Applied Mathematics Masters Examination Fall 2016, August 18, 1 4 pm.

Transcription

1 Applied Mathematics Masters Examination Fall 16, August 18, 1 4 pm. Each of the fifteen numbered questions is worth points. All questions will be graded, but your score for the examination will be the sum of your scores on your eight best questions. Problem 1. Let Show that (, ) exists but Real Analysis xy f (x, y) = x if (x, y) = (, ) + y if (x, y) = (, ). is not continuous at (, ). Solution: To compute (, ), we consider, for h > : lim h 1 ( ) f (h, ) f (, ) = lim. h h Hence (, ) exists and is equal to zero. On the other hand, for (x, y) = (, ), we compute: (x, y) = y (x + y ) x y (x + y ) = y4 x y (x + y ). We see, that along the y-axis: (, y) = 1, whereas, along the x-axis: (x, ) =. Thus is not continuous at the point (, ). Problem. Define f : R 3 R f (x, y, z) = x y + e x + z. a) Prove that there exists a differentiable, realvalued function g defined in a neighborhood of (1, 1) R such that b) Compute g g y (1, 1) and z (1, 1). g(1, 1) = and f ( g(y, z), y, z ) =. Solution: (a) The Implicit Function Theorem yields the existence of such a differentiable function g, provided (, 1, 1) =. This can easily be checked, since = xy + ex = 1 =, 1 at (, 1, 1).

2 (b) In order to g y (1, 1) we use the chain rule in the expression f ( g(y, z), y, z ) =, to obtain g (, 1, 1) (1, 1) + (, 1, 1) =. y y This yields g g y (1, 1) =. Likewise we obtain z (1, 1) = 1. Problem 3. Let H be the parallelogram in R whose vertices are (1, 1), (3, ), (4, 5), (, 4). Find the affine map T : R R which sends (, ) to (1, 1), (1, ) to (3, ), (, 1) to (, 4). Show that the Jacobian of this map is J T = 5. Use T to convert the integral α = e x y dx dy to an integral over unit-square, and thus compute α. H Solution: The affine map T : (u, v) (x, y), mapping the unit-square into H, is easily found to be: ( ) ( ( ) ( x 1 u 1 = +, y 1 3) v 1) and the corresponding Jacobian is: J T (x, y) (u, v) = det ( ) 1 = Using the change of variables formula, we find 1 1 α = e x y dx dy = e u e v J T du dv = 5 (e 1)(e 1). H Problem 4. Evaluate the following integral Complex Analysis sin(x) x dx. Solution: sin(x) x dx = 1 i e ix = π Res z= x dx [ e iz ] z 1 i lim e R CR iz z dz, where C R is the upper half of the circle z = R from z = R to z = R. Note that [ e iz ] Res z= = 1, z

3 and by Jordan s lemma. Therefore, e lim R CR iz z dz = sin(x) x dx = π. Problem 5. Consider the function f (z) = cos z z + 1. (1) Find all singular points. () Classify the singular points as essential, removable, or poles. (3) Find the residues of all singular points. Solution: z = i, z = i are simple poles. Res z=i f (z) = cos(i) i, Res z= i f (z) = cos( i). i Problem 6. Determine the number of zeros, counting multiplicities, of the polynomial in the disk {z : z < 1}. p(z) = z 7 4z 3 + z 1 Solution: Since 4z 3 = 4 > 3 = z 7 + z + 1 z 7 + z 1 on the circle {z : z = 1}, p(z) has 3 zeros by Rouché s theorem. Applied Differential Equations Problem 7. Find a 1-parameter family of solutions of the equation (x 1) dx + (x + y) dy =. Solution: Let u = x 1 and v = x + y, then du = dx and dv = dx + dy, which gives dx = du, dy = dv du. Thus the equation can be rewritten as udu + v(dv du) = = (u v)du + vdv =. 3

4 Take w = u v. Thus, v = uw and dv = udw + wdu. Rewriting the equation above gives (u uw)du + uw(udw + wdu) = = u(1 w) du + u wdw =. The last equation is separable. If u = and w = 1, we have 1 u du = w (1 w) dw = 1 u du = [ 1 w (1 w) 1 (1 w) Integrating yields [ ] 1 1 w u du = (1 w) 1 (1 w) dw. It follows that for a constant C log u + C = log 1 w + 1 w 1, and substituting the variables gives log x 1 + C = log which is a 1-parameter family of solutions. x + y + 1 x 1 + x 1 x + y + 1, ] dw. Problem 8. Find the general solution to the system: x + 4x = 3 sin t, x y + y = cos t. Solution: The first equation does not contain y, so we solve for x by this equation. The corresponding homogeneous equation has solutions x h1 = cos t, x h = sin t. A particular solution has the form x p = A sin t + B cos t; inserting x p to the first equation yields A sin t B cos t + 4A sin t + 4B cos t = 3 sin t = A = 1, B =. Thus, the solution of x is x(t) = c 1 sin t + c cos t + sin t. Inserting x(t) to the second equation, we obtain y y = c 1 cos t c sin t cos t. The corresponding homogeneous equation has solutions y h1 = e t, y h = e t. A particular solution has the form y p = C cos t + D sin t + E sin t + F cos t; inserting y p to the second equation yields 5C cos t 5D sin t E sin t F cos t = c 1 cos t c sin t cos t 4

