A e 1 f and g x y 0 so we use 2 1 4 1

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Unformatted text preview: s 36ˆ + 3 (⇠ ⌘ )ˆ + 4 (⇠ ⌘ )ˆ + sin( 4 ⇠ 1 ⌘ )ˆ = 0. u 3 3u 1 1 1 4 1 Canonical form: u ˆ (⇠ ⌘ )u (⇠ ⌘ )ˆ u sin( 3 ⇠ 3 ⌘ )ˆ = 0. u 108 27 36 6. a) Given uxx +Dux +Euy +F u = g (x, y ), (where E 6= 0), we let u (x, y ) = eax+by v (x, y ), as instructed. Then ux = aeax+by v + eax+by vx uy = beax+by v + eax+by vy uxx = a2 eax+by v + 2aeax+by vx + eax+by vxx . Substituting into the DE gives eax+by a2 v + 2avx + vxx +Deax+by (av + vx )+Eeax+by (bv + vy )+F eax+by v = g (x, y ) . Collecting terms, this is vxx + (2a + D) vx + Evy + a2 + Da + Eb + F v = e ax by g (x, y ) . b) Choosing a = D/2 and b = a2 Da F /E = D2 /4 F /E will eliminate the second and fourth terms, giving the following DE for v (x, y ): vxx + Evy = e c) Here we have D = 1 ( a= and b = 2 the PDE will be ax by g (x, y ) . a, E = 1, F = , and g (x, y ) = 0, so we use 2 1) /4 + = (1 + )2 . The simplified form of 1 vxx vy = 0. AMATH 350 Page 6 Assignment #8 Solutions - Fall 2013 7. Solve the IVP ut = xux , u(x, 0) = 5 , x3 x>0 by the method of separation of variables. Solution: Assume that u(x, y ) = F (x)G(t). Then ut = F (x)G0 (t) and ux = F 0 (x)G(t), so ut = xux =) F (x)G0 (t) = xF 0 (x)G(t). Isolating functions of x and t, we have xF 0 (x) G0 ( t ) = =. F ( x) G( t ) Here we have also drawn the conclusion that the expressions in each variable must equal a constant, since they are equal for every value of x and every value of t. We have thus separated the problem into two ODEs: xF 0 (x) = F (x), Z Z dF =) = dx, F x =) ln |F | = =) =) l n x + C1 , | F | = e C1 e G 0 ( t ) = G( t ) Z Z dG = dt G ln |G| = t + C2 ln x | G | = e C2 e F = C3 x , =) G( t) = C4 e u(x, t) = C5 x e t t t Applying the initial condition, we have 5 = C5 x , x3 so we must have C5 = 5 and = 3; t...
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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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