设f（u，v）具有二阶连续偏导数，且满足∂2f∂u2+∂2f∂v2=1，g（x，y）=f[xy，12（x2-y2）]，求∂2g∂x2+∂2g∂y2．

Question

设f（u，v）具有二阶连续偏导数，且满足(∂2f)/(∂u2)+(∂2f)/(∂v2)=1，g（x，y）=f[xy，(1)/(2)（x²-y²）]，求(∂2g)/(∂x2)+(∂2g)/(∂y2)．

Answer 1

设u＝xy，v＝(1)/(2)(x2−y2)，则z=f（u，v）
∴(∂g)/(∂x)＝y(∂f)/(∂u)+x(∂f)/(∂v)，(∂g)/(∂y)＝x(∂f)/(∂u)−y(∂f)/(∂v)．
∴(∂2g)/(∂x2)＝y2(∂2f)/(∂u2)+2xy(∂2f)/(∂u∂v)+x2(∂2f)/(∂v2)+(∂f)/(∂v)，
(∂2g)/(∂y2)＝x2(∂2f)/(∂u2)−2xy(∂2f)/(∂v∂u)+y2(∂2f)/(∂v2)−(∂f)/(∂v)．
∴(∂2g)/(∂x2)+(∂2g)/(∂y2)＝(x2+y2)(∂2f)/(∂u2)+(x2+y2)(∂2f)/(∂v2)＝x2+y2．