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[朱长江阮立志偏微分方程简明教程第2版] 7.1调和函数习题参考解答

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发表于 2023-2-11 07:53:53 | 显示全部楼层 |阅读模式
# 调和函数习题参考解答 --- 1、 对二维的 Laplace 算子, 在极坐标变换 \begin\{aligned\} x=x\_0+r\cos\theta, y=y\_0+r\sin\theta \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 变换下, 验证 \begin\{aligned\} \Delta u=\frac\{\partial^2 u\}\{\partial r^2\}+\frac\{1\}\{r\}\frac\{\partial u\}\{\partial r\} +\frac\{1\}\{r^2\}\frac\{\partial^2u\}\{\partial \theta^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题设, \begin\{aligned\} \frac\{\partial u\}\{\partial r\}=\frac\{\partial u\}\{\partial x\}\cos\theta+\frac\{\partial u\}\{\partial y\}\sin \theta, \frac\{\partial u\}\{\partial \theta\}=-\frac\{\partial u\}\{\partial x\}r\sin\theta+\frac\{\partial u\}\{\partial y\}r\cos\theta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 解得 \begin\{aligned\} \frac\{\partial u\}\{\partial x\}=\frac\{\partial u\}\{\partial r\}\cos\theta -\frac\{\partial u\}\{\partial\theta\}\frac\{1\}\{r\}\sin\theta, \frac\{\partial u\}\{\partial y\}=\frac\{\partial u\}\{\partial r\}\sin\theta+\frac\{\partial u\}\{\partial\theta\}\frac\{1\}\{r\}\cos\theta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 换言之, 我们有 \begin\{aligned\} \frac\{\partial\}\{\partial x\}=\cos\theta\frac\{\partial\}\{\partial r\}-\frac\{1\}\{r\}\sin\theta\frac\{\partial\}\{\partial\theta\}, \frac\{\partial\}\{\partial y\}=\sin\theta\frac\{\partial\}\{\partial r\}+\frac\{1\}\{r\}\cos\theta\frac\{\partial\}\{\partial\theta\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 \begin\{aligned\} &\frac\{\partial^2u\}\{\partial x^2\} =\frac\{\partial\}\{\partial x\}\left(\frac\{\partial u\}\{\partial x\}\right) = \cos\theta\frac\{\partial\}\{\partial r\}-\frac\{1\}\{r\}\sin\theta\frac\{\partial\}\{\partial\theta\}\left(\frac\{\partial u\}\{\partial x\}\right)\\\\ =&\cos\theta \frac\{\partial\}\{\partial r\}\left(\frac\{\partial u\}\{\partial x\}\right) -\frac\{1\}\{r\}\sin \theta\frac\{\partial\}\{\partial \theta\}\left(\frac\{\partial u\}\{\partial x\}\right)\\\\ =&\cos\theta\frac\{\partial\}\{\partial r\}\left( \frac\{\partial u\}\{\partial y\}r\cos\theta-\frac\{\partial u\}\{\partial x\}r\sin\theta\right) -\frac\{1\}\{r\}\sin\theta\frac\{\partial\}\{\partial \theta\} \left( \frac\{\partial u\}\{\partial y\}r\cos\theta-\frac\{\partial u\}\{\partial x\}r\sin\theta\right)\\\\ =&\frac\{\partial^2u\}\{\partial r^2\}\cos^2\theta -\frac\{\partial^2u\}\{\partial r\partial\theta\} \frac\{1\}\{r\}\sin\theta\cos\theta+\frac\{\partial u\}\{\partial\theta\}\frac\{1\}\{r^2\}\sin\theta\cos\theta\\\\ &-\frac\{\partial^2u\}\{\partial\theta\partial r\}\frac\{1\}\{r\}\cos\theta\sin\theta+\frac\{\partial u\}\{\partial r\}\frac\{1\}\{r\}\sin^2\theta +\frac\{\partial^2u\}\{\partial \theta^2\}\frac\{1\}\{r^2\}\sin^2\theta+\frac\{\partial u\}\{\partial\theta\}\frac\{1\}\{r^2\}\sin\theta\cos\theta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 同理, \begin\{aligned\} \frac\{\partial^2u\}\{\partial