教材:[DLI-BSc-Mathematics-Documents/Ordinary Differential Equatrions.pdf at main · JavaZeroo/DLI-BSc-Mathematics-Documents (github.com)](https://github.com/JavaZeroo/DLI-BSc-Mathematics-Documents/blob/main/y3s2/常微分方程/Ordinary Differential Equatrions.pdf)
第一章
Explicit First Order Equations
这种形式的
y ′ = f ( x , y ) y' = f(x, y) y ′ = f ( x , y )
称为 'Explicit First Order Equations' 。
y ′ = f ( x )
1. Equations with Separated Variables
y ′ = f ( x ) g ( y ) y' = f(x)g(y) y ′ = f ( x ) g ( y )
这种可以直接变成 d y g ( y ) = f ( x ) d x \frac{dy}{g(y)}=f(x)dx g ( y ) d y = f ( x ) d x
积分完后
∫ d y g ( y ) = ∫ f ( x ) d x \int\frac{dy}{g(y)}=\int f(x)dx ∫ g ( y ) d y = ∫ f ( x ) d x
IVP: y ( ξ ) = η y(\xi)=\eta y ( ξ ) = η
∫ η y d y g ( y ) = ∫ ξ x f ( x ) d x \int^y_{\eta}\frac{dy}{g(y)}=\int^x_{\xi} f(x)dx ∫ η y g ( y ) d y = ∫ ξ x f ( x ) d x
这里需要注意的是如果g ( y ( ξ ) ) = g ( η ) = 0 g(y(\xi))=g(\eta)=0 g ( y ( ξ )) = g ( η ) = 0 那么,直接就有 y ′ = 0 y'=0 y ′ = 0 因此y = η y=\eta y = η ;
2. 普通的替换
y ′ = f ( a x + b y + c ) y'=f(ax+by+c) y ′ = f ( a x + b y + c )
这种情况用u ( x ) = a x + b y + c u(x) = ax+by+c u ( x ) = a x + b y + c 去替换掉变量x x x 。原理是u ′ = a + b y ′ ( x ) = a + b f ( u ) u'=a+by'(x)=a+bf(u) u ′ = a + b y ′ ( x ) = a + b f ( u ) 。很关键的一点是,u u u 和y y y 都是一次的。求导后刚好是线性关系。
因此 最后得出的u ( x ) u(x) u ( x ) 后可以直接利用u ( x ) = a x + b y + c u(x)=ax+by+c u ( x ) = a x + b y + c 得到y y y 。
3. 普通的Homogeneous Differential Equation
y ′ = f ( y x ) y'=f\left(\frac{y}{x}\right) y ′ = f ( x y )
同样的道理用u ( x ) = y ( x ) x , ( x ≠ 0 ) u(x) = \frac{y(x)}{x}, (x\neq0) u ( x ) = x y ( x ) , ( x = 0 ) 替换掉变量y y y 。有y ′ = u + x u ′ = f ( y x ) y'=u+xu'=f(\frac{y}{x}) y ′ = u + x u ′ = f ( x y ) ,可得u ′ = f ( u ) − u x u'=\frac{f(u)-u}{x} u ′ = x f ( u ) − u
因此 最后得出的u ( x ) u(x) u ( x ) 后可以直接利用u ( x ) = y ( x ) x u(x)=\frac{y(x)}{x} u ( x ) = x y ( x ) 得到y y y 。
4. 高级的 Homogeneous Differential Equation
y ′ = f ( a x + b y + c α x + β y + γ ) y'=f\left(\frac{ax+by+c}{\alpha x+\beta y+ \gamma}\right) y ′ = f ( α x + β y + γ a x + b y + c )
这个的核心思想是转换成“普通的Homogeneous Differential Equation”。
首先分析行列式
∣ a b α β ∣ \left | \begin{matrix}
a &b \\
\alpha &\beta \\
\end{matrix} \right | a α b β
行列式为零时,说明a = λ a α , b = λ b β a=\lambda_a \alpha, b=\lambda_b \beta a = λ a α , b = λ b β 此时 可以直接转换成“普通的Homogeneous Differential Equation”
行列式不为零的时候,说明方程组有唯一解。
首先解方程组
{ a x + b y + c = 0 α x + β y + γ = 0 \left\{\begin{align}
