FM5002-HW2-2.1.12

# Dt 4t 3 t 2 38 19 t 4t 39 t 1 dz after converting

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Unformatted text preview: ly, 1 1, dt 8x 1 6 0 dy dx 3 0. 26 4t 4t . 2 FM5002− HW2− 2.1.12.nb 0038 −3. Let V be the linear vector field defined by V (x, y) = (−8 x −18 y + 6,6 x +13 y − 5). Find the flowline of V footed at (2, 6). This is the same problem as 0038 − 2, stated in terms of flowlines. 10 x2 0038 4. Find the reverse gradient flow for f x, y 2 y2 6x y 2x 5y 7, footed at 1, 2 . We find the reverse gradient, f x, y 20 x 6 y 2, 6 x 4 y 5 . This is inhomogenous, so let z 1, dz so that 0. We can restate the problem as a homogenous problem by asking for the reverse gradient flow for f x, y, z dt 10 x2 6 x y 2 y2 2 x z 5 y z 7 , footed at 1, 2, 1 . The reverse gradient is, by construction , f x, y, z 20 x 6 y 2 z, 6 x 4 y 5 z, 0 . As in 0038 1, xt the solution is given by y t zt tM To compute tM 20 6 C, M 6 4 0 0 2 5 , and C 1 2. 0 1 , we have to diagonalize M. The eigenvalues for M are 22, 2, 3 1 1 4 . We let A and 0. The corresponding eigenvectors are 1 , 3 , 0 2 0 3 1 0 1 3 0 1 4 , so that A 1 t M A 2 22 t 0 A 0 0 0 2t 0 0 0 xt we conclude that y t zt 22 0 0 0 2t A0 0 0 2t 10 C 3 10 22 t 3 10 2t 10 0 2t 10 1 0A 1 22 t 9 tM 22 t 9 22 t A1 0 2 0 t. We therefore have that 00 0 0 3 22 t 3 3 10 22 t 3 10 2t 22 t 2t 3 10 10 22 t 9 10 22 t 0 1 2 3 2t 9 1 2 3 22 t 22 t . Finally, 20 1 2 1 2t 39 20 1 2t 39 20 2t 13 20 20 2t 13 1 20 2 A1 20 22 t 2 10 22 t A1t MA A 20 0 2t 10 10 10 0 2t 3 10 10 A A 1 t M AA 1 tM 2 1 2 9 22 t 20 3 22 t 20 1 2t 20 3 2t 20 ....
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## This note was uploaded on 01/19/2014 for the course MATH 5002 taught by Professor Adams during the Spring '08 term at Minnesota.

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