Unformatted text preview: ) e y G(1) = e2y ) G(1) = e3y .
This is impossible, so the BVP has no solutions.
This is impossible, so the BVP has no solutions.
2
5. (a) yuxx 4uxy + 2yuyy + exy ux x3u = xyex 2
xy
3
x
5. (a) yuxx 4uxy + 2yuyy + e u2 x u = xye 2
a = y, b = 4, c = 2y ) b x 4ac = 16 8y 2.
a = y, b = 4, c = 2y ) b2 4ac = 16 8y .
Equation is hyperbolic if 16 8y 2 > 0
Equation is hyperbolic if 16 8y 2 > 0
Equation is parabolic if 16 8y 2 = 0 ) y = ± 2
Equation is parabolic if 16 8y 2 = 0 ) y = ± 2
Equation is elliptic if 16 8y 2 < 0
Equation is elliptic if 16 8y 2 < 0 (b) 2uxx + cos(x)uxy + (1 sin2(x))uyy + cos2(x)u + sin(xy )ux
(b) 2uxx + cos(x)uxy + (1 sin2(x))uyy + cos2(x)u + sin(xy )ux
a = 2, b = cos(x), c = (1 sin2(x))
a = 2, b = cos(x), c = (1 sin2(x))
) b2 4ac = cos2 (x) 8(1 sin2 (x)) = 7(sin2 (x) 1).
) b2 4ac = cos2 (x) 8(1 sin2 (x)) = 7(sin2 (x) 1). uy = tan(x)
uy = tan(x) AMATH 350 Assignment #7 Solutions - Fall 2012 Page 8 8. 2ux + uy + u = 0; u(x, 0) = cos x.
This is a first-order linear PDE, and we cannot solve i...
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- Fall '14
- Trigraph, general solution, BVP
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