ax+by+c&=0\\
\alpha x+\beta y+ \gamma&=0\\
\end{align}\right. { a x + b y + c α x + β y + γ = 0 = 0
可以解出一组( x 0 , y 0 ) (x_0, y_0) ( x 0 , y 0 ) 利用这组解将**“高级的 Homogeneous Differential Equation”转换成“普通的Homogeneous Differential Equation”**。
原理是新建坐标系得x ˉ : = x − x 0 , y ˉ : = y − y 0 \bar{x}:=x-x_0, \bar{y}:=y-y_0 x ˉ := x − x 0 , y ˉ := y − y 0 那么在这个坐标系下面原方程就变成了y ˉ ( x ˉ ) : = y ( x ˉ + x 0 ) − y 0 \bar{y}(\bar{x}):=y(\bar{x}+x_0)-y_0 y ˉ ( x ˉ ) := y ( x ˉ + x 0 ) − y 0 。对这个方程求导可以将y y y 消掉,将原问题变y ˉ \bar{y} y ˉ 和x x x 的关系。这样做的目的就是将**“高级的 Homogeneous Differential Equation”转换成“普通的Homogeneous Differential Equation”**。
对y ˉ \bar{y} y ˉ 求导后可以发现(利用方程y ˉ ( x ˉ ) : = y ( x ˉ + x 0 ) − y 0 , a x 0 + b y 0 + c = 0 , α x 0 + β y 0 + γ = 0 \bar{y}(\bar{x}):=y(\bar{x}+x_0)-y_0, ax_0+by_0+c=0, \alpha x_0+\beta y_0+ \gamma=0 y ˉ ( x ˉ ) := y ( x ˉ + x 0 ) − y 0 , a x 0 + b y 0 + c = 0 , α x 0 + β y 0 + γ = 0 )
y ˉ ( x ) d x ˉ = y ′ ( x ˉ + x 0 ) = f ( a x ˉ + b y ˉ ( x ˉ ) α x ˉ + β y ˉ ( x ˉ ) ) \frac{\bar{y}(x)}{d\bar{x}} = y'(\bar{x}+x_0)=f\left(\frac{a\bar{x}+b\bar{y}(\bar{x})}{\alpha\bar{x}+\beta\bar{y}(\bar{x})}\right) d x ˉ y ˉ ( x ) = y ′ ( x ˉ + x 0 ) = f ( α x ˉ + β y ˉ ( x ˉ ) a x ˉ + b y ˉ ( x ˉ ) )
这里就可以像刚才的普通的Homogeneous Differential Equation 一样去做了。值得注意的是,这样解出来的y ˉ \bar{y} y ˉ 需要转换成原来的y y y 。这里可以用
y ( x ) : = y 0 + y ˉ ( x − x 0 ) y(x):=y_0+\bar{y}(x-x_0) y ( x ) := y 0 + y ˉ ( x − x 0 )
得到最后的y y y 。
The Linear Differential Equation
这种形式的
y ′ + g ( x ) y = h ( x ) y' + g(x)y=h(x) y ′ + g ( x ) y = h ( x )
称为 'The Linear Differential Equation' 。
这时有两种 情况:h ( x ) = 0 h(x)=0 h ( x ) = 0 和h ( x ) ≠ 0 h(x)\neq0 h ( x ) = 0 , 分别称为"homogeneous" 和"nonhomogeneous"
事实上当h ( x ) = 0 h(x)=0 h ( x ) = 0 也就是"homogeneous"时,就是上面的"Explicit First Order Equations",这里就不在赘述了。
对于h ( x ) ≠ 0 h(x)\neq0 h ( x ) = 0 的情况,也就是"nonhomogeneous "时,我们需要用到"Method of variation of constants"。
**Method of variation of constants:**这个方法首先计算出齐次的时候的通解。对于方程y ′ + g ( x ) y = h ( x ) y'+g(x)y=h(x) y ′ + g ( x ) y = h ( x ) 他的齐次方程的通解是通过解y ′ + g ( x ) y = 0 y'+g(x)y=0 y ′ + g ( x ) y = 0 ,可得
y = C ⋅ e − ∫ g ( x ) d x y=C\cdot e^{-\int g(x)dx} y = C ⋅ e − ∫ g ( x ) d x
此时我们将常数C C C 作为一个与x x x 的函数C ( x ) C(x) C ( x ) 。这个时候,只需要解出一个C ( x ) C(x) C ( x ) 就是非齐次情况下的答案 。
那么现在的问题就变成了,如何得到一个C ( x ) C(x) C ( x ) 使得y ′ + g ( x ) y = h ( x ) y' + g(x)y=h(x) y ′ + g ( x ) y = h ( x ) ?
