MATH 31BH Homework 5 Solutions

Transcription

1 MATH 3BH Homework 5 Solutions February 4, 204 Problem.8.2 (a) Let x t f y = x 2 + y 2 + 2z 2 and g(t) = t 2. z t 3 Then by the chain rule a a a D(g f) b = Dg f b Df b c c c = [Dg(a 2 + b 2 + 2c 2 )] [ 2a 2b 4c ] = 2a 2 + 2b 2 + 4c 2 [ 2a 2b 4c ] 3(a 2 + b 2 + 2c 2 ) 2 2a 2b 4c = 4a + 4ab 2 + 8ac 2 4a 2 b + 4b 3 + 8bc 2 8a 2 c + 8b 2 c + 6c 3. 6a(a 2 + b 2 + 2c 2 ) 2 6b(a 2 + b 2 + 2c 2 ) 2 2c(a 2 + b 2 + 2c 2 ) 2 (b) Let x f y = z ( x 2 ) + z yz and g ( ) a = a 2 + b 2. b Then by the chain rule x x x D(g f) y = Dg f y Df y z z z [ ( x = Dg 2 )] [ ] + z 2x 0 yz 0 z y = [ 2x 2 + 2z 2yz ] [ ] 2x 0 0 z y = [ 4x 3 + 4xz 2yz 2 2x 2 + 2z + 2y 2 z ].

2 Problem.8.0 (a) If a function f : R 2 R can be written as ϕ(x 2 + y 2 ) for some differentiable function ϕ : R R, then f satisfies xd 2 f yd f = 0. Proof. By the chain rule, we have D f = D (ϕ(x 2 + y 2 )) = 2xϕ (x 2 + y 2 ). Similarly, Thus D 2 f = D 2 (ϕ(x 2 + y 2 )) = 2yϕ (x 2 + y 2 ). xd 2 f yd f = 2xyϕ (x 2 + y 2 ) 2xyϕ (x 2 + y 2 ) = 0. Problem 2 (i) The function f : R n R defined by is homogeneous of degree d. f(x, x 2,..., x n ) = x d + x d x d n Proof. Let t R. Then f(tx, tx 2,..., tx n ) = (tx ) d + (tx 2 ) d (tx n ) d = t d x d + t d x d t d x d n = t d (x d + x d x d n) = t d f(x, x 2,..., x n ). So f is homogeneous. (ii) If f is differentiable and homogeneous of degree d then x f x x n f = d f(x,..., x n ). Proof. Since f is homogeneous, for all t R we have f(tx, tx 2,..., tx n ) = t d f(x, x 2,..., x n ). 2

3 We will differentiate both sides of this equation with respect to t. By the chain rule, the lefthand side becomes f(tx,..., tx n ) x x t f(tx,..., tx n ) t = x f(tx,..., tx n ) x x n f(tx,..., tx n ). The righthand side becomes dt d f(x,..., x n ). It follows that x f(tx,..., tx n ) x x n f(tx,..., tx n ) = dt d f(x,..., x n ) for all t R. In particular, if t =, then x f x x n f = d f(x,..., x n ). Problem 3 Find the tangent planes to the following level surfaces. (i) xy 2 + xyz = 2 at (,, ) Let F (x, y, z) = xy 2 + xyz 2. Then F x = y 2 + yz F y = 2xy + xz F z = xy. Thus F x (,, ) = 2, F y (,, ) = 3, F z (,, ) =. So the tangent plane to this level surface at (,, ) is 2(x ) + 3(y ) + (z ) = 0. (ii) xyz 2 + 2x 3 y = 3 at (,, ) Let G(x, y, z) = xyz 2 + 2x 3 y 3. Then G x = yz 2 + 6x 2 y G y = xz 2 + 2x 3 G z = 2xyz. 3

