A 2 b cosx c 1 sin x b2 4ac cos2 x 81 sin2

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Unformatted text preview: = 2, b = cos(x), c = (1 sin (x)) ) b2 4ac = cos2 (x) 8(1 sin2 (x)) = 7(sin2 (x) 1). Equation is hyperbolic if sin2 (x) > 1. Never occurs. Equation is parabolic if sin2 (x) = 1 ) x = 2 + k ⇡ , k Equation is elliptic if sin2 (x) < 1 6. uxx 10uxy + 16uyy (a) a = 1, b = Z Z xux + sin(y )u = 0. 10, c = 16 ) b2 4ac = 36 > 0. PDE is hyperbolic. (b) Characteristic equations: p b + b2 4ac dy = dx 2a =2 Integrating: y = 2 x + k1 dy b = dx =8 y= p b2 2a 8 x + k2 4ac AMATH 350 Page 3 Assignment #8 Solutions - Fall 2013 3. With the transformation ⇠ = p ax, ⌘ = y , we have p @u @u @ u d⇠ ˆ ˆ = =a @x @⇠ dx @⇠ ✓◆ ✓ ◆ ✓◆ p @ @ u d⇠ @ 2u @ @u @ p @u ˆ ˆ @ 2u ˆ and = = a =a = a 2. 2 @x @x @x @x @⇠ @⇠ @⇠ dx @⇠ @u @u ˆ @ 2u @ 2u ˆ = , and = 2 . Hence the given equation uxx + auyy + 2 @y @⌘ @y @⌘ bux + cuy + du = f becomes p au⇠⇠ + au⌘⌘ + abu⇠ + cu⌘ + du = f, ˆ ˆ ˆ ˆ ˆ Meanwhile, that is, b c d f u⇠⇠ + u⌘⌘ + p u⇠ + u⌘ + u = . ˆ ˆ ˆ ˆ ˆ a a a a 4. uxx + 2uxy 15uyy = 1 Notice that b2 4ac = 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 p p dy b ± b2 4ac 2 ± 4 + 60 = = = 3, 5. dx 2a 2 Solving the first of these, we find dy = dx 3 =) y = 3x + C =) 3x + y = C1 . The second one gives dy = 5 =) y = 5x + C, =) 5x dx y = C2 . Therefore a transformation which will work is ⇠ = 3x + y, ⌘ = 5x y (we could also use the reverse, or use multiples of these functions). Next, we need several applications of the chain rule to rewrite the PDE: ux = u⇠ ⇠x + u⌘ ⌘x = 3ˆ⇠ + 5ˆ⌘ . ˆ ˆ u u u y...
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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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