5 = C = c 1 5, D = c 5, E =, F = 1. Thus, the solution of y is y(t) = c 3 e t + c 4 e t c 1 5 cos t + c 5 sin t + 1 cos t. Problem 9. Find the Laplace transform of x, y, i.e. L[x] and L[y] for the system: d x dt dy dt = 1 t, dx dt + dy dt = 4et + x, with x() =, y() =, x () = 1. Solution: Take Laplace transform on the two equations to obtain s L[x] sl[y] = s 1 s, (1) (s 1)L[x] + L[y] = 4 s 1. Multiplying the first equation by and second one by s, adding them yields (3s s)l[x] = + s s + 4s s 1 [ 1 = L[x] = 3s + s s s + 4s ]. s 1 Inserting L[x] to the second equation in (1) gives L[y] = s 1 s 1 [ (3s + s) s s + 4s ]. s 1 Applied Partial Differential Equations Problem 1. Consider the following eigenvalue problem: d φ (x) = λφ(x), < x < L, dx subject to the boundary conditions φ() = dφ (L) =. dx Find all eigenvalues and the associated eigenfunctions under the assumption that λ R. 5

6 Solution: Case 1) λ =. We have φ(x) = c 1 + c x, c 1, c R. Using the boundary conditions, we find φ =. Case ) λ <. For s = λ, we have φ(x) = c 1 e sx + c e sx = A cosh( λx) + B sinh( λx). Using the boundary condition at x =, since sinh() =, cosh() = 1, we find A =. For the boundary condition at x = L, we derive that φ (L) = B λ cosh( λl) =, which implies B =. Hence, φ is trivial. Case 3) λ >. We have φ(x) = a cos( λx) + b sin( λx). Using boundary conditions, φ() = = a =, φ (L) = b λ cos( λl) =. Since λ =, b ==, we deduce that ( (n + 1)π ), λ = λ n = n =, 1,,... L Hence, ( ) (n + 1)π φ = b n sin x. L Problem 11. Consider the D Laplace equation u = u rr + r 1 u r + r u θθ =, in a half disc Ω = {(r, θ) : < r < R, < θ < π}. Assume that u is zero on the diameter {(x, y) : R x R, y = } and that Find the solution. Solution: Since u = on the diameter, Setting u(r, θ) = ψ(r)φ(θ), we find u(r, θ) = sin θ. u(r, ) = u(r, π) = u(, θ) =. φ (θ) = λφ(θ), φ() = φ(π) =. Hence, putting λ = n, we find φ n (θ) = sin(nθ). For the radial direction, we deduce A general solution is r ψ n(r) + rψ n(r) = n ψ(r). φ n (r) = ar n + br n. 6

7 Since u(, θ) =, we have ψ() = = b =. Hence, u(r, θ) = Since u(r, θ) = sin θ, we derive that c n r n sin(nθ). n=1 u(r, θ) = r sin θ. R Problem 1. Use the Fourier transform to derive d Alembert s formula u t c u =, < x <, t >, u(x, ) = f (x), u (x, ) =. t Solution: Using Fourier transform, tt û + c ξ û = Then, we find û(t, ξ) = a(ξ) cos(cξt) + b(ξ) sin(cξt). Using the initial condition and boundary condition, Then, we deduce that û(t, ξ) = ˆf (ξ) cos(cξt). u(ξ, t) = F 1 ( ˆf (ξ) cos(cξt) ) = 1 ( ) F 1 ˆf (ξ)(e icξt + e icξt ) = 1 ( f (x ct) + f (x + ct)). Numerical Analysis Problem 13. The forward first-order derivative approximation of a function f (x) at the point x is given by f f (x + h) f (x) (x) = + O(h). h Using Richardson s extrapolation, compute the forward second-order derivative approximation of f (x). 7

8 Solution: Denote the first-order approximation as A(h), then, for the second-order approximation B(h) Richardson s extrapolation yields B(h) = A(h) A(h) 1 + O(h ) = 1 [ ] 3 f (x) + 4 f (x + h) f (x + h) + O(h ). h Problem 14. Consider the function f (x) = x (x + 1) 3. Obviously, it has two real roots: x 1 = and x = 1. Without actually running the calculation, answer the following questions: a) Which root-finding method will converge faster to x 1, the Bisection method or the Newton s method? Why? b) Which root-finding method will converge faster to x, the Bisection method or the Newton s method? Why? Solution: a) The bisection method will not converge to x 1, since the root cannot be bracketed. The Newton s method will converge as a first-order method, due to root multiplicity. b) The bisection method will converge faster to x, because it halves the distance to the root after each iteration, while the Newton method reduces the distance by one-third, due to the root multiplicity. Problem 15. For the first order differential equation the midpoint method is given by x n+1 = x n + h f dx dt = f (x), ( x n + h ) f (x n), where h is the length of the time discretization step. Assume that the midpoint method is used to solve the ordinary differential equation dx dt = αx(t), where α > is a constant parameter. Find the range of values for the discretization step h, for which the method retains stability (that is, x n remains bounded as n ). Solution: Substituting αx as f (x) into the midpoint method yields ( x n+1 = x n αh x n αh ) ) x n = (1 αh + (αh) x n. 8

9 For stability, we need to choose (αh) > so that the expression in brackets is < 1. The latter yields αh + (αh) <, or h < α. 9