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分离变数法||数理方程

2021-05-08 10:19 作者:湮灭的末影狐  | 我要投稿

//上一章还有一点点内容需要收尾。

//这一篇基本上是关于分离变数法这一章的内容。

//前面有一次专栏做到最后他提示我图片超过一百张投不了,我这才了解到那些公式都是以特殊图片形式插入的。因此这一次为了尽量避免这一问题有一些简短公式就不专门用公式格式打了,略有不规范请谅解...

我们考虑以下问题,并以此为例介绍一些微分方程解法。

考虑长度为l的弦上的波动方程

u_%7Btt%7D-a%5E2%20u_%7Bxx%7D%3D0

有两端固定边界条件u(0%2Ct)%3Du(l%2Ct)%3D0

和初始条件u(x%2C0)%3D%5Cphi(x)%2C%20u_t(x%2C0)%3D%5Cpsi(x).

I 达朗贝尔公式

我们在上一章中对于相关方程的分析提示我们,可以考虑变量代换:

%5Cleft%5C%7B%0A%5Cbegin%7Barray%7D%7Bc%7D%0A%5Cxi%3Dx%2Bat%5C%5C%0A%5Ceta%3Dx-at%0A%5Cend%7Barray%7D%5Cright.

并且我们也证明了微分方程有通解u%3Df(x%2Bat)%2Bg(x-at)

这里我们直接代入初始条件得到:

%5Cleft%20%5C%7B%5Cbegin%7Balign%7D%0Af(x)%2Bg(x)%3D%5Cphi%20(x)%5C%5C%0Aaf'(x)-bg'(x)%3D%5Cpsi%20(x)%0A%5Cend%7Balign%7D%5Cright%20.

后一式积分,结合前式即可证明:

u%3D%5Cfrac12%5B%5Cphi(x%2Bat)%2B%5Cphi(x-at)%5D%2B%5Cfrac1%7B2a%7D%5Cint_%7Bx-at%7D%5E%7Bx%2Bat%7D%5Cpsi(x)%5Cmathrm%20d%20x

上式被称为达朗贝尔公式。

II 分离变数法

事实上,我们知道固定端点会引起波的反射,这让我们联想到驻波。猜驻波解:

u(x%2Ct)%3DX(x)T(t)

代入微分方程得

XT''-a%5E2X''T%3D0

%5CRightarrow%20%5Cfrac%7BT%5E%7B%5Cprime%20%5Cprime%7D%7D%7Ba%5E%7B2%7D%20T%7D%3D%5Cfrac%7BX%5E%7B%5Cprime%20%5Cprime%7D%7D%7BX%7D%3D-%5Clambda

一边关于时间,一边关于空间,总是相等,因此必为常数%5Clambda.

%5CRightarrow%20X''%2B%5Clambda%20X%3D0

又有边界条件:X(0)%3DX(l)%3D0

因为任何指数发散的解都是无意义的,这里只能%5Clambda%3E0,得三角函数解:

X_k(x)%3DC_k%5Csin%20%5Cfrac%7Bk%5Cpi%20x%7D%7Bl%7D%20

%5Clambda%3D%5Cfrac%7Bk%5E2%5Cpi%5E2%7D%7Bl%5E2%7D,这种特定取值称为本征值,以上对X的求解为本征值问题。

同理不难解得

T(t)%3DA%20%5Ccos%20%5Cfrac%7Bk%20%5Cpi%20a%20t%7D%7Bl%7D%2BB%20%5Csin%20%5Cfrac%7Bk%20%5Cpi%20a%20t%7D%7Bl%7D

u_k(x%2Ct)%3D(A_k%5Ccos%20%5Cfrac%7Bk%5Cpi%20at%7D%7Bl%7D%20%2BB_k%5Csin%20%5Cfrac%7Bk%5Cpi%20at%7D%7Bl%7D)%5Csin%5Cfrac%7Bk%5Cpi%20x%7D%7Bl%7D%20

以上就是所有可能的驻波,每一个k值均对应一个一种驻波,或称本征振动。k=1称为基波,而其他为k次谐波。

分离变数法原理

考虑到微分方程是线性的,各个本征值对应的解可以叠加起来:

u(x%2Ct)%3D%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%20%7D%20(A_k%5Ccos%20%5Cfrac%7Bk%5Cpi%20at%7D%7Bl%7D%20%2BB_k%5Csin%20%5Cfrac%7Bk%5Cpi%20at%7D%7Bl%7D)%5Csin%5Cfrac%7Bk%5Cpi%20x%7D%7Bl%7D%20%20

这个形式让我们想到傅里叶级数。事实上,将初值条件:

u(x%2C0)%3D%5Cphi(x)%2Cu_t(x%2C0)%3D%5Cpsi(x)

代入,即发现它们分别展开为傅里叶正弦级数,即可求得A_k%2CB_k.

%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D%0AA_%7Bk%7D%3D%5Cfrac%7B2%7D%7Bl%7D%20%5Cint_%7B0%7D%5E%7Bl%7D%20%5Cphi(%5Cxi)%20%5Csin%20%5Cfrac%7Bk%20%5Cpi%20%5Cxi%7D%7Bl%7D%20%5Cmathrm%7B~d%7D%20%5Cxi%20%5C%5C%0AB_%7Bk%7D%3D%5Cfrac%7B2%7D%7Bk%20%5Cpi%20a%7D%20%5Cint_%7B0%7D%5E%7Bl%7D%20%5Cpsi(%5Cxi)%20%5Csin%20%5Cfrac%7Bk%20%5Cpi%20%5Cxi%7D%7Bl%7D%20%5Cmathrm%7B~d%7D%20%5Cxi%0A%5Cend%7Barray%7D%5Cright.

