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AMath350.F13.A8.Sol - AMath 350 Assignment#8 Fall 2013...

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AMath 350 Assignment #8 Fall 2013 Solutions 1. Following the hint, we have Z 1 0 e ( x - x 2 ) dx = Z 1 0 e - ( x 2 - x ) dx = Z 1 0 e - h ( x - 1 2 ) 2 - 1 4 i dx = e 1 4 Z 1 0 e - ( x - 1 2 ) 2 dx. Now, letting u = x - 1 2 ,th isbecomes e 1 4 Z 1 2 - 1 2 e - u 2 du and since the integrand is even, this is 2 e 1 4 Z 1 2 0 e - u 2 du. Using the defnition o± the error ±unction, this can be expressed as 2 e 1 4 p 2 er± 1 2 ◆◆ = p e 1 4 er± 1 2 1 . 1846 . 2. For each o± the ±ollowing second order, linear PDEs, determine in which regions R 2 the region is hyperbolic, parabolic, and elliptic. Sketch and label the regions. a) yu xx - 4 u xy +2 yy + e xy u x - x 3 u = xye x 2
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AMATH 350 Assignment #8 Solutions - Fall 2013 Page 2 ) yu xx - 4 u xy +2 yy + e xy u x - x 3 u = xye x 2 a = y,b = - 4 ,c =2 y ) b 2 - 4 ac =16 - 8 y 2 . Equation is hyperbolic if 16 - 8 y 2 > 0 Equation is parabolic if 16 - 8 y 2 =0 ) y = ± p 2 Equation is elliptic if 16 - 8 y 2 < 0 b) 2 u xx +cos( x ) u xy + ( 1 - sin 2 ( x ) ) u +cos 2 ( x ) u +sin( xy ) u x - u y =tan( x ) 2 u xx + cos( x ) u xy +(1 - sin 2 ( x )) u + cos 2 ( x ) u xy ) u x - u y x ) a ,b = cos( x ) =(1 - sin 2 ( x )) ) b 2 - 4 ac = cos 2 ( x ) - 8(1 - sin 2 ( x )) = 7(sin 2 ( x ) - 1). Equation is hyperbolic if sin 2 ( x ) > 1. Never occurs. Equation is parabolic if sin 2 ( x )=1 ) x = π 2 + k ,k ZZ Equation is elliptic if sin 2 ( x ) < 1
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AMATH 350 Assignment #8 Solutions - Fall 2013 Page 3 3. With the transformation = p ax , = y ,wehave @ u @ x = @ ˆ u @⇠ d dx = p a @ ˆ u and @ 2 u @ x 2 = @ @ x @ u @ x = @ @ x p a @ ˆ u = p a @ @ ˆ u d dx = a @ 2 ˆ u 2 . Meanwhile, @ u @ y = @ ˆ u @⌘ ,and @ 2 u @ y 2 = @ 2 ˆ u 2 .Hencetheg iv enequa t ion u xx + au yy + bu x + cu y + du = f becomes a ˆ u ⇠⇠ + a ˆ u ⌘⌘ + p ab ˆ u + c ˆ u + d ˆ u = f, that is, ˆ u u + b p a ˆ u + c a ˆ u + d a ˆ u = f a . 4. u xx +2 u xy - 15 u =1 Notice that b 2 - 4 ac =64 > 0 , so the equation is hyperbolic. This means that we can solve it by the method of characteristics. To Fnd the characteristics, we need to solve the equations dy dx = b ± p b 2 - 4 ac 2 a = 2 ± p 4+60 2 = - 3 , 5 .
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AMath350.F13.A8.Sol - AMath 350 Assignment#8 Fall 2013...

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