Civil Engineering Reference
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at x = x 0 . This means that
(U T A(x 0 )U) ii u ξ i ξ i 0 ,
Lu(x 0 ) =−
i
in contradiction with Lu(x 0 ) = f(x 0 )< 0.
(2) Now suppose that f(x)
0 and that there exists x = x with
u(x) > sup x u(x) . The auxiliary function h(x) :
= (x 1 x 1 ) 2
+ (x 2 x 2 ) 2
+
x d ) 2
···+
(x d −¯
is bounded on .Nowif δ> 0 is chosen sufficiently small,
then the function
w :
= u + δh
attains its maximum at a point x 0 in the interior. Since h x i x k =
2 δ ik ,wehave
Lw(x 0 ) = Lu(x 0 ) + δLh(x 0 )
2 δ
i
= f(x 0 )
a ii (x 0 )< 0 .
This leads to a contradiction just as in the first part of the proof.
Examples
The maximum principle has interesting interpretations for the equations (1.1)-
(1.3). If the charge density vanishes in a domain , then the potential is determined
by the potential equation. Without any charge, the potential in the interior cannot
be larger than its maximum on the boundary. The same holds if there are only
negative charges.
Next we consider the variational problem 1.3. Let c : = max x u(x) .Ifthe
solution u does not attain its maximum on the boundary, then
w(x) :
=
min
{
u(x), c
}
defines an admissible function which is different from u . Now the integral D(w, w)
exists in the sense of Lebesgue, and
(u x + u y )dxdy <
(u x + u y )dxdy,
D(w, w) =
1
where 1 :
. Thus, w leads to a smaller (generalized)
surface than u . We can smooth w to get a differentiable function which also
provides a smaller surface. This means that the minimal solution must satisfy the
maximum principle.
={ (x, y) ; u(x) < c }
 
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