Noting that d x e 0 f c let y 2x 1 8x

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Unformatted text preview: rabolic, or elliptic. a) 2 (a) 1 u = 10, x x + sin(y 4 0 6. uxx a =01,xyb + 16uyy c =u16 ) b )u =ac. = 36 > 0. PDE is hyperbolic. (b) a = , b Characteristic equations: (a) Find1the = 10, c = 16 ) b2 4ac = 36 > 0. PDE is hyperbolic. b) characteristics. p p b2 4ac dy b dy (b) Characteristic equations:b + b2 4ac = = p2a p2a dx dx dy = b 8 b2 4ac dy = b + b2 4ac =2 = dx 2a dx 2a Integrating: =2 =8 y = 2 x + k1 y = 8 x + k2 Integrating: y + 2 x = k1 y + 8 x = k2 y = 2 x + k1 y = 8 x + k2 y + ⇠y = k (c) Let ⇠ = y + y x, 2x == +1 8x, then ⇠x = 2, 8x = 1, 2⌘x = 8, ⌘y = 1 and 2+ ⌘ yk all second partial derivatives are zero. Noting that d = x, e = 0, f = (c) Let ⇠ = y + 2x, ⌘ = 1 + 8x, then ⇠x = 2, ⇠y = 1, ⌘x = 8, ⌘y = 1 and y sin(y ), g = 0, and x = 6 (⌘ ⇠ ), y = 1 (4⇠ ⌘ ), we obtain g = 0 and ˆ 3 all second partial derivatives are zero. Noting that d = x, e = 0, f = 1 B 0, 2(1)(2)(8) + sin(y ), g = = and x = 6 (⌘ 10 ), y = 1 (4⇠ ⌘ ),+ 2(16)(1)(1)= 0 and ⇠ ((2)(1) 3 (8)(1)) we obtain g = 36 ˆ 1 1 1 (⌘ ⇠ )(2) 0 ⇠ ⌘ D = 2(1)(2)(8) 1=((2)(1) + (8)(1)) + 2(16)(1)(1) = 36 B 6 3 3 1 4 4 1 (⌘ ⇠ )(2) = 1 ⇠ 1 ⌘ D= (⌘ ⇠ )(8) = ⇠ ⌘ E= dx = 2a 2 dx = 8 2a Assignment #8 Solutions - Fall 2013 Page 5 Integrating: y = 2 x + k1 y = 8 x + k2 y + 2 x = k1 y + 8 x = k2 c) Put the equation in its canonical form. AMATH 350 (c) Let ⇠ = y + 2x, ⌘ = y + 8x, then ⇠x = 2, ⇠y = 1, ⌘x = 8, ⌘y = 1 and all second partial derivatives are zero. Noting that d = x, e = 0, f = 1 sin(y ), g = 0, and x = 6 (⌘ ⇠ ), y = 1 (4⇠ ⌘ ), we obtain g = 0 and ˆ 3 B = 2(1)(2)(8) 10 ((2)(1) + (8)(1)) + 2(16)(1)(1) = 1 1 1 (⌘ ⇠ )(2) = ⇠ ⌘ D= 6 3 3 4 4 1 (⌘ ⇠ )(8) = ⇠ ⌘ E= 6 3 3 4 1 ˆ f = sin(y (⇠ , ⌘ )) = sin( ⇠ ⌘) 3 3 36 1 u3 u Thus the PDE become...
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This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

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