题目

2.设u=f(x,y),x=rcosθ,y=rsinθ,证明：(∂^2u)/(∂r^2)+(1)/(r)(∂u)/(∂r)+(1)/(r^2)(∂^2u)/(∂θ^2)=(∂^2u)/(∂x^2)+(∂^2u)/(∂y^2).

2.设u=f(x,y),x=rcosθ,y=rsinθ,证明： $\frac{∂^{2}u}{∂r^{2}}+\frac{1}{r}\frac{∂u}{∂r}+\frac{1}{r^{2}}\frac{∂^{2}u}{∂θ^{2}}=\frac{∂^{2}u}{∂x^{2}}+\frac{∂^{2}u}{∂y^{2}}.$

题目解答

答案

设 $ u = f(x, y) $，其中 $ x = r \cos \theta $，$ y = r \sin \theta $。

  1. 由链式法则，得
    $\frac{\partial u}{\partial r} = u_x \cos \theta + u_y \sin \theta,$
    $\frac{\partial^2 u}{\partial r^2} = u_{xx} \cos^2 \theta + 2u_{xy} \sin \theta \cos \theta + u_{yy} \sin^2 \theta.$
  2. 同理，
    $\frac{\partial u}{\partial \theta} = -r u_x \sin \theta + r u_y \cos \theta,$
    $\frac{\partial^2 u}{\partial \theta^2} = r^2(u_{xx} \sin^2 \theta - 2u_{xy} \sin \theta \cos \theta + u_{yy} \cos^2 \theta) - r \frac{\partial u}{\partial r}.$
  3. 代入原式，化简得
    $\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = u_{xx} + u_{yy}.$
    结论：
    $\boxed{\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}}$