Posted by haifeng on 2023-03-25 08:03:12 last update 2023-03-25 08:03:12 | Edit | Answers (1)
设 r=x2+y2+z2, 证明: ∂2(lnr)∂x2+∂2(lnr)∂y2+∂2(lnr)∂z2=1r2. 即 Δ(lnr)=1r2.
1
Posted by haifeng on 2023-03-25 08:11:39
∂∂x(lnr)=1r⋅∂r∂x=1r⋅xx2+y2+z2=1r⋅xr=xr2
类似的,
∂∂y(lnr)=yr2,∂∂z(lnr)=zr2
∂2∂x2(lnr)=∂∂x(xr2)=1⋅r2−x⋅2r⋅rx′r4=r2−2rx⋅xrr4=r2−2x2r4.
类似的, 有
∂2∂y2(lnr)=r2−2y2r4,∂2∂z2(lnr)=r2−2z2r4.
于是
Δ(lnr)=(∂2∂x2+∂2∂y2+∂2∂z2)(lnr)=r2−2x2r4+r2−2y2r4+r2−2z2r4=3r2−2r2r4=1r2.