§2.1达朗贝尔公式 定解问题: l1n=ax,-0 l10=0(x),-0 0=V(x),2-∞
§2.1 达朗贝尔公式 一、定解问题： ï î ï í ì = -¥ = -¥ = -¥ = = | ( ), 3 | ( ) , 2 , 1 0 0 2 u x x u x x u a u x t t t tt xx y j
二、求解: 1、分析:对于常微分方穆/(0)=0 y(0)=0 y(O)=1/3 先求通解: y(t)=At+B A.0+B=0 再用初始条件求特解: →B=0 A=1/ y()=(1/3)t
y(t) = At + B 0 1/ 3 0 0 ® = î í ì = × + = B A A B 二、求解： 1、分析：对于常微分方程 ï î ï í ì ¢ = = ¢¢ = ( ) 1/3 (0) 0 ( ) 0 y o y y t 先求通解： 再用初始条件求特解： y(t) = (1/ 3)t
启示:简化偏微分方程,先求通解,再求其特解 若将方程化为型如: 则通解简单易求,故引入变换 首先,将变为=0 22a?? 02 a=0 即(+ax)( a u=o ot ox ot Ox
启示:简化偏微分方程,先求通解,再求其特解 若将方程化为型如： uxh = 0 ( ) 0 2 2 2 2 2 = ¶ ¶ - ¶ ¶ u x a t 则通解简单易求，故引入变换 首先，将变为 ( )( ) = 0 ¶ ¶ - ¶ ¶ ¶ ¶ + ¶ ¶ u x a x t a t 即
若引入 t=l(, 使得:=+00x=2+a at ax 00t0x =A(+a n at an ax an a t dx 则方程1>化为 0
î í ì = = ( , ) ( , ) x h x h t t x x 若引入 ( ) x a t A x x t t ¶ ¶ + ¶ ¶ = ¶ ¶ ¶ ¶ + ¶ ¶ ¶ ¶ = ¶ ¶ x x x 使得： ( ) x a t A x x t t ¶ ¶ + ¶ ¶ = ¶ ¶ ¶ ¶ + ¶ ¶ ¶ ¶ = ¶ ¶ h h h 则方程 uxh = 0 化为
t Ox 为此需要:0 t x an an 故可令: x=a(2+m) E=(x+ at)/2a In=(xr-at)/2a
ïïîïïíì = ¶¶ = ¶¶ 11 hxtt ïïîïïíì = - ¶¶ = ¶¶ a x a xh 为此需要： x 故可令： îíì = - = + x hx h tx a ( ) îíì = - = + x at a x at a ( ) / 2 ( ) / 2 h 即： x
则方程<1化为: l(,n)=0 50n 2、求通解 选择 5 (x+at) [n=(x-ar 即:x=012X5+n) t=(1/2a)(-m) 0vdn=C1(2)
则方程化为： ( , ) 0 2 = ¶ ¶ ¶ x h x h u 2、求通解： 选择 î í ì = - = + ( ) ( ) x at x at h x î í ì = - = + (1/ 2 )( ) (1/ 2)( ) x h x h t a x 即： ò = ¶ ¶ = ¶ ¶ \ 0 ( ) 1 h x h h x x d C u u
a5=c1(5) 0l=C1(5)+f2(n) 即(2,n)=f(2)+(m) 故通解为 u(x,y)=f(x+ at)+f2(x-at
ò = + ¶ ¶ = ¶ ¶ ( ) ( ) ( ) 1 x 1 x 2 h x x x d C f u C u 即 u(x ,h) = f 1 (x ) + f 2 (h) 故通解为： ( , ) ( ) ( ) u x y = f 1 x + at + f 2 x - at
3、用初始条件定特解: 由方程可得: f1(x)+f2(x)=(x) 由方程〈3>可得: dfi(+at)d(x+at d(x+ at) dt df,(x-at d(x-at) y(x) d(x-at) dt t=0
3、用初始条件定特解： 由方程可得： f 1 (x) + f 2 (x) = j (x) 由方程可得： ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 0 1 x dt d x at d x at df x at dt d x at d x at df x at t t = y - - - + + + + = =
0f1(x)-af2(x)=V(x) 即 f(x)-(x)= V()dc+C 5式加上<6式再除2,可得: f(x)=70(m、r y(ada 2a
( ) ( ) ( ) 1 2 af x af x = y x ¢ - ¢ 即 + - = ò ( ) 6 1 ( ) ( ) 0 1 2 x x d C a f x f x y a a 式加上式再除2，可得： 2 ( ) 2 1 ( ) 2 1 ( ) 0 1 C d a f x x x x = + + ò j y a a
5>式减去<6》式再除2,可得: f2(x)=q(x) La x y(ada 2 f1(x+a)=(x+a) x+at y(a)da+ f,(x-at)=o(x-at) 2a y(ada+
式减去式再除2，可得： 2 ( ) 2 1 ( ) 2 1 ( ) 0 2 C d a f x x x x = - - ò j y a a 2 ( ) 2 1 ( ) 2 1 ( ) 0 1 C d a f x at x at x at x + + \ + = + ò + y a a j 2 ( ) 2 1 ( ) 2 1 ( ) 0 2 C d a f x at x at x at x + + - = - ò - y a a j