y^2\}=&\frac\{\partial^2u\}\{\partial r^2\}\sin^2\theta +\frac\{\partial^2u\}\{\partial r\partial\theta\} \frac\{1\}\{r\}\sin\theta\cos\theta -\frac\{\partial u\}\{\partial\theta\}\frac\{1\}\{r^2\}\sin\theta\cos\theta\\\\ &+\frac\{\partial^2u\}\{\partial\theta\partial r\}\frac\{1\}\{r\}\cos\theta\sin\theta +\frac\{\partial u\}\{\partial r\}\frac\{1\}\{r\}\cos^2\theta +\frac\{\partial^2u\}\{\partial \theta^2\}\frac\{1\}\{r^2\}\cos^2\theta -\frac\{\partial u\}\{\partial\theta\}\frac\{1\}\{r^2\}\sin\theta\cos\theta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 相加即得结论. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 2、 对 $\displaystyle n\ (n\geq 3)$ 维的 Laplace 算子, 验证 (1.6) 式成立. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 在球坐标 \begin\{aligned\} x\_1-x\_1^0=&r\cos\theta\_1,\\\\ x\_2-x\_2^0=&r\sin\theta\_1\cos\theta\_2,\\\\ \cdots=&\cdots,\\\\ x\_\{n-1\}-x\_\{n-1\}^0=&r\sin\theta\_1\cdots\sin \theta\_\{n-2\}\cos \theta\_\{n-1\},\\\\ x\_n-x\_n^0=&r\sin\theta\_1\cdots \sin \theta\_\{n-1\}\sin\theta\_\{n-1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 变换下, 验证 \begin\{aligned\} \Delta\_nu=&\frac\{1\}\{r^\{n-1\}\}\frac\{\partial\}\{\partial r\}\left(r^\{n-1\}\frac\{\partial u\}\{\partial r\}\right) +\frac\{1\}\{r^2\}\left(\frac\{\partial^2u\}\{\partial \theta\_1^2\}+\frac\{(n-2)\cos\theta\_1\}\{\sin\theta\_1\}\frac\{\partial u\}\{\partial \theta\_1\}\right)\\\\ &+\frac\{1\}\{r^2\}\sum\_\{i=2\}^n \frac\{1\}\{\sin^2\theta\_1\cdots \sin^2\theta\_\{i-1\}\} \left(\frac\{\partial^2u\}\{\partial \theta\_i^2\}+\frac\{(n-1-i)\cos\theta\_i\}\{\sin\theta\_i\}\frac\{\partial u\}\{\partial \theta\_i\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 不妨设 $\displaystyle x\_i^0=0$. 对 $\displaystyle n$ 作数学归纳法. 当 $\displaystyle n=2$ 时, 由题 1 知结论成立. 设结论对 $\displaystyle n$ 成立, 则对 $\displaystyle n+1$, 令 \begin\{aligned\} x\_2^2+\cdots+x\_\{n+1\}^2=s^2, x\_1=r\cos \theta\_1, s=r\sin \theta\_1, x\_1^2+s^2=r^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由 $\displaystyle n=2$ 的情形知 \begin\{aligned\} u\_\{x\_1x\_1\}+u\_\{ss\}=u\_\{rr\}+\frac\{1\}\{r\}u\_r+\frac\{1\}\{r^2\}u\_\{\theta\_1\theta\_1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而由归纳假设知 \begin\{aligned\} &u\_\{x\_2x\_2\}+\cdots+u\_\{x\_\{n+1\}x\_\{n+1\}\}\\\\ =&\frac\{n-2\}\{s\}u\_s+u\_\{ss\} +\frac\{1\}\{s^2\}\sum\_\{i=1\}^\{n-1\}\frac\{1\}\{\sin^2\theta\_1 \cdots \sin^2\theta\_i\}\left\[ \frac\{\partial^2u\}\{\partial\theta\_\{i+1\}^2\} +\frac\{(n-1-i)\cos\theta\_\{i+1\}\}\{\sin\theta\_\{i+1\}\} \frac\{\partial u\}\{\partial\theta\_\{i+1\}\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 上两式相加得 \begin\{aligned\} \Delta\_\{n+1\}u=&u\_\{rr\}+\frac\{1\}\{r\}u\_r+\frac\{1\}\{r^2\}u\_\{\theta\_1\theta\_1\} +\frac\{n-2\}\{s\}u\_s\\\\ &+\frac\{1\}\{s^2\}\sum\_\{i=1\}^\{n-1\}\frac\{1\}\{\sin^2\theta\_1 \cdots \sin^2\theta\_i\}\left\[ \frac\{\partial^2u\}\{\partial\theta\_\{i+1\}^2\} +\frac\{(n-1-i)\cos\theta\_\{i+1\}\}\{\sin\theta\_\{i+1\}\} \frac\{\partial u\}\{\partial\theta\_\{i+1\}\}\right\].