Euler’s Theorem Statement: If [katex]u=f(x,y)[/katex] is a homogeneous function of [katex]x,y[/katex] of degree [katex]n[/katex], then
[katex]x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial u}=nu[/katex].
Proof: we have
[katex]u=f(x,y)=x^n\;g\left(\frac yx\right)\;\;[/katex],Therefore
[katex]\frac{\partial u}{\partial x}=n\;x^{n-1\;}g\left(\frac yx\right)\;+\;x^{n\;}g’\left(\frac yx\right)\left(\frac{-y}{x^2}\right)[/katex].
[katex]\frac{\partial u}{\partial x}=n\;x^{n-1\;}g\left(\frac yx\right)\;-y\;x^{n-2\;}g’\left(\frac yx\right)…………………(1)[/katex].
and
[katex]\frac{\partial u}{\partial y}=\;x^{n\;}g’\left(\frac yx\right)\;\frac1x[/katex].
[katex]\frac{\partial u}{\partial y}=\;x^{n-1\;}g’\left(\frac yx\right)…………………(2)[/katex].
Thus from (1) and (2), we have
[katex]\frac{\displaystyle\partial u}{\displaystyle\partial x}+\frac{\displaystyle\partial u}{\displaystyle\partial y}=n\;x^{n-1\;}g\left(\frac{\displaystyle y}{\displaystyle x}\right)\;-y\;x^{n-2\;}g’\left(\frac{\displaystyle y}{\displaystyle x}\right)+\;x^{n-1\;}g’\left(\frac{\displaystyle y}{\displaystyle x}\right)[/katex].
[katex]\frac{\displaystyle\partial u}{\displaystyle\partial x}+\frac{\displaystyle\partial u}{\displaystyle\partial y}=n\;x^n\;g\left(\frac yx\right)[/katex].
[katex]\frac{\displaystyle\partial u}{\displaystyle\partial x}+\frac{\displaystyle\partial u}{\displaystyle\partial y}=nu[/katex].
[katex]\frac{\displaystyle\partial u}{\displaystyle\partial x}+\frac{\displaystyle\partial u}{\displaystyle\partial y}=nf(x,y)[/katex]
Hence proved.
Example:
If [katex]u=arc\tan\left(\frac{x^3+y^3}{x-y}\right)[/katex], show that
[katex]x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin 2u[/katex].
Solution:
we have
[katex]z =tan u[/katex].
[katex]\Rightarrow\;\;z(x,y)=\frac{x^3+y^3}{x-y}[/katex].
[katex]\Rightarrow\;\;z(tx,ty)=\frac{\left(tx\right)^3+\left(ty\right)^3}{tx-ty}[/katex].
[katex]\Rightarrow\;\;z(tx,ty)=\frac{t^3\left(x^3+y^3\right)}{t\left(x-y\right)}[/katex].
[katex]\Rightarrow\;\;z(tx,ty)=\frac{t^2\left(x^3+y^3\right)}{\left(x-y\right)}[/katex].
Thus [katex]z[/katex] is homogeneous function of [katex]x,y[/katex] of degree [katex]2[/katex].
Therefore, by Euler’s theorem, we have
[katex]x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2z \;\;\;……………..(1)[/katex].
But
[katex]\frac{\partial z}{\partial x}=sec^2u\;\frac{\partial u}{\partial x}\;\;\;\;\;and\;\;\;\;\frac{\partial z}{\partial y}=sec^2u\;\frac{\displaystyle\partial u}{\displaystyle\partial y}[/katex].
Substituting into (1), we get
[katex]sec^2u\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2z[/katex].
[katex]\Rightarrow \;\;\sec^2u\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2\;\tan\;u[/katex].
[katex]or\;\;\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2\;\tan\;u\;.\frac{\displaystyle1}{sec^2u}[/katex].
[katex]\Rightarrow\;\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2\;\frac{\displaystyle\sin\;u}{\cos{\displaystyle\;}{\displaystyle u}}\;.\cos^2u[/katex].
[katex]\Rightarrow\;\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=2\;\sin\;u\;.\;\cos u[/katex].
[katex]\Rightarrow\;\;\left(x\frac{\partial u}{\partial x}\;+y\;\frac{\displaystyle\partial u}{\displaystyle\partial y}\right)=\sin\;2u[/katex].
Hence Euler’s theorem verified.
To read about Euler click here.