早稲田大学 創造理工学研究科 経営システム工学専攻 2024年8月実施 微积分 Author 思齐塾 , 祭音Myyura
Description y = f ( x ) , x = s 2 + t 2 + u 2 y = f(x), x = \sqrt{s^2 + t^2 + u^2} y = f ( x ) , x = s 2 + t 2 + u 2 ∂ 2 y ∂ s 2 + ∂ 2 y ∂ t 2 + ∂ 2 y ∂ u 2 \frac{\partial^2 y}{\partial s^2} + \frac{\partial^2 y}{\partial t^2} + \frac{\partial^2 y}{\partial u^2} ∂ s 2 ∂ 2 y + ∂ t 2 ∂ 2 y + ∂ u 2 ∂ 2 y x , f ′ ( x ) , f ′ ′ ( x ) x, f'(x), f''(x) x , f ′ ( x ) , f ′′ ( x )
题目描述
设
y = f ( x ) , x = s 2 + t 2 + u 2 . y=f(x),\qquad x=\sqrt{s^2+t^2+u^2}. y = f ( x ) , x = s 2 + t 2 + u 2 . 用 x , f ′ ( x ) , f ′ ′ ( x ) x,f'(x),f''(x) x , f ′ ( x ) , f ′′ ( x )
∂ 2 y ∂ s 2 + ∂ 2 y ∂ t 2 + ∂ 2 y ∂ u 2 . \frac{\partial^2y}{\partial s^2}
+\frac{\partial^2y}{\partial t^2}
+\frac{\partial^2y}{\partial u^2}. ∂ s 2 ∂ 2 y + ∂ t 2 ∂ 2 y + ∂ u 2 ∂ 2 y . 原题没有说明原点处的光滑性；因此一般公式应在 x > 0 x>0 x > 0 x = 0 x=0 x = 0
Kai Let x = s 2 + t 2 + u 2 x = \sqrt{s^2 + t^2 + u^2} x = s 2 + t 2 + u 2 x 2 = s 2 + t 2 + u 2 x^2 = s^2 + t^2 + u^2 x 2 = s 2 + t 2 + u 2
∂ x ∂ s = s s 2 + t 2 + u 2 = s x \frac{\partial x}{\partial s} = \frac{s}{\sqrt{s^2 + t^2 + u^2}} = \frac{s}{x} ∂ s ∂ x = s 2 + t 2 + u 2 s = x s ∂ x ∂ t = t s 2 + t 2 + u 2 = t x \frac{\partial x}{\partial t} = \frac{t}{\sqrt{s^2 + t^2 + u^2}} = \frac{t}{x} ∂ t ∂ x = s 2 + t 2 + u 2 t = x t ∂ x ∂ u = u s 2 + t 2 + u 2 = u x \frac{\partial x}{\partial u} = \frac{u}{\sqrt{s^2 + t^2 + u^2}} = \frac{u}{x} ∂ u ∂ x = s 2 + t 2 + u 2 u = x u
Now, we compute the second derivatives:
∂ 2 y ∂ s 2 = ∂ ∂ s ( ∂ y ∂ s ) = ∂ ∂ s ( f ′ ( x ) ∂ x ∂ s ) = ∂ ∂ s ( f ′ ( x ) s x ) \frac{\partial^2 y}{\partial s^2} = \frac{\partial}{\partial s} \left( \frac{\partial y}{\partial s} \right) = \frac{\partial}{\partial s} \left( f'(x) \frac{\partial x}{\partial s} \right) = \frac{\partial}{\partial s} \left( f'(x) \frac{s}{x} \right) ∂ s 2 ∂ 2 y = ∂ s ∂ ( ∂ s ∂ y ) = ∂ s ∂ ( f ′ ( x ) ∂ s ∂ x ) = ∂ s ∂ ( f ′ ( x ) x s )
∂ 2 y ∂ s 2 = f ′ ′ ( x ) ∂ x ∂ s s x + f ′ ( x ) x − s ∂ x ∂ s x 2 = f ′ ′ ( x ) s 2 x 2 + f ′ ( x ) x − s 2 x x 2 = f ′ ′ ( x ) s 2 x 2 + f ′ ( x ) x 2 − s 2 x 3 \frac{\partial^2 y}{\partial s^2} = f''(x) \frac{\partial x}{\partial s} \frac{s}{x} + f'(x) \frac{x - s \frac{\partial x}{\partial s}}{x^2} = f''(x) \frac{s^2}{x^2} + f'(x) \frac{x - \frac{s^2}{x}}{x^2} = f''(x) \frac{s^2}{x^2} + f'(x) \frac{x^2 - s^2}{x^3} ∂ s 2 ∂ 2 y = f ′′ ( x ) ∂ s ∂ x x s + f ′ ( x ) x 2 x − s ∂ s ∂ x = f ′′ ( x ) x 2 s 2 + f ′ ( x ) x 2 x − x s 2 = f ′′ ( x ) x 2 s 2 + f ′ ( x ) x 3 x 2 − s 2
