Answer to Question #199529 in Differential Equations for Raj kumar

Question

Let x=e^r cosθ, y=e^r sinθ and f be a continuously differentible function of x and y, whose partial derivatives are also continuously differentible. show that ∂²f/∂r² + ∂²f/∂θ² = (x²+y²)(∂²f/∂x² + ∂²f/∂y²)

Answer 1

$\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial f}{\partial x}e^r cosθ+\frac{\partial f}{\partial y}e^r sinθ$

$\frac{\partial^2 f}{\partial r^2}=\frac{\partial }{\partial r}(\frac{\partial f}{\partial x}e^r cosθ)+\frac{\partial }{\partial r}(\frac{\partial f}{\partial y}e^r sinθ)=(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial r}+\frac{\partial^2 f}{\partial x\partial y }\frac{\partial y}{\partial r})e^r cosθ+\frac{\partial f}{\partial x}e^r cosθ+$

$+(\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial r}+\frac{\partial^2 f}{\partial x\partial y }\frac{\partial x}{\partial r})e^r sinθ+\frac{\partial f}{\partial y}e^r sinθ=$

$=(\frac{\partial^2 f}{\partial x^2}e^r cosθ+\frac{\partial^2 f}{\partial x\partial y }e^r sinθ)e^r cosθ+\frac{\partial f}{\partial x}e^r cosθ+$

$+(\frac{\partial^2 f}{\partial y^2}e^r sinθ+\frac{\partial^2 f}{\partial x\partial y }e^r cosθ)e^r sinθ+\frac{\partial f}{\partial y}e^r sinθ$

$\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}=-\frac{\partial f}{\partial x}e^r sinθ+\frac{\partial f}{\partial y}e^r cosθ$

$\frac{\partial^2 f}{\partial θ^2}=-\frac{\partial }{\partial θ}(\frac{\partial f}{\partial x}e^r sinθ)+\frac{\partial }{\partial θ}(\frac{\partial f}{\partial y}e^r cosθ)=-(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial θ}+\frac{\partial^2 f}{\partial x\partial y }\frac{\partial y}{\partial θ})e^r sinθ-\frac{\partial f}{\partial x}e^r cosθ+$

$+(\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial θ}+\frac{\partial^2 f}{\partial x\partial y }\frac{\partial x}{\partial θ})e^r cosθ-\frac{\partial f}{\partial y}e^r sinθ=$

$=-(-\frac{\partial^2 f}{\partial x^2}e^r sinθ+\frac{\partial^2 f}{\partial x\partial y }e^r cosθ)e^r sinθ-\frac{\partial f}{\partial x}e^r cosθ+$

$+(\frac{\partial^2 f}{\partial y^2}e^r cosθ-\frac{\partial^2 f}{\partial x\partial y }e^r sinθ)e^r cosθ-\frac{\partial f}{\partial y}e^r sinθ$

then:

$\frac{\partial^2 f}{\partial r^2}+\frac{\partial^2 f}{\partial θ^2}=(\frac{\partial^2 f}{\partial x^2}e^r cosθ+\frac{\partial^2 f}{\partial x\partial y }e^r sinθ)e^r cosθ+\frac{\partial f}{\partial x}e^r cosθ+$

$+(\frac{\partial^2 f}{\partial y^2}e^r sinθ+\frac{\partial^2 f}{\partial x\partial y }e^r cosθ)e^r sinθ+\frac{\partial f}{\partial y}e^r sinθ-$

$-(-\frac{\partial^2 f}{\partial x^2}e^r sinθ+\frac{\partial^2 f}{\partial x\partial y }e^r cosθ)e^r sinθ-\frac{\partial f}{\partial x}e^r cosθ+$

$+(\frac{\partial^2 f}{\partial y^2}e^r cosθ-\frac{\partial^2 f}{\partial x\partial y }e^r sinθ)e^r cosθ-\frac{\partial f}{\partial y}e^r sinθ=$

$=f_{xx}x^2+f_{xy}xy+f_{yy}y^2+f_{xy}xy+f_{xx}y^2-f_{xy}xy+f_{yy}x^2-f_{xy}xy=$

$=f_{xx}(x^2+y^2)+f_{yy}(x^2+y^2)=(x^2+y^2)(f_{xx}+f_{yy})$