首先先计算C ′ ( x ) C'(x) C ′ ( x )
y = C ( x ) ⋅ e − ∫ g ( x ) d x y ′ = ( C ( x ) ) ′ ⋅ e − ∫ g ( x ) + C ( x ) ⋅ ( e − ∫ g ( x ) d x ) ′ y ′ = C ′ ( x ) ⋅ e − ∫ g ( x ) + C ( x ) ⋅ ( e − ∫ g ( x ) d x ⋅ ( − ∫ g ( x ) ) ′ ) y ′ = C ′ ( x ) ⋅ e − ∫ g ( x ) + C ( x ) ⋅ ( e − ∫ g ( x ) d x ⋅ − g ( x ) ) ⇓ y ′ = C ′ ⋅ e − ∫ g ( x ) − g C ⋅ e − ∫ g ( x ) d x \begin{align}
y &=C(x)\cdot e^{-\int g(x)dx}\\
y'&=\left( C(x) \right)'\cdot e^{-\int g(x)}+C(x)\cdot \left( e^{-\int g(x)dx }\right)' \\
y'&=C'(x)\cdot e^{-\int g(x)}+C(x)\cdot \left( e^{-\int g(x)dx }\cdot\left(-\int g(x)\right)'\right) \\
y'&=C'(x)\cdot e^{-\int g(x)}+C(x)\cdot \left( e^{-\int g(x)dx }\cdot -g(x)\right) \\
&\Downarrow \\
y'&=C'\cdot e^{-\int g(x)}-gC\cdot e^{-\int g(x)dx}
\end{align} y y ′ y ′ y ′ y ′ = C ( x ) ⋅ e − ∫ g ( x ) d x = ( C ( x ) ) ′ ⋅ e − ∫ g ( x ) + C ( x ) ⋅ ( e − ∫ g ( x ) d x ) ′ = C ′ ( x ) ⋅ e − ∫ g ( x ) + C ( x ) ⋅ ( e − ∫ g ( x ) d x ⋅ ( − ∫ g ( x ) ) ′ ) = C ′ ( x ) ⋅ e − ∫ g ( x ) + C ( x ) ⋅ ( e − ∫ g ( x ) d x ⋅ − g ( x ) ) ⇓ = C ′ ⋅ e − ∫ g ( x ) − g C ⋅ e − ∫ g ( x ) d x
然后代入y ′ + g ( x ) y = h ( x ) y' + g(x)y=h(x) y ′ + g ( x ) y = h ( x )
L y ≡ y ′ + g ( x ) y = h ( x ) L y = C ′ ⋅ e − ∫ g ( x ) d x − g C ⋅ e − ∫ g ( x ) d x + g C ⋅ e − ∫ g ( x ) d x L y = C ′ ⋅ e − ∫ g ( x ) d x = h ( x ) ⇓ C ′ ( x ) = h ( x ) ⋅ e ∫ g ( x ) d x ⇓ C ( x ) = ∫ h ( x ) ⋅ e ∫ g ( x ) d x d x + C 0 \begin{align}
L_y \equiv y' + g(x)y&=h(x)\\
L_y&=C'\cdot e^{-\int g(x)dx}-gC\cdot e^{-\int g(x)dx}+gC\cdot e^{-\int g(x)dx}\\
L_y&=C'\cdot e^{-\int g(x)dx}=h(x) \\
&\Downarrow \\
C'(x)&=h(x)\cdot e^{\int g(x)dx} \\
&\Downarrow \\
C(x)&=\int h(x)\cdot e^{\int g(x)dx}dx + C_0
\end{align} L y ≡ y ′ + g ( x ) y L y L y C ′ ( x ) C ( x ) = h ( x ) = C ′ ⋅ e − ∫ g ( x ) d x − g C ⋅ e − ∫ g ( x ) d x + g C ⋅ e − ∫ g ( x ) d x = C ′ ⋅ e − ∫ g ( x ) d x = h ( x ) ⇓ = h ( x ) ⋅ e ∫ g ( x ) d x ⇓ = ∫ h ( x ) ⋅ e ∫ g ( x ) d x d x + C 0
现在我们可以知道,当y = C ( x ) ⋅ e − ∫ g ( x ) d x y=C(x)\cdot e^{-\int g(x)dx} y = C ( x ) ⋅ e − ∫ g ( x ) d x 且C ( x ) = ∫ h ( x ) ⋅ e ∫ g ( x ) d x + C 0 C(x)=\int h(x)\cdot e^{\int g(x)dx} + C_0 C ( x ) = ∫ h ( x ) ⋅ e ∫ g ( x ) d x + C 0 时。,有y ′ + g ( x ) y = h ( x ) y' + g(x)y=h(x) y ′ + g ( x ) y = h ( x ) 。这里有个二级结论
If y , y ˉ y, \bar{y} y , y ˉ , yare two solutions to the nonhomogeneous equation L y = h L_y = h L y = h , then L ( y − y ) = L y − L y ˉ = 0 , i . e . , z ( x ) = y − y ˉ L(y - y) = L_y - L_{\bar{y}} = 0, i.e., z(x) = y - \bar{y} L ( y − y ) = L y − L y ˉ = 0 , i . e . , z ( x ) = y − y ˉ is a solution of the homogeneous equation L y = 0 L_y = 0 L y = 0 . Thus all solutions y ( x ) y(x) y ( x ) of the nonhomogeneous equation can be written in the form
