Question

Determine \( f:R->R \) twice differentiable such that for function  \( \varphi \) defined by 

\( \varphi(x,y)=f(\frac{x}{y}) \) satisfies \( \frac{\partial^2\varphi}{\partial x^2}+\frac{\partial^2\varphi}{\partial y^2}=0 \)

Answer 1

solution

Let \( u=\frac{x}{y}=> \varphi(x,y)=f(u) \)

\( (*)\frac{ \partial\varphi }{\partial x}(x,y)=f'(u)u_x=\frac{1}{y}f'(u) \)

\( (**)\frac{\partial^2\varphi}{\partial x^2}(x,y)=\frac{1}{y^2}f''(u) \)

Then , \( \frac{\partial ^2\varphi}{\partial x^2}+\frac{\partial ^2\varphi}{\partial y^2}=\frac{1}{y^2}f''(u)+\frac{2x}{y^3}f'(u)+\frac{x^2}{y^4}f''(u)=0 \)

\( <=>\frac{1}{y^2}[(1+\frac{x^2}{y^2})f''(u)+2\frac{x}{y}f'(u)]=0 \)

\( <=>(1+\frac{x^2}{y^2})f''(u)+2\frac{x}{y}f'(u)=0 \)

\( <=>(1+u^2)f''(u)+2uf'(u)=0 \)

\( <=>[(1+u^2)f'(u)]'=0 \)

\( <=>(1+u^2)f'(u)=k,k\in R \)

\( <=>\int f'(u) du=k\int \frac{1}{1+u^2}du \)

\( <=>f(u)=karctan(u)+c, \)  \( k,c\in R \)

Therefore, \( \varphi(x,y)=f(\frac{x}{y}) =karctan(u)+c, \)  \( k,c\in R \)

answer 

Therfore , \( \varphi(x,y)=f(\frac{x}{y}) =karctan(u)+c, \)  \( k,c\in R \)