非数专题四 多元函数积分学 (4)
4.4 与重积分有关的不等式证明问题
4.9 (清华大学1985年竞赛题)
设函数
f(x)在
[0,1]上连续且单调递减,又
f(x) > 0,求证:
\frac{\displaystyle\int_{0}^{1}xf^{2}(x)dx}{\displaystyle \int_{0}^{1}xf(x)dx}\leq \frac{\displaystyle\int_{0}^{1}f^{2}(x)dx}{\displaystyle \int_{0}^{1}f(x)dx}
【解析】:由题意知,
f(x)单调递减,所以
(f(x)-f(y))(x-y) \leq 0,所以两边乘以
f(x)f(y) ,即
f(x)f(y)(f(x)-f(y))(x-y) \leq 0;展开得
xf^{2}(x)f(y)+yf^{2}(y)f(x) \leq xf(x)f^{2}(y)+yf(y)f^{2}(x)
构造二重积分
D:\{(x,y)|0 \leq x \leq 1,0 \leq y \leq 1\},对上面不等式进行二重积分,则
\begin{align*}\displaystyle \underset{D}{\iint}[xf^{2}(x)f(y)+yf^{2}(y)f(x)]dxdy&=\int_{0}^{1}xf^{2}(x)dx\int_{0}^{1}f(y)dy+\int_{0}^{1}yf^{2}(y)dy\int_{0}^{1}f(x)dx\\&=2\int_{0}^{1}xf^{2}(x)dx\int_{0}^{1}f(x)dx\end{align*}
\begin{align*}\displaystyle \underset{D}{\iint}[xf(x)f^{2}(y)+yf(y)f^{2}(x)]dxdy&=\int_{0}^{1}xf(x)dx\int_{0}^{1}f^{2}(y)dy+\int_{0}^{1}yf(y)dy\int_{0}^{1}f^{2}(x)dx\\&=2\int_{0}^{1}xf(x)dx\int_{0}^{1}f^{2}(x)dx\end{align*}
所以
\begin{align*}\displaystyle \int_{0}^{1}xf^{2}(x)dx\int_{0}^{1}f(x)dx \leq \int_{0}^{1}xf(x)dx\int_{0}^{1}f^{2}(x)dx\end{align*}
即
\frac{\displaystyle\int_{0}^{1}xf^{2}(x)dx}{\displaystyle \int_{0}^{1}xf(x)dx}\leq \frac{\displaystyle\int_{0}^{1}f^{2}(x)dx}{\displaystyle \int_{0}^{1}f(x)dx}.
4.10 (广东省1991年竞赛题)
设
D域是
x^2+y^2 \leq 1,试着证明不等式
\dfrac{61}{165}\pi \leq \displaystyle \underset{D}{\iint}\sin\sqrt{(x^2+y^2)^3}dxdy \leq \dfrac{2}{5}\pi
【解析】:先利用极坐标,有
\displaystyle \underset{D}{\iint}\sin\sqrt{(x^2+y^2)^3}dxdy =\int_{0}^{2\pi}d\theta\int_{0}^{1}\rho\sin(\rho^3)d\rho=2\pi\rho\sin(\rho^3)d\rho
当
x \geq 0时,
\sin x \leq x;又泰勒公式展开可知,
\sin x \geq x-\dfrac{1}{6}x^3,所以
\sin(\rho^3)\leq \rho^3,
\sin(\rho^3) \geq \rho^3-\dfrac{1}{6}\rho^9,记原二重积分为
I,即
\displaystyle I \leq 2\pi\int_{0}^{1}\rho^4d\rho =\dfrac{2}{5}\pi,I \geq 2\pi\int_{0}^{1}\rho(\rho^3-\frac{1}{6}\rho^9)d\rho=\frac{61}{165}\pi
故原不等式得证。
4.12 (江苏省2004年竞赛题)
已知
\Omega为
x^2+y^2+z^2 \leq 1,求证:
\displaystyle \frac{3}{2}\pi < \underset{\Omega}{\iiint}\sqrt[3]{x+2y-2z+5}dV < 3\pi【解析】:令
f=x+2y-2z+5,当
f处于
x^2+y^2+z^2 < 1的内部时,
f^{'}_{x}=1 \neq 0,
f^{'}_{y}=2 \neq 0,
f^{'}_{z}=-2\neq 0,无驻点,即
f在内部无极大值和极小值。当处于边界
x^2+y^2+z^2=1时,利用拉格朗日乘数法令
F=x+2y-2z+5\lambda(x^2+y^2+z^2-1),
\begin{matrix}&F^{'}(x)=1+2\lambda x=0,F^{'}(y)=2+2\lambda y=0\\&F^{'}(z)=-2+2\lambda z=0,F^{'}(\lambda)=x^2+y^2+z^2-1=0\end{matrix}
解得极值点为
(\dfrac{1}{3},\dfrac{2}{3},-\dfrac{2}{3})和
(-\dfrac{1}{3},-\dfrac{2}{3},\dfrac{2}{3}),带入,求得
F_{\min}=2,
F_{\max}=8,所以
\displaystyle\underset{\Omega}{\iiint}\sqrt[3]{x+2y-2z+5}dV >\sqrt[3]{2}\underset{\Omega}{\iiint}dV=\dfrac{4\sqrt[3]{2}}{3}\pi > \frac{3}{2}\pi
\displaystyle\underset{\Omega}{\iiint}\sqrt[3]{x+2y-2z+5}dV < 2\underset{\Omega}{\iiint}dV=\dfrac{8}{3}\pi < 3\pi
所以得证。
4.13 (全国大学生2014年决赛题)
设
\displaystyle I=\underset{D}{\iint} f(x,y)dxdy,其中
D:\{(x,y)|0\leq x\leq 1,0\leq y\leq 1\},函数
f(x,y)在
D上有连续的二阶偏导数。若多任意的
x,y均有