y ( x ) = y ˉ + z ( x ) y(x)=\bar{y}+z(x) y ( x ) = y ˉ + z ( x )
这里面z ( x ) z(x) z ( x ) 就是刚刚的y = C ⋅ e − ∫ g ( x ) d x y=C\cdot e^{-\int g(x)dx} y = C ⋅ e − ∫ g ( x ) d x ,y ˉ \bar{y} y ˉ 就是C ( x ) = ∫ h ( x ) ⋅ e ∫ g ( x ) d x + C 0 C(x)=\int h(x)\cdot e^{\int g(x)dx} + C_0 C ( x ) = ∫ h ( x ) ⋅ e ∫ g ( x ) d x + C 0 和y = C ⋅ e − ∫ g ( x ) d x y=C\cdot e^{-\int g(x)dx} y = C ⋅ e − ∫ g ( x ) d x 的结合中的y y y ,也就是y ˉ = ( ∫ h ( x ) ⋅ e ∫ g ( x ) d x d x + C 0 ) ⋅ e − ∫ g ( x ) d x \bar{y}=\left(\int h(x)\cdot e^{\int g(x)dx}dx + C_0\right)\cdot e^{-\int g(x)dx} y ˉ = ( ∫ h ( x ) ⋅ e ∫ g ( x ) d x d x + C 0 ) ⋅ e − ∫ g ( x ) d x
Bernoulli's Equation
这种形式的
y ′ + g ( x ) y + h ( x ) y α = 0. α ≠ 1 y'+g(x)y+h(x)y^{\alpha}=0.\alpha \neq 1 y ′ + g ( x ) y + h ( x ) y α = 0. α = 1
非常的简单,只需要把y α y^{\alpha} y α 解决了就可以了。等式去除y α y^{\alpha} y α 有
y ′ y − α + g ( x ) y ( 1 − α ) + h ( x ) = 0 y'y^{-\alpha}+g(x)y^{(1-\alpha)}+h(x)=0 y ′ y − α + g ( x ) y ( 1 − α ) + h ( x ) = 0
利用z = y ( 1 − α ) ⟹ z ′ = ( 1 − α ) y − α ⋅ y ′ z=y^{(1-\alpha)} \implies z'=(1-\alpha)y^{-\alpha}\cdot y' z = y ( 1 − α ) ⟹ z ′ = ( 1 − α ) y − α ⋅ y ′ 替换原式得
1 1 − α z ′ + g ( x ) z + h ( x ) = 0 \frac{1}{1-\alpha}z'+g(x)z+h(x)=0 1 − α 1 z ′ + g ( x ) z + h ( x ) = 0
现在,就变成了nonhomogeneous的"The Linear Differential Equation"。最后解出z z z ,别忘了替换回y y y 。
Exact differential equations
这种形式的
M ( x , y ) d x + N ( x , y ) d y = 0 , ∃ U ( x , y ) s . t . U x ( x , y ) = M ( x , y ) , U y ( x , y ) = N ( x , y ) M(x,y)dx+N(x,y)dy=0,\\ \exists\ U(x, y)\ s.t.\ U_x(x,y)=M(x,y),U_y(x,y)=N(x,y) M ( x , y ) d x + N ( x , y ) d y = 0 , ∃ U ( x , y ) s . t . U x ( x , y ) = M ( x , y ) , U y ( x , y ) = N ( x , y )
x d x + y d y = 0 xdx+ydy=0 x d x + y d y = 0 is an exact equation, and U ( x , y ) = 1 / 2 ( x 2 + y 2 ) U(x,y)=1/2 (x^2+y^2 ) U ( x , y ) = 1/2 ( x 2 + y 2 ) is a potential function.
Integrating Factors
Integrating Factors是用来让非 Exact 变成Exact differential equations。
E.g . y d x + 2 x d y = 0 ydx + 2xdy = 0 y d x + 2 x d y = 0 is not exact. However, it can easily be made an exact differential equation (in the domain x > 0 x > 0 x > 0 ) by multiplying the equation by 1 / x 1/\sqrt{x} 1/ x . The resulting differential equation
y x d x + 2 x d y = 0 \frac{y}{\sqrt{x}}dx+2\sqrt{x}dy=0 x y d x + 2 x d y = 0
is exact, and a potential function is given by
F ( x , y ) = 2 y x = 0 ( x > 0 ) F(x,y)=2y\sqrt{x}=0\ (x>0) F ( x , y ) = 2 y x = 0 ( x > 0 )
