第七章：消除某些限制。微分方程 G = 0 {\displaystyle G=0}
96 曲線可能由有限個正則軌跡組成，而不是單個正則軌跡。97 F {\displaystyle F} x ′ {\displaystyle x'} y ′ {\displaystyle y'} G = 0 {\displaystyle G=0} 98 前一篇文章中給出的結果的解釋。99 總結。100 第一章問題 I 的微分方程解。101 ,102 不連續解。103 對於問題 II，文章 9 中求解的方程 G = 0 {\displaystyle G=0} 104 兩個固定點必須位於擺線的同一個環上。105 經過兩點可以畫出一條且只有一條擺線環，它不包含尖點。106 問題 III. 表面上的最短線問題。107 用不同的方法推匯出相同的結果。108 問題 IV. 提供最小阻力的旋轉曲面。109 ,110 對於第一章問題 IV，方程 G = 0 {\displaystyle G=0} 第 96 條 x ′ {\displaystyle x'} y ′ {\displaystyle y'} G = 0 {\displaystyle G=0} A C {\displaystyle AC} C B {\displaystyle CB} A B {\displaystyle AB} C B {\displaystyle CB} A C {\displaystyle AC}
如上所述，我們得出結論，這部分曲線必須滿足微分方程 G = 0 {\displaystyle G=0} C B {\displaystyle CB}
我們現在可以取消曲線由一個正則軌跡組成的限制，並假設它由有限個正則軌跡組成。
第 97 條 F {\displaystyle F} x {\displaystyle x} ∂ F ∂ x = 0 {\displaystyle {\frac {\partial F}{\partial x}}=0} G = 0 {\displaystyle G=0} G 1 = 0 {\displaystyle G_{1}=0}
∂ F ∂ x − d d t ∂ F ∂ x ′ = 0 {\displaystyle {\frac {\partial F}{\partial x}}-{\frac {\text{d}}{{\text{d}}t}}{\frac {\partial F}{\partial x'}}=0} 由此可得
∂ F ∂ x ′ = constant {\displaystyle {\frac {\partial F}{\partial x'}}={\text{ constant}}} 此常數與 t {\displaystyle t} ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} x ′ {\displaystyle x'} y ′ {\displaystyle y'} G = 0 {\displaystyle G=0}
即使 x ′ {\displaystyle x'} y ′ {\displaystyle y'} G = 0 {\displaystyle G=0} ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} ∂ F ∂ y ′ {\displaystyle {\frac {\partial F}{\partial y'}}}
如果曲線中存在間斷點 t ′ {\displaystyle t'} t ′ {\displaystyle t'} τ {\displaystyle \tau } τ ′ {\displaystyle \tau '} τ … t ′ {\displaystyle \tau \ldots t'} t ′ … τ ′ {\displaystyle t'\ldots \tau '} t 0 … τ {\displaystyle t_{0}\ldots \tau } τ ′ … t 1 {\displaystyle \tau '\ldots t_{1}} τ … τ ′ {\displaystyle \tau \ldots \tau '} τ … τ ′ {\displaystyle \tau \ldots \tau '} τ {\displaystyle \tau } τ ′ {\displaystyle \tau '} t ′ {\displaystyle t'}
積分的變化
I = ∫ t 0 t 1 F ( x , y , x ′ , y ′ ) d t {\displaystyle I=\int _{t_{0}}^{t_{1}}F(x,y,x',y')~{\text{d}}t} 然後僅取決於積分總和的變化
∫ τ t ′ F ( x , y , x ′ , y ′ ) d t + ∫ t ′ τ ′ F ( x , y , x ′ , y ′ ) d t {\displaystyle \int _{\tau }^{t'}F(x,y,x',y')~{\text{d}}t+\int _{t'}^{\tau '}F(x,y,x',y')~{\text{d}}t} 由於此表示式的第一個變分必須消失，我們必然有（第 79 節）
0 = ∫ τ t ′ G ( y ′ ξ − x ′ η ) d t + ∫ t ′ τ ′ G ( y ′ ξ − x ′ η ) d t + [ ∂ F ∂ x ′ ξ + ∂ F ∂ y ′ η ] τ t ′ + [ ∂ F ∂ x ′ ξ + ∂ F ∂ y ′ η ] t ′ τ {\displaystyle 0=\int _{\tau }^{t'}G(y'\xi -x'\eta )~{\text{d}}t+\int _{t'}^{\tau '}G(y'\xi -x'\eta )~{\text{d}}t+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{\tau }^{t'}+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{t'}^{\tau }} 由於整個曲線上的 G = 0 {\displaystyle G=0}
[ ∂ F ∂ x ′ ξ + ∂ F ∂ y ′ η ] τ t ′ + [ ∂ F ∂ x ′ ξ + ∂ F ∂ y ′ η ] t ′ τ = 0 {\displaystyle \left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{\tau }^{t'}+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{t'}^{\tau }=0} 量 ξ {\displaystyle \xi } η {\displaystyle \eta } τ {\displaystyle \tau } τ ′ {\displaystyle \tau '}
[ ∂ F ∂ x ′ ] t ′ − {\displaystyle \left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{-}} [ ∂ F ∂ x ′ ] t ′ + {\displaystyle \left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{+}} τ {\displaystyle \tau } τ ′ {\displaystyle \tau '} t ′ {\displaystyle t'} ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} 上面的表示式變為
( [ ∂ F ∂ x ′ ] t ′ − − [ ∂ F ∂ x ′ ] t ′ + ) ( ξ ) ′ + ( [ ∂ F ∂ y ′ ] t ′ − − [ ∂ F ∂ y ′ ] t ′ + ) ( η ) ′ = 0 {\displaystyle \left(\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{-}-\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{+}\right)(\xi )'+\left(\left[{\frac {\partial F}{\partial y'}}\right]_{t'}^{-}-\left[{\frac {\partial F}{\partial y'}}\right]_{t'}^{+}\right)(\eta )'=0} 其中 ( ξ ) ′ {\displaystyle (\xi )'} ( η ) ′ {\displaystyle (\eta )'} ξ {\displaystyle \xi } η {\displaystyle \eta } t ′ {\displaystyle t'} ( ξ ) ′ {\displaystyle (\xi )'} ( η ) ′ {\displaystyle (\eta )'}
[ ∂ F ∂ x ′ ] t ′ − = [ ∂ F ∂ x ′ ] t ′ + {\displaystyle \left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{-}=\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{+}} [ ∂ F ∂ y ′ ] t ′ − = [ ∂ F ∂ y ′ ] t ′ + {\displaystyle \left[{\frac {\partial F}{\partial y'}}\right]_{t'}^{-}=\left[{\frac {\partial F}{\partial y'}}\right]_{t'}^{+}} 也就是說，量 ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} ∂ F ∂ y ′ {\displaystyle {\frac {\partial F}{\partial y'}}} x ′ {\displaystyle x'} y ′ {\displaystyle y'}
這是一個新的必要條件，用於積分 I {\displaystyle I} G = 0 {\displaystyle G=0}