\ (I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 在题 2 中作替换 $\displaystyle x\to x\_1, y\to s, \theta\to\theta\_1$, 得 \begin\{aligned\} u\_s=\sin\theta u\_r+\frac\{\cos\theta\_1\}\{r\}u\_\{\theta\_1\} \Rightarrow \frac\{1\}\{s\}u\_s=\frac\{1\}\{r\}u\_r+\frac\{\cos\theta\_1\}\{r^2\sin\theta\_1\} u\_\{\theta\_1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 而 \begin\{aligned\} &u\_\{rr\}+\frac\{1\}\{r\}u\_r+\frac\{n-2\}\{s\}u\_s\\\\ =&u\_\{rr\}+\frac\{n-1\}\{r\}u\_r+\frac\{(n-2)\cos\theta\_1\}\{r^2\sin\theta\_1\}u\_\{\theta\_1\}\\\\ =&\frac\{1\}\{r^n\}\frac\{\partial\}\{\partial r\}\left(r^n\frac\{\partial u\}\{\partial r\}\right) +\frac\{(n-2)\cos\theta\_1\}\{r^2\sin\theta\_1\}u\_\{\theta\_1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 代入 $\displaystyle (I)$ 即知归纳步证毕. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 3、 证明下列函数是调和函数: (1)、 $\displaystyle x^3-3xy^2$ 和 $\displaystyle 3x^2y-y^3$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u=x^3-3xy^2, v=3x^2y-y^3$, 则 \begin\{aligned\} &u\_x=3x^2-3y^2, u\_y=-6xy\Rightarrow u\_\{xx\}=6x, u\_y=-6x\Rightarrow \Delta\_2u=0,\\\\ &v\_x=6xy, v\_y=3x^2-3y^2\Rightarrow v\_\{xx\}=6y, v\_\{yy\}=-6y\Rightarrow \Delta\_2v=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \sinh (my) \sin (mx)$ 和 $\displaystyle \sinh (my) \cos (mx)$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u=\sinh(my)\sin (mx)$, 则 \begin\{aligned\} &u\_x=m\sinh(my)\cos(mx), u\_y=m\cosh(my)\sin (mx)\\\\ \Rightarrow&u\_\{xx\}=-m^2\sinh(my)\sin (mx), u\_\{yy\}=m\sinh(my)\sin(mx) \Rightarrow \Delta\_2u=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 同理, $\displaystyle \Delta\_2\left\[\sinh(my)\cos(mx)\right\]=0$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle \cosh (my) \sin (mx)$ 和 $\displaystyle \cosh (my) \cos (mx)$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u=\cosh(my)\sin (mx)$, 则 \begin\{aligned\} &u\_x=n\cosh(my)\cos(mx), u\_y=n\sinh(my)\sin (mx)\\\\ \Rightarrow&u\_\{xx\}=-n^2\cosh(my)\sin(mx), u\_\{yy\}=n\cosh(my)\sin(mx)\Rightarrow \Delta\_2u=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 同理, $\displaystyle \Delta\_2\left\[\cosh(my)\cos(mx)\right\]=0$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (4)、 $\displaystyle \sinh x (\cosh x+\cos y)^\{-1\}$ 和 $\displaystyle \sin y (\cosh x+\cos y)^\{-1\}$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u=\frac\{\sinh x\}\{\cosh x+\cos y\}$, 则 \begin\{aligned\} u\_x=&\frac\{\cosh x(\cosh x+\cos y)-\sinh^2x\}\{(\cosh x+\cos y)^2\},\\\\ u\_\{xx\}=&\frac\{\boxed\{\begin\{array\}\{c\}\sinh x\cos y(\cosh x+\cos y)^2\\\\ -(1+\cosh x \cos y)\cdot 2(\cosh x+\cos y) \sinh x\end\{array\}\}\}\{(\cosh x+\cos y)^4\}\\\\ =&\frac\{\sin hx\cos y(\cosh x+\cos y)-2(1+\cosh x\cos y)\sinh x\}\{(\cosh x+\cos y)^3\}\\\\ =&\frac\{\sinh x\}\{(\cosh x+\cos y)^3\} \left\[\cos^2y-2-\cosh x\cos y\right\], \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} u\_y=&\frac\{-\sinh x(-\sin y)\}\{(\cosh x+\cos y)^2\} =\frac\{\sin hx\sin y\}\{(\cosh x+\cos y)^2\},\\\\ u\_\{yy\}=&\frac\{\sinh x\cos y(\cosh x+\cos y)^2-\sinh x\sin y\cdot(\cosh x+\cos y)(-\sin y)\}\{(\cosh x+\cos y)^4\}\\\\ =&\frac\{\sinh x\cos y(\cosh x+\cos y)+2\sinh x\sin^2y\}\{(\cosh x+\cos y)^3\}\\\\ =&\frac\{\sinh x\}\{(\cosh x+\cos y)^3\}\left\[\cos^2y+\cosh x\cos y+2\sin^2y\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle \Delta\_2u=0$. 