题目
13.17 设函数 =f(ln sqrt ({x)^2+(y)^2}) 有二阶连续偏导数,满足 dfrac ({partial )^2u}(partial {x)^2}+dfrac ({partial )^2u}(partial {y)^2}=(({x)^2+(y)^2)}^dfrac (3{2)}, 且-|||-极限 lim _(xarrow 0)dfrac ({int )_(0)^1f(xt)dt}(x)=-1 求函数f(x)的表达式.
题目解答
答案
解析
步骤 1：计算 $\dfrac {\partial u}{\partial x}$ 和 $\dfrac {\partial u}{\partial y}$
设 $t=\ln \sqrt {{x}^{2}+{y}^{2}}$ ,则 ${x}^{2}+{y}^{2}={e}^{2t}$ , $u=f(t)$ ,可知 $\dfrac {\partial u}{\partial x}=\dfrac {df}{dt}\cdot \dfrac {\partial t}{\partial x}=f'(t)\cdot \dfrac {x}{{x}^{2}+{y}^{2}}$ ,同理 $\dfrac {\partial u}{\partial y}=f'(t)\cdot \dfrac {y}{{x}^{2}+{y}^{2}}$ .
步骤 2：计算 $\dfrac {{\partial }^{2}u}{\partial {x}^{2}}$ 和 $\dfrac {{\partial }^{2}u}{\partial {y}^{2}}$
$\dfrac {{\partial }^{2}u}{\partial {x}^{2}}=\dfrac {\partial }{\partial x}\left(f'(t)\cdot \dfrac {x}{{x}^{2}+{y}^{2}}\right)=f''(t)\cdot \dfrac {{x}^{2}}{{({x}^{2}+{y}^{2})}^{2}}+f'(t)\cdot \dfrac {{y}^{2}-{x}^{2}}{{({x}^{2}+{y}^{2})}^{2}}$ ,同理 $\dfrac {{\partial }^{2}u}{\partial {y}^{2}}=f''(t)\cdot \dfrac {{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}+f'(t)\cdot \dfrac {{x}^{2}-{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}$ .
步骤 3：利用给定条件求解 $f(t)$
由 $\dfrac {{\partial }^{2}u}{\partial {x}^{2}}+\dfrac {{\partial }^{2}u}{\partial {y}^{2}}={({x}^{2}+{y}^{2})}^{\dfrac {3}{2}}$ ,得 $f''(t)={({x}^{2}+{y}^{2})}^{\dfrac {5}{2}}={e}^{5t}$ . 积分两次得 $f(t)=\dfrac {1}{25}{e}^{5t}+{C}_{1}t+{C}_{2}$ ,即 $f(x)=\dfrac {1}{25}{e}^{5x}+{C}_{1}x+{C}_{2}$ .
步骤 4：利用极限条件求解 $C_1$ 和 $C_2$
由 $\lim _{x\rightarrow 0}\dfrac {{\int }_{0}^{1}f(xt)dt}{x}=-1$ ,得 $f(0)=0$ 和 $f'(0)=-2$ . 将其代入 $f(x)$ 表达式中,得 ${C}_{1}=-\dfrac {11}{5}$ 和 ${C}_{2}=-\dfrac {1}{25}$ .
设 $t=\ln \sqrt {{x}^{2}+{y}^{2}}$ ,则 ${x}^{2}+{y}^{2}={e}^{2t}$ , $u=f(t)$ ,可知 $\dfrac {\partial u}{\partial x}=\dfrac {df}{dt}\cdot \dfrac {\partial t}{\partial x}=f'(t)\cdot \dfrac {x}{{x}^{2}+{y}^{2}}$ ,同理 $\dfrac {\partial u}{\partial y}=f'(t)\cdot \dfrac {y}{{x}^{2}+{y}^{2}}$ .
步骤 2：计算 $\dfrac {{\partial }^{2}u}{\partial {x}^{2}}$ 和 $\dfrac {{\partial }^{2}u}{\partial {y}^{2}}$
$\dfrac {{\partial }^{2}u}{\partial {x}^{2}}=\dfrac {\partial }{\partial x}\left(f'(t)\cdot \dfrac {x}{{x}^{2}+{y}^{2}}\right)=f''(t)\cdot \dfrac {{x}^{2}}{{({x}^{2}+{y}^{2})}^{2}}+f'(t)\cdot \dfrac {{y}^{2}-{x}^{2}}{{({x}^{2}+{y}^{2})}^{2}}$ ,同理 $\dfrac {{\partial }^{2}u}{\partial {y}^{2}}=f''(t)\cdot \dfrac {{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}+f'(t)\cdot \dfrac {{x}^{2}-{y}^{2}}{{({x}^{2}+{y}^{2})}^{2}}$ .
步骤 3：利用给定条件求解 $f(t)$
由 $\dfrac {{\partial }^{2}u}{\partial {x}^{2}}+\dfrac {{\partial }^{2}u}{\partial {y}^{2}}={({x}^{2}+{y}^{2})}^{\dfrac {3}{2}}$ ,得 $f''(t)={({x}^{2}+{y}^{2})}^{\dfrac {5}{2}}={e}^{5t}$ . 积分两次得 $f(t)=\dfrac {1}{25}{e}^{5t}+{C}_{1}t+{C}_{2}$ ,即 $f(x)=\dfrac {1}{25}{e}^{5x}+{C}_{1}x+{C}_{2}$ .
步骤 4：利用极限条件求解 $C_1$ 和 $C_2$
由 $\lim _{x\rightarrow 0}\dfrac {{\int }_{0}^{1}f(xt)dt}{x}=-1$ ,得 $f(0)=0$ 和 $f'(0)=-2$ . 将其代入 $f(x)$ 表达式中,得 ${C}_{1}=-\dfrac {11}{5}$ 和 ${C}_{2}=-\dfrac {1}{25}$ .