对于一个not excat differential equation我们需要找到一个Factor U ( x , y ) U(x,y) U ( x , y ) 使得U ( x , y ) ⋅ M ( x , y ) d x + U ( x , y ) ⋅ N ( x , y ) d y = 0 U(x,y)\cdot M(x,y)dx+U(x,y)\cdot N(x,y)dy=0 U ( x , y ) ⋅ M ( x , y ) d x + U ( x , y ) ⋅ N ( x , y ) d y = 0 变成一个Exact differential equations。
这里有一个"Theorem on potential functions "保证可以找到U ( x , y ) U(x,y) U ( x , y )
现在的问题就是,如何去找?首先令M ′ = U ⋅ M , N ′ = U ⋅ N M' = U\cdot M,N'=U\cdot N M ′ = U ⋅ M , N ′ = U ⋅ N 如果F x = M ′ , F y = N ′ F_x=M',F_y=N' F x = M ′ , F y = N ′ 则有M y ′ = N x ′ M'_y=N'_x M y ′ = N x ′ 。利用这个关系可以知道
( U ⋅ M ) y = ( U ⋅ N ) x ⟹ U y ⋅ M + U ⋅ M y = U x ⋅ N + U ⋅ N x \begin{align}
&(U\cdot M)_y=(U\cdot N)_x\\
\implies &U_y\cdot M+U\cdot M_y=U_x\cdot N+U\cdot N_x \\
\end{align} ⟹ ( U ⋅ M ) y = ( U ⋅ N ) x U y ⋅ M + U ⋅ M y = U x ⋅ N + U ⋅ N x
此时需要考虑,Integrating Factors是只与x x x 有关还是只与y y y 有关(只需要选一个)
假如只与x x x 有关则U y = 0 U_y=0 U y = 0 则有
U y ⋅ M + U ⋅ M y = U x ⋅ N + U ⋅ N x U ⋅ M y = U ′ ⋅ N + U ⋅ N x 1 U U ′ = M y − N x N ( ln U ) ′ = M y − N x N U = e ∫ M y − N x N d x \begin{align}
U_y\cdot M+U\cdot M_y&=U_x\cdot N+U\cdot N_x \\
U\cdot M_y &= U' \cdot N+U \cdot N_x \\
\frac{1}{U}U'&=\frac{M_y-N_x}{N} \\
(\ln U)'&=\frac{M_y-N_x}{N} \\
U&=e^{\int \frac{M_y-N_x}{N}dx}
\end{align} U y ⋅ M + U ⋅ M y U ⋅ M y U 1 U ′ ( ln U ) ′ U = U x ⋅ N + U ⋅ N x = U ′ ⋅ N + U ⋅ N x = N M y − N x = N M y − N x = e ∫ N M y − N x d x
这里的答案不是U = C ⋅ e ∫ M y − N x N d x U=C\cdot e^{\int \frac{M_y-N_x}{N}dx} U = C ⋅ e ∫ N M y − N x d x 的原因是,我们只需要找到一个U U U ,因此你可以认为我们选择C = 1 C=1 C = 1 作为答案。下面的情况同理。
假如只与y y y 有关则U x = 0 U_x=0 U x = 0 则有
U y ⋅ M + U ⋅ M y = U x ⋅ N + U ⋅ N x U ′ ⋅ M + U ⋅ M y = U ⋅ N x 1 U U ′ = N x − M y M ( ln U ) ′ = N x − M y M U = e ∫ N x − M y M \begin{align}
U_y\cdot M+U\cdot M_y&=U_x\cdot N+U\cdot N_x \\
U'\cdot M+U\cdot M_y &= U \cdot N_x \\
\frac{1}{U}U'&=\frac{N_x-M_y}{M} \\
(\ln U)'&=\frac{N_x-M_y}{M} \\
U &=e^{\int \frac{N_x-M_y}{M}}
\end{align} U y ⋅ M + U ⋅ M y U ′ ⋅ M + U ⋅ M y U 1 U ′ ( ln U ) ′ U = U x ⋅ N + U ⋅ N x = U ⋅ N x = M N x − M y = M N x − M y = e ∫ M N x − M y
Implicit First Order Differential Equations
这种形式的
F ( x , y , y ′ ) = 0 F(x, y, y')=0 F ( x , y , y ′ ) = 0
一般来说有两种解决办法。要么通过一些方法获得explicit differential equation,要么就用参数化。
在这里我们只讨论两种情况:
F ( x , y ′ ) = 0 F(x, y')=0 F ( x , y ′ ) = 0 , F ( y , y ′ ) = 0 F(y, y')=0 F ( y , y ′ ) = 0
y = f ( x , y ′ ) y=f(x,y') y = f ( x , y ′ ) , x = f ( y , y ′ ) x=f(y,y') x = f ( y , y ′ )
第一种情况
对于F ( x , y ′ ) = 0 F(x, y')=0 F ( x , y ′ ) = 0 我们使用参数化:
{ x = ϕ ( t ) y ′ = ψ ( t ) \left\{\begin{array}{l}
x=\phi(t) \\
y'=\psi(t)
\end{array}\right. { x = ϕ ( t ) y ′ = ψ ( t )
此时方程变为F ( ϕ ( t ) , ψ ( t ) ) = 0 F(\phi(t), \psi(t))=0 F ( ϕ ( t ) , ψ ( t )) = 0 ,同时我们有