同理, $\displaystyle \Delta\_2\left\[\sin (\cosh x+\cos y)^\{-1\} \right\]=0$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 4、 证明用极坐标表示的下列函数满足调和方程: (1)、 $\displaystyle \ln r$ 和 $\displaystyle \theta$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u=\ln r, v=\theta$, 则 \begin\{aligned\} \Delta\_2u=u\_\{rr\}+\frac\{1\}\{r\}u\_r+\frac\{1\}\{r^2\}u\_\{\theta\theta\}=0, \Delta\_2v=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle r^m\cos m\theta$ 和 $\displaystyle r^m\sin m\theta$, $\displaystyle m$ 为常数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u=r^m\cos m\theta$, 则 \begin\{aligned\} \Delta\_2u=&\frac\{1\}\{r\}(ru\_r)\_r+u\_\{\theta\theta\} =\frac\{1\}\{r\}(mr^m\cos m\theta)\_r-m^2r^m \sin m\theta=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 同理, $\displaystyle \Delta\_2\left(r^m\sin m\theta\right)=0$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (3)、 $\displaystyle r\ln r\cos\theta-r\theta\sin\theta$ 和 $\displaystyle r\ln r\sin\theta+r\theta\cos\theta$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u=r\ln r\cos \theta-r\theta\sin \theta$, 则 \begin\{aligned\} u\_r&=(\ln r\cos \theta+\cos\theta)-\theta\sin \theta, u\_\{rr\}=\frac\{1\}\{r\}\cos\theta,\\\\ u\_\theta&=-r\ln r\sin \theta-r\sin\theta-r\theta\cos\theta,\\\\ u\_\{\theta\theta\}&=-r\ln r\cos\theta-r\cos\theta-r\cos\theta+r\theta\sin\theta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 整理发现 $\displaystyle \Delta\_2u=u\_\{rr\}+\frac\{1\}\{r\}u\_r+\frac\{1\}\{r^2\}u\_\{\theta\theta\}=0$. 同理, $\displaystyle r\ln r\sin\theta+r\theta\cos\theta$ 也是调和函数. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 5、 求下列 $\displaystyle n\ (n\geq 3)$ 维方程的球对称解 $\displaystyle u(r)$, 其中 $\displaystyle r=\left(\sum\_\{i=1\}^n x\_i^2\right)^\frac\{1\}\{2\}$: (1)、 $\displaystyle \Delta\_nu=\sum\_\{i=1\}^n x\_i^2$; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 2 知 \begin\{aligned\} &\frac\{1\}\{r^\{n-1\}\}(r^\{n-1\}u\_r)\_r=r^2 \Rightarrow (r^\{n-1\}u\_r)\_r=r^\{n+1\} \Rightarrow r^\{n-1\}u\_r=\frac\{r^\{n+2\}\}\{n+2\}+C\\\\ \Rightarrow& u\_r=\frac\{r^3\}\{n+2\}+Cr^\{1-n\} \Rightarrow u=\frac\{r^4\}\{4(n+2)\}+\frac\{A\}\{r^\{n-2\}\}+B. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 $\displaystyle \Delta\_nu-u=0$. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由题 2 知 \begin\{aligned\} \frac\{1\}\{r^\{n-1\}\}(r^\{n-1\}u\_r)\_r=u \Leftrightarrow u''+\frac\{n-1\}\{r\}u'-u=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 抱歉, 算不下去的. 会涉及特殊函数. