固體物理中擴(kuò)散方程兩種邊界條件的求解

2023-08-19 10:39 作者:回憶早已抹去6 0人讀過 | 我要投稿

固體物理中一維擴(kuò)散方程的泛定方程為

$%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20t%7D%3DD%5Cfrac%7B%5Cpartial%5E2%20C%7D%7B%5Cpartial%20x%5E2%7D%2C$

第一類邊界條件為在單位面積上有 $Q$ 個粒子欲向晶體內(nèi)部單方向擴(kuò)散，即

$%5Cbegin%7Balign%7Dt%26%3D0%2Cx%3D0%2CC_0%3DQ%2C%5C%5Ct%26%3D0%2Cx%3E0%2CC(x)%3D0%3B%5C%5C%5Cend%7Balign%7D$

而且時間足夠長時，晶體內(nèi)部的擴(kuò)散粒子總數(shù)為 $Q$ ，滿足這些條件的擴(kuò)散方程的解為

$C(x%2Ct)%3D%5Cfrac%7BQ%7D%7B%5Csqrt%7B%5Cpi%20Dt%7D%7D%5Cmathrm%7Be%7D%5E%7B-x%5E2%2F4Dt%7D%2C$

第二類邊界條件為擴(kuò)散粒子在晶體表面維持一個不變濃度 $C_0$ ，即

$%5Cbegin%7Balign%7D%20%26t%5Cgeq0%2Cx%3D0%2CC%3DC_0%3B%5C%5C%26t%3D0%2Cx%3E0%2CC%3D0.%20%20%20%20%5Cend%7Balign%7D$

在此條件下擴(kuò)散方程的解為

$C(x%2Ct)%3DC_0%5B1-%5Cfrac%7B2%7D%7B%5Csqrt%5Cpi%7D%5Cint_%7B0%7D%5E%7Bx%2F2%7D%5Cmathrm%7Be%7D%5E%7B-%5Cxi%5E2%7D%5Cmathrm%7Bd%7D%5Cxi%5D.$

在許多固體物理教材中都給出了以上結(jié)果，但它們都沒有給出具體的證明過程，本文給出在這兩個條件下對擴(kuò)散方程的求解過程.

第一類邊界條件的求解

由泛定方程和邊界條件可得到定解問題：

$%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20t%7D%3DD%5Cfrac%7B%5Cpartial%5E2%20C%7D%7B%5Cpartial%20x%5E2%7D%2C%20%5C%5CC%7C_%7Bx%3D0%2Ct%3D0%7D%3DQ%2C%20%20%5C%5CC%7C_%7Bt%3D0%7D%3D0%2C(x%3E0)%20%20%5Cend%7Bmatrix%7D%5Cright.$ ?

這是半無界問題，最方便的方法是傅里葉積分變換法. $F%5BC(x%2Ct)%5D%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7DC(x%2Ct)%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx%3D%5Ctilde%7BC%7D(k%2Ct)%2C$

將泛定方程左右兩邊同乘 $%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dkx%7D$ 并對 $x$ 積分

$%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20t%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx%3DD%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Cpartial%5E2%20C%7D%7B%5Cpartial%20x%5E2%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx%2C$ ?

其中

$%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20t%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx%3D%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20t%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7DC%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx%3D%5Cfrac%7B%5Cpartial%20%5Ctilde%7BC%7D%7D%7B%5Cpartial%20t%7D%2C$

$%20%5Cbegin%7Balign%7D%20%26D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Cpartial%5E2%20C%7D%7B%5Cpartial%20x%5E2%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx%5C%5C%26%3DD%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7D(%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D)%5C%5C%20%26%3DD(%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D-%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%5Cmathrm%7Bd%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D)%5C%5C%20%26%3DD(%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%2B%5Cmathrm%7Bi%7Dk%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx)%5C%5C%20%26%3DD(%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%2B%5Cmathrm%7Bi%7Dk%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7DC)%5C%5C%26%3DD%5B%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%2B%5Cmathrm%7Bi%7Dk(%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7DC%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D-%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7DC%5Cmathrm%7Bd%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D)%5D%5C%5C%26%3DD(%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%2B%5Cmathrm%7Bi%7Dk%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7DC%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%2B%5Cmathrm%7Bi%7Dk%5Ccdot%5Cmathrm%7Bi%7Dk%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7DC%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx)%5C%5C%20%26%3DD(%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%2B%5Cmathrm%7Bi%7Dk%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7DC%7C_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D-k%5E2%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7DC%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx)%2C%20%20%5Cend%7Balign%7D$ ?

