- LINEAR AND INTERPOLATED DIFFERENTIAL HARNACK ESTIMATES 121
Next we verify (2.133). Using (2.132) and (2.21), we compute( :t -.6.) (\713 div (h)a)
= \713 (!-.6.) div(h)a + (\713.6.-.6.\713) div(h)a
= \7 j3 ( Rμfi \7 vhap, + hμv \7 aRμfi - ~Rafi div ( h) v)
1 1
+ 2\713RJa div (h)5 - 2R/38 \73 div (h)a + R13ry8a \7 ;y div (h) 8
= \713Rμfi \7 vhap, + \713hμv \7 aRμfi+ Rμfi \7 /3 \7 vha;;; + hμv \7 j3 \7 aRμfi - ~Rafi \713 div ( h) v
1- 2R135\73div (h)a + R13ryJa \7;y div (h) 8.
Tracing and cancelling terms, we have(:t -.6.) (ga/3\713div(h)a)
=ga/3 (:t -.6.) (\713div(h)a) +R,aa\713div(h)a
= \7 aRμfi \7 vhap, + \7 ahp,v \7 aRμfi + R,aa \713 div ( h) a
+ Rμfi \7 a \7 vhap, + hμv \7 a \7 aRμfi,which is (2.133). DEXERCISE 2.100. Write down the formulas for the complex conjugate
equations to (2.132) and (2.133).Now let for c: > 0
and
ha/3 ~ ha/3 + c:ga/3·Since Z is of the form Z = A+ Ba Va+ Ba Va+ ha/3 Va V,a and ha/3 ;:=:: c:ga/3 > 0,
at each (x, t) we have that Z attains its minimum for some V. By taking the
first variation, we immediately see that
(2.134)Differentiating this, we have(2.135)
(2.136)\7 ,6 div (h )a+ \7 ,aha;y Vry + ha;y \7 ,6 Vry = 0,
\713div(h)a + \713ha;yVry + ha;y\713Vry = 0.