Euler's Theorem

The theorem #

If u is a homogeneous function of degree ’n’ in x,y: u(x,y)=$x^nf(y/x)$ (OR) u(tx,ty)=tⁿu(x,y) $$ x\frac{\partial u(x,y)}{\partial x}+y\frac{\partial u(x,y)}{\partial y}=n\cdot u(x,y) $$

Proof #

$$ \begin{align} u=x^nf\left(\frac{y}{x}\right) \\ \frac{\partial u}{\partial x}= nx^{n-1}f\left(\frac{y}{x}\right) + x^nf’\left(\frac{y}{x}\right)\frac{y}{-x^2}\\ \frac{\partial u}{\partial y} = x^{n-1}f’\left(\frac{y}{x}\right)\\ \\ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=n\left(x^nf\left(\frac{y}{x}\right)\right) = nu\\ =g(u) \end{align} $$

Modified Euler’s Theorem #

If u = Φ(v(x,y)) where v(x,y) is a homogeneous function of x,y of degree n: $$ x\left(\frac{\partial u}{\partial x}\right)+y\left(\frac{\partial u}{\partial y}\right)=n\cdot \frac{\Phi^{-1}(u)}{\frac{\partial \Phi^{-1}(u)}{\partial u}} $$

Proof #

$$ \begin{align} x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y} = nv\\ \Phi^{-1}(u)=v(x,y)\\ \frac{\partial v}{\partial x}=\left(\frac{\partial \Phi^{-1}(u)}{\partial u}\right)\left(\frac{\partial u}{\partial x}\right),\\ \frac{\partial v}{\partial y}=\left(\frac{\partial \Phi^{-1}(u)}{\partial u}\right)\left(\frac{\partial u}{\partial y}\right)\\ \\ \text{Substituting these in (7)}\\ x\left(\frac{\partial \Phi^{-1}(u)}{\partial u}\right)\left(\frac{\partial u}{\partial x}\right)+y\left(\frac{\partial \Phi^{-1}(u)}{\partial u}\right)\left(\frac{\partial u}{\partial y}\right) = nv = n\Phi^{-1}(u)\\ x\left(\frac{\partial u}{\partial x}\right)+y\left(\frac{\partial u}{\partial y}\right)=n\cdot \frac{\Phi^{-1}(u)}{\frac{\partial \Phi^{-1}(u)}{\partial u}}\\ =g(u) \end{align} $$

Euler’s deduction II #

$$ x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x \partial y}+y^2\frac{\partial^2u}{\partial y^2} = g(u)[\frac{\partial g(u)}{\partial u}-1]=g(u)[g’(u)-1] $$ For u being homogenous the RHS of this expression would be $nu(n-1)$