Euler’s Theorem for Homogeneous Functions

Euler’s theorem for homogeneous functions states that for a homogeneous function of degree n, the sum of each variable multiplied by its partial derivative equals n times the function itself.

Homogeneous Function

Definition: A function f(x,y) is said to be homogeneous of degree n if

f(tx,ty) = tⁿ f(x,y)

for all real numbers t.

Euler’s Theorem

Statement: Let f(x,y) be a homogeneous function of degree n in variables x and y. Then we have that

$\boxed{x \dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}=nf}$.

Questions Answers

$\boxed{\color{blue}\textbf{Q 1}:}$ Verify Euler’s theorem for f(x,y) = (ax+by)^1/3.

Answer:

We have

f(tx,ty) = (atx+bty)^1/3

⇒ f(tx,ty) = t^1/3(ax+by)^1/3.

⇒ f(tx,ty) = t^1/3 f(x,y).

This shows that f(x,y) is a homogeneous function of degree 3. So according to Euler’s theorem, we need to verify that $x \dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}=\dfrac{f}{3}$.

Now,

$\dfrac{\partial f}{\partial x}=\dfrac{1}{3}$ (ax+by)^-2/3 ⋅ a

$\dfrac{\partial f}{\partial y}=\dfrac{1}{3}$ (ax+by)^-2/3 ⋅ b

Therefore,

$x \dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}$ $=\dfrac{1}{3}$ (ax+by)^-2/3 ax + $\dfrac{1}{3}$ (ax+by)^-2/3 by

= $\dfrac{1}{3}$ (ax+by)^-2/3 (ax + by)

= $\dfrac{1}{3}$ (ax+by)^1/3

= f/3.

So, Euler’s theorem is verified.

$\boxed{\color{blue}\textbf{Q 2}:}$ If $u=\log\dfrac{x^2+y^2}{x+y}$, then prove that

xu_x+yu_y = 1.

Answer:

Write f=e^u.

So, f(x,y) = $\dfrac{x^2+y^2}{x+y}$.

Now, f(tx,ty) = $\dfrac{t^2x^2+t^2y^2}{tx+ty}$ = $t \dfrac{x^2+y^2}{x+y}$ = t¹ f(x,y).

∴ f(x,y) is a homogeneous function of degree 1.

So by Euler’s theorem applying to f, we have that

$xf_x+yf_y=f$ as the degree n=1.

Now,

$f_x=\dfrac{\partial}{\partial x}(e^u)=e^u u_x$ and $f_x=\dfrac{\partial}{\partial y}(e^u)=e^u u_y$.

Thus,

$xf_x+yf_y=f$ ⇒ $xe^u u_x+y e^u u_y=f=e^u$ ⇒ $xu_x+y u_y=1$ as eu ≠ 0.

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