Similarly,
∂ 2 y ∂ t 2 = f ′ ′ ( x ) t 2 x 2 + f ′ ( x ) x 2 − t 2 x 3 \frac{\partial^2 y}{\partial t^2} = f''(x) \frac{t^2}{x^2} + f'(x) \frac{x^2 - t^2}{x^3} ∂ t 2 ∂ 2 y = f ′′ ( x ) x 2 t 2 + f ′ ( x ) x 3 x 2 − t 2 ∂ 2 y ∂ u 2 = f ′ ′ ( x ) u 2 x 2 + f ′ ( x ) x 2 − u 2 x 3 \frac{\partial^2 y}{\partial u^2} = f''(x) \frac{u^2}{x^2} + f'(x) \frac{x^2 - u^2}{x^3} ∂ u 2 ∂ 2 y = f ′′ ( x ) x 2 u 2 + f ′ ( x ) x 3 x 2 − u 2
Thus,
∂ 2 y ∂ s 2 + ∂ 2 y ∂ t 2 + ∂ 2 y ∂ u 2 = f ′ ′ ( x ) s 2 + t 2 + u 2 x 2 + f ′ ( x ) 3 x 2 − ( s 2 + t 2 + u 2 ) x 3 = f ′ ′ ( x ) x 2 x 2 + f ′ ( x ) 3 x 2 − x 2 x 3 = f ′ ′ ( x ) + f ′ ( x ) 2 x 2 x 3 = f ′ ′ ( x ) + 2 x f ′ ( x ) \frac{\partial^2 y}{\partial s^2} + \frac{\partial^2 y}{\partial t^2} + \frac{\partial^2 y}{\partial u^2} = f''(x) \frac{s^2 + t^2 + u^2}{x^2} + f'(x) \frac{3x^2 - (s^2 + t^2 + u^2)}{x^3} = f''(x) \frac{x^2}{x^2} + f'(x) \frac{3x^2 - x^2}{x^3} = f''(x) + f'(x) \frac{2x^2}{x^3} = f''(x) + \frac{2}{x} f'(x) ∂ s 2 ∂ 2 y + ∂ t 2 ∂ 2 y + ∂ u 2 ∂ 2 y = f ′′ ( x ) x 2 s 2 + t 2 + u 2 + f ′ ( x ) x 3 3 x 2 − ( s 2 + t 2 + u 2 ) = f ′′ ( x ) x 2 x 2 + f ′ ( x ) x 3 3 x 2 − x 2 = f ′′ ( x ) + f ′ ( x ) x 3 2 x 2 = f ′′ ( x ) + x 2 f ′ ( x )
Therefore,
For x > 0 x>0 x > 0
∂ 2 y ∂ s 2 + ∂ 2 y ∂ t 2 + ∂ 2 y ∂ u 2 = f ′ ′ ( x ) + 2 x f ′ ( x ) . \boxed{
\frac{\partial^2 y}{\partial s^2}
+\frac{\partial^2 y}{\partial t^2}
+\frac{\partial^2 y}{\partial u^2}
=f''(x)+\frac{2}{x}f'(x)
}. ∂ s 2 ∂ 2 y + ∂ t 2 ∂ 2 y + ∂ u 2 ∂ 2 y = f ′′ ( x ) + x 2 f ′ ( x ) . At x = 0 x=0 x = 0 x x x f ( s 2 + t 2 + u 2 ) f(\sqrt{s^2+t^2+u^2}) f ( s 2 + t 2 + u 2 ) f ′ ( 0 ) = 0 f'(0)=0 f ′ ( 0 ) = 0
( ∂ 2 y ∂ s 2 + ∂ 2 y ∂ t 2 + ∂ 2 y ∂ u 2 ) ∣ ( 0 , 0 , 0 ) = 3 f ′ ′ ( 0 ) . \left.
\left(
\frac{\partial^2 y}{\partial s^2}
+\frac{\partial^2 y}{\partial t^2}
+\frac{\partial^2 y}{\partial u^2}
\right)\right|_{(0,0,0)}
=3f''(0). ( ∂ s 2 ∂ 2 y + ∂ t 2 ∂ 2 y + ∂ u 2 ∂ 2 y ) ( 0 , 0 , 0 ) = 3 f ′′ ( 0 ) . Without these regularity conditions, the second partial derivatives at the origin need not exist.