4 So G x (,, ) = 7, G y (,, ) = 3, G z (,, ) = 2. Therefore the tangent plane to this level surface at (,, ) is 7(x ) + 3(y ) + 2(z ) = 0. Problem 4 (i) The function T (x, y) = ln(x 2 + y 2 ) satisfies Laplace s equation, T xx + T yy = 0. Proof. We have and T x = 2x x 2 + y 2 T xx = 2(x2 + y 2 ) 4x 2 (x 2 + y 2 ) 2 = 2y2 2x 2 (x 2 + y 2 ) 2 T y = 2y x 2 + y 2 T yy = 2(x2 + y 2 ) 4y 2 (x 2 + y 2 ) 2 = 2x2 2y 2 (x 2 + y 2 ) 2. So T xx + T yy = 2y2 2x 2 (x 2 + y 2 ) 2 + 2x2 2y 2 (x 2 + y 2 ) 2 = 0. (ii) The function T (x, y) = e x cos(y) is harmonic. Proof. We have T x = e x cos(y) T xx = e x cos(y) and T y = e x sin(y) T yy = e x cos(y), 4

5 so T xx + T yy = e x cos(y) e x cos(y) = 0. (iii) If T (x, y) = h(x 2 + y 2 ) is harmonic for some function h : R + R, then h(z) = A ln z + B for some constants A, B R. Proof. Suppose T is harmonic. Then T xx + T yy = 0. Now, by the chain rule, T x = 2xh (x 2 + y 2 ) T xx = 4x 2 h (x 2 + y 2 ) + 2h (x 2 + y 2 ) and T y = 2yh (x 2 + y 2 ) T yy = 4y 2 h (x 2 + y 2 ) + 2h (x 2 + y 2 ), so 0 = T xx + T yy = 4(x 2 + y 2 )h (x 2 + y 2 ) + 4h (x 2 + y 2 ). Now let z = x 2 + y 2 and g(z) = h (z). Then we have 4zg (z) + 4g(z) = 0. This gives us a differential equation with the solution g(z) = C z for some constant C R. By integrating g, we find that h(z) = A ln z + B for some constants A, B R. (iv) We might define a harmonic function of n variables to be a function f : R n R which satisfies Problem 5 f = 2 f x f x 2 n 5 = 0.

6 The d Alambertian of the function u(x,..., x n, t) of (n + ) variables is defined to be u = 2 u x u x 2 2 u n c 2 t 2. A function u(x, t) satisfies the wave equation if its d Alambertian is zero: u(x, t) = 0. (i) Assume F and F 2 are arbitrary twice differentiable functions of a single variable, and let n =. Then satisfy the wave equation. Proof. By the chain rule we have u (x, t) = F (x ct) and u 2 (x, t) = F 2 (x + ct) and So u x = F (x ct) 2 u x 2 = F (x ct) u t = cf (x ct) 2 u t 2 = c2 F (x ct). u = F (x ct) c 2 c2 F (x ct) = 0. Similarly, and So u 2 x = F 2(x + ct) 2 u 2 x 2 = F 2 (x + ct) u 2 t = cf 2(x + ct) 2 u 2 t 2 = c2 F 2 (x + ct). u 2 = F 2 (x + ct) c 2 c2 F 2 (x + ct) = 0. 6

7 (ii) For all n, the sinusoidal wave satisfies the wave equation with constant c = Proof. For each i =,..., n we have We also have Then the d Alembertian of u with c = u = u(x,..., x n, t) = A sin( k x ωt) ω k. u xi = k i A cos( k x ωt) u xi x i = k 2 i A sin( k x ωt). u t = ωa cos( k x ωt) u tt = ω 2 A sin( k x ωt). ω k is n ki 2 A sin( k x ωt) + w 2 k 2 A sin( k x ωt) i= = k 2 A sin( k x ωt) + k 2 A sin( k x ωt) = 0. ω 2 Problem 6 The surface z = ln(cos(x)) ln(cos(y)) is a minimal surface; i.e., it satisfies the partial differential equation ( + z 2 y)z xx + ( + z 2 x)z yy = zz x z y z xy. Proof. We have the following partial derivatives of z: z x = sin(x) cos(x) = tan(x) z y = sin(y) cos(y) = tan(x) z xx = sec 2 (x) z yy = sec 2 (y) z xy = 0. 7

