin 2 t cos t]

7 Another Formulation of the Chain Rule 7 = sin t cos t (b 3 sin t a 3 cos t). Problem 2. Use the chain rule to fin the rate of change of f(x, y, z) = x 2 y + z cos x with respect to t along the twiste cubic r(t) = ti + t 2 j + t 3 k. Solution. Once again we use the relation t [f(r(t)] = f(r(t)) r (t). This time f = (2xy z sin x, x 2, cos x). With x(t) = t, y(t) = t 2, z(t) = t 3, we have f(r(t)) = (2t 3 t 3 sin t, t 2, cos t). Since r (t) = (1, 2t, 3t 2 ), we have t [f(r(t)] = f(r(t)) r (t). = (2t 3 t 3 sin t, t 2, cos t) (1, 2t, 3t 2 ) = 2t 3 t 3 sin t + 2t 3 + 3t 2 cos t = 4t 3 t 3 sin t + 3t 2 cos t. Another Formulation of the Chain Rule The chain rule for functions of one variable, t [u(x(t))] = u (x(t))x (t), can be written In a similar manner, the relation u t = u x x t. t [f(r(t)] = f(r(t)) r (t)

8 Another Formulation of the Chain Rule 8 can be written With this equation takes the form u = ( x, u t y, = u r t. ) an r t = (x t, y t, z t ), u t = x x t + y y t + z t. In the two-variable case, the z-term rops out an we have Problem 3. Fin u/t if u t = x x t + y y t. u = x 2 y 2 an x = t 2 1, y = 3 sin πt. Solution. Here we are in the two-variable case u t = x x t + y y t. Since we have x = 2x, y u t = 2y an x t = 2t, y t = (2x)(2t) + ( 2y)(3π cos πt) = 4t 3 4t 18π sin πt cos πt. = 3π cos πt, This same result may be obtaine by first writing u irectly as a function of t an then ifferentiating: so that u = x 2 y 2 = (t 1) 2 (3 sin πt) 2 u t = 4t3 4t 18π sin πt cos πt. Problem 4. The upper raius of the frustum of a cone is 10 inches, the lower raius is 12 inches, an the height is 18 inches. At what rate is the volume changing if the upper raius ecreases at the rate of 2 inches per minute, the lower raius increases

9 Other Chain Rules 9 at the rate of 3 inches per minute, an the height ecreases at the rate of 4 inches per minute? Solution. Let x be the upper raius, y the lower raius, an z the height. Then V = 1 3 πz(x2 + xy + y 2 ) so that V x = 1 V πz(2x + y), 3 y = 1 V πz(x + 2y), 3 = 1 3 π(x2 + xy + y 2 ). Since we have here V t = V x x t + V y y t + V z t, Set we fin that V t = 1 πz(2x + y)x 3 t + 1 πz(x + 2y)y 3 t π(x2 + xy + y 2 ) z t. x = 10, y = 12, z = 18, x t = 2, y t V t = π. = 4, z t = 4 The volume ecreases at the rate of 772/3 cubic inches per minute. Other Chain Rules In the setting of functions of several variables there are numerous chain rules. They can all be euce from the theorem prove above an its corollaries. If, for example, u = u(x, y), where x = x(s, t) an y = y(s, t), then s = x x s + y y s an t = x x t + y To obtain the first equation, keep t fixe an ifferentiate u with respect to s accoring to the formula in the chain rule; to obtain the secon equation, keep s fixe an ifferentiate u with respect to t. y t.

10 Other Chain Rules 10 In the figure below we have rawn a tree iagram for the formula. We construct such a tree by branching at each stage from a function to all the variables that irectly etermine it. Each path starting at u an ening at a variable etermines a prouct of (partial) erivatives. The partial erivative of u with respect to each variable is the sum of the proucts generate by all the irect paths to that variable. Suppose, for example, that u = u(x, y, z), where x = x(s, t), y = y(s, t), z = z(s, t). The partials of u with respect to s an t can be rea from the iagram: s = x x s + y y s + s an t = x x t + y y t + Problem 5. (Implicit ifferentiation) Let u = u(x, y, z) be a continuously ifferentiable function, an suppose that the equation u(x, y, z) = 0 efines z implicitly as a ifferentiable function of x an y. Show that if 0, then x = / x / Solution. To be able to apply the chain rule, we write Then Since u(s, t, z(s, t)) = 0 for all s an t, an y = / y /. u = u(x, y, z) with x = s, y = t, z = z(s, t). s = x x s + y y s + s. s = 0. t.