第 98 條 函式 ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} ∂ F ∂ y ′ {\displaystyle {\frac {\partial F}{\partial y'}}} x ′ {\displaystyle x'} y ′ {\displaystyle y'} x ′ {\displaystyle x'} y ′ {\displaystyle y'} 為了回答這個問題，我們可以說這些函式的組合具有特殊性質，即，包含 x ′ {\displaystyle x'} y ′ {\displaystyle y'} ∂ F ∂ x ′ = {\displaystyle {\frac {\partial F}{\partial x'}}=}
第 99 條 I {\displaystyle I}
1) 提供最大值或最小值的曲線必須滿足微分方程
G ≡ ∂ 2 F ∂ x ∂ y ′ − ∂ 2 F ∂ y ∂ x ′ + F 1 ( x ′ d y ′ d t − y ′ d x ′ d t ) = 0 {\displaystyle G\equiv {\frac {\partial ^{2}F}{\partial x\partial y'}}-{\frac {\partial ^{2}F}{\partial y\partial x'}}+F_{1}\left(x'{\frac {{\text{d}}y'}{{\text{d}}t}}-y'{\frac {{\text{d}}x'}{{\text{d}}t}}\right)=0} 或者，等價地，這兩個方程
G 2 ≡ ∂ F ∂ x − d d t ( ∂ F ∂ x ′ ) = 0 {\displaystyle G_{2}\equiv {\frac {\partial F}{\partial x}}-{\frac {\text{d}}{{\text{d}}t}}\left({\frac {\partial F}{\partial x'}}\right)=0} G 2 ≡ ∂ F ∂ y − d d t ( ∂ F ∂ y ′ ) = 0 {\displaystyle G_{2}\equiv {\frac {\partial F}{\partial y}}-{\frac {\text{d}}{{\text{d}}t}}\left({\frac {\partial F}{\partial y'}}\right)=0} 2) 函式 F {\displaystyle F} x ′ {\displaystyle x'} y ′ {\displaystyle y'}
為了確定透過方程 G = 0 {\displaystyle G=0} Δ I {\displaystyle \Delta I}
\begin{center}第一章問題中微分方程 G=0 的解
第 100 條
S 2 π = ∫ t 0 t 1 y x ′ 2 + y ′ 2 d t {\displaystyle {\frac {S}{2\pi }}=\int _{t_{0}}^{t_{1}}y{\sqrt {x'^{2}+y'^{2}}}~{\text{d}}t} [1] {\displaystyle \qquad {\text{[1]}}} 因此
F = y x ′ 2 + y ′ 2 {\displaystyle F=y{\sqrt {x'^{2}+y'^{2}}}} [2] {\displaystyle \qquad {\text{[2]}}} 因此
∂ F ∂ x ′ = y x ′ x ′ 2 + y ′ 2 {\displaystyle {\frac {\partial F}{\partial x'}}={\frac {yx'}{\sqrt {x'^{2}+y'^{2}}}}} ∂ F ∂ y ′ = y y ′ x ′ 2 + y ′ 2 {\displaystyle \qquad {\frac {\partial F}{\partial y'}}={\frac {yy'}{\sqrt {x'^{2}+y'^{2}}}}} [3] {\displaystyle \qquad {\text{[3]}}} 由此可見， ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} ∂ F ∂ y ′ {\displaystyle {\frac {\partial F}{\partial y'}}} x ( t ) {\displaystyle x(t)} y ( t ) {\displaystyle y(t)} ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} ∂ F ∂ y ′ {\displaystyle {\frac {\partial F}{\partial y'}}} y = 0 {\displaystyle y=0} x ′ 2 + y ′ 2 {\displaystyle {\sqrt {x'^{2}+y'^{2}}}} x ′ {\displaystyle x'} y ′ {\displaystyle y'} y {\displaystyle y}
由於 F {\displaystyle F} x {\displaystyle x}
G 1 = 0 {\displaystyle G_{1}=0} ∂ F ∂ x ′ = y x ′ x ′ 2 + y ′ 2 = β {\displaystyle {\frac {\partial F}{\partial x'}}={\frac {yx'}{\sqrt {x'^{2}+y'^{2}}}}=\beta } [4] {\displaystyle \qquad {\text{[4]}}} 其中 β {\displaystyle \beta }
y 2 ( d x d t ) 2 = β 2 [ ( d x d t ) 2 + ( d y d t ) 2 ] {\displaystyle y^{2}\left({\frac {{\text{d}}x}{{\text{d}}t}}\right)^{2}=\beta ^{2}\left[\left({\frac {{\text{d}}x}{{\text{d}}t}}\right)^{2}+\left({\frac {{\text{d}}y}{{\text{d}}t}}\right)^{2}\right]} [5] {\displaystyle \qquad {\text{[5]}}} 該方程的解是懸鏈線
x = α + β t {\displaystyle x=\alpha +\beta t} y = β 2 ( e t + e − t ) , {\displaystyle \qquad y={\frac {\beta }{2}}(e^{t}+e^{-t}),} [6] {\displaystyle \qquad {\text{[6]}}} 其中 α {\displaystyle \alpha }
第 101 條 不連續解 。如果我們取弧 s {\displaystyle s} t {\displaystyle t}
y d x d s = β {\displaystyle y{\frac {{\text{d}}x}{{\text{d}}s}}=\beta } Suppose that β = 0 {\displaystyle \beta =0} t 0 … t 1 {\displaystyle t_{0}\ldots t_{1}} y ≠ 0 {\displaystyle y\neq 0} P 0 {\displaystyle P_{0}} d x d s = cos ( ϕ ) = 0 {\displaystyle {\frac {{\text{d}}x}{{\text{d}}s}}=\cos(\phi )=0} ϕ {\displaystyle \phi } X {\displaystyle X} cos ( ϕ ) {\displaystyle \cos(\phi )} y = 0 {\displaystyle y=0} P 0 M 0 {\displaystyle P_{0}M_{0}} M 0 {\displaystyle M_{0}} d x d s {\displaystyle {\frac {{\text{d}}x}{{\text{d}}s}}} P 1 {\displaystyle P_{1}} M 0 {\displaystyle M_{0}} M 1 {\displaystyle M_{1}} P 0 M 0 M 1 P 1 {\displaystyle P_{0}M_{0}M_{1}P_{1}} β = 0 {\displaystyle \beta =0} P 0 M 0 {\displaystyle P_{0}M_{0}} P 1 M 1 {\displaystyle P_{1}M_{1}} i.e. , x = x 0 {\displaystyle x=x_{0}} x = x 1 {\displaystyle x=x_{1}} G = 0 {\displaystyle G=0} y ′ G = G 1 {\displaystyle y'G=G_{1}} G 1 = 0 {\displaystyle G_{1}=0} y ≠ 0 {\displaystyle y\neq 0} G ≠ 0 {\displaystyle G\neq 0} M 0 M 1 {\displaystyle M_{0}M_{1}}
第 102 條 X {\displaystyle X} 正 的時，這一點就一目瞭然。這個問題是第 79 條的特殊情況。
第一變分可以分解成幾個部分（參見第 79 條和第 81 條）
δ I = − ∫ M 0 P 0 G w N d s + [ ∂ F ∂ x ′ ξ + ∂ F ∂ y ′ η ] M 0 P 0 − ∫ M 1 M 0 G w N d s + [ ∂ F ∂ x ′ ξ + ∂ F ∂ y ′ η ] M 1 M 0 − ∫ P − 1 M 1 G w N d s + [ ∂ F ∂ x ′ ξ + ∂ F ∂ y ′ η ] P 1 M 1 {\displaystyle \delta I=-\int _{M_{0}}^{P_{0}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{M_{0}}^{P_{0}}-\int _{M_{1}}^{M_{0}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{M_{1}}^{M_{0}}-\int _{P-1}^{M_{1}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{P_{1}}^{M_{1}}} 因為，現在所有邊界項都為零
