f(u,x)具有二阶连续偏导数,f(u,x)具有二阶连续偏导数,且满足δ^2f/δu^2+δ^2f/δv^2=1,又g(x,y)=_数学解答

f(u,x)具有二阶连续偏导数,
f(u,x)具有二阶连续偏导数,且满足δ^2f/δu^2+δ^2f/δv^2=1,
又g(x,y)=f[xy,(x^2-y^2)/2],
求δ^2g/δx^2+δ^2g/δy^2.
我做到δg/δx=y(δf/δu)+x(δf/δv),
符号是偏导数的符号啊，不是全微分，δ^2g/δx^2这个是2阶x偏导数

人气：480 ℃　时间：2019-11-15 07:01:11

解答

以下以df/dx表示一阶偏导数,d2f/dx2表示二阶偏导数
g(x,y)看作是由f(u,v),u＝xy,v＝1/2×(x^2－y^2)复合而成
则
dg/dx＝df/du×y＋df/dv×x
d2g/dx2
＝y×d(df/du)/dx＋df/dv＋x×d(df/dv)/dx
＝y×[d2f/du2×y＋d2f/dudv×x]＋df/dv＋x×[d2f/dudv×y＋d2f/dv2×x]
＝y^2×d2f/du2＋2xy×d2f/dudv＋df/dv＋x^2×d2f/dv2
dg/dy＝df/du×x－df/dv×y
d2g/dx2
＝x×d(df/du)/dy－df/dv－y×d(df/dv)/dy
＝x×[d2f/du2×x－d2f/dudv×y]－df/dv－y×[d2f/dudv×x－d2f/dv2×y]
＝x^2×d2f/du2－2xy×d2f/dudv－df/dv＋y^2×d2f/dv2
所以,
d2g/dx2＋d2g/dy2＝(x^2＋y^2)×[d2f/du2＋d2f/dv2]＝0