MATH323 EXAM SOLUTIONS

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1 MATH EXAM SOLUTIONS These are mostly solutions to the 006 eam with the eception of Qu.5, which was covered in class, and therefore I have included the solution of the 004 eam question instead.. 4 d y 8 d d + 9y = 4 ln For solution of homogeneous equation try y = n d =nn d y = nn n d 4nn n 8n n + 9 n =0 4n n + 9 = 0 n = 0 n = repeated. We need two independent solutions; the second is y = ln. So the general solution of the homogeneous equation is y = c + c ln. Using the method of variation of abitrary constants, we try for a solution of the full equation Now impose L + L ln =0 y =L + L ln d =L + L + L ln + L =L + L + L ln + L d = L + L d y d = L + L + L = L + L 4 ln + + L L + L ln + ln + ln + + L ln + + L ln + + L 4 ln + + L ln + + L ln + ] 8 L + L ln + ] ] + 9 L + L ln = 4 ln. 4

2 This simplifies to 6L 5 + L 6 ln =4 ln 6L + 6L ln + 4L =4 ln. But the first condition was so we have L + L ln = 0 L = ln L = ln d = ln + c using the integral provided in the question. Then we have So the solution is y =L. The differential equation is In this case F = y + y so L = L ln = ln L = ln + c. + L ln = ln + c + ln + c ln = 6 ln + c + c ln. y d d y =0 d y d y =0 F y d F d y = 0. y =A cosh + B sinh. y = =A cosh + B sinh y = =A cosh + sinh = A cosh B sinh Adding A =0 Subtracting B = sinh y et = sinh sinh.

3 Then Iy et ] = 4 cosh = 4 sinh = 4 sinh sinh + 4sinh sinh cosh d sinh ] d = 4 sinh sinh sinh = 4 sinh sinh = 4 sinh sinh cosh =8 coth. The straight line joining the endpoints is y = m + c with = m + c, = m + c so c = 0 and m =, i.e. y =. For the straight line, I = ]d = ] = = Now Iy et ] = 8 coth 0.50 < 0.67 so the etremum must be a minimum.. Define a new functional I + λj where λ is a Lagrange multiplier then apply the Euler- Lagrange equation to F + λg. This determines y up to two arbitrary constants and λ which are found by the endpoint conditions and the constraint. E-L gives d d 4 y ] 8 y =λ 4 4 y + 8 y 8 y =λ 4 y + 4y 4y = λ For the homogeneous equation try y = n, then For a particular integral try nn + 4n 4 = 0 n + n 4 = 0 n n + 4 = 0 n =, n = 4 y = a + b + c d =a + b d y d = a a + 4a + b 4a + b + c = λ a = λ, c = 0.

4 b can be anything since is alrea in the homogeneous solution. So the general solution is y = A + B 4 + λ. We require 6 4 A + B 4 + λ d = 6 A6 + B + 84 λ7] = A64 + B λ = Also A + B + λ =, A + 6 B + λ =0 I omit the details of the solution; I suspect that it wasn t intended to be quite this messy. The answer is A = 80, B = 8 99, λ = 5. is the etremal curve. 4. The equations may be written so y = Au + Bu y = 0 where A = 0, B = u uy, u 0 =, u v y =. v y The eigenvalue equation is detλa B = 0 λ + λ λ = 0 λ + λ λ = 0 λ λ 4 λ 4λ + = 0 λ = 4,. So we need to solve as given. We have = 4 y = 4 + constant = ν, say d = y = + constant = η, say, d η =, η y =, ν = 4, ν y = u =u η η + u ν ν = u η 4u ν, u y =u η η y + u ν ν y = u η + u ν, 4

5 The original equations now become u η 4u ν u η + u ν + v η + v ν =0 u η 4u ν + v η 4v ν + v η + v ν =0 u η + v η 6u ν + v ν =0 u η + v η + u ν v ν =0 + : 5u η + 5v η =0 u η v η = 0 6u ν v ν =0 u v =fν, 6u v = gη 4 for some functions f and g. If u, 0 = 6 and v, 0 =, we have Writing X = 4 = 4X we have 6 = f 4, = g. Then 84 X + 4X = fx fν = 84 ν + 4 ν = 84 ν96 ν. 4 : 5u = gη fν u = 5 gη fν] = 5 η η + 84 νν 96 ] = 5 + y + y ] + 84 y 4y 4 96] ] 4 6 : 5v = gη 6fν v = 5 gη 6fν] = 5 η η + 64 νν 96 ] + y + y ] + 64 y 4y 4 96] ] = 5 5. Jan. 004 The equation is of the form Au + Bu y + Cu yy = F with A =, B = 0, C = y and so B 4AC = 4 y > 0. The associarted quadratic is λ y = 0 λ = ± y. 5