∂ F ∂ x ′ = y ′ x x ′ 2 + y ′ 2 {\displaystyle {\frac {\partial F}{\partial x'}}={\frac {y'x}{\sqrt {x'^{2}+y'^{2}}}}} ∂ F ∂ y ′ = y y ′ x ′ 2 + y ′ 2 {\displaystyle \qquad {\frac {\partial F}{\partial y'}}={\frac {yy'}{\sqrt {x'^{2}+y'^{2}}}}} 因此，兩者在點 M 0 {\displaystyle M_{0}} M 1 {\displaystyle M_{1}} ξ {\displaystyle \xi } η {\displaystyle \eta } P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} G = 0 {\displaystyle G=0} d s {\displaystyle {\text{d}}s} w N {\displaystyle w_{N}} δ I {\displaystyle \delta I} 正 。
當任意常數 < β ≠ 0 {\displaystyle <\beta \neq 0} X {\displaystyle X}
第 103 條
T = ∫ t 0 t 1 x ′ 2 + y ′ 2 4 g y + α 2 d t {\displaystyle T=\int _{t_{0}}^{t_{1}}{\frac {\sqrt {x'^{2}+y'^{2}}}{\sqrt {4gy+\alpha ^{2}}}}~{\text{d}}t} [1] {\displaystyle \qquad {\text{[1]}}} 為了使此表示式實際上表達下降時間（時間和因此增量 d t {\displaystyle {\text{d}}t} 4 g y + α 2 {\displaystyle {\sqrt {4gy+\alpha ^{2}}}} x ′ 2 + y ′ 2 {\displaystyle {\sqrt {x'^{2}+y'^{2}}}} T 0 … t 1 {\displaystyle T_{0}\ldots t_{1}}
然而，如果我們用 x {\displaystyle x} y {\displaystyle y} t {\displaystyle t} x ′ {\displaystyle x'} y ′ {\displaystyle y'} t 0 … t 1 {\displaystyle t_{0}\ldots t_{1}} t {\displaystyle t} t {\displaystyle t} x {\displaystyle x} y {\displaystyle y}
假設這種情況發生在 t = t ′ {\displaystyle t=t'} x 0 {\displaystyle x_{0}} y 0 {\displaystyle y_{0}}
x = x 0 + a ( t − t ′ ) m + ⋯ {\displaystyle x=x_{0}+a(t-t')^{m}+\cdots } y = y 0 + ( t − t ′ ) m + ⋯ {\displaystyle \qquad y=y_{0}+(t-t')^{m}+\cdots } 其中 m ≥ 2 {\displaystyle m\geq 2} a {\displaystyle a} b {\displaystyle b}
那麼有
x ′ 2 + y ′ 2 = m 2 ( a 2 + b 2 ) ( t − t ′ ) 2 ( m − 1 ) + ⋯ {\displaystyle x'^{2}+y'^{2}=m^{2}(a^{2}+b^{2})(t-t')^{2(m-1)}+\cdots } 以及
x ′ 2 + y ′ 2 = m a 2 + b 2 ( t − t ′ ) m − 1 + ⋯ {\displaystyle {\sqrt {x'^{2}+y'^{2}}}=m{\sqrt {a^{2}+b^{2}}}(t-t')^{m-1}+\cdots } 這裡我們假設 a 2 + b 2 {\displaystyle {\sqrt {a^{2}+b^{2}}}}
如果現在 m {\displaystyle m} t − t ′ {\displaystyle t-t'} x ′ 2 + y ′ 2 {\displaystyle {\sqrt {x'^{2}+y'^{2}}}}
相反，如果 m {\displaystyle m} x 0 {\displaystyle x_{0}} y 0 {\displaystyle y_{0}} 尖點 ，因為這裡 x ′ 2 + y ′ 2 {\displaystyle {\sqrt {x'^{2}+y'^{2}}}} t > t ′ {\displaystyle t>t'} t < t ′ {\displaystyle t<t'}
因此，如果上面的積分要表示時間， x ′ 2 + y ′ 2 {\displaystyle {\sqrt {x'^{2}+y'^{2}}}} t {\displaystyle t} 尖點 後被設為級數的相反值。因此，我們只考慮沒有奇點的曲線的一部分。
在問題中，我們經常需要做出這樣的限制，否則積分就沒有確切的意義。因此，有了這個假設， x ′ 2 + y ′ 2 {\displaystyle {\sqrt {x'^{2}+y'^{2}}}}
我們可以寫成
F = x ′ 2 + y ′ 2 4 g y + α 2 {\displaystyle F={\frac {\sqrt {x'^{2}+y'^{2}}}{\sqrt {4gy+\alpha ^{2}}}}} [2] {\displaystyle \qquad {\text{[2]}}} 因此
∂ F ∂ x ′ = 1 4 g y + α 2 x ′ x ′ 2 + y ′ 2 {\displaystyle {\frac {\partial F}{\partial x'}}={\frac {1}{\sqrt {4gy+\alpha ^{2}}}}{\frac {x'}{\sqrt {x'^{2}+y'^{2}}}}} ∂ F ∂ y ′ = 1 4 g y + α 2 y ′ x ′ 2 + y ′ 2 {\displaystyle \qquad {\frac {\partial F}{\partial y'}}={\frac {1}{\sqrt {4gy+\alpha ^{2}}}}{\frac {y'}{\sqrt {x'^{2}+y'^{2}}}}} [3] {\displaystyle \qquad {\text{[3]}}} 從這裡我們可以得出結論，與第一個例子類似， ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} ∂ F ∂ y ′ {\displaystyle {\frac {\partial F}{\partial y'}}} x {\displaystyle x} y {\displaystyle y} ∂ F ∂ x ′ {\displaystyle {\frac {\partial F}{\partial x'}}} ∂ F ∂ y ′ {\displaystyle {\frac {\partial F}{\partial y'}}} 4 g y + α 2 {\displaystyle {\sqrt {4gy+\alpha ^{2}}}}
同樣在這裡 F {\displaystyle F} x {\displaystyle x} G 1 = 0 {\displaystyle G_{1}=0}
∂ F ∂ x ′ = 1 4 g y + α 2 x ′ x ′ 2 + y ′ 2 = C {\displaystyle {\frac {\partial F}{\partial x'}}={\frac {1}{\sqrt {4gy+\alpha ^{2}}}}{\frac {x'}{\sqrt {x'^{2}+y'^{2}}}}=C} [4] {\displaystyle \qquad {\text{[4]}}} 其中 C {\displaystyle C}
如果 C {\displaystyle C} C {\displaystyle C} C {\displaystyle C} 4 g y + α 2 {\displaystyle {\sqrt {4gy+\alpha ^{2}}}} ∞ {\displaystyle \infty } x ′ x ′ 2 + y ′ 2 = cos ( α ) {\displaystyle {\frac {x'}{\sqrt {x'^{2}+y'^{2}}}}=\cos(\alpha )} C {\displaystyle C}
從 [4] 可以看出
d x 2 = C 2 ( 4 g y + α 2 ) ( d x 2 + d y 2 ) {\displaystyle {\text{d}}x^{2}=C^{2}(4gy+\alpha ^{2})({\text{d}}x^{2}+{\text{d}}y^{2})} 或者，如果我們將 4 g {\displaystyle 4g}
α 2 4 g = a {\displaystyle {\frac {\alpha ^{2}}{4g}}=a} 4 g C 2 = c 2 {\displaystyle 4gC^{2}=c^{2}} 我們有
d x 2 = c 2 ( y + a ) ( d x 2 + d y 2 ) {\displaystyle {\text{d}}x^{2}=c^{2}(y+a)({\text{d}}x^{2}+{\text{d}}y^{2})} 因此
d x = c ( y + a ) d y ( y + a ) [ 1 − c 2 ( y + 1 ) ] ] {\displaystyle {\text{d}}x={\frac {c(y+a){\text{d}}y}{\sqrt {(y+a)[1-c^{2}(y+1)]]}}}} [5] {\displaystyle \qquad {\text{[5]}}} 為了進行最後的積分，寫出
d ϕ = c d y ( y + a ) [ 1 − c 2 ( y + a ) ] {\displaystyle {\text{d}}\phi ={\frac {c{\text{d}}y}{\sqrt {(y+a)[1-c^{2}(y+a)]}}}} [6] {\displaystyle \qquad {\text{[6]}}} 因此
d x = ( y + a ) d ϕ {\displaystyle {\text{d}}x=(y+a){\text{d}}\phi } [ 5 a ] {\displaystyle \qquad [5^{\text{a}}]} 在 d ϕ {\displaystyle {\text{d}}\phi }
