數學 中的特徵線法 是求解偏微分方程 的一種方法,適用於準線性偏微分方程的求解。只要初始值不是沿著特徵線給定,即可通過特徵線法獲得偏微分方程的精確解。 其基本思想是通過把雙曲線型的準線性偏微分方程轉化為兩組常微分方程,再對常微分方程進行求解。兩組常微分方程中的一組用於定義特徵線,另一組用以描述解沿給定特徵線變化。
設所需求解的準線性偏微分方程為
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)}
。
取某變量
s
{\displaystyle s}
,令
u
(
x
)
{\displaystyle u({\boldsymbol {x}})}
對
s
{\displaystyle s}
求導數,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
若定義
∂
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 )的過程變作對聯立的常微分方程組作積分
{
∂
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
積分過程需要給定初始條件。一般初始條件給定的形式為
x
{\displaystyle x}
空間中的流形
g
(
x
,
u
)
=
0
{\displaystyle g\left({\boldsymbol {x}},u\right)=0}
5
將此曲面對應為
s
=
0
{\displaystyle s=0}
。
設想
x
{\displaystyle {\boldsymbol {x}}}
和
u
{\displaystyle u}
依賴於變量
{
s
,
t
1
,
t
2
,
.
.
.
,
t
N
−
1
}
{\displaystyle \{s,t_{1},t_{2},...,t_{N-1}\}}
,則
{
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
從方程組(4 )和初始條件(6 )確定
x
{\displaystyle {\boldsymbol {x}}}
和
u
{\displaystyle u}
後,可以得到解的隱式形式。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}}
,則可獲得顯式形式的解。
沿著一階偏微分方程的特徵線, 偏微分方程簡化為一個常微分方程 . 沿著特徵線求出對應常微分方程的解就可以得到偏微分方程的解.
為了更好地解釋這一方法, 考慮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
假設解 u 已知, 考慮R 3 中的曲面 z = u (x ,y ). 曲面的法向量 為
(
u
x
(
x
,
y
)
,
u
y
(
x
,
y
)
,
−
1
)
.
{\displaystyle (u_{x}(x,y),u_{y}(x,y),-1).\,}
那麼,[ 1] 方程 (1 ) 表示向量場
(
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))\,}
與曲面 z = u (x ,y ) 在任意點處相切. 換句話說, 解函數的圖像必定是該向量場的積分曲線 的並. 這些積分曲線被稱作偏微分方程的特徵線.
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9
t
h
{\displaystyle 9^{th}}
Revised, McGraw-Hill Higher Education, 1998