粒子擴(kuò)散不到無窮遠(yuǎn)處，即

$%5Cbegin%7Balign%7D%26%5Clim_%7Bx%5Crightarrow%5Cpm%5Cinfty%7DC%3D0%2C(%E6%97%A0%E7%A9%B7%E8%BF%9C%E5%A4%84%E6%B5%93%E5%BA%A6%E4%B8%BA%E9%9B%B6)%5C%5C%26%5Clim_%7Bx%5Crightarrow%5Cpm%5Cinfty%7D%5Cfrac%7B%5Cpartial%20C%7D%7B%5Cpartial%20x%7D%3D0%2C(%E6%97%A0%E7%A9%B7%E8%BF%9C%E5%A4%84%E6%B5%93%E5%BA%A6%E8%BF%91%E4%BC%BC%E4%B8%BA%E9%9B%B6%EF%BC%8C%E4%B8%80%E9%98%B6%E5%AF%BC%E6%95%B0%E4%BA%A6%E8%BF%91%E4%BC%BC%E4%B8%BA%E9%9B%B6)%5Cend%7Balign%7D.$

且

$%5Ctilde%7BC%7D(k%2Ct)%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7DC(x%2Ct)%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx%2C$

$%5CRightarrow%20D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%5Cpartial%5E2C%7D%7B%5Cpartial%20x%5E2%7D%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dx%3D-Dk%5E2%5Ctilde%7BC%7D%2C$

即泛定方程在傅里葉變換下變?yōu)榱顺Ｎ⒎址匠?/p>

$%5Cfrac%7B%5Cmathrm%7Bd%7D%5Ctilde%7BC%7D%7D%7B%5Cmathrm%7Bd%7Dt%7D%2BDk%5E2%5Ctilde%7BC%7D%3D0%2C$

由邊界條件，不妨對邊界條件作偶延拓，邊界條件可表示為（ $0%5E%2B$ 貢獻(xiàn)一個 $Q$ ， $0%5E-$ 貢獻(xiàn)一個 $Q$ ,一共為 $2Q$

$C%7C_%7Bt%3D0%7D%3D2Q%5Cdelta(x)%2C$ ?

對上式作傅里葉變換

$%5Ctilde%7BC%7D%7C_%7Bt%3D0%7D%3D%5CPhi(k)%2C$

其中 $%5CPhi(k)$ 為 $2Q%5Cdelta(x)$ 的傅里葉變換，則 $%5Ctilde%7BC%7D$ 的微分方程與上式組成的定解問題的通解為

$%5Ctilde%7BC%7D(k%2Ct)%3D%5CPhi(k)%5Cmathrm%7Be%7D%5E%7B-k%5E2Dt%7D%2C$

再進(jìn)行傅里葉逆變換

$%5Cbegin%7Balign%7D%20%20C(x%2Ct)%26%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5CPhi(k)%5Cmathrm%7Be%7D%5E%7B-k%5E2Dt%7D%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dk%5C%5C%20%26%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5B%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D2Q%5Cdelta(%5Cxi)%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dk%5Cxi%7D%5Cmathrm%7Bd%7D%5Cxi%5D%5Cmathrm%7Be%7D%5E%7B-k%5E2Dt%7D%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dkx%7D%5Cmathrm%7Bd%7Dk%2C%20%20%20%20%20%20%20%5Cend%7Balign%7D$ ?

交換積分次序

$C(x%2Ct)%3D%5Cfrac%7BQ%7D%7B%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cdelta(%5Cxi)%5B%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B-k%5E2Dt%7D%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dk(x-%5Cxi)%7D%5Cmathrm%7Bd%7Dk%5D%5Cmathrm%7Bd%7D%5Cxi%2C$

為了求出上述積分，我們先考察如下形式的積分

$%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B%5Calpha%5E2k%5E2%2B%5Cbeta%20k%7D%5Cmathrm%7Bd%7Dk%3D%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Cbeta%5E2%7D%7B4%5Calpha%5E2%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-%5Calpha%5E2(k-%5Cfrac%7B%5Cbeta%7D%7B2%5Calpha%5E2%7D)%5E2%7D%5Cmathrm%7Bd%7Dk%2C$

