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Partial Differential Equations: An Introduction With by Ioannis P. Stavroulakis, Stepan A. Tersian

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By Ioannis P. Stavroulakis, Stepan A. Tersian

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Additional resources for Partial Differential Equations: An Introduction With Mathematica and Maple, 2nd Edition

Example text

Solution . The discriminant is b2-ac c = x2y2 and the equation is hyperbolic in R2\{(x,y) : x = 0, y = 0). On the lines x = 0 and y = 0 the equation is parabolic. Let us consider the hyperbolic damain. We apply MAPLE procedures of ScientifiCWorkPlacE to realize the algorithm of canonization. Namely 1. Solution of characteristic equation. 18) we derive the equation aX2 - 2bX + c = 0. For u = x 2, b = 0 , 2 c=-y applying Solve to the last equation we get the solutions 2. Applying Solve O D E we find solutions of equations -dy= dx Y, x Y = CI exact solution is X and 3 - -Y, dx x exact solution is xy = ~ 3.

Next we will show that we can find new coordinates and 7 so that in terms of the new coordinates the form of Eq. 11) is such that its principal part is particularly simple. Then we say that the equation is in canonical fomn. 2. Assume that Eq. 13) to uaa - upp + 0 (a,p, u,U a ,u p ) = 0, called the second canonical form for hyperbolic equations. 2. (i) Let PO( X O ,yo) be a hyperbolic point. We choose and q in order to have A = a<: C = aq2 So + 2b<,<, + <: = 0, + 2bqxqy + cqy2 = 0. < and q are solutions of the first-order nonlinear equation a(p2,+ 2b(p,cpy + ccp2y = 0.

26). We have for the functions V = v (x,y, u (x,y ) ) and W = 20 (2, y, u (2, y)) vx IVY wx WYl = = 1 21%- vy + + v,uy VUUX + wx W U U X wy i- w,uy + (vuwy - 'uywu)u, (vxwu - V u W x ) ~ y (vywx - vxwy) = X (au, buy - c) = 0. e. 6. Find the general solution of the equation (u - y)ux + yuy = x + y. and solve the initial value problem u ( s ,1) = 2 + s. Solution. The characteristic system is dx -- dY Y (ii) - du - -= U-Y (4 Z+Y' (iii) + Using the proportion property ( i ) (iii)= (ii)we have Then v=-- u t x Y - c1 + (ii)= (iii) is a first integral.

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