高専の数学解説

第6章 6.23

(1)

(∂z∂x)2+(∂z∂y)2=(∂z∂r)2+1r2(∂z∂θ)2\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2=\left(\frac{\partial z}{\partial r}\right)^2+\frac{1}{r^2}\left(\frac{\partial z}{\partial \theta}\right)^2

右辺=(∂z∂x⋅∂x∂r+∂z∂y⋅∂y∂r)2+1r2(∂z∂x⋅∂x∂θ+∂z∂y⋅∂y∂θ)2=\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial r}\right)^2+\frac{1}{r^2}\left(\frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial \theta}\right)^2

=cos⁡2θ(∂z∂x)2+2cos⁡θ⋅sin⁡θ∂z∂x⋅∂z∂y+sin⁡2θ(∂z∂y)2=\cos^2\theta\left(\frac{\partial z}{\partial x}\right)^2+2\cos\theta\cdot\sin\theta\frac{\partial z}{\partial x}\cdot\frac{\partial z}{\partial y}+\sin^2\theta\left(\frac{\partial z}{\partial y}\right)^2

+(1r)2{r2sin⁡2θ(∂z∂x)2−2r2cos⁡θ⋅sin⁡θ∂z∂x⋅∂z∂y+r2cos⁡2θ(∂z∂y)2}+\left(\frac{1}{r}\right)^2\left\{r^2\sin^2\theta\left(\frac{\partial z}{\partial x}\right)^2-2r^2\cos\theta\cdot\sin\theta\frac{\partial z}{\partial x}\cdot\frac{\partial z}{\partial y}+r^2\cos^2\theta\left(\frac{\partial z}{\partial y}\right)^2\right\}

=(∂z∂x)2+(∂z∂y)2=左辺=\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2=\text{左辺}

(2)

∂2z∂x2+∂2z∂y2=∂2z∂r2+1r∂z∂r+1r2∂2z∂θ2\frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial y^2}=\frac{\partial^2 z}{\partial r^2}+\frac{1}{r}\frac{\partial z}{\partial r}+\frac{1}{r^2}\frac{\partial^2 z}{\partial \theta^2}

∂z∂r=cos⁡θ∂z∂x+sin⁡θ∂z∂y∂z∂θ=r(−sin⁡θ∂z∂x+cos⁡θ∂z∂y)\frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}\qquad\qquad \frac{\partial z}{\partial \theta}=r\left(-\sin\theta\frac{\partial z}{\partial x}+\cos\theta\frac{\partial z}{\partial y}\right)

∂2z∂r2=cos⁡2θ∂2z∂x2+2sin⁡θ⋅cos⁡θ∂2z∂x∂y+sin⁡2θ∂2z∂y2①\frac{\partial^2 z}{\partial r^2}=\cos^2\theta\frac{\partial^2 z}{\partial x^2}+2\sin\theta\cdot\cos\theta\frac{\partial^2 z}{\partial x \partial y}+\sin^2\theta\frac{\partial^2 z}{\partial y^2}\qquad ①

∂2z∂θ2=r2(sin⁡2θ∂2z∂x2−2sin⁡θ⋅cos⁡θ∂2z∂x∂y+cos⁡2θ∂2z∂y2)−r∂z∂r\frac{\partial^2 z}{\partial \theta^2}=r^2\left(\sin^2\theta\frac{\partial^2 z}{\partial x^2}-2\sin\theta\cdot\cos\theta\frac{\partial^2 z}{\partial x \partial y}+\cos^2\theta\frac{\partial^2 z}{\partial y^2}\right)-r\frac{\partial z}{\partial r}

1r2∂2z∂θ2+1r∂z∂r=sin⁡2θ∂2z∂x2−2sin⁡θ⋅cos⁡θ∂2z∂x∂y+cos⁡2θ∂2z∂y2②\frac{1}{r^2}\frac{\partial^2 z}{\partial \theta^2}+\frac{1}{r}\frac{\partial z}{\partial r}=\sin^2\theta\frac{\partial^2 z}{\partial x^2}-2\sin\theta\cdot\cos\theta\frac{\partial^2 z}{\partial x \partial y}+\cos^2\theta\frac{\partial^2 z}{\partial y^2}\qquad ②

右辺＝①+②

=cos⁡2θ∂2z∂x2+2sin⁡θ⋅cos⁡θ∂2z∂x∂y+sin⁡2θ∂2z∂y2+sin⁡2θ∂2z∂x2−2sin⁡θ⋅cos⁡θ∂2z∂x∂y+cos⁡2θ∂2z∂y2=\cos^2\theta\frac{\partial^2 z}{\partial x^2}+2\sin\theta\cdot\cos\theta\frac{\partial^2 z}{\partial x \partial y}+\sin^2\theta\frac{\partial^2 z}{\partial y^2}+\sin^2\theta\frac{\partial^2 z}{\partial x^2}-2\sin\theta\cdot\cos\theta\frac{\partial^2 z}{\partial x \partial y}+\cos^2\theta\frac{\partial^2 z}{\partial y^2}

=∂2z∂x2+∂2z∂y2=左辺=\frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial y^2}=\text{左辺}

解説: r31bn1z