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Unformatted text preview: u is speciﬁed, so the second integral disappears. Also, we notice that the domain Ω is a line, so d Ω = dx , and the integral over its boundary (or any part thereof) is just the evaluation at a point, so Z Ωx 2 dw dx du dx + 5 wuwe x ! dx + wx 2 du dx n x ± ± ± ± ± Γ q = L = , where the normal to the boundary on the right hand side is n x  L = 1. Thus, the weak form is as follows: Z Ωx 2 dw dx du dx + 5 wuwe x ! dx14 L 2 w ( L ) = ....
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This note was uploaded on 08/01/2008 for the course ME 180 taught by Professor Zohdi during the Spring '08 term at Berkeley.
 Spring '08
 ZOHDI

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