11 A first-orer partial ifferential equation with constant coefficients 11 Since we have If 0, then The formula for y 0 = x 1 + x s y 0 + = 1 an y s = 0, s = x + x = / x / can be obtaine in a similar manner. s = x + x. A first-orer partial ifferential equation with constant coefficients Consier the first-orer partial ifferential equation f(x, y) f(x, y) x y All the solutions of this equation can be foun by geometric consierations. We express = 0. the left member as a ot prouct, an write the equation in the form (3, 2) f(x, y) = 0. This tells us that the graient vector f(x, y) is orthogonal to the vector 3i + 2j at each point (x, y). But we also know that f(x, y) is orthogonal to the level curves of f. Hence these level curves must be straight lines parallel to 3i + 2j. In other wors, the level curves of f are the lines 2x 3y = c. Therefore f(x, y) is constant when 2x 3y is constant. This suggests that f(x, y) = g(2x 3y) for some function g.

12 A first-orer partial ifferential equation with constant coefficients 12 Now we verify that, for each ifferentiable function g, the scalar fiel f efine by this equation oes, inee, satisfy the given PDE. Using the chain rule to compute the partial erivatives of f we fin f x = 2g (2x 3y), f y = 3g (2x 3y), 3 f x + 2 f y = 6g (2x 3y) 6g (2x 3y) = 0. Therefore, f satisfies the PDE as require. Conversely, we can show that every ifferentiable f which satisfies the PDE must necessarily have the form f(x, y) = g(2x 3y) for some g. To o this, we introuce a linear change of variables, x = Au + Bv, y = Cu + Dv. This transforms f(x, y) into a function of u an v, say h(u, v) = f(au + Bv, Cu + Dv). We shall choose the constants A, B, C, D so that h satisfies the simpler equation h(u, v) = 0. Then we shall solve this equation an show that f has the require form. Using the chain rule we fin Since f satisfies h = f x x + f y y = f x A + f y C. f(x, y) f(x, y) x y = 0. we have f/ y = (3/2)( f/ x), so the equation for h/ becomes Therefore, h will satisfy h(u,v) fin h = f x (A 3 2 C). = 0 if we choose A = 3 C. Taking A = 3 an C = 2 we 2 x = 3u + Bv, y = 2u + Dv.

13 A first-orer partial ifferential equation with constant coefficients 13 For this choice of A an C, the function h satisfies h(u,v) of v alone, say h(u, v) = g(v) = 0, so h(u, v) is a function for some function g. To express v in terms of x an y we eliminate u from x = 3u + Bv, y = 2u + Dv an obtain 2x 3y = (2B 3D)v. Now we choose B an D to make 2B 3D = 1, say B = 2, D = 1. For this choice the transformation x = Ax + Bv, y = Cu + Dv. is nonsingular; we have v = 2x 3y, an hence f(x, y) = h(u, v) = g(v) = g(2x 3y). This shows that every ifferentiable solution for has the form f(x, y) f(x, y) x y f(x, y) = g(2x 3y). Exactly the same type of argument proves the following theorem for first-orer equations with constant coefficients. THEOREM Let g be ifferentiable on R, an let f be the scalar fiel efine on R 2 by the equation f(x, y) = g(bx ay), where a an b are constants, not both zero. ifferential equation f(x, y) f(x, y) a + b x y = 0 Then f satisfies the first-orer partial everywhere in R 2. Conversely, every ifferentiable solution of this PDE necessarily has the form for some g. = 0

14 Alembert s Solution of the Wave Equation 14 The one-imensional wave equation Imagine a string of infinite length stretche along the x-axis an allowe to vibrate in the xy-plane. We enote by y = f(x, t) the vertical isplacement of the string at the point x at time t. We assume that, at time t = 0, the string is isplace along a prescribe curve, y = F (x). We regar the isplacement f(x, t) as an unknown function of x an t to be etermine. A mathematical moel for this problem (suggeste by physical consierations) is the partial ifferential equation 2 f t 2 = c2 2 f x 2, where c is a positive constant epening on the physical characteristics of the string. This equation is calle the one-imensional wave equation. We will solve this equation subject to certain auxiliary conitions. Since the initial isplacement is the prescribe curve y = F (x), we seek a solution satisfying the conition f(x, 0) = F (x). We also assume that y/ t, the velocity of the vertical isplacement, is prescribe at time t = 0, say D 2 f(x, 0) = G(x), where G is a given function. It seems reasonable to expect that this information shoul suffice to etermine the subsequent motion of the string. We will show that, inee, this is true by etermining the function f in terms of F an G. The solution is expresse in a form given by Jean Alembert ( ), a French mathematician an philosopher. Alembert s Solution of the Wave Equation Theorem. Let F an G be given functions such that G is ifferentiable an F is twice ifferentiable on R. Then the function f given by the formula f(x, t) = F (x + ct) + F (x ct) c x+ct x ct G(s) s