2 c 2 ( y + a ) = 1 − ξ {\displaystyle 2c^{2}(y+a)=1-\xi } [ 7 ] {\displaystyle \qquad [7]} 那麼有
2 [ 1 − c 2 ( y + a ) ] = 1 + ξ {\displaystyle 2[1-c^{2}(y+a)]=1+\xi } [ 8 ] {\displaystyle \qquad [8]} 以及
2 c 2 d y = − d ξ {\displaystyle 2c^{2}{\text{d}}y=-{\text{d}}\xi } [ 9 ] {\displaystyle \qquad [9]} 因此
d ϕ = − d ξ 1 − ξ 2 {\displaystyle {\text{d}}\phi =-{\frac {{\text{d}}\xi }{\sqrt {1-\xi ^{2}}}}} [ 10 ] {\displaystyle \qquad [10]} 因此
ξ = cos ( ϕ ) {\displaystyle \xi =\cos(\phi )} [ 11 ] {\displaystyle \qquad [11]} 這裡可以省略積分常數，因為 ϕ {\displaystyle \phi }
因此，
y + a = 1 2 c 2 ( 1 − cos ( ϕ ) ) {\displaystyle y+a={\frac {1}{2c^{2}}}(1-\cos(\phi ))} 5 a {\displaystyle 5^{\text{a}}} x + x 0 = 1 2 c 2 ( ϕ − sin ( ϕ ) ) {\displaystyle x+x_{0}={\frac {1}{2c^{2}}}(\phi -\sin(\phi ))} [ 12 ] {\displaystyle \qquad [12]} {\displaystyle } 這些方程表示一個擺線 。
積分常數 x 0 {\displaystyle x_{0}} c {\displaystyle c} A {\displaystyle A} B {\displaystyle B} x {\displaystyle x} y {\displaystyle y} ϕ {\displaystyle \phi } y {\displaystyle y} ϕ 2 {\displaystyle \phi ^{2}} x {\displaystyle x} ϕ {\displaystyle \phi } ϕ = 0 {\displaystyle \phi =0} ϕ = 2 π , 4 π , … {\displaystyle \phi =2\pi ,4\pi ,\ldots }
A {\displaystyle A} B {\displaystyle B}
如果我們透過點 − x 0 {\displaystyle -x_{0}} − a {\displaystyle -a} 1 / ( 2 c 2 ) {\displaystyle 1/(2c^{2})} − x 0 {\displaystyle -x_{0}} − a {\displaystyle -a} X {\displaystyle X} A {\displaystyle A}
第 104 條 A {\displaystyle A} B {\displaystyle B} α {\displaystyle \alpha }
則擺線的方程為
x = r ( ϕ − sin ( ϕ ) ) {\displaystyle x=r(\phi -\sin(\phi ))} y r ( 1 − cos ( ϕ ) ) {\displaystyle \qquad yr(1-\cos(\phi ))} 其中我們用 r {\displaystyle r} 1 / ( 2 c 2 ) {\displaystyle 1/(2c^{2})}
從附圖中可以看到擺線弧。取兩個位於不同環上的點，它們非常靠近並且關於頂點對稱，讓我們比較從一個點到另一個點經過頂點所需的時間和沿連線這兩個點的直線所需的時間。這兩個點的引數可以表示為
ϕ 0 = 2 π − ψ 0 {\displaystyle \phi _{0}=2\pi -\psi _{0}} ϕ 1 = 2 π + ψ 0 {\displaystyle \qquad \phi _{1}=2\pi +\psi _{0}} 經過頂點所需的時間為
T = 1 2 g ∫ t 0 t 1 x ′ 2 + y ′ 2 y d t = 1 2 g ∫ s 0 s 1 d s y {\displaystyle T={\frac {1}{\sqrt {2g}}}\int _{t_{0}}^{t_{1}}{\sqrt {\frac {x'^{2}+y'^{2}}{y}}}~{\text{d}}t={\frac {1}{\sqrt {2g}}}\int _{s_{0}}^{s_{1}}{\frac {{\text{d}}s}{\sqrt {y}}}} 現在
d x = r ( 1 − cos ( ϕ ) ) d ϕ {\displaystyle {\text{d}}x=r(1-\cos(\phi ))~{\text{d}}\phi } 以及
d y = r sin ( ϕ ) d ϕ {\displaystyle {\text{d}}y=r\sin(\phi )~{\text{d}}\phi } 所以
d s = d x 2 + d y 2 = 2 r sin ( ϕ / 2 ) d ϕ {\displaystyle {\text{d}}s={\sqrt {{\text{d}}x^{2}+{\text{d}}y^{2}}}=2r\sin(\phi /2)~{\text{d}}\phi } 因此
T = 1 2 g ∫ ϕ 0 ϕ 1 2 r sin ( ϕ / 2 ) r 1 − cos ( ϕ ) d ϕ = r g ∫ ϕ 0 ϕ 1 d ϕ {\displaystyle T={\frac {1}{\sqrt {2g}}}\int _{\phi _{0}}^{\phi _{1}}{\frac {2r\sin(\phi /2)}{{\sqrt {r}}{\sqrt {1-\cos(\phi )}}}}{\text{d}}\phi ={\sqrt {\frac {r}{g}}}\int _{\phi _{0}}^{\phi _{1}}{\text{d}}\phi } r g ( ϕ 1 − ϕ 0 ) = r g [ 2 π + ψ 0 − 2 π + ψ 0 ] = 2 r g ψ 0 {\displaystyle {\sqrt {\frac {r}{g}}}(\phi _{1}-\phi _{0})={\sqrt {\frac {r}{g}}}[2\pi +\psi _{0}-2\pi +\psi _{0}]=2{\sqrt {\frac {r}{g}}}\psi _{0}} 從 ϕ 0 {\displaystyle \phi _{0}} ϕ 1 {\displaystyle \phi _{1}} [ v d x d s ] {\displaystyle \left[v{\frac {{\text{d}}x}{{\text{d}}s}}\right]} d x d s = sin ϕ 2 {\displaystyle {\frac {{\text{d}}x}{{\text{d}}s}}=\sin {\frac {\phi }{2}}} v 2 = 2 g y {\displaystyle v^{2}=2gy}
[ 2 g r 1 − cos ( ϕ ) sin ϕ 2 ] ϕ 0 ϕ 1 = 2 g r sin 2 ψ 0 2 {\displaystyle \left[{\sqrt {2gr}}{\sqrt {1-\cos(\phi )}}\sin {\frac {\phi }{2}}\right]_{\phi _{0}}^{\phi _{1}}=2{\sqrt {gr}}\sin ^{2}{\frac {\psi _{0}}{2}}} 從 ϕ 0 {\displaystyle \phi _{0}} ϕ 1 {\displaystyle \phi _{1}}
x 1 − x 0 = r [ ϕ 1 − ϕ 0 − sin ϕ 1 + sin ϕ 0 ] = 2 r [ ψ 0 − sin ψ 0 ] {\displaystyle x_{1}-x_{0}=r[\phi _{1}-\phi _{0}-\sin \phi _{1}+\sin \phi _{0}]=2r[\psi _{0}-\sin \psi _{0}]} 因此，所需的時間為
T 1 = 2 r ( ψ 0 − sin ψ 0 ) 2 g r sin 2 ( ψ 0 / 2 ) {\displaystyle T_{1}={\frac {2r(\psi _{0}-\sin \psi _{0})}{2{\sqrt {gr}}\sin ^{2}(\psi _{0}/2)}}} 因此
T 1 T = 2 r ( ψ 0 − sin ψ 0 ) 2 g r sin 2 ψ 0 2 ⋅ 2 r g ψ 0 = ψ 0 − sin ψ 0 2 ψ 0 sin 2 ψ 0 2 = ψ 0 3 3 ! − ψ 0 5 5 ! + ⋯ 2 ψ 0 [ ψ 0 2 − 1 3 ! ( ψ 0 2 ) 3 + ⋯ ] 2 {\displaystyle {\frac {T_{1}}{T}}={\frac {2r(\psi _{0}-\sin \psi _{0})}{2{\sqrt {gr}}\sin ^{2}{\frac {\psi _{0}}{2}}\cdot 2{\sqrt {\frac {r}{g}}}\psi _{0}}}={\frac {\psi _{0}-\sin \psi _{0}}{2\psi _{0}\sin ^{2}{\frac {\psi _{0}}{2}}}}={\frac {{\frac {\psi _{0}^{3}}{3!}}-{\frac {\psi _{0}^{5}}{5!}}+\cdots }{2\psi _{0}\left[{\frac {\psi _{0}}{2}}-{\frac {1}{3!}}\left({\frac {\psi _{0}}{2}}\right)^{3}+\cdots \right]^{2}}}} 因此，
T 1 T < ψ 0 3 3 ! 2 ψ 0 [ ψ 0 2 − 1 6 ( ψ 0 2 ) 3 + ⋯ ] 2 {\displaystyle {\frac {T_{1}}{T}}<{\frac {\frac {\psi _{0}^{3}}{3!}}{2\psi _{0}\left[{\frac {\psi _{0}}{2}}-{\frac {1}{6}}\left({\frac {\psi _{0}}{2}}\right)^{3}+\cdots \right]^{2}}}} 或者
