现在考虑一般的非齐次混合问题
{ L t u + L x u = f ( t , x ) , t > 0 , a < x < b ( α 1 u − β 1 ∂ u ∂ x ) ∣ x = a = g 1 ( t ) , ( a 2 u + β 2 ∂ u ∂ x ) ∣ x = b = g 2 ( t ) u ∣ t = 0 = φ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) \begin{cases} L_tu+L_xu=f(t,x), \quad t>0,a
边界条件出现非齐次项 g 1 ( t ) , g 2 ( t ) g_1(t),g_2(t) g1(t),g2(t),不能再用分离变量法(Fourier展开法)或冲量原理法求解,但是可以通过寻找满足非齐次边界条件的特解把边界条件齐次化。
设 v ( t , x ) v(t,x) v(t,x)满足非齐次边界条件
( a 1 v − β 1 ∂ v ∂ x ) ∣ x = a = g 1 ( t ) , ( a 2 v + β 2 ∂ v ∂ x ) ∣ x = b = g 2 ( t ) (a_1v-\beta_1\frac{\partial v}{\partial x})|_{x=a}=g_1(t),\quad (a_2v+\beta_2\frac{\partial v}{\partial x})|_{x=b}=g_2(t) (a1v−β1∂x∂v)∣x=a=g1(t),(a2v+β2∂x∂v)∣x=b=g2(t)
这里的 v ( t , x ) v(t,x) v(t,x)显然不唯一,最简单地,可取为x的线性函数
v ( t , x ) = A ( t ) x + B ( t ) v(t,x)=A(t)x+B(t) v(t,x)=A(t)x+B(t)
其中, A ( t ) , B ( t ) A(t),B(t) A(t),B(t)待定,将此 v ( t , x ) v(t,x) v(t,x)代入边界条件,整理得 A ( t ) , B ( t ) A(t),B(t) A(t),B(t)的线性方程组
{ ( α 1 a − β 1 ) A ( t ) + α 1 B ( t ) = g 1 ( t ) ( α 2 b + β 2 ) A ( t ) + α 2 B ( t ) = g 2 ( t ) \begin{cases} (\alpha_1a-\beta_1)A(t)+\alpha_1B(t)=g_1(t) \\ (\alpha_2b+\beta_2)A(t)+\alpha_2B(t)=g_2(t) \end{cases} {(α1a−β1)A(t)+α1B(t)=g1(t)(α2b+β2)A(t)+α2B(t)=g2(t)
解出 A ( t ) , B ( t ) A(t),B(t) A(t),B(t)便可得满足非齐次边界的 v ( t , x ) v(t,x) v(t,x)。若上述线性方程组无解,则可设 v ( t , x ) v(t,x) v(t,x)为x的二次函数。
找到这样的 v ( t , x ) v(t,x) v(t,x),令 u ( t , x ) = v ( t , x ) + w ( t , x ) u(t,x)=v(t,x)+w(t,x) u(t,x)=v(t,x)+w(t,x),由叠加原理知, w ( t , x ) w(t,x) w(t,x)满足非齐次方程齐次边界条件的定解问题
{ L t w + L x w = f ( t , x ) − L t v − L x v , t > 0 , a < x < b ( α 1 w − β 1 ∂ w ∂ x ) ∣ x = a = 0 , ( a 2 w + β 2 ∂ w ∂ x ) ∣ x = b = 0 w ∣ t = 0 = φ ( x ) − v ∣ t = 0 , ∂ w ∂ t ∣ t = 0 = ψ ( x ) − ∂ v ∂ t ∣ t = 0 \begin{cases} L_tw+L_xw=f(t,x)-L_tv-L_xv, \quad t>0,a
这在齐次边界条件下非齐次发展方程的混合问题已解决。