y ′ = d y d x and ϕ ′ ( t ) = d ϕ ( t ) d t d y = y ′ d x and d ϕ ( t ) = ϕ ′ ( t ) d t y = ∫ y ′ d x + C y = ∫ ψ ( t ) d ϕ ( t ) + C y = ∫ ψ ( t ) ϕ ′ ( t ) d t + C \begin{align}
y'&=\frac{dy}{dx} \ \text{and} \ \phi'(t)=\frac{d\phi(t)}{dt} \\
dy&=y'dx \ \ \text{and} \ d\phi(t)=\phi'(t)dt\\
y&=\int y'dx + C\\
y&=\int \psi(t)d\phi(t) + C \\
y&=\int \psi(t)\phi'(t)dt +C
\end{align} y ′ d y y y y = d x d y and ϕ ′ ( t ) = d t d ϕ ( t ) = y ′ d x and d ϕ ( t ) = ϕ ′ ( t ) d t = ∫ y ′ d x + C = ∫ ψ ( t ) d ϕ ( t ) + C = ∫ ψ ( t ) ϕ ′ ( t ) d t + C
最后得到:
{ x = ϕ ( t ) y = ∫ ψ ( t ) ϕ ′ ( t ) d t + C \left\{\begin{array}{l}
x=\phi(t) \\
y=\int \psi(t)\phi'(t)dt +C
\end{array}\right. { x = ϕ ( t ) y = ∫ ψ ( t ) ϕ ′ ( t ) d t + C
第一种情况2
对于F ( y , y ′ ) = 0 F(y,y')=0 F ( y , y ′ ) = 0 我们仍然参数化:
{ y = ϕ ( t ) y ′ = ψ ( t ) \left\{\begin{array}{l}
y=\phi(t) \\
y'=\psi(t)
\end{array}\right. { y = ϕ ( t ) y ′ = ψ ( t )
此时有F ( ϕ ( t ) , ψ ( t ) = 0 F(\phi(t),\psi(t)=0 F ( ϕ ( t ) , ψ ( t ) = 0 ,同时有:
y ′ = d y d x and ϕ ′ ( t ) = d y d t d x = d y ψ ( t ) and d y = ϕ ′ ( t ) d t d x = ϕ ′ ( t ) d t ψ ( t ) ∫ d x = ∫ ϕ ′ ( t ) d t ψ ( t ) x = ∫ ϕ ′ ( t ) d t ψ ( t ) \begin{align}
y'&=\frac{dy}{dx} \ \text{and} \ \phi'(t)=\frac{dy}{dt} \\
dx&=\frac{dy}{\psi(t)} \ \text{and} \ dy=\phi'(t)dt \\
dx&=\frac{\phi'(t)dt}{\psi(t)} \\
\int dx&= \int \frac{\phi'(t)dt}{\psi(t)} \\
x&=\int \frac{\phi'(t)dt}{\psi(t)}
\end{align} y ′ d x d x ∫ d x x = d x d y and ϕ ′ ( t ) = d t d y = ψ ( t ) d y and d y = ϕ ′ ( t ) d t = ψ ( t ) ϕ ′ ( t ) d t = ∫ ψ ( t ) ϕ ′ ( t ) d t = ∫ ψ ( t ) ϕ ′ ( t ) d t
An Existence and Uniqueness Theorem
中文名是存在唯一性定理 。首先我们介绍Lipschitz condition :
We consider the following initial value problem
y ′ = f ( x , y ) , for ξ ≤ x ≤ ξ + a , y ( ξ ) = η y' = f(x,y),\ \text{for}\ \xi \leq x\leq \xi+a,\ y(\xi)=\eta y ′ = f ( x , y ) , for ξ ≤ x ≤ ξ + a , y ( ξ ) = η
The main assumptions in the following theorem are that f f f is continuous in the strip S = J × R S=J\times\mathbb{R} S = J × R with J = [ ξ , ξ + a ] J=[\xi,\xi+a] J = [ ξ , ξ + a ] and satisfies a Lipschitz condition with respect to y y y in S S S
∣ f ( x , y ) − f ( x , y ˉ ) ∣ ≤ L ∣ y − y ˉ ∣ |f(x,y)-f(x, \bar{y})|\leq L|y-\bar{y}| ∣ f ( x , y ) − f ( x , y ˉ ) ∣ ≤ L ∣ y − y ˉ ∣
No restrictions are placed on the value of the Lipschitz constant L ≥ 0 L\geq 0 L ≥ 0
然后引出存在唯一性定理:
Let f ∈ C ( S ) f \in C(S) f ∈ C ( S ) satisfy the Lipschitz condition . Then the IVP has exactly one solution y ( x ) y(x) y ( x ) . The solution exists in the interval J : ξ ≤ x ≤ ξ + a J: \xi\leq x \leq\xi +a J : ξ ≤ x ≤ ξ + a
The extension of solutions.
以下三个定理是用于 The extension of solutions.
