合成関数の2次偏導関数

■問題

$z=f\left(x,y\right),x=u\mathrm{cos}\theta -v\mathrm{sin}\theta ,$ $y=u\mathrm{sin}\theta +v\mathrm{cos}\theta$ のとき

$\frac{{\partial }^{2}z}{\partial {x}^{2}}+\frac{{\partial }^{2}z}{\partial {y}^{2}}=\frac{{\partial }^{2}z}{\partial {u}^{2}}+\frac{{\partial }^{2}z}{\partial {v}^{2}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

となることを示せ．

■解説

$x$$u$偏微分すると

$\frac{\partial x}{\partial u}=cos\theta$

これを更に$u$偏微分すると

$\frac{{\partial }^{2}x}{\partial {u}^{2}}=\frac{\partial }{\partial u}\left(\frac{\partial x}{\partial u}\right)=\frac{\partial }{\partial u}\left(cos\theta \right)=0$

$\frac{\partial y}{\partial u}=sin\theta$

$\frac{{\partial }^{2}y}{\partial {u}^{2}}=\frac{\partial }{\partial u}\left(\frac{\partial y}{\partial u}\right)=\frac{\partial }{\partial u}\left(sin\theta \right)=0$

よって，$\frac{{\partial }^{2}z}{\partial {u}^{2}}$合成関数の2次偏導関数の公式より

$\frac{{\partial }^{2}z}{\partial {u}^{2}}={f}_{xx}{\left(\frac{\partial x}{\partial u}\right)}^{2}+2{f}_{xy}\frac{\partial x}{\partial u}\frac{\partial y}{\partial u}+{f}_{yy}{\left(\frac{\partial y}{\partial u}\right)}^{2}+{f}_{x}\frac{{\partial }^{2}x}{\partial {u}^{2}}+{f}_{y}\frac{{\partial }^{2}y}{\partial {u}^{2}}$

$={f}_{xx}{\left(cos\theta \right)}^{2}+2{f}_{xy}cos\theta sin\theta +{f}_{yy}{\left(sin\theta \right)}^{2}+{f}_{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}0+{f}_{y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}0$

$={f}_{xx}{cos}^{2}\theta +2{f}_{xy}sin\theta cos\theta +{f}_{yy}{sin}^{2}\theta$　･･････(1)

$\frac{\partial x}{\partial v}=-sin\theta$

$\frac{{\partial }^{2}x}{\partial {v}^{2}}=\frac{\partial }{\partial v}\left(\frac{\partial x}{\partial v}\right)=\frac{\partial }{\partial v}\left(-sin\theta \right)=0$

$\frac{\partial y}{\partial v}=cos\theta$

$\frac{{\partial }^{2}y}{\partial {u}^{2}}=\frac{\partial }{\partial u}\left(\frac{\partial y}{\partial u}\right)=\frac{\partial }{\partial v}\left(cos\theta \right)=0$

よって，$\frac{{\partial }^{2}z}{\partial {v}^{2}}$合成関数の2次偏導関数の公式より

$\frac{{\partial }^{2}z}{\partial {v}^{2}}={f}_{xx}{\left(\frac{\partial x}{\partial v}\right)}^{2}+2{f}_{xy}\frac{\partial x}{\partial v}\frac{\partial y}{\partial v}$$+{f}_{yy}{\left(\frac{\partial y}{\partial v}\right)}^{2}$$+{f}_{x}\frac{{\partial }^{2}x}{\partial {v}^{2}}$$+{f}_{y}\frac{{\partial }^{2}y}{\partial {v}^{2}}$

$={f}_{xx}{\left(-sin\theta \right)}^{2}$$+2{f}_{xy}\left(-sin\theta \right)cos\theta$$+{f}_{yy}{\left(cos\theta \right)}^{2}$$+{f}_{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}0$$+{f}_{y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}0$

$={f}_{xx}{sin}^{2}\theta -2{f}_{xy}sin\theta cos\theta$$+{f}_{yy}{cos}^{2}\theta$　･･････(2)

(1)，(2)より

$\frac{{\partial }^{2}z}{\partial {u}^{2}}+\frac{{\partial }^{2}z}{\partial {v}^{2}}$

$=\left({f}_{xx}{cos}^{2}\theta +2{f}_{xy}sin\theta cos\theta +{f}_{yy}{sin}^{2}\theta \right)$$+\left({f}_{xx}{sin}^{2}\theta -2{f}_{xy}sin\theta cos\theta +{f}_{yy}{cos}^{2}\theta \right)$

$={f}_{xx}{cos}^{2}\theta +2{f}_{xy}sin\theta cos\theta$$+{f}_{yy}{sin}^{2}\theta$$+{f}_{xx}{sin}^{2}\theta -2{f}_{xy}sin\theta cos\theta$$+{f}_{yy}{cos}^{2}\theta$

$=\left({cos}^{2}\theta +{sin}^{2}\theta \right){f}_{xx}$$+\left({sin}^{2}\theta +{cos}^{2}\theta \right){f}_{yy}$

$={f}_{xx}+{f}_{yy}$

$=\frac{{\partial }^{2}z}{\partial {x}^{2}}+\frac{{\partial }^{2}z}{\partial {y}^{2}}$

となる．したがって

$\frac{{\partial }^{2}z}{\partial {x}^{2}}+\frac{{\partial }^{2}z}{\partial {y}^{2}}=\frac{{\partial }^{2}z}{\partial {u}^{2}}+\frac{{\partial }^{2}z}{\partial {v}^{2}}$

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