T 1 T < 1 3 1 ( 1 − ψ 0 2 24 + ⋯ ) 3 {\displaystyle {\frac {T_{1}}{T}}<{\frac {1}{3}}{\frac {1}{\left(1-{\frac {\psi _{0}^{2}}{24}}+\cdots \right)^{3}}}} 因此，對於 ψ 0 {\displaystyle \psi _{0}}
T 1 < T {\displaystyle T_{1}<T} 由此可見，包含頂點的粒子路徑不能給出最小值。
第 105 條 G = 0 {\displaystyle G=0} r {\displaystyle r} X {\displaystyle X} G = 0 {\displaystyle G=0}
現在我們將證明，只有一個這樣的擺線能夠包含這兩個點 A {\displaystyle A} B {\displaystyle B} A {\displaystyle A} B {\displaystyle B} D B > A C {\displaystyle DB>AC} r {\displaystyle r} X {\displaystyle X} O {\displaystyle O} O {\displaystyle O} A B {\displaystyle AB} A ′ C ′ {\displaystyle A'C'} A ′ B ′ {\displaystyle A'B'} A ′ B ′ : A ′ C ′ {\displaystyle A'B':A'C'} A ′ {\displaystyle A'}
那麼，對於某個位置，我們必須有
A ′ B ′ A ′ C ′ = A B A C {\displaystyle {\frac {A'B'}{A'C'}}={\frac {AB}{AC}}} 由於點 A {\displaystyle A} B {\displaystyle B} A B {\displaystyle AB} A B {\displaystyle AB}
如果 A ′ B ′ = A B {\displaystyle A'B'=AB} A ′ C ′ = A C {\displaystyle A'C'=AC} A {\displaystyle A} B {\displaystyle B} A ′ B ′ ≠ A B {\displaystyle A'B'\neq AB}
接下來選擇一個數量 r ′ {\displaystyle r'}
r : r ′ = A C : A ′ C ′ {\displaystyle r:r'=AC:A'C'} 以 O {\displaystyle O} r : r ′ {\displaystyle r:r'}
x = r ′ ( ϕ − sin ϕ ) {\displaystyle x=r'(\phi -\sin \phi )} y = r ′ ( 1 − cos ϕ ) {\displaystyle \qquad y=r'(1-\cos \phi )} 這是新的擺線的座標。
後面的擺線與第一個相似，因為變換使縱座標 A ′ C ′ {\displaystyle A'C'} A ′ B ′ {\displaystyle A'B'}
r ′ r A ′ C ′ = A C {\displaystyle {\frac {r'}{r}}A'C'=AC\qquad } r ′ r A ′ B ′ = A B {\displaystyle \qquad {\frac {r'}{r}}A'B'=AB} 這給了我們一個擺線，它對縱座標 A C {\displaystyle AC} A B {\displaystyle AB}
此外，只有一個擺線滿足要求的條件。因為，如果我們已經有了 A ′ B ′ = A B {\displaystyle A'B'=AB} A ′ C ′ = A C {\displaystyle A'C'=AC} r ′ r A ′ B ′ = A B {\displaystyle {\frac {r'}{r}}A'B'=AB} r ′ {\displaystyle r'} r ′ = r {\displaystyle r'=r} 因此，透過兩點 A {\displaystyle A} B {\displaystyle B} X {\displaystyle X} [ 1]
第 106 條 曲面上最短線問題 。這個問題一般來說是無法解決的，因為微分方程中的變數無法分離，積分也無法進行。只有在少數情況下，人們才能成功地進行積分，從而表示滿足微分方程的曲線。
例如，這在平面、球面和所有其他二次曲面情況下已經完成。
作為一個簡單的例子，我們將討論球面上的兩點之間的最短線問題。球體的半徑被設為 1，球面的方程以以下形式給出：
x 2 + y 2 + z 2 = 1 {\displaystyle x^{2}+y^{2}+z^{2}=1} 現在寫成
x = cos u {\displaystyle x=\cos u} y = sin u cos v {\displaystyle y=\sin u\cos v} z =∈ u sin v {\displaystyle z=\in u\sin v} [ 1 ] {\displaystyle \qquad [1]} 那麼 u = {\displaystyle u=} v = {\displaystyle v=}
弧微元是
d s = d u 2 + sin 2 u d v 2 {\displaystyle {\text{d}}s={\sqrt {{\text{d}}u^{2}+\sin ^{2}u{\text{d}}v^{2}}}} [ 2 ] {\displaystyle \qquad [2]} 因此，要使最小化的積分是
L = ∫ t 0 t 1 u ′ 2 + v ′ 2 sin 2 u d t {\displaystyle L=\int _{t_{0}}^{t_{1}}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}~{\text{d}}t} [ 3 ] {\displaystyle \qquad [3]} 所以這裡我們有
F = u ′ 2 + v ′ 2 sin 2 u {\displaystyle F={\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}} 以及
∂ F ∂ u ′ = u ′ u ′ 2 + v ′ 2 sin 2 u {\displaystyle {\frac {\partial F}{\partial u'}}={\frac {u'}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}}} ∂ F ∂ v ′ = v ′ sin 2 u u ′ 2 + v ′ 2 sin 2 u {\displaystyle \qquad {\frac {\partial F}{\partial v'}}={\frac {v'\sin ^{2}u}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}}} [ 5 ] {\displaystyle \qquad [5]} 由於 F {\displaystyle F} v {\displaystyle v} G 1 = 0 {\displaystyle G_{1}=0}
∂ F ∂ v ′ = v ′ sin 2 u u ′ 2 + v ′ 2 sin 2 u = c {\displaystyle {\frac {\partial F}{\partial v'}}={\frac {v'\sin ^{2}u}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}}=c} 其中 c {\displaystyle c}
如果曲線的初始點 A {\displaystyle A} u ≠ 0 {\displaystyle u\neq 0} v ′ = 0 {\displaystyle v'=0} c {\displaystyle c} v {\displaystyle v} A {\displaystyle A} B {\displaystyle B}
如果不是這種情況，那麼總是 c ≠ 0 {\displaystyle c\neq 0} c < 1 {\displaystyle c<1} sin c {\displaystyle \sin c} c {\displaystyle c}
v ′ sin 2 u u ′ 2 + v ′ 2 sin 2 u = sin c {\displaystyle {\frac {v'\sin ^{2}u}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}}=\sin c} [ 6 ] {\displaystyle \qquad [6]} 或者
d v = sin c d u sin u sin 2 u − sin c {\displaystyle {\text{d}}v={\frac {\sin c~{\text{d}}u}{\sin u{\sqrt {\sin ^{2}u-\sin c}}}}} [ 7 ] {\displaystyle \qquad [7]} 如果我們寫
cos u = cos c cos w {\displaystyle \cos u=\cos c\cos w} [ 8 ] {\displaystyle \qquad [8]} 那麼是
d v = sin c d w 1 − cos 2 c cos 2 w {\displaystyle {\text{d}}v={\frac {\sin c~{\text{d}}w}{1-\cos ^{2}c\cos ^{2}w}}} 由於 1 可以被 sin 2 w + cos 2 w {\displaystyle \sin ^{2}w+\cos ^{2}w}
d v = sin c d w sin 2 w + cos 2 w sin 2 c = sin c d w cos 2 w sin 2 c + tan 2 w = d tan w sin c 1 + tan 2 w sin 2 c {\displaystyle {\text{d}}v={\frac {\sin c~{\text{d}}w}{\sin ^{2}w+\cos ^{2}w\sin ^{2}c}}={\frac {\sin c{\frac {{\text{d}}w}{\cos ^{2}w}}}{\sin ^{2}c+\tan ^{2}w}}={\frac {{\text{d}}{\frac {\tan w}{\sin c}}}{1+{\frac {\tan ^{2}w}{\sin ^{2}c}}}}} 因此
