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AMath350.F12.A8.Sol

# Equation is hyperbolic if sin2 x 1 never occurs

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Unformatted text preview: f sin2 (x) > 1. Never occurs. Equation is parabolic if sin2 (x) = 1 ) x = 2 + k ⇡ , k Equation is elliptic if sin2 (x) < 1 uy = tan(x) Z Z 6. uxx 10uxy + 16uyy xux + sin(y )u = 0. 3 (a) 2 = , uyy 1 , 3. uxx + a uxy1+ b = =00 c = 16 ) b2 4ac = 36 > 0. PDE is hyperbolic. 4 (b) Characteristic equations: the equation is hyperbolic. This means that 2 Notice that b 4ac = 1 > 0, so p we can solve it by the dy method bof characteristics. To ﬁnd b pb2 4ac the characteristics, + b2 4ac dy = = we need to solve the equations dx 2a dx 2a p p =22 =8 dy b± b 4ac 2± 4 3 31 Integrating: = = =,. dx 2 2 y = 2a 2 x + k 1 y = 2 8 x + k2 Solving the ﬁrst of y + 2xwe ﬁnd these, = k1 y + 8 x = k2 dy ⇠ 3 + x, (c) Let = = y ) 2y =⌘3= + C 8x, then=⇠x x +2, , yi.e. 3x ⌘x2y = ,K⌘. = 1 and =8 y = x y + =) 2y 3 = C1 ⇠ = 1, 1 dx 2 2 all second partial derivatives are zero. Noting that d = x, e = 0, f = 1 1 sin(y ) g = 0, and ˆ The second ,one gives x = 6 (⌘ ⇠ ), y = 3 (4⇠ ⌘ ), we obtain g = 0 and B = 2(1)(2)(8) 10 x ((2)(1) + (8)(1)) + 2(16)(1)(1)...
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