15 Alembert s Solution of the Wave Equation 15 satisfies the one-imensional wave equation 2 f t 2 = c2 2 f x 2 an the initial conitions f(x, 0) = F (x), D 2 f(x, 0) = G(x). Conversely, any function f with equal mixe partials which satisfies these conitions necessarily has this form. Proof. It is a straightforwar to verify that the function f given this way satisfies the wave equation an the given initial conitions. We shall prove the converse. One way to procee is to assume that f is a solution of the wave equation, introuce a linear change of variables, x = Au + Bv, t = Cu + Dv which transforms f(x, t) into a function of u an v, say g(u, v) = f(au + Bv, Cu + Dv), an choose the constants A, B, C, D so that g satisfies the simpler equation 2 g v = 0. Solving this equation for g we fin that g(u, v) = ϕ 1 (u) + ϕ 2 (v), where ϕ 1 (u) is a function of u alone an ϕ 2 (v) is a function of v alone. The constants A, B, C, D can be chosen so that u = x + ct, v = x ct, from which we obtain f(x, t) = ϕ 1 (x + ct) + ϕ 2 (x ct). Then we use the initial conitions to etermine the functions ϕ 1 (x) an ϕ 2 (x) in terms the given functions F an G. We will obtain this form by another metho which makes use of the previous theorem an avois the change of variables. First we rewrite the wave equation in the form L 1 (L 2 f) = 0,

16 Alembert s Solution of the Wave Equation 16 where L 1 an L 2 are the first-orer linear ifferential operators given by L 1 = t c x, L 2 = t + c x. Let f be a solution of L 1 (L 2 f) = 0 an let u(x, t) = L 2 f(x, t). Equation L 1 (L 2 f) = 0 states that u satisfies the first-orer equation L 1 (u) = 0. Hence, by the previous theorem we have u(x, t) = ϕ(x + ct) for some function ϕ. Let Φ be any primitive of ϕ, say Φ(y) = y 0 We will show that L 2 (v) = L 2 (f). We have v(x, t) = 1 Φ(x + ct). 2c ϕ(s) s, an let v x = 1 2c Φ (x + ct) an v t = 1 2 Φ (x + ct), so L 2 v = v t + c v x = Φ (x + ct) = u(x, t) = L 2 f. In other wors, the ifference f v satisfies the first-orer equation L 2 (f v) = 0. By the previous theorem we must have f(x, t) v(x, t) = ψ(x ct) for some function ψ. Therefore f(x, t) = v(x, t) + ψ(x ct) = 1 Φ(x + ct) + ψ(x ct). 2c This proves that f(x, t) = ϕ 1 (x + ct) + ϕ 2 (x ct) with ϕ 1 = 1 2c Φ an ϕ 2 = ψ. Now we use the initial conitions f(x, 0) = F (x), D 2 f(x, 0) = G(x).

17 Alembert s Solution of the Wave Equation 17 (9.13) to etermine the functions ϕ 1 an ϕ 2 in terms of the given functions F an G. The relation f(x, 0) = F (x) implies ϕ 1 (x) + ϕ 2 (x) = F (x). The other initial conition, D 2 f(x, 0) = G(x), implies cϕ 1(x) cϕ 2(x) = G(x). Differentiating we obtain Solving these we fin ϕ 1(x) + ϕ 2(x) = F (x). ϕ 1(x) = 1 2 F (x) + 1 2c G(x), ϕ 2(x) = 1 2 F (x) 1 2c G(x). Integrating these relations we get ϕ 1 (x) ϕ 1 (0) = F (x) F (0) c x 0 G(s) s, ϕ 2 (x) ϕ 2 (0) = F (x) F (0) 2 1 2c x 0 G(s) s. In the first equation we replace x by x+ct; in the secon equation we replace x by x ct. Then we a the two resulting equations an use the fact that ϕ 1 (0) + ϕ 2 (0) = F (0) to obtain f(x, t) = ϕ 1 (x + ct) + ϕ 2 (x + ct) = F (x + ct) + F (x ct) c x+ct x ct G(s) s This completes the proof. EXAMPLE. Assume the initial isplacement is given by the formula 1 + cos πx for 1 x 1 F (x) = 0 for x > 1. Suppose that the initial velocity G(x) = 0 for all x. Then the resulting solution of the wave equation is given by the formula f(x, t) = F (x + ct) + F (x ct). 2