mirror of
https://github.com/Didnelpsun/Math.git
synced 2026-02-06 20:14:34 +08:00
更新微分
This commit is contained in:
Binary file not shown.
@@ -56,9 +56,19 @@
|
||||
|
||||
解:这个函数是复合函数$u=u(r)$和$r=\sqrt{x^2+y^2}$而成。根据复合函数求导法则:
|
||||
|
||||
$\dfrac{\partial u}{\partial x}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{\partial r}{\partial x}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{x}{\sqrt{x^2+y^2}}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{x}{r}$。
|
||||
$\dfrac{\partial u}{\partial x}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{\partial r}{\partial x}=\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{x}{\sqrt{x^2+y^2}}=\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{x}{r}$,$\dfrac{1}{x}\cdot\dfrac{\partial u}{\partial x}=\dfrac{1}{r}\cdot\dfrac{\textrm{d}u}{\textrm{d}r}$。
|
||||
|
||||
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}\right)=\dfrac{\partial}{\partial x}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\dfrac{x}{r}\right)=\dfrac{x}{r}\cdot\dfrac{\partial}{\partial x}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{\partial}{\partial x}\left(\dfrac{x}{r}\right)=$。
|
||||
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}\right)=\dfrac{\partial}{\partial x}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{x}{r}\right)=\dfrac{x}{r}\cdot\dfrac{\partial}{\partial x}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{\partial}{\partial x}\left(\dfrac{x}{r}\right)=\dfrac{x}{r}\cdot\dfrac{\partial}{\partial r}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)\dfrac{\partial r}{\partial x}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{r-x\cdot(\partial r/\partial x)}{r^2}=\dfrac{x^2}{r^2}\cdot\dfrac{\textrm{d}^2u}{\textrm{d}r^2}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{r^2-x^2}{r^3}$。
|
||||
|
||||
$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial y}\right)=\dfrac{\partial}{\partial y}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{y}{r}\right)=\dfrac{y}{r}\cdot\dfrac{\partial}{\partial y}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{\partial}{\partial y}\left(\dfrac{y}{r}\right)=\dfrac{y}{r}\cdot\dfrac{\partial}{\partial r}\left(\dfrac{\textrm{d}u}{\textrm{d}r}\right)\dfrac{\partial r}{\partial y}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{r-y\cdot(\partial r/\partial y)}{r^2}=\dfrac{y^2}{r^2}\cdot\dfrac{\textrm{d}^2u}{\textrm{d}r^2}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{r^2-x^2}{r^3}$
|
||||
|
||||
代入不等式:$\dfrac{x^2+y^2}{r^2}\cdot\dfrac{\textrm{d}u^2}{\textrm{d}r^2}+\dfrac{\textrm{d}u}{\textrm{d}r}\cdot\dfrac{2r^2-x^2-y^2}{r^3}-\dfrac{1}{r}\cdot\dfrac{\textrm{d}u}{\textrm{d}r}+u=x^2+y^2$。
|
||||
|
||||
代入$x^2+y^2=r^2$:$\dfrac{\textrm{d}^2u}{\textrm{d}r^2}+u=r^2$,为二阶线性常系数微分方程。
|
||||
|
||||
通解为$u=C_1\cos r+C_2\sin r+r^2-2$。
|
||||
|
||||
即$u(\sqrt{x^2+y^2})=C_1\cos\sqrt{x^2+y^2}+C_2\sin\sqrt{x^2+y^2}+x^2+y^2-2$。
|
||||
|
||||
\subsubsection{特殊值反代}
|
||||
|
||||
|
||||
Reference in New Issue
Block a user