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 6、 证明三维空间中的调和函数在球心 $\displaystyle P\_0(x^0)$ 的函数值等于球体 $\displaystyle K\_R$ 上的平均值, 即 \begin\{aligned\} u(x^0)=\frac\{3\}\{4\pi R^3\}\iiint\_\{K\_R\}u\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由调和函数的平均值定理知对 $\displaystyle \forall\ 0 < r < R$, \begin\{aligned\} &4\pi r^2u(P\_0)=\iint\_\{x^2+y^2=r^2\}u(P)\mathrm\{ d\} S\\\\ =&\int\_0^\{2\pi\}\mathrm\{ d\} \theta\int\_0^\pi u(x\_0+r\sin \phi \cos\theta, y\_0+r\sin\phi \sin\theta, z\_0+r\cos\phi)r^2\sin \phi \mathrm\{ d\} \phi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 两边关于 $\displaystyle r$ 在 $\displaystyle [0,R]$ 上积分得 \begin\{aligned\} \frac\{4\pi R^3\}\{3\}u(P\_0) =&\int\_0^R \mathrm\{ d\} r \int\_0^\{2\pi\}\mathrm\{ d\} \theta\int\_0^\pi u\left(\cdots\right)r^2\sin \phi \mathrm\{ d\} \phi\\\\ =&\iiint\_\{K\_R\}u(P)\mathrm\{ d\} \varOmega. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 7、 证明 Neumann 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_nu=g(x), x\in \varOmega\subset \mathbb\{R\}^n, n\geq 3,\\\\ \left.\frac\{\partial u\}\{\partial \nu\}\right|\_\{\varGamma\}=f(y), y\in\varGamma\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 有解的必要条件是 \begin\{aligned\} \int\_\varOmega g(x)\mathrm\{ d\} \varOmega=\int\_\varGamma f(y)\mathrm\{ d\} S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / \begin\{aligned\} &\int\_\varOmega g(x)\mathrm\{ d\} \varOmega= \int\_\varOmega \Delta\_nu\mathrm\{ d\} \varOmega \xlongequal[\tiny\mbox\{定理\}]\{\tiny\mbox\{散度\}\} \int\_\{\varGamma\}\frac\{\partial u\}\{\partial \nu\}\mathrm\{ d\} S =\int\_\varGamma f(y)\mathrm\{ d\} S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 8、 证明区域 $\displaystyle \varOmega$ 内的下 (上) 调和函数 $\displaystyle u$ 满足 (1.15). [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由书第 161 页基本积分公式 \begin\{aligned\} u(P\_0)=&\frac\{1\}\{(n-2)\omega\_n\}\int\_\{\varGamma\_R\} \left\[ \frac\{1\}\{r^\{n-2\}\}\frac\{\partial u\}\{\partial \nu\} -u\frac\{\partial\}\{\partial\nu\}\left(\frac\{1\}\{r^\{n-2\}\}\right)\right\]\mathrm\{ d\} S\\\\ &-\frac\{1\}\{(n-2)\omega\_n\}\int\_\{K\_R\}\frac\{\Delta\_nu\}\{r^\{n-2\}\}\mathrm\{ d\} \varOmega. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 对下调和函数 $\displaystyle u: -\Delta u\leq 0$, 我们有 \begin\{aligned\} u(P\_0)\leq&\frac\{1\}\{(n-2)\omega\_n\}\int\_\{\varGamma\_R\} \left\[ \frac\{1\}\{r^\{n-2\}\}\frac\{\partial u\}\{\partial \nu\} -u\frac\{\partial\}\{\partial\nu\}\left(\frac\{1\}\{r^\{n-2\}\}\right)\right\]\mathrm\{ d\} S\\\\ =&\frac\{1\}\{(n-2)\omega\_nR^\{n-2\}\}\int\_\{\varGamma\_R\}\frac\{\partial u\}\{\partial \nu\}\mathrm\{ d\} S -\frac\{1\}\{(n-2)\omega\_n\} \int\_\{\varGamma\_R\} u\frac\{\partial\}\{\partial r\} (r^\{2-n\})\mathrm\{ d\} S\\\\ \xlongequal[\tiny\mbox\{定理\}]\{\tiny\mbox\{散度\}\}&\frac\{1\}\{(n-2)\omega\_nR^\{n-2\}\} \int\_\{K\_R\}\Delta\_nu\mathrm\{ d\} \varOmega +\frac\{1\}\{\omega\_n\} \int\_\{\varGamma\_R\}ur^\{1-n\}\mathrm\{ d\} S \leq \frac\{1\}\{\omega\_nR^\{n-1\}\}\int\_\{\varGamma\_R\}u\mathrm\{ d\} S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 对上调和函数, 上述 $\displaystyle \leq$ 改为 $\displaystyle \geq$ 即可. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 9、 证明下 (上) 调和函数的强最大 (小) 值原理. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 也即要证: 若非常数的 $\displaystyle u\in C^2(\varOmega)\cap C(\bar\{\varOmega\})$ 是有界区域 $\displaystyle \varOmega$ 上的下调和函数, 则 $\displaystyle u$ 在 $\displaystyle \bar\varOmega$ 上的最大值只能在边界 $\displaystyle \varGamma$ 上达到. 用反证法. 若不然, $\displaystyle u$ 在 $\displaystyle \varOmega$ 内某一点 $\displaystyle P\_0$ 取得最大值 $\displaystyle M$, 则由题 8 知对 $\displaystyle \forall\ 0 < R < \mathrm\{dist\}(P\_0,\varGamma)$, \begin\{aligned\} M=u(P\_0)\leq\frac\{1\}\{\omega\_nR^\{n-1\}\}\int\_\{\varGamma\_R\}u\mathrm\{ d\} S \leq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle \forall\ P: |PP\_0| < \mathrm\{dist\}(P\_0,\varGamma), u(P)=M$. 设 \begin\{aligned\} \varOmega'=\left\\{P\in\varOmega; u(P)=M\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 $\displaystyle P\_0\in\varOmega'\Rightarrow \varOmega'\neq \varnothing$. 由上述推理知 $\displaystyle \varOmega'$ 是开集. 又由 $\displaystyle u$ 是连续的知 $\displaystyle \varOmega'$ (相对于 $\displaystyle \varOmega$ 而言) 是闭集. 由 $\displaystyle \varOmega$ 是区域 (而是连通的) 知 $\displaystyle \varOmega'=\varOmega$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 10、 设 $\displaystyle u$ 是区域 $\displaystyle \varOmega\subset \mathbb\{R\}^n$ 内的调和函数, 证明: (1)、 若 $\displaystyle \varPhi: \mathbb\{R\}\to \mathbb\{R\}$ 是光滑的凸函数, 则函数 $\displaystyle v=\varPhi(u)$ 是下调和函数; [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 \begin\{aligned\} v\_\{x\_i\}=\varPhi'(u)u\_\{x\_i\}, v\_\{x\_ix\_i\}=\varPhi''(u)u\_\{x\_i\}^2+\varPhi'(u)u\_\{x\_ix\_i\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} \Delta\_nv=\varPhi''(u)|\nabla u|^2+\varPhi'(u)\Delta\_nu =\varPhi''(u)|\nabla u|^2\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle v$ 是下调和函数. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ (2)、 函数 $\displaystyle v=|Du|^2$ 是下调和函数. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 \begin\{aligned\} v\_\{x\_i\}=2Du\cdot Du\_\{x\_i\}, v\_\{x\_ix\_i\}=2Du\_\{x\_i\}\cdot Du\_\{x\_i\} +2Du\cdot Du\_\{x\_ix\_i\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 \begin\{aligned\} \Delta\_nv=2|D^2u|^2+2Du\cdot D\Delta\_nu =2|D^2u|^2\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle v$ 是下调和函数. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 11、 若 $\displaystyle u(x,y)$ 是二维调和函数, 证明 $\displaystyle xu(x,y)$ 是调和函数的充要条件是: $\displaystyle u(x,y)=ay+b$, 其中 $\displaystyle a,b$ 为常数. 试举例说明两个调和函数的乘积不一定是调和函数. [纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 设 $\displaystyle v=xu$, 则 \begin\{aligned\} v\_x=u+xu\_x, v\_\{xx\}=2u\_x+xu\_\{xx\}, v\_\{yy\}=xu\_\{yy\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle \Delta\_2v=2u\_x+x\Delta\_2u=2u\_x$. 故 $\displaystyle v$ 调和 \begin\{aligned\} \Leftrightarrow u\_x=0\stackrel\{\Delta\_2u=0\}\{\Leftrightarrow\}u\_x=0, u\_\{yy\}=0 \Leftrightarrow u=ay+b, a,b\in\mathbb\{R\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、 取 $\displaystyle u=x^3-3xy^2$, $\displaystyle v=x$, 则由题 3 (1) 知 $\displaystyle u$ 调和. 而显然 $\displaystyle v$ 调和. 但由第 1 步知 $\displaystyle uv$ 不是调和函数. 这就说明了调和函数的乘积未必调和. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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