1️⃣Local Lipschitz condition. The function f ( x , y ) f(x,y) f ( x , y ) is said to satisfy a local Lipschitz condition with respect to y y y in D ⊂ R 2 D\subset R^2 D ⊂ R 2 if for every ( x 0 , y 0 ) ∈ D (x_0,y_0 )\in D ( x 0 , y 0 ) ∈ D there exists a neighborhood U = U ( x 0 , y 0 ) U=U(x_0,y_0 ) U = U ( x 0 , y 0 ) and an L = L ( x 0 , y 0 ) L=L(x_0,y_0 ) L = L ( x 0 , y 0 ) such that in U ∪ D U\cup D U ∪ D the function f f f satisfies the Lipschitz condition ∣ f ( x , y ) − f ( x , y ˉ ) ∣ ≤ L ∣ y − y ˉ ∣ |f(x,y)-f(x,\bar{y})|\leq L|y-\bar{y}| ∣ f ( x , y ) − f ( x , y ˉ ) ∣ ≤ L ∣ y − y ˉ ∣ .
注意!我们一般通过连续性来判断Local Lipschitz condition
If D D D is open and if f ∈ C ( D ) f \in C(D) f ∈ C ( D ) has a continuous derivative f y f_y f y in D D D , then f satisfies a local Lipschitz condition in this set.
2️⃣Theorem on local solvability If D D D is open and f ∈ C ( D ) f\in C(D) f ∈ C ( D ) satisfies a local Lipschitz condition in D D D , then the IVP is locally uniquely solvable for( x 0 , y 0 ) ∈ D (x_0,y_0 )∈D ( x 0 , y 0 ) ∈ D ; i.e., there is a neighborhood I I I ofx 0 x_0 x 0 such that exactly one solution exists in I I I .
3️⃣Theorem on the extension of solutions Let f ∈ C ( D ) f \in C (D) f ∈ C ( D ) satisfy a local Lipschitz condition with respect to y y y in D D D . Then for every ( x 0 , y 0 ) ∈ D (x_0, y_0)\in D ( x 0 , y 0 ) ∈ D the initial value problem y ′ = f ( x , y ) , y ( x 0 ) = y 0 y' = f (x, y), y(x_0) = y_0 y ′ = f ( x , y ) , y ( x 0 ) = y 0 has a solution that can be extended to the left and to the right comes arbitrarily close to the boundary of D D D .
最后我们有:
The Peano existence theorem. If f ( x , y ) f(x,y) f ( x , y ) is continuous in a domain D D D and ( ξ , η ) (\xi,\eta) ( ξ , η ) is a point in D D D , then at least one solution of the differential equation y ′ = f ( x , y ) y′=f(x,y) y ′ = f ( x , y ) goes through ( ξ , η ) (\xi,\eta) ( ξ , η ) . Every solution can be extended to the left and to the right up to the boundary of D D D .
Linear System
这里开始就是在讨论,常微分方程组。
Systems of n Linear Differential Equations
给出常微分方程组的形式:
y 1 ′ = a 11 ( t ) y 1 + ⋯ + a 1 n ( t ) y n + b 1 ( t ) ⋮ y 1 ′ = a 11 ( t ) y 1 + ⋯ + a 1 n ( t ) y n + b 1 ( t ) \begin{align}
y_1' &= a_{11}(t)y_1+\cdots+ a_{1n}(t)y_n+b_1(t) \\
&\ \ \vdots \\
y_1' &= a_{11}(t)y_1+\cdots+a_{1n}(t)y_n+b_1(t) \\
\end{align} y 1 ′ y 1 ′ = a 11 ( t ) y 1 + ⋯ + a 1 n ( t ) y n + b 1 ( t ) ⋮ = a 11 ( t ) y 1 + ⋯ + a 1 n ( t ) y n + b 1 ( t )
或者:
y ′ = A ( t ) y + b ( t ) where \hfill A ( t ) = ( a i j ( t ) ) , b ( t ) = ( b 1 ( t ) , … , b n ( t ) ) ⊤ \mathbf{y}'=A(t)\mathbf{y}+\mathbf{b}(t)\\
\text{where} \hfill \\
A(t)=(a_{ij}(t)),\ \mathbf{b}(t)=(b_1(t), \dots, b_n(t))^\top y ′ = A ( t ) y + b ( t ) where \hfill A ( t ) = ( a ij ( t )) , b ( t ) = ( b 1 ( t ) , … , b n ( t ) ) ⊤
值得一提的是,存在唯一性定理在常微分方程组也同样适用。
Homogeneous Linear Systems
对于齐次的形式,常微分方程组就变成了:
y ′ = A ( x ) y \mathbf{y}'=A(x)\mathbf{y} y ′ = A ( x ) y
此时,根据存在唯一性定理有:
∃ exactly one solution y = y ( t ; τ , η ) ∀ τ ∈ J , η ∈ R n or \C n \exist \text{ exactly one solution } \mathbf{y}=\mathbf{y}(t;\tau,\boldsymbol{\eta})\ \forall \tau \in J, \boldsymbol{\eta}\in \R^n\text{ or }\C^n ∃ exactly one solution y = y ( t ; τ , η ) ∀ τ ∈ J , η ∈ R n or \C n
当然,齐次常微分方程组有一些重要的性质:
y ≡ 0 \mathbf{y} \equiv 0 y ≡ 0 in J J J is a solution of the homogeneous linear systems .