v − β = tan − 1 ( tan w sin c ) {\displaystyle v-\beta =\tan ^{-1}\left({\frac {\tan w}{\sin c}}\right)} 其中 β {\displaystyle \beta }
由此可得
tan ( v − β ) = tan w sin c {\displaystyle \tan(v-\beta )={\frac {\tan w}{\sin c}}} [ 9 ] {\displaystyle \qquad [9]} 透過 [8] 消去 w {\displaystyle w}
tan u cos ( v − β ) = tan c {\displaystyle \tan u\cos(v-\beta )=\tan c} [ 10 ] {\displaystyle \qquad [10]} 這是我們正在尋找的曲線方程，用球座標 u {\displaystyle u} v {\displaystyle v}
為了更仔細地研究它們的含義，我們可以透過弧長 s {\displaystyle s} u {\displaystyle u} v {\displaystyle v} s {\displaystyle s}
透過 [7]，表示式 [2] 變為
d s = sin u d u sin 2 u − sin 2 c {\displaystyle {\text{d}}s={\frac {\sin u~{\text{d}}u}{\sqrt {\sin ^{2}u-\sin ^{2}c}}}} 並且，由於替換了 [8]，它變為
d s = d w {\displaystyle {\text{d}}s={\text{d}}w} 因此，如果 b {\displaystyle b}
s − b = w {\displaystyle s-b=w} [ 11 ] {\displaystyle \qquad [11]} 因此，從公式 [8] 和 [9] 中，我們得到以下公式
cos u = cos c cos ( s − b ) {\displaystyle \cos u=\cos c\cos(s-b)} cot ( v − β ) = sin c cot ( s − b ) {\displaystyle \qquad \cot(v-\beta )=\sin c\cot(s-b)} [ 12 ] {\displaystyle \qquad [12]} 但這些關係存在於直角球面三角形的邊和角之間。
如果我們考慮從北極點畫出的子午線，它與我們正在尋找的曲線垂直相交，那麼這條子午線與曲線和任何其他子午線形成一個三角形，上述關係可以應用於這個三角形。
因此，滿足微分方程的曲線本身必須是大圓的一部分。積分常數 c {\displaystyle c} b {\displaystyle b} β {\displaystyle \beta } A {\displaystyle A} B {\displaystyle B}
幾何解釋是： c {\displaystyle c} u = 0 {\displaystyle u=0} s − b {\displaystyle s-b} v − β {\displaystyle v-\beta }
因此，如果我們假設零子午線穿過 A {\displaystyle A} b {\displaystyle b} A {\displaystyle A} β {\displaystyle \beta }
第 107 條 G = 0 {\displaystyle G=0}
由於
F 1 u ′ 2 = ∂ 2 F ∂ v ′ 2 {\displaystyle F_{1}u'^{2}={\frac {\partial ^{2}F}{\partial v'^{2}}}} 我們有
F 1 = sin 2 u ( u ′ 2 + v ′ 2 sin 2 u ) 3 / 2 {\displaystyle F_{1}={\frac {\sin ^{2}u}{(u'^{2}+v'^{2}\sin ^{2}u)^{3/2}}}} 將此值代入
G ≡ ∂ 2 F ∂ v ∂ u ′ − ∂ 2 F ∂ u ∂ v ′ + F 1 ( v ′ u ″ − u ′ v ″ ) = 0 {\displaystyle G\equiv {\frac {\partial ^{2}F}{\partial v\partial u'}}-{\frac {\partial ^{2}F}{\partial u\partial v'}}+F_{1}(v'u''-u'v'')=0} 使此表示式變為
− [ 2 cos u u ′ 2 v ′ + sin 2 u cos u v ′ 3 ] + sin u ( v ′ u ″ − u ′ v ″ ) = 0 {\displaystyle -[2\cos uu'^{2}v'+\sin ^{2}u\cos uv'^{3}]+\sin u(v'u''-u'v'')=0} 或者
d v d u [ 2 + sin 2 u ( d v d u ) 2 ] cos u + sin u d 2 v d u 2 = 0 {\displaystyle {\frac {{\text{d}}v}{{\text{d}}u}}\left[2+\sin ^{2}u\left({\frac {{\text{d}}v}{{\text{d}}u}}\right)^{2}\right]\cos u+\sin u{\frac {{\text{d}}^{2}v}{{\text{d}}u^{2}}}=0} 在此方程中寫
1 ) w = sin u d v d u {\displaystyle 1)\qquad w=\sin u{\frac {{\text{d}}v}{{\text{d}}u}}} 我們有
cot u ( w + w 3 ) + d w d u = 0 {\displaystyle \cot u(w+w^{3})+{\frac {{\text{d}}w}{{\text{d}}u}}=0} 或者
2 ) d w w + w 3 + cot u d u = 0 {\displaystyle 2)\qquad {\frac {{\text{d}}w}{w+w^{3}}}+\cot u~{\text{d}}u=0} 對最後一個方程進行積分，得到
ln ( w 1 + w 2 sin u ) = c {\displaystyle \ln \left({\frac {w}{\sqrt {1+w^{2}}}}\sin u\right)=c} 因此
3 ) w 2 sin 2 u = C 2 ( 1 + w 2 ) {\displaystyle 3)\qquad w^{2}\sin ^{2}u=C^{2}(1+w^{2})} 假設 A {\displaystyle A} u {\displaystyle u} A {\displaystyle A} v {\displaystyle v} B {\displaystyle B}
因此，對於所有透過 A {\displaystyle A} G = 0 {\displaystyle G=0} C = 0 {\displaystyle C=0} sin u = 0 {\displaystyle \sin u=0} u = 0 {\displaystyle u=0} w = 0 {\displaystyle w=0}
sin u d v d u = 0 {\displaystyle \sin u{\frac {{\text{d}}v}{{\text{d}}u}}=0} 或者
v = {\displaystyle v=} 因此，如上所述， A {\displaystyle A} B {\displaystyle B} A {\displaystyle A} C ≠ 0 {\displaystyle C\neq 0}
d s 2 = d u 2 + sin 2 u d v 2 = d u 2 [ 1 + w 2 ] {\displaystyle {\text{d}}s^{2}={\text{d}}u^{2}+\sin ^{2}u{\text{d}}v^{2}={\text{d}}u^{2}[1+w^{2}]} 或者
d s = ± sin u d u sin 2 u − sin 2 C {\displaystyle {\text{d}}s=\pm {\frac {\sin u~{\text{d}}u}{\sqrt {\sin ^{2}u-\sin ^{2}C}}}} sin 2 C {\displaystyle \sin ^{2}C} C 2 {\displaystyle C^{2}} 以及
d v = sin C d u sin ( u ) sin 2 u − sin 2 C {\displaystyle {\text{d}}v={\frac {\sin C~{\text{d}}u}{\sin(u){\sqrt {\sin ^{2}u-\sin ^{2}C}}}}} 寫成 cos u = cos C cos t {\displaystyle \cos u=\cos C\cos t}
cos u = cos C cos ( s − b ) {\displaystyle \cos u=\cos C\cos(s-b)} cot ( v − β ) = sin C cot ( s − b ) {\displaystyle \qquad \cot(v-\beta )=\sin C\cot(s-b)}
第108條 提供最小阻力的旋轉曲面。 要解決這個問題，我們看到（第12條）積分
I = ∫ t 0 t 1 x x ′ 3 x ′ 2 + y ′ 2 d t {\displaystyle I=\int _{t_{0}}^{t_{1}}{\frac {xx'^{3}}{x'^{2}+y'^{2}}}~{\text{d}}t} 必須是最小值。
我們這裡有
F = x x ′ 3 x ′ 2 + y ′ 2 {\displaystyle F={\frac {xx'^{3}}{x'^{2}+y'^{2}}}} 我們看到 F {\displaystyle F} x ′ {\displaystyle x'} y ′ {\displaystyle y'}
我們可以從以下關係式確定函式 F − 1 {\displaystyle F-1}
∂ 2 F ∂ x ′ ∂ y ′ = − x ′ y ′ F 1 {\displaystyle {\frac {\partial ^{2}F}{\partial x'\partial y'}}=-x'y'F_{1}} 可以看出
