# hw1a_98W - Homework#1 MEAM 502 Differential Equation...

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Homework #1 MEAM 502 Differential Equation Methods in Mechanical Engineering 1998 Winter, Kikuchi Noting that the functional given provides a nonlinear problem, it is better to find out a linearized problem that yields an iteration method to find a solution. Taking the first variation of the functional F v dv dx dx x vdx v b g = + F H G I K J - - F H G I K J - F H G I K J z z 1 2 1 1 2 3 4 2 0 1 0 1 we have δ δ δ δ δ δ δ δ δ δ F v dv dx dx x vdx v dv dx dv dx dv dx dx x vdx v dv dx dv dx d v dx dx x vdx v b g = + F H G I K J - - F H G I K J - F H G I K J = + F H G I K J - - F H G I K J - F H G I K J = + F H G I K J - - F H G I K J - F H G I K J z z z z z z 1 1 2 3 4 1 2 1 1 2 1 2 3 4 1 2 1 1 2 1 2 3 4 2 0 1 0 1 2 0 1 0 1 2 0 1 0 1 After decomposing the interval (0,1) into two sub-domains (0,3/4) and (3/4,1), we apply the integration by parts rule :

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δ δ δ δ δ δ δ F v dv dx dv dx d v dx dx dv dx dv dx d v dx dx x vdx v dv dx dv dx v d dx dv dx dv dx vdx dv dx dv dx v d dx dv dx dv dx b g = + F H G I K J + + F H G I K J - - F H G I K J - F H G I K J = + F H G I K J - + F H G I K J F H G G G G I K J J J J + + F H G I K J - + F H G I K J F H G G G z z z z 1 1 1 1 1 2 3 4 1 1 1 1 1 1 1 1 2 0 3 4 2 3 4 1 0 1 2 0 3 4 2 0 3 4 2 3 4 1 2 G I K J J J J - - F H G I K J - F H G I K J = + F H G I K J + + F H G I K J - + F H G I K J - F H G G G G G I K J J J J J F H G I K J + - + F H G I K J F H G G G G I K J J J J - - F H G I K J R S | | T | | U V | | W | | z z - + δ δ δ δ δ δ vdx x vdx v dv dx dv dx v dv dx dv dx dv dx dv dx v d dx dv dx dv dx x v 3 4 1 0 1 2 0 1 2 3 4 2 3 4 2 0 1 2 3 4 1 1 1 1 1 1 1 3 4 1 1 1 2 1 z dx we have the Euler equation
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hw1a_98W - Homework#1 MEAM 502 Differential Equation...

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