There exist n n n linearly independent solutions 1 , … , y n _1,\dots,\mathbf{y}_n 1 , … , y n . Every such set of n n n linearly independent solutions is called a fundamental system of solutions . If y 1 , … , y n \mathbf{y}_1,\dots,\mathbf{y}_n y 1 , … , y n is a fundamental system, then every solution y \mathbf{y} y can be written in a unique way as a linear combination y = C 1 y 1 + ⋯ + C n y n \mathbf{y}=C_1 \mathbf{y}_1+\dots+C_n \mathbf{y}_n y = C 1 y 1 + ⋯ + C n y n .
A system of n n n solutions y 1 , … , y n \mathbf{y}_1,…,\mathbf{y}_n y 1 , … , y n can be assembled into an n × n n\times n n × n solution matrix Φ ( x ) = ( y 1 , … , y n ) \Phi(x)=(\mathbf{y}_1,\dots,\mathbf{y}_n ) Φ ( x ) = ( y 1 , … , y n ) . If n n n solutions y 1 , … , y n \mathbf{y}_1,\dots,\mathbf{y}_n y 1 , … , y n are linearly independent, then Φ ( x ) \Phi(x) Φ ( x ) is a system of n n n solutions y 1 , … , y n \mathbf{y}_1,\dots,\mathbf{y}_n y 1 , … , y n can be assembled into an n × n n\times n n × n solution matrix Φ ( x ) = ( y 1 , … , y n ) \Phi(x)=(\mathbf{y}_1,\dots,\mathbf{y}_n ) Φ ( x ) = ( y 1 , … , y n ) . If n n n solutions y 1 , … , y n \mathbf{y}_1,\dots,\mathbf{y}_n y 1 , … , y n are linearly independent, then Φ ( x ) \Phi(x) Φ ( x ) is a Fundamental Matrix.
The Wronskian
现在讨论一下,齐次常微分方程的解,是线性无关还是线性相关。
The Wronskian. If Φ ( x ) = ( y 1 , … , y n ) \Phi(x)=(\mathbf{y}_1,\dots,\mathbf{y}_n ) Φ ( x ) = ( y 1 , … , y n ) is a solution matrix of y ′ = A ( x ) y \mathbf{y}^′=A(x)\mathbf{y} y ′ = A ( x ) y , then its determinant W ( x ) = ∣ Φ ( x ) ∣ W(x)=|\Phi(x)| W ( x ) = ∣Φ ( x ) ∣ is called the Wronskian determinant.
Theorem If y 1 , … , y n \mathbf{y}_1,\dots,\mathbf{y}_n y 1 , … , y n are linearly dependent in J J J , then the Wronskian W ( x ) ≡ 0 W(x)\equiv0 W ( x ) ≡ 0 .
Theorem If y 1 , … , y n \mathbf{y}_1,…,\mathbf{y}_n y 1 , … , y n is a fundamental system of equation y ′ = A ( x ) y \mathbf{y}'=A(x)\mathbf{y} y ′ = A ( x ) y , then the Wronskian W ( x ) ≠ 0 W(x)\neq0 W ( x ) = 0 in J J J .
Theorem. There exists a fundamental system of solutions for equation y ′ = A ( x ) y \mathbf{y}'=A(x)\mathbf{y} y ′ = A ( x ) y .
因此,我们先求出n n n 个解,然后再去判断这n n n 个解的Fundamental Matrix 的行列式,也就是The Wronskian ,是否为零。
Inhomogeneous Systems
对于非齐次的常微分方程组,就是最开始样子:
y ′ = A ( t ) y + b ( t ) \mathbf{y}'=A(t)\mathbf{y}+\mathbf{b}(t) y ′ = A ( t ) y + b ( t )
下面这个定理类似线性代数,非齐次方程组的通解是,齐次方程组的通解+非齐次方程组的特解:
Theorem. Let y ~ ( x ) \tilde{\mathbf{y}}(x) y ~ ( x ) be a fixed solution of the inhomogeneous equation (1). If y 0 ( x ) \mathbf{y}_0 (x) y 0 ( x ) is an arbitrary solution of the homogeneous equation, then y ( x ) = y ~ ( x ) + y 0 ( x ) \mathbf{y}(x)=\tilde{\mathbf{y}}(x)+\mathbf{y}_0 (x) y ( x ) = y ~ ( x ) + y 0 ( x ) is a solution of the inhomogeneous equation, and all solutions of the inhomogeneous equation are obtained in this way.