F 1 = 2 x x ′ ( 3 y ′ 2 − x ′ 2 ) ( x ′ 2 + y ′ 2 ) 3 {\displaystyle F_{1}={\frac {2xx'(3y'^{2}-x'^{2})}{(x'^{2}+y'^{2})^{3}}}} 我們可以取 x {\displaystyle x} x {\displaystyle x} t {\displaystyle t} x ′ {\displaystyle x'}
因此， F 1 {\displaystyle F_{1}} 3 y ′ 2 − x ′ 2 {\displaystyle 3y'^{2}-x'^{2}} 3 sin 2 λ − cos 2 λ {\displaystyle 3\sin ^{2}\lambda -\cos ^{2}\lambda } λ {\displaystyle \lambda } X {\displaystyle X}
F 1 {\displaystyle F_{1}} 正數 ，如果 | tan λ | > 1 3 {\displaystyle |\tan \lambda |>{\frac {1}{\sqrt {3}}}} 負數 ，如果 | tan λ | < 1 3 {\displaystyle |\tan \lambda |<{\frac {1}{\sqrt {3}}}}
我們將在後面（第 117 條）看到， F 1 {\displaystyle F_{1}} | tan λ | {\displaystyle |\tan \lambda |} 1 3 {\displaystyle {\frac {1}{\sqrt {3}}}} x ′ {\displaystyle x'} X {\displaystyle X} 30 ∘ {\displaystyle 30^{\circ }}
第 109 條 G = 0 {\displaystyle G=0}
由於 F {\displaystyle F} y {\displaystyle y}
− x ′ G = G 2 ≡ ∂ F ∂ y − d d t ∂ F ∂ y ′ = 0 {\displaystyle -x'G=G_{2}\equiv {\frac {\partial F}{\partial y}}-{\frac {\text{d}}{{\text{d}}t}}{\frac {\partial F}{\partial y'}}=0} 我們立刻有
∂ F ∂ y ′ = {\displaystyle {\frac {\partial F}{\partial y'}}=} 或者
− 2 x x ′ 3 y ′ ( x ′ 2 + y ′ 2 ) 2 = C {\displaystyle -{\frac {2xx'^{3}y'}{(x'^{2}+y'^{2})^{2}}}=C} 現在，如果有任何提供阻力的表面部分無限接近於旋轉軸，則該常數必須為零，因為 x = 0 {\displaystyle x=0} C = 0 {\displaystyle C=0}
如果 C = 0 {\displaystyle C=0}
x ′ 3 y ′ = 0 {\displaystyle x'^{3}y'=0} 因此
x ′ = 0 {\displaystyle x'=0} y ′ = 0 {\displaystyle y'=0} 由此得出
x = {\displaystyle x=} y = {\displaystyle y=} 在第一種情況下，該表面將是一個無限長的圓柱體，以 Y {\displaystyle Y} Y {\displaystyle Y} Y {\displaystyle Y}
第 110 條
− 2 x x ′ 3 y ′ ( x ′ 2 + y ′ 2 ) 2 = C {\displaystyle -{\frac {2xx'^{3}y'}{(x'^{2}+y'^{2})^{2}}}=C} 其中 C {\displaystyle C} x {\displaystyle x} C {\displaystyle C} x ′ y ′ {\displaystyle x'y'}
不保留變數 t {\displaystyle t}
t = − y {\displaystyle t=-y} 以及
d x d y = u {\displaystyle {\frac {{\text{d}}x}{{\text{d}}y}}=u} 則微分方程為
− 2 x u 3 ( u 2 + 1 ) 2 = C {\displaystyle {\frac {-2xu^{3}}{(u^{2}+1)^{2}}}=C} 我們可以將 − y {\displaystyle -y} t {\displaystyle t} x ′ y ′ {\displaystyle x'y'} y {\displaystyle y} x {\displaystyle x} y {\displaystyle y} x {\displaystyle x}
然後我們有
x = − C ( u 2 + 1 ) 2 2 u 3 = − C 2 ( u + 2 u − 1 + u − 3 ) {\displaystyle x=-{\frac {C(u^{2}+1)^{2}}{2u^{3}}}=-{\frac {C}{2}}(u+2u^{-1}+u^{-3})} 以及
d x d u = d x d y d y d u = u d y d u = − C 2 ( 1 − 2 u − 2 − 3 u − 4 ) {\displaystyle {\frac {{\text{d}}x}{{\text{d}}u}}={\frac {{\text{d}}x}{{\text{d}}y}}{\frac {{\text{d}}y}{{\text{d}}u}}=u{\frac {{\text{d}}y}{{\text{d}}u}}=-{\frac {C}{2}}(1-2u^{-2}-3u^{-4})} 或者
d y d u = − C 2 ( u − 1 − 2 u − 3 − 4 u − 5 ) {\displaystyle {\frac {{\text{d}}y}{{\text{d}}u}}=-{\frac {C}{2}}(u^{-1}-2u^{-3}-4u^{-5})} 因此
y = − C 2 [ ln u + u − 2 + 3 4 u − 4 ] + C 1 {\displaystyle y=-{\frac {C}{2}}\left[\ln u+u^{-2}+{\frac {3}{4}}u^{-4}\right]+C_{1}} 方程
x = − C 2 ( u + 2 u − 1 + u − 3 ) {\displaystyle x=-{\frac {C}{2}}(u+2u^{-1}+u^{-3})} y = − C 2 [ ln u + u − 2 + 3 4 u − 4 ] + C 1 {\displaystyle \qquad y=-{\frac {C}{2}}\left[\ln u+u^{-2}+{\frac {3}{4}}u^{-4}\right]+C_{1}} 確定一族曲線，其中之一是弧，該弧產生旋轉曲面，如果存在最小值，則該旋轉曲面產生最小值。對於這樣的曲線，如果我們給 u {\displaystyle u} + ∞ {\displaystyle +\infty } 3 {\displaystyle {\sqrt {3}}}
d y d x = 1 u > 1 3 {\displaystyle {\frac {{\text{d}}y}{{\text{d}}x}}={\frac {1}{u}}>{\frac {1}{\sqrt {3}}}} 或者
1 u < 1 3 {\displaystyle {\frac {1}{u}}<{\frac {1}{\sqrt {3}}}} 換句話說，如果在弧線上任意一點處的切線與 X {\displaystyle X} 30 ∘ {\displaystyle 30^{\circ }} 30 ∘ {\displaystyle 30^{\circ }} 30 ∘ {\displaystyle 30^{\circ }} 30 ∘ {\displaystyle 30^{\circ }} P {\displaystyle P} X {\displaystyle X} 30 ∘ {\displaystyle 30^{\circ }} P {\displaystyle P} 30 ∘ {\displaystyle 30^{\circ }} 30 ∘ {\displaystyle 30^{\circ }}
↑ 該證明由 Schwarz 教授提出。 
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![{\displaystyle \qquad {\text{[4]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd8a7658c47a27cccb1415bdcbf70acbfd7dd9b8)
![{\displaystyle y^{2}\left({\frac {{\text{d}}x}{{\text{d}}t}}\right)^{2}=\beta ^{2}\left[\left({\frac {{\text{d}}x}{{\text{d}}t}}\right)^{2}+\left({\frac {{\text{d}}y}{{\text{d}}t}}\right)^{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8867f84b944ff09c66e049cb1dd7ce71ffb3b66)
![{\displaystyle \qquad {\text{[5]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0196c8b891ab87d7f54d182078421ca80dfa51b8)
![{\displaystyle \qquad {\text{[6]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a2c550975cc4ab8462417d8c5f425cb5ad3433e)
![{\displaystyle \delta I=-\int _{M_{0}}^{P_{0}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{M_{0}}^{P_{0}}-\int _{M_{1}}^{M_{0}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{M_{1}}^{M_{0}}-\int _{P-1}^{M_{1}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{P_{1}}^{M_{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb6bc4b1cb23f31a295f647e29c933eac16d552d)
