[求指教] 工程數學研究所題目?

令 u_bar = μ , ψ( x , y , t ) = X(x)Y(y)T(t)
則 ∂ψ / ∂x = X'YT , ∂²ψ / ∂x² = X''YT , ∂²ψ / ∂y² = XY''T

原式即 ( ∂/∂t + μ*∂/∂x )( X''YT + XY''T ) + βX'YT = 0
∂/∂t( X''YT + XY''T ) + μ*∂/∂x( X''YT + XY''T ) + βX'YT = 0
X''YT' + XY''T' + μX'''YT + μX'Y''T + βX'YT = 0
T' ( X''Y + XY'' ) = - T( μX'''Y + μX'Y'' + βX'Y )
T' / T = - ( μX'''Y + μX'Y'' + βX'Y ) / ( X''Y + XY'' )
等式左邊由變數 t 決定, 右邊由 x, y 決定, 故可令等式等於一常數值 λ , 即 :
T' / T = - ( μX'''Y + μX'Y'' + βX'Y ) / ( X''Y + XY'' ) = λ

T' - λT = 0
T = c1*e^(λt)

μX'''Y + μX'Y'' + βX'Y = - λ( X''Y + XY'' )
μX'''Y + μX'Y'' + βX'Y + λX''Y + λXY'' = 0
Y( μX''' + βX' + λX'' ) = - Y'' ( μX' + λX )
Y'' / Y = - ( μX''' + βX' + λX'' ) / ( μX' + λX )
令 Y'' / Y = - ( μX''' + βX' + λX'' ) / ( μX' + λX ) = - k²

Y'' + k²Y = 0
Y = c2*cos ky + c3*sin ky

μX''' + βX' + λX'' = k²( μX' + λX )
μX''' + βX' + λX'' - μk²X' - λk²X = 0
μX''' + λX'' + ( β - μk² )X' - λk²X = 0 , 此為常係數三階齊次ODE
其特徵方程式為 :
μm³ + λm² + ( β - μk² )m - λk² = 0

ψ( x , y , t )
= X(x)Y(y)T(t)
= X(x) * ( c2*cos ky + c3*sin ky ) * c1*e^(λt)
= c1 * X(x) * ( c2*cos ky + c3*sin ky ) * e^(λt)

Ans :
ψ( x , y , t ) = c1 * X(x) * ( c2*cos ky + c3*sin ky ) * e^(λt)
其中 X(x) 為常係數三階齊次ODE之解,
其特徵方程式為 μm³ + λm² + ( β - μk² )m - λk² = 0

係數部份 :
μ = u_bar
λ , k , c1 , c2 , c3 為常數, 需另提供初始值或邊界值方能決定

請問工數高手 : u_bar是指....?
謝謝你為我解惑了工數最後一題~~~ ^__^b
我做了考古題5年 > <