1846 the bvp has the unique solution and 3 exx 2 ux

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Unformatted text preview: unique solution: , and 3 ex+x 2 u(x, order,xlinear PDEs, determine in which regions y ) = + 2 (1 e x y ). 2. For each of the following second x of R2 the region is hyperbolic, parabolic, and elliptic. Sketch and label the (iii) u(x, 0) = x2 , uy (1, y ) = e2y . As above H (x) = x2 G(x). regions. Apply second BC: uy (1, y ) = e2y ) e y G(1) = e2y ) G(1) = e3y . This uxy + 2yuyy + the x x u = xyex2 a) yuxx is4impossible, so exy uBVP 3has no solutions. 2 2 5. (a) yuxx 4uxy + 2yuyy + exy ux x3u = xyex a = y, b = 4, c = 2y ) b2 4ac = 16 8y 2 . Equation is hyperbolic if 16 8y 2 > 0 p Equation is parabolic if 16 8y 2 = 0 ) y = ± 2 Equation is elliptic if 16 8y 2 < 0 (b) 2uxx + cos(x)uxy + (1 sin2(x))uyy + cos2(x)u + sin(xy )ux a = 2, b = cos(x), c = (1 sin2(x)) ) b2 4ac = cos2 (x) 8(1 sin2 (x)) = 7(sin2 (x) 1). uy = tan(x) AMATH 350 Page 2 Assignment #8 Solutions - Fall 2012 b) 2uxx + cos(x)uxy + 1 sin2 (x) uyy + cos2 (x)u + sin (xy ) ux uy = tan(x) (b) 2uxx + cos(x)uxy + (1 sin2(x))uyy + cos2(x)u + sin(xy )ux a = 2, b = cos(x), c = (1 sin2(x)) ) b2 4ac = cos2 (x) 8(1 sin2 (x)) = 7(sin2 (x) 1). Equation is hyperbolic i...
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This document was uploaded on 02/09/2014.

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