![{\displaystyle {\text{d}}x={\frac {c(y+a){\text{d}}y}{\sqrt {(y+a)[1-c^{2}(y+1)]]}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97216d28368da84d39099bdc239b4262bdf710fd)
![{\displaystyle {\text{d}}\phi ={\frac {c{\text{d}}y}{\sqrt {(y+a)[1-c^{2}(y+a)]}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3db50e6105bea051360dd7fd69826d8977741b82)
![{\displaystyle \qquad [5^{\text{a}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8316b1f878cf1f7e2662beac418b6a059b8f503)
![{\displaystyle \qquad [7]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7a76b74167cf784fb8ea94a80925fa8e30f6c6d)
![{\displaystyle 2[1-c^{2}(y+a)]=1+\xi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/49dadd639d8590cf3d04c6bc909400115007b51b)
![{\displaystyle \qquad [8]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2409f65d77283ea98f0ecb79a5a6455b7531813)
![{\displaystyle \qquad [9]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb5d3946d4cecfe2ebc3a8444f5dc7ba9adbefb1)
![{\displaystyle \qquad [10]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14c913f5e542d1f928c5e824b8eadfc8e2e929cd)
![{\displaystyle \qquad [11]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc2aeb4e6f12d05d34d893e304b8bfc8ee23be34)
![{\displaystyle \qquad [12]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/062121e770d7e2f486fd8b5bd56f3c28189ced41)
![{\displaystyle {\sqrt {\frac {r}{g}}}(\phi _{1}-\phi _{0})={\sqrt {\frac {r}{g}}}[2\pi +\psi _{0}-2\pi +\psi _{0}]=2{\sqrt {\frac {r}{g}}}\psi _{0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b8a20f4a1499b01271ca92fb0cb757d357242a4)
![{\displaystyle \left[v{\frac {{\text{d}}x}{{\text{d}}s}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c77f15b55cf67814e9f293dbaea4b89a56ea1b6)
![{\displaystyle \left[{\sqrt {2gr}}{\sqrt {1-\cos(\phi )}}\sin {\frac {\phi }{2}}\right]_{\phi _{0}}^{\phi _{1}}=2{\sqrt {gr}}\sin ^{2}{\frac {\psi _{0}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e553311597332a37540f5ab6b37c5ea1e01bbe33)
![{\displaystyle x_{1}-x_{0}=r[\phi _{1}-\phi _{0}-\sin \phi _{1}+\sin \phi _{0}]=2r[\psi _{0}-\sin \psi _{0}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3b281662a6c6cef707b95a7d4a2a2b75f9f793a)
![{\displaystyle {\frac {T_{1}}{T}}={\frac {2r(\psi _{0}-\sin \psi _{0})}{2{\sqrt {gr}}\sin ^{2}{\frac {\psi _{0}}{2}}\cdot 2{\sqrt {\frac {r}{g}}}\psi _{0}}}={\frac {\psi _{0}-\sin \psi _{0}}{2\psi _{0}\sin ^{2}{\frac {\psi _{0}}{2}}}}={\frac {{\frac {\psi _{0}^{3}}{3!}}-{\frac {\psi _{0}^{5}}{5!}}+\cdots }{2\psi _{0}\left[{\frac {\psi _{0}}{2}}-{\frac {1}{3!}}\left({\frac {\psi _{0}}{2}}\right)^{3}+\cdots \right]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5678092f4f38f80df589c1da2ca95126bc5a5cc9)
![{\displaystyle {\frac {T_{1}}{T}}<{\frac {\frac {\psi _{0}^{3}}{3!}}{2\psi _{0}\left[{\frac {\psi _{0}}{2}}-{\frac {1}{6}}\left({\frac {\psi _{0}}{2}}\right)^{3}+\cdots \right]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c128344908d1a03125de140c1e03866ec1fae83f)
![{\displaystyle \qquad [1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9844ea2f4fc15e13afd45b24fb41f6aaf744dbb7)
![{\displaystyle \qquad [2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e8f4f1dd2ae6b9aa3bfdc6b1bb102e663b266f67)
![{\displaystyle \qquad [3]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d30ee95572bf3617b4f921cc67d7a0cb5a5dddc2)
![{\displaystyle \qquad [5]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd4a7683798777cff440a306b1f4f16a1188272f)
![{\displaystyle \qquad [6]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f5cc94e473f6aeba5103d6c87b25484f65c7b9e)
![{\displaystyle -[2\cos uu'^{2}v'+\sin ^{2}u\cos uv'^{3}]+\sin u(v'u''-u'v'')=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12e4537d13f43d650cdcc25536ea462a8e00bf53)
![{\displaystyle {\frac {{\text{d}}v}{{\text{d}}u}}\left[2+\sin ^{2}u\left({\frac {{\text{d}}v}{{\text{d}}u}}\right)^{2}\right]\cos u+\sin u{\frac {{\text{d}}^{2}v}{{\text{d}}u^{2}}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/caf1da4e5184ec293f31b4880b99ecda6b395db0)
![{\displaystyle {\text{d}}s^{2}={\text{d}}u^{2}+\sin ^{2}u{\text{d}}v^{2}={\text{d}}u^{2}[1+w^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4aec5cbc1ef6f93633fa61a00eec17ebab641ad6)
![{\displaystyle y=-{\frac {C}{2}}\left[\ln u+u^{-2}+{\frac {3}{4}}u^{-4}\right]+C_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/431362fa13e558a57038f861c204a89014b417eb)
![{\displaystyle \qquad y=-{\frac {C}{2}}\left[\ln u+u^{-2}+{\frac {3}{4}}u^{-4}\right]+C_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ad6c7ce78ce45915d00fe3c637679f7524b841d)
