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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 , this becomes 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 definition of the error function, this can be expressed as 2 e 1 4 p 2 erf 1 2 ◆◆ = p e 1 4 erf 1 2 1 . 1846 . 2. For each of the following second order, linear PDEs, determine in which regions of R 2 the region is hyperbolic, parabolic, and elliptic. Sketch and label the regions. a) yu xx - 4 u xy + 2 yu 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 yu 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 yy + cos 2 ( x ) u + sin ( xy ) u x - u y = tan( x ) 2 u xx + cos( x ) u xy + (1 - sin 2 ( x )) u yy + cos 2 ( x ) u + sin( xy ) u x - u y = tan( x ) a = 2 , b = cos( x ) , c = (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 , we have @ 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 . Hence the given equation 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 ⇠⇠ + ˆ u ⌘⌘ + b p a ˆ u + c a ˆ u + d a ˆ u = f a . 4. u xx + 2 u xy - 15 u yy = 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 find the characteristics, we need to solve the equations
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