6 So we need to solve d = y d y = ln y = ln + constant ln y = constant y = constant = η d = y d y = ln y = ln + constant ln y = constant y = constant = ν η = y, η y =, ν = y, ν y =. u = u η η + u ν ν = y u η + yu ν u = y u η y u ηηη + u ην ν + y u νη η + u νν ν = y u η y y u ηη + yu ην + y y u νη + yu νν = y u η + y 4 u ηη y u ην + y u νν u y =u η η y + u ν ν y = u η + u ν u yy = u ηηη y + u ην ν y + u νη η y + u νν ν y = u ηη + u ην + u νη + u νν = u ηη + u ην + u νν. So the PDE becomes y u η 4y u νη = F. But y = η and νη = y, so we have 4ηνu ην + ηu η = F. For F = u + yu y = y u η + yu ν + y u η + u ν = y u η = ηu η 6

7 we have 4ηνu ην + ηu η = ηu η u ην =0 u η = fη u = fηdη + gν u =fη + gν y i.e. u, y =f + gy. 6a If Φ is independent of y then we have Φ =0 Φ = P Φ = P + Q. Φ0 = Q = Φa = P a + Q = P = a Φ = a +. b w = z z = w so + iy = u + iv = u iv u + v u = u + v, y = v u + v. The line through AB is the line = a so we have u a = u + v u + v u a = 0 u + v = a 4a which is a circle of radius a with centre at a, 0. From w = z u = + y, v = y + y we obtain so Aa, 0 with = a, y = 0 maps to A a, 0 and Ba, b with = a, y = b maps to B a a +b, b a +b. The edge AB maps to the arc A B of the circle. The edge OA with y = 0 is mapped to the part of the line v = 0 with u > a. The line through BC is y = b and so b = v u + v u + v + v b = 0 u + v + = b 4b 7

8 which is a circle with radius b and centre 0, b. C0, b maps to C 0, b and so the edge BC maps to the arc B C of this circle. The edge OC with = 0 is mapped to the part of the line u = 0 with v < b. c Since Φ = a + is harmonic in the z-plane with Φ = on = 0 and Φ = on = a, and since w = z is a conformal map, Φ = a + = u au + v is harmonic in the w-plane with Φ = on u = 0 and Φ = on the circle of radius a with centre at a, Suppose that the result is true for a particular k < n, i.e. F f k ; ω = iω k fω. Then F f k+ ; ω = f k+ e iω d d = d = f k e iω] {f k e iω} + iωf k e iω d =iω f k e iω d = iωf =iωiω k fω = iω k+ fω, + iω f k e iω d f k ; ω where we have used the fact that f r 0 as ± for 0 r n. So the result is also true for k +. It is clearly true for k = 0. So the result follows for n by induction. For f = e a, Laplace s equation: 0 fω = e a iω = a iω e a e iω d + ] 0 + = a iω + a + iω = a a + ω. φ + φ y = 0. e a e iω d 0 ] e a+iω a + iω 0 8

9 Solve by setting φω, y = φ, y = π φ, ye iω d φω, ye iω d. i.e. F.T. in one variable. So Laplace s equation becomes Note: the solution is y iω φω, y + φ y =0 i.e. is at constant ω; could write d φ y =ω φ. φω, y = c ωe ωy + c ωe ωy. as there are no other derivatives. The Note: c and c are not constants they depend on ω. φ, 0 = Ae a = Af. Require a solution valid in y 0 such that φ 0 as y. This does not imply c = 0 since < ω <. Therefore to obtain a result that 0 as y we need to take Then At y = 0 from the definition. Thus φ, y = π φω, y = c ωe ω y. φ, y = c ωe ω y+iω dω. π Af = c ωe iω dω π c ω =Afω = Aa a + ω dωafωe ω y e iω = aa dω e ω y e iω π a + ω. However this involves a product of Fourier transforms. But the convolution theorem says F f g; ω =fωgω where f g = and so f g = π With gω = e ω y, g = φ, y = π = y π duguf u = y π +y dωfωgωe iω. dufug u, taking b = y in the equation in the question so dωafωgωe iω = Af g = A Afu du u + y = Ay π e a u du u + y. dufug u 9

Final Exam May 4, 2016

1 Math 425 / AMCS 525 Dr. DeTurck Final Exam May 4, 2016 You may use your book and notes on this exam. Show your work in the exam book. Work only the problems that correspond to the section that you prepared.