8 So ( + z 2 y)z xx + ( + z 2 x)z yy = ( + tan 2 (y))( sec 2 (x)) + ( + tan 2 (x))(sec 2 (y)) = sec 2 (y) sec 2 (x) + sec 2 (x) sec 2 (y) = 0. Moreover, since z xy = 0, we have zz x z y z xy = 0. It follows that ( + z 2 y)z xx + ( + z 2 x)z yy = zz x z y z xy. Problem 7 For any twice continuously differentiable function f : R 2 R, f = f rr + r f r + r 2 f θθ. Proof. We recall that in polar coordinates, x = r cos(θ) and y = r sin(θ). Therefore, by the chain rule, x f r = f x r + f y y r = cos(θ)f x + sin(θ)f y x f θ = f x θ + f y y θ = r sin(θ)f x + r cos(θ)f y. Further differentiating these two functions according to the chain rule, we see f rr = cos(θ) (cos(θ)f xx + sin(θ)f xy ) + sin(θ) (cos(θ)f yx + sin(θ)f yy ) = cos 2 (θ)f xx + 2 sin(θ) cos(θ)f xy + sin 2 (θ)f yy f θθ = r cos(θ)f x r sin(θ) ( r sin(θ)f xx + r cos(θ)f xy ) r sin(θ)f y + r cos(θ) ( r sin(θ)f yx + r cos(θ)f yy ) = r cos(θ)f x r sin(θ)f y + r 2 sin 2 (θ)f xx + r 2 cos 2 (θ)f yy 2r 2 sin(θ) cos(θ)f xy. Thus f rr = cos 2 (θ)f xx + 2 sin(θ) cos(θ)f xy + sin 2 (θ)f yy r f r = cos(θ) f x + sin(θ) f y r r r 2 f θθ = cos(θ) f x sin(θ) f y + sin 2 (θ)f xx 2 sin(θ) cos(θ)f xy + cos 2 (θ)f yy. r r 8

9 Adding like terms, one can see that frr + r f r + r 2 f θθ = ( cos 2 (θ) + sin 2 (θ) ) f xx + ( sin 2 (θ) + cos 2 (θ) ) f yy = f xx + f yy = f. Problem 8 Let f : R 2 R be a harmonic function and let A be an orthogonal 2 2 matrix. Then the function g defined for all x R 2 by g(x) = f(ax) is a harmonic function. Proof. Suppose Then we have By the chain rule, we have [ ] a a A = 2 and x = a 2 a 22 g ( ) x = f y [ ] x. y ( ) a x + a 2 y. a 2 x + a 22 y g x = a f x + a 2 f y g xx = a 2 f xx + a a 2 f xy + a 2 a f yx + a 2 2f yy g y = a 2 f x + a 22 f y f yy = a 2 2f xx + a 2 a 22 f xy + a 22 a 2 f yx + a 2 22f yy. Therefore g = g xx + g yy = a 2 f xx + 2a a 2 f xy + a 2 2f yy + a 2 2f xx + 2a 2 a 22 f xy + a 2 22f yy = ( a 2 + a 2 2) fxx + 2(a a 2 + a 2 a 22 )f xy + ( a a 2 22) fyy. Now since A is orthogonal, it has orthonormal rows. Therefore, the length of each row is and the dot product of the two distinct rows is 0. In other words, we have a 2 + a 2 2 =, a a 2 + a 2 a 22 = 0, a a 2 22 =. 9

10 Therefore the Laplacian of g simplifies to g = f xx + f yy = f. But f is harmonic, so f = 0. It follows that g is also harmonic. Problem 9 A pair of functions u, v : R 2 R in C 2 is said to satisfy the Cauchy-Riemann equations if v y = u x, u y = v x. (i) If u and v satisfy the Cauchy-Riemann equations, then u and v are harmonic functions. Proof. We first show that u is harmonic. Indeed, differentiating the first equation with respect to x and the second equation with respect to y, we see that u xx = v yx u yy = v xy. So u xx + u yy = v yx v xy = v xy v xy = 0. Now let us differentiate the first equation with respect to y and the second equation with respect to x. We find v yy = u xy v xx = u yx. Thus v xx + v yy = ( v xx ) + v yy = u yx + u xy = 0. (ii) We will show that u(x, y) = e x cos(y) and v(x, y) = e x sin(y) are harmonic conjugates. We have u x = e x cos(y) u y = e x sin(y) v x = e x sin(y) v y = e x cos(y). One can verify that indeed u x = v y and u y = v x. 0