題目
7. (10 Points) Let
f:R2→R be a function. Assume that all second partial derivatives of
f exist and continuous and
fxx+fyy=0,for all (x,y)∈R2.
Define g:R2→R by g(u,v)=f(u2−v2,2uv). Find guu+gvv.
解答
解法一
令 X=u2−v2、Y=2uv。兩次套用鏈鎖律可得
guu+gvv=4(u2+v2)(fxx+fyy)(X,Y).
題設 fxx+fyy=0,因此 guu+gvv=0。