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[zingzuize]

$\Delta u= u_{xx} + u _{yy} =0$

đặt $x= r.cos(\varphi )$

       $y= r.sin(\varphi )$

xây dựng cách chuyển $\Delta u$ sang tọa độ cực

[funcalys]

Kí hiệu

 

$u(x(r,\phi),y(r,\phi))$

 

Ta có:

 

$\frac{\partial x}{\partial r}=\cos\phi, \frac{\partial x}{\partial \phi}=-r\sin \phi$

 

$\frac{\partial y}{\partial r}=\sin \phi, \frac{\partial y}{\partial \phi}=r\cos \phi$

 

Từ đây, có công thức của:

 

$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}=\cos \phi \frac{\partial u}{\partial x}+\sin \phi \frac{\partial u}{\partial y}$

 

Ta lại có:

 

$\frac{\partial^2 u}{\partial r^2}=\cos \phi \frac{\partial }{\partial r}\frac{\partial u}{\partial x}+\sin \phi \frac{\partial }{\partial r}\frac{\partial u}{\partial y}$

 

$=\cos \phi \frac{\partial }{\partial x}\frac{\partial x}{\partial r}\frac{\partial u}{\partial x}+\cos \phi \frac{\partial }{\partial y}\frac{\partial y}{\partial r}\frac{\partial u}{\partial x}+\sin \phi \frac{\partial }{\partial x}\frac{\partial x}{\partial r}\frac{\partial u}{\partial y}+\sin \phi \frac{\partial }{\partial y}\frac{\partial y}{\partial r}\frac{\partial u}{\partial y}=\cos ^2\phi \frac{\partial^2 u}{\partial x^2}+2\cos \phi \sin \phi \frac{\partial^2 u}{\partial x\partial y}+\sin^2 \phi \frac{\partial^2 u}{\partial y^2}$

 

Ta có:

 

 

$\frac{\partial u}{\partial \phi}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \phi}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \phi}=r\cos \phi \frac{\partial u}{\partial y}-r\sin \phi \frac{\partial u}{\partial x}$

 

$\Rightarrow \frac{\partial^2 u}{\partial \phi^2}= -r \cos \phi \frac{\partial u}{\partial x}-r\sin \phi \frac{\partial^2 u}{\partial x \partial \phi}-r\sin \phi \frac{\partial u}{\partial y}+ r\cos \phi \frac{\partial^2 u}{\partial y \partial \phi}$

$= \frac{\partial^2 u}{\partial \phi^2}= -r \cos \phi \frac{\partial u}{\partial x}-r\sin \phi \frac{\partial^2 u}{\partial x \partial \phi}-r\sin \phi \frac{\partial u}{\partial y}+ r\cos \phi \frac{\partial^2 u}{\partial y \partial \phi}$

$= -r \cos \phi \frac{\partial u}{\partial x} - r\sin \phi \frac{\partial u}{\partial y} + r \cos \phi\left (r\cos \phi \frac{\partial^2 u}{\partial y^2}+ -r\sin \phi\frac{\partial^2 u}{\partial x \partial y} \right )-r \sin \phi \left ( -r \sin \phi \frac{\partial^2 u}{\partial x^2}+ r\cos \phi \frac{\partial^2 u}{\partial x \partial y} \right )$

$=r^2\left ( \sin^2 \phi \frac{\partial^2 u}{\partial x^2}-2\cos \phi \sin \phi \frac{\partial^2 u}{\partial x\partial y}+\cos^2 \phi \frac{\partial^2 u}{\partial y^2} \right )-r\left ( \cos \phi \frac{\partial u}{\partial x} +\sin \phi \frac{\partial u}{\partial y}\right )$

Chia 2 vế cho $r^2$, ta có:

$\frac{1}{r^2}\frac{\partial^2 u}{\partial x^2}+ \frac{1}{r}\frac{\partial u}{\partial r}= \sin ^2\phi \frac{\partial^2 u}{\partial x^2}-2\cos \phi \sin \phi \frac{\partial^2 u}{\partial x\partial y}+\cos^2 \phi \frac{\partial^2 u}{\partial y^2}$

Vậy:

$\Delta u= \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial \phi^2}= \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$

chính là pt Laplace cần tìm trong tọa độ cực.

Edited by Ispectorgadget, 19-07-2013 - 21:49.