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Midterm

# Midterm - ACM95c Midterm Exam 2004 Problem 0(2 points Write...

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ACM95c Midterm Exam 2004 Problem 0 (2 points) Write down your grading section number. Problem 1 ( 25 points) (a) (10 points) (1) x 2 u x + y 2 u y = u 2 u H x, 2 x L = 1 The characteristic equations are (2) x t = x 2 y t = y 2 u t = u 2 with initial data (3) x H 0, s L = s y H 0, s L = 2 s u H 0, s L = 1 An initial value problem of the form (4) z' = z 2 z H 0 L = a has solution (found using the fact that the ODE is separable) given by (5) z H t L = a ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 - a t Since each of our characteristic ODE is of this form we conclude (6) x H t , s L = s ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 - s t y H t , s L = 2 s ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅ Å 1 - 2 s t u H t , s L = 1 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 - t The Jacobian of the transformation is (7) J = ¢ x t y t x s y s = ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ s 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅ Å H 1 - s t L 2 4 s 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅ H 1 - 2 s t L 2 1 ÅÅÅÅÅÅÅÅÅÅÅÅÅ Å H 1 - s t L 2 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅ H 1 - 2 s t L 2 ƒ ƒ ƒ ƒ ƒ § ƒ ƒ ƒ ƒ ƒ = - 2 s 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅ H 1 - s t L 2 H 1 - 2 s t L 2 This vanishes for s=0 and is infinite for s t =1 and 2s t =1. We expect that these zeros and singularities of the Jacobian will cause the solution to be badly behaved. Using the characteristic equations to invert the coordinate transform gives Printed by Mathematica for Students

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(8) s H x, y L = x y ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ 2 H y - x L t H x, y L = y - 2 x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x y We see from this the following (9) s = 0 Ø xy = 0 t s = 1 Ø y = 0 t s = 1 ê 2 Ø x = 0 So the transformation is poorly defined on the x and y axes. Plugging t into the expression for u gives (10) u H x, y L = x y ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ x H 2 + y L - y which is undefined on (11) y = 2 x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 1 - x We see by plotting this, together with the initial curve y=2x, that the solution to the Cauchy problem can only be defined in the region between the two parts of the graph of y=2x/(1-x). (2 points off if they only say that the solution fails to exist on the curves and don't mention what happens outside the region) Figure 1 -4 -2 2 4 6 -6 -4 -2 2 4 (b) (15 points) (12) u t + u u x = 0 u H x, 0 L = f H x L (13) f H x L = 0 x § 0 x 0 § x § 1 1 x ¥ 1 The characteristic equations are (14) x t = u t t = 1 u t = 0 with initial data Printed by Mathematica for Students
(15 x H 0, s L = s t H 0, s L = 0 u H 0, s L = f H s L Solving the t and u problems gives (16) t = t u = f H s L Plugging this expression for u into the ODE for x gives (17) x t = f H s L x H 0, s L = s Which we solve to find (18) x = t f H s L + s The Jacobian of the transformation is (19) J = ¢ x t t t x s t s = ¢ f H s L 1 t f ' H s L + 1 0 = - 1 - t f ' H s L = - 1 0 s § 0 -t 0 § s § 1 0 s ¥ 1 So the Jacobian is non-zero and finite for all t >0 (i.e. t>0). We now find an explicit expression for u(x,t). Notice (20) t = t u = f H s L = 0 s § 0 s 0 § s § 1 1

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