III 极坐标系的处理

匀强电场中,放置垂直导体圆柱。自由空间的静电场满足拉普拉斯方程:

%5Cnabla%5E2%5Cphi%3D0%5C%3B%20(r%3ER)

边界条件:(假设匀强电场沿x轴)

%5Cphi(R%2C%5Ctheta)%3D0%2C%20%5C%3B%5Clim_%7Br%5Crightarrow%20%5Cinfty%7D%5Cphi%5Csim%20-E_0r%5Ccos%5Cphi

我们仍然分类变量%5Cphi%3DR(r)%5CPhi(%5Ctheta%20%20)

极坐标中的拉普拉斯方程为%5Cfrac1r%5Cfrac%7B%5Cpartial%20%7D%7B%5Cpartial%20r%7D%20(r%5Cfrac%7B%5Cpartial%20%5Cphi%20%7D%7B%5Cpartial%20r%7D)%2B%5Cfrac1%7Br%5E2%7D%5Cfrac%7B%5Cpartial%5E2%20%5Cphi%20%7D%7B%5Cpartial%20%5Ctheta%20%5E2%7D%3D0

%5CRightarrow%20%5Cfrac%7Br%7D%7BR%7D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20r%7D(rR')%20%20%3D-%5Cfrac%7B%5CPhi''%7D%7B%5CPhi%7D%3D%5Clambda

%5CPhi''%2B%5Clambda%5CPhi%3D0

我们注意到电势必然是单值函数,这意味着存在自然边界条件:%5CPhi(%5Ctheta%2B2%5Cpi)%3D%5CPhi(%5Ctheta)

以上构成了本征值问题,从而

%5CPhi(%5Ctheta)%3DA_m%5Ccos%20m%5Ctheta%2BB_m%5Csin%20m%5Ctheta%2C%5C%3B%5Clambda%3Dm%5E2%2C%5C%3Bm%3D0%2C1%2C2%2C...

把本征值m²代入关于R的常微方程

r%5E2R''%2BrR'-m%5E2R%3D0

这是欧拉型常微方程,令r%3De%5Et可得到%5Cfrac%7B%5Cmathrm%7Bd%7D%5E2R%20%7D%7B%5Cmathrm%7Bd%7D%20t%5E2%7D-m%5E2R%3D0

%5CRightarrow%20...%5CRightarrow%20R(r)%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bc%7D%0AC%2BD%5Cln%20r%20%2C%20%5C%3Bm%3D0%5C%5C%0AC%20r%5Em%2BDr%5E%7B-m%7D%2C%5C%3Bm%3D1%2C2%2C...%0A%5Cend%7Barray%7D%5Cright.

对结果进行整理,我们得到极坐标下的静电场通解:

%5Cphi(r%2C%5Ctheta)%3DC_0%2BD_0%5Cln%20r%20%2B%5Csum_%7Bm%3D1%7D%5E%5Cinfty%20%5B%5Crho%5Em(A_m%5Ccos%20m%5Ctheta%2BB_m%5Csin%20m%5Ctheta)%2B%5Crho%5E%7B-m%7D(C_m%5Ccos%20m%5Ctheta%2BD_m%5Csin%20m%5Ctheta)%5D

非齐次方程的处理

例如,初始条件不变,范定方程改为

u_%7Bt%20t%7D-a%5E%7B2%7D%20u_%7Bx%20x%7D%3DA%20%5Ccos%20%5Cfrac%7B%5Cpi%20x%7D%7Bl%7D%20%5Csin%20%5Comega%20t

而边界条件改为u_x%7C_%7Bx%3D0%7D%3Du_x%7C_%7Bx%3Dl%7D%3D0 (即将u改为对x的偏导)

我们考虑将解展开成傅里叶级数,在当前边界条件下应考虑余弦级数:

u(x%2Ct)%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty%20T_k(t)%5Ccos%7B%5Cfrac%7Bk%5Cpi%20x%20%7D%7Bl%7D%7D

代入范定方程,会得到

%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7D%5Cleft%5BT_%7Bk%7D%5E%7B%5Cprime%20%5Cprime%7D%2B%5Cfrac%7Bk%5E%7B2%7D%20%5Cpi%5E%7B2%7D%20a%5E%7B2%7D%7D%7Bl%5E%7B2%7D%7D%20T_%7Bk%7D%5Cright%5D%20%5Ccos%20%5Cfrac%7Bk%20%5Cpi%20x%7D%7Bl%7D%3DA%20%5Ccos%20%5Cfrac%7B%5Cpi%20x%7D%7Bl%7D%20%5Csin%20%5Comega%20t

而等式右边恰好也是x的傅里叶余弦级数。比较系数得

T_k''%2B%5Cfrac%7Bk%5E2%5Cpi%5E2a%5E2%7D%7Bl%5E2%7DT_k%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D%0A0%2Ck%5Cneq1%5C%5C%0AA%5Csin%5Comega%20t%2C%20k%3D1%0A%5Cend%7Barray%7D%5Cright.

可以看到,非齐次项仅影响了T1.

我们再将初始条件也展开为傅里叶级数,就可以解得最终解了。

...先更这么多吧...最近太忙了...


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