f(0,y)=f(x,0)=0,且有
\dfrac{\partial ^2 f}{\partial x\partial y} \leq A,证明
I \leq \dfrac{A}{4}【解析】:先将二重积分化为二次积分,再利用分部积分有
\begin{align*}I&=\int_{0}^{1}dx\int_{0}^{1}f(x,y)dy=\int_{0}^{1}dx\int_{0}^{1}f(x,y)d(y-1)\\&=\int_{0}^{1}dx\left((y-1)f(x,y)\bigg|_{y=0}^{y=1}-\int_{0}^{1}(y-1)f^{'}_{y}(x,y)dy\right)\\&=-\int_{0}^{1}dx\int_{0}^{1}(y-1)f^{'}_{y}(x,y)dy=-\int_{0}^{1}dy\int_{0}^{1}(y-1)f^{'}_{y}(x,y)dx\\&=-\int_{0}^{1}dy\int_{0}^{1}(y-1)f^{'}_{y}(x,y)d(x-1)\\&=-\int_{0}^{1}[(x-1)(y-1)f^{'}_{y}(x,y)\bigg|_{x=0}^{x=1}-\int_{0}^{1}(x-1)(y-1)f^{''}_{xy}(x,y)dx]\\&=\int_{0}^{1}dx\int_{0}^{1}f^{''}_{xy}(x,y)dx=\underset{D}{\iint}(x-1)(y-1)f^{''}_{xy}(x,y)dxdy\end{align*}\begin{align*}\displaystyle \underset{D}{\iint}(x-1)(y-1)f^{''}_{xy}(x,y)dxdy& \leq A\underset{D}{\iint}(x-1)(y-1)dxdy\\&=A\times \frac{1}{2}(x-1)^2\Big|_{0}^{1}\cdot\frac{1}{2}(y-1)^2\Big|_{0}^{1}=\frac{A}{4}\end{align*}故得证。
4.14 ( 广东省1991年竞赛题)
设二元函数
f(x,y)在区域
D:\{0 \leq x \leq 1,0 \leq y\leq 1\}上具有连续的四阶偏导数,并且
f(x,y)在区域
D的边界上恒为
0,且
\left|\dfrac{\partial^4 f}{\partial x^2 \partial y^2}\right| \leq 3,试着证明:
\displaystyle\left|\underset{D}{\iint}f(x,y)dxdy\right| \leq \frac{1}{48}【解析】:可以考虑一下上述的二重积分,利用分部积分法,有
\begin{align*}\displaystyle \underset{D}{\iint}xy(x-1)(y-1)\frac{\partial^4 f}{\partial x^2 \partial y^2}dxdy&=\int_{0}^{1}x(1-x)dx\int_{0}^{1}y(y-1)\frac{\partial^4 f}{\partial x^2 \partial y^2}dy \\&=\int_{0}^{1}x(1-x)\left[y(1-y)\frac{\partial^3 f}{\partial x^2 \partial y}\Bigg|_{0}^{1}+\int_{0}^{1}(2y-1)\frac{\partial^3 f}{\partial x^2 \partial y}dy\right]dx\\&=\int_{0}^{1}x(1-x)dx\int_{0}^{1}(2y-1)\frac{\partial^3 f}{\partial x^2\partial y}dy\\&=\int_{0}^{1}x(1-x)\left[(2y-1)\frac{\partial^2 f}{\partial x^2}\Bigg|_{0}^{1}-2\int_{0}^{1}\frac{\partial^2 f}{\partial x^2}dy\right]dx\\&=\int_{0}^{1}x(1-x)\left[(2y-1)\frac{\partial^2 f(x,1)}{\partial x^2}+\frac{\partial^2 f(x,0)}{\partial x^2}\right]dx-2\int_{0}^{1}\left[\int_{0}^{1}x(1-x)\frac{\partial^2 f}{\partial x^2}dx\right]dy\\&=x(1-x)\left[f^{'}_{x}(x,1)+f^{'}_{x}(x,0)\right]\bigg|_{0}^{1}+\int_{0}^{1}(2x-1)\left[f^{'}_{x}(x,1)+f_{x}^{'}(x,0)\right]dx\\&-2\int_{0}^{1}\left[x(1-x)f^{'}_{x}(x,y)\bigg|_{0}^{1}+\int_{0}^{1}(2x-1)f^{'}_{x}(x,y)dx\right]dy\\&=(2x-1)\left[f(x,1)+f(x,0)]\bigg|_{0}^{1}-2\int_{0}^{1}[f(x,1)+f(x,0)\right]dx-2\int_{0}^{1}\left[(2x-1)f(x,y)\bigg|_{0}^{1}-2\int_{0}^{1}f(x,y)dx\right]dy\\&=4\underset{D}{\iint}f(x,y)d\sigma\end{align*}
所以
\begin{align*}\displaystyle \left|\underset{D}{\iint}f(x,y)d\sigma\right|&=\frac{1}{4}\left|\underset{D}{\iint}xy(1-x)(1-y)\dfrac{\partial^4 f}{\partial x^2 \partial y^2}d\sigma\right|\\&\leq\frac{3}{4}\underset{D}{\iint}xy(1-x)(1-y)d\sigma\\&=\frac{3}{4}(\frac{x^2}{2}-\frac{x^3}{3})\bigg|_{0}^{1}\times(\frac{y^2}{2}-\frac{y^3}{3})\bigg|_{0}^{1}\\&=\frac{3}{4}\times(\frac{1}{6})^2=\frac{1}{48}\end{align*}
故得证。
作者:小熊
写作日期:8.5