数学 中的特征线法 是求解偏微分方程 的一种方法,适用于准线性偏微分方程的求解。只要初始值不是沿着特征线给定,即可通过特征线法获得偏微分方程的精确解。 其基本思想是通过把双曲线型的准线性偏微分方程转化为两组常微分方程,再对常微分方程进行求解。两组常微分方程中的一组用于定义特征线,另一组用以描述解沿给定特征线变化。
设所需求解的准线性偏微分方程为
a 1 ( x , u ) ∂ u ∂ x 1 + a 2 ( x , u ) ∂ u ∂ x 2 + ⋯ + a N ( x , u ) ∂ u ∂ x N = b ( x , u ) {\displaystyle a_{1}\left({\boldsymbol {x}},u\right){\frac {\partial u}{\partial x_{1}}}+a_{2}\left({\boldsymbol {x}},u\right){\frac {\partial u}{\partial x_{2}}}+\cdots +a_{N}\left({\boldsymbol {x}},u\right){\frac {\partial u}{\partial x_{N}}}=b\left({\boldsymbol {x}},u\right)} 1
其中 u ( x ) = u ( x 1 , x 2 , … , x N ) {\displaystyle u\left({\boldsymbol {x}}\right)=u\left(x_{1,}x_{2},\ldots ,x_{N}\right)} 。
取某变量 s {\displaystyle s} ,令 u ( x ) {\displaystyle u({\boldsymbol {x}})} 对 s {\displaystyle s} 求导數,可得
d u d s = ( ∂ x 1 ∂ s ) ∂ u ∂ x 1 + ( ∂ x 2 ∂ s ) ∂ u ∂ x 2 + … + ( ∂ x N ∂ s ) ∂ u ∂ x N {\displaystyle {\frac {{\text{d}}u}{{\text{d}}s}}=\left({\frac {\partial x_{1}}{\partial s}}\right){\frac {\partial u}{\partial x_{1}}}+\left({\frac {\partial x_{2}}{\partial s}}\right){\frac {\partial u}{\partial x_{2}}}+\ldots +\left({\frac {\partial x_{N}}{\partial s}}\right){\frac {\partial u}{\partial x_{N}}}} 2
若定义 ∂ x k ∂ s = a k ( x , u ) {\displaystyle {\frac {\partial x_{k}}{\partial s}}=a_{k}\left({\boldsymbol {x}},u\right)} ,可知
d u d s = a 1 ( x , u ) ∂ u ∂ x 1 + a 2 ( x , u ) ∂ u ∂ x 2 + … + a N ( x , u ) ∂ u ∂ x N = b ( x , u ) {\displaystyle {\frac {{\text{d}}u}{{\text{d}}s}}=a_{1}\left({\boldsymbol {x}},u\right){\frac {\partial u}{\partial x_{1}}}+a_{2}\left({\boldsymbol {x}},u\right){\frac {\partial u}{\partial x_{2}}}+\ldots +a_{N}\left({\boldsymbol {x}},u\right){\frac {\partial u}{\partial x_{N}}}=b\left({\boldsymbol {x}},u\right)} 3
即,求解的偏微分方程(1 )的过程变作对联立的常微分方程组作积分
{ ∂ x k ∂ s = a k ( x , u ) d u d s = b ( x , u ) {\displaystyle \left\{{\begin{array}{rcl}{\dfrac {\partial x_{k}}{\partial s}}&=&a_{k}\left({\boldsymbol {x}},u\right)\\[1em]{\dfrac {{\text{d}}u}{{\text{d}}s}}&=&b\left({\boldsymbol {x}},u\right)\end{array}}\right.} 4
积分过程需要给定初始条件。一般初始条件给定的形式为 x {\displaystyle x} 空间中的流形
g ( x , u ) = 0 {\displaystyle g\left({\boldsymbol {x}},u\right)=0} 5
将此曲面对应为 s = 0 {\displaystyle s=0} 。
设想 x {\displaystyle {\boldsymbol {x}}} 和 u {\displaystyle u} 依赖于变量 { s , t 1 , t 2 , . . . , t N − 1 } {\displaystyle \{s,t_{1},t_{2},...,t_{N-1}\}} ,则 { t 1 , t 2 , . . . , t N − 1 } {\displaystyle \{t_{1},t_{2},...,t_{N-1}\}} 可作方程(5 )中的初始值,即
x 1 ( s = 0 ) = h 1 ( t 1 , t 2 , … , t N − 1 ) x 2 ( s = 0 ) = h 2 ( t 1 , t 2 , … , t N − 1 ) ⋮ u ( s = 0 ) = v ( t 1 , t 2 , … , t N − 1 ) {\displaystyle {\begin{array}{rcl}x_{1}\left(s=0\right)&=&h_{1}\left(t_{1},t_{2},\ldots ,t_{N-1}\right)\\x_{2}\left(s=0\right)&=&h_{2}\left(t_{1},t_{2},\ldots ,t_{N-1}\right)\\\vdots \\u\left(s=0\right)&=&v\left(t_{1},t_{2},\ldots ,t_{N-1}\right)\end{array}}} 6
从方程组(4 )和初始条件(6 )确定 x {\displaystyle {\boldsymbol {x}}} 和 u {\displaystyle u} 后,可以得到解的隐式形式。如果可以解析消掉 s , t 1 , t 2 , . . . , t N − 1 {\displaystyle s,t_{1},t_{2},...,t_{N-1}} ,则可获得显式形式的解。
沿着一阶偏微分方程的特征线, 偏微分方程简化为一个常微分方程 . 沿着特征线求出对应常微分方程的解就可以得到偏微分方程的解.
为了更好地解释这一方法, 考虑具有以下形式的偏微分方程
a ( x , y , u ) ∂ u ∂ x + b ( x , y , u ) ∂ u ∂ y = c ( x , y , u ) . {\displaystyle a(x,y,u){\frac {\partial u}{\partial x}}+b(x,y,u){\frac {\partial u}{\partial y}}=c(x,y,u).} 1
假设解 u 已知, 考虑R 3 中的曲面 z = u (x ,y ). 曲面的法向量 为
( u x ( x , y ) , u y ( x , y ) , − 1 ) . {\displaystyle (u_{x}(x,y),u_{y}(x,y),-1).\,} 那么,[ 1] 方程 (1 ) 表示向量场
( a ( x , y , z ) , b ( x , y , z ) , c ( x , y , z ) ) {\displaystyle (a(x,y,z),b(x,y,z),c(x,y,z))\,} 与曲面 z = u (x ,y ) 在任意点处相切. 换句话说, 解函数的图像必定是该向量场的积分曲线 的并. 这些积分曲线被称作偏微分方程的特征线.
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