為了求解上述積分，令 $%5Cxi%3Dk-%5Cfrac%7B%5Cbeta%7D%7B2%5Calpha%5E2%7D$ ，有

$%5Cbegin%7Balign%7D%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-%5Calpha%5E2%5Cxi%5E2%7D%5Cmathrm%7Bd%7D%5Cxi%26%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-%5Calpha%5E2%5Cxi%5E2%7D%5Cmathrm%7Bd%7D(%5Calpha%20%5Cxi)%5C%5C%26%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-x%5E2%7D%5Cmathrm%7Bd%7Dx%5C%5C%26%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%5Ccdot%20I%2C%20%20%20%20%5Cend%7Balign%7D$

其中

$%5Cbegin%7Balign%7D%20I%5E2%26%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20%5Cmathrm%7Be%7D%5E%7B-x%5E2%7D%5Cmathrm%7Bd%7Dx%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20%5Cmathrm%7Be%7D%5E%7B-y%5E2%7D%5Cmathrm%7Bd%7Dy%5C%5C%26%3D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-(x%5E2%2By%5E2)%7D%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dy%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B2%5Cpi%7D%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B-r%5E2%7Dr%5Cmathrm%7Bd%7Dr%5Cmathrm%7Bd%7D%5Ctheta%5C%5C%26%3D2%5Cpi%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B-r%5E2%7Dr%5Cmathrm%7Bd%7Dr%5C%5C%26%3D%5Cpi%2C%20%5Cend%7Balign%7D$

$%5Cbegin%7Balign%7D%20%26%5CRightarrow%20I%3D%5Csqrt%7B%5Cpi%7D%2C%5C%5C%26%5CRightarrow%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cmathrm%7Be%7D%5E%7B-%5Calpha%5E2%5Cxi%5E2%7D%5Cmathrm%7Bd%7D%5Cxi%3D%5Cfrac%7B%5Csqrt%7B%5Cpi%7D%7D%7B%5Calpha%7D%2C%5C%5C%26%5CRightarrow%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-%5Calpha%5E2k%5E2%2B%5Cbeta%20k%7D%5Cmathrm%7Bd%7Dk%3D%5Cfrac%7B%5Csqrt%7B%5Cpi%7D%7D%7B%5Calpha%7D%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Cbeta%5E2%7D%7B4%5Calpha%5E2%7D%7D%2C%20%5Cend%7Balign%7D$

運(yùn)用上述公式，我們就能得到前面 $C(x%2Ct)$ 積分的結(jié)果，令 $C(x%2Ct)$ 積分中的 $%5Calpha%3D%5Csqrt%7BDt%7D%2C%5Cbeta%3D%5Cmathrm%7Bi%7D(x-%5Cxi)$ 積分化為?

$%5Cbegin%7Balign%7D%20C(x%2Ct)%26%3D%5Cfrac%7BQ%7D%7B%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cdelta(%5Cxi)%5B%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cmathrm%7Be%7D%5E%7B-k%5E2%5Calpha%5E2%2B%5Cbeta%20k%7D%5Cmathrm%7Bd%7Dk%5D%5Cmathrm%7Bd%7D%5Cxi%5C%5C%26%3D%5Cfrac%7BQ%7D%7B%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cdelta(%5Cxi)%5Cfrac%7B%5Csqrt%7B%5Cpi%7D%7D%7B%5Calpha%7D%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B%5Cbeta%5E2%7D%7B4%5Calpha%5E2%7D%7D%5Cmathrm%7Bd%7D%5Cxi%5C%5C%26%3D%5Cfrac%7BQ%7D%7B%5Cpi%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cdelta(%5Cxi)%5Csqrt%7B%5Cfrac%7B%5Cpi%7D%7BDt%7D%7D%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7B(x-%5Cxi)%5E2%7D%7B4Dt%7D%7D%5Cmathrm%7Bd%7D%5Cxi%5C%5C%26%3D%5Cfrac%7BQ%7D%7B%5Cpi%7D%5Csqrt%7B%5Cfrac%7B%5Cpi%7D%7BDt%7D%7D%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7Bx%5E2%7D%7B4Dt%7D%7D%5C%5C%26%3D%5Cfrac%7BQ%7D%7B%5Csqrt%7B%5Cpi%20Dt%7D%7D%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7Bx%5E2%7D%7B4Dt%7D%7D.%20%5Cend%7Balign%7D$