Unformatted text preview: t of the last (boundary) term will need greater
attention. For the moment, let v satisfy the same trivial boundary conditions (1.2.1b) as
0 6 Introduction u. In this case, the boundary term vanishes and (1.2.8) becomes
A(v u) ; (v f ) = 0
where A(v u) = Z 1 (v pu + vqu)dx:
0 0 0 (1.2.9a)
(1.2.9b) The integration by parts has eliminated second derivative terms from the formulation.
Thus, solutions of (1.2.9) might have less continuity than those satisfying either (1.2.1) or
(1.2.2). For this reason, they are called weak solutions in contrast to the strong solutions
of (1.2.1) or (1.2.2). Weak solutions may lack the continuity to be strong solutions, but
strong solutions are always weak solutions. In situations where weak and strong solutions
di er, the weak solution is often the one of physical interest.
Since we've added a derivative to v by the integration by parts, v must be restricted
to a space where functions have more continuity than those in L2 . Having symmetry in
mind, we will select functions...
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