Euler’s Theorem: Definition, Proof & Applications

Euler’s theorem for Homogeneous Functions is used to derive a relationship between the product of the function with its degree and partial derivatives of it.

Euler’s Theorem

In this article, we will first discuss the statement of the theorem followed by the mathematical expression of Euler’s theorem and prove the theorem. We will also discuss the things for which Euler’s Theorem is used and is applicable. A brief history of mathematician Leonhard Euler will also be discussed after whom the theorem is named so.

History of Leonhard Euler

Image:

Leonhard Euler,

• Name: Leonhard Euler

• Born: 15 April 1707

• Died: 18 September 1783

Statement of Euler’s Theorem

Euler's theorem states that if $(f$) is a homogeneous function of the degree $n$ of $k$ variables $x_{1}, x_{2}, x_{3}, \ldots \ldots, x_{k}$, then

$x_{1} \dfrac{\partial f}{\partial x_{1}}+x_{2} \dfrac{\partial f}{\partial x_{2}}+x_{3} \dfrac{\partial f}{\partial x_{3}}+\ldots \ldots+x_{k} \dfrac{\partial f}{\partial x_{k}}=n f$

Here, we will be discussing 2 variables only. So, if $f$ is a homogeneous function of degree $n$ of variables $x$ and $y$, then from Euler's Theorem, we get

$x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=n f$

Proof of Euler’s Theorem

Proof:

Let $f=u[x, y]$ be a homogenous function of degree $n$ of the variables $x, y$.

$f=u[x, y] \ldots \ldots \ldots[1]$

Now, we know that

$u[X, Y]=t^{n} u[x, y] \ldots \ldots \ldots[2]$

This is because when $u$ is a function of $X, Y$, then it becomes a function of $x, y, t$ because $X, Y$ are a function of $t$.

This means that

$X=x t^{n} Y=y t^{n}$

Now, we will consider the above equations for degree 1.

$X=x t \ldots \ldots \ldots[3] \\$

$Y=y t \ldots \ldots \ldots[4]$

where $t$ is an arbitrary parameter.

Now, we will differentiate the equation [3] partially w.r.t. $t$.

$X=x t \\$

$\Rightarrow \dfrac{\partial}{\partial t} X=\dfrac{\partial}{\partial t} x t \\$

$\Rightarrow \dfrac{\partial X}{\partial t}=x \dfrac{\partial t}{\partial t} \\$

$\Rightarrow \dfrac{\partial X}{\partial t}=x \ldots \ldots \ldots[5]$

Again, we will differentiate the equation [4] partially w.r.t. $t$.

$Y=y t \\$

$\Rightarrow \dfrac{\partial}{\partial t} Y=\dfrac{\partial}{\partial t} y t \\$

$\Rightarrow \dfrac{\partial Y}{\partial t}=y \ldots \ldots \ldots[6]$

Now, we will substitute $t=1$ in equation [3].

$X=x t \\$

$\Rightarrow X=x[1]$

Multiplying the terms, we get

$X=x \ldots \ldots \ldots \ldots \text { [7] }$

Now, we will substitute $t=1$ in equation [4].

$Y=y t \\$

$\Rightarrow Y=y[1]$

Multiplying the terms, we get

$Y=y \ldots \ldots \ldots \ldots[8]$

We will now differentiate the equation [1] partially w.r.t. $x$.

$f=u[x, y] \\$

$\Rightarrow \dfrac{\partial}{\partial x} f=\dfrac{\partial}{\partial x} u[x, y] \\$

$\Rightarrow \dfrac{\partial f}{\partial x}=\dfrac{\partial u}{\partial x} \ldots \ldots \ldots \ldots .[9]$

Partial differentiation of $y$ w.r.t. $x$ is $\underline{\underline{0} \text {. }}$

Now, we will differentiate the equation [1] partially w.r.t. $y$.

$f=u[x, y] \\$

$\Rightarrow \dfrac{\partial}{\partial y} f=\dfrac{\partial}{\partial y} u[x, y] \\$

$\Rightarrow \dfrac{\partial f}{\partial y}=\dfrac{\partial u}{\partial y} \ldots \cdots \cdots[10]$

Partial differentiation of $x$ w.r.t. $y$ is 0 .

Now, we will differentiate the equation [2] partially w.r.t. $t$.

Since $u$ is a function of $X, Y$ and $X, Y$ are a function of $x, y$, and $t$, thus apply the chain rule of partial differentiation to differentiate.

$u[X, Y]=t^{n} u[x, y] \\$

$\Rightarrow \dfrac{\partial u}{\partial x} \cdot \dfrac{\partial X}{\partial t}+\dfrac{\partial u}{\partial y} \cdot \dfrac{\partial Y}{\partial t}=n t^{n-1} u[x, y] \ldots \ldots \ldots[\text { [11] }$

We will substitute $x$ for $\dfrac{\partial x}{\partial t}, y$ for $\dfrac{\partial Y}{\partial t}, \dfrac{\partial f}{\partial x}$ for $\dfrac{\partial u}{\partial x}, \dfrac{\partial f}{\partial y}$ for $\dfrac{\partial u}{\partial y}$ in equation [11]. Thus, the equation [11] becomes:

$\Rightarrow \dfrac{\partial u}{\partial x} \cdot \dfrac{\partial X}{\partial t}+\dfrac{\partial u}{\partial y} \cdot \dfrac{\partial Y}{\partial t}=n t^{n-1} u[x, y]$

Using the result of equations [7] and [8], we can write the above equation as

$\Rightarrow \dfrac{\partial f}{\partial x} \cdot x+\dfrac{\partial f}{\partial y} \cdot y=n t^{n-1} u[x, y] \ldots \ldots \ldots \text { [12] }$

We will again substitute $t=1$ and $f$ for $u[x, y]$ in the equation [12], we get

$\Rightarrow \dfrac{\partial f}{\partial x} \cdot x+\dfrac{\partial f}{\partial y} \cdot y=n[1]^{n-1} f \\$

$\Rightarrow \dfrac{\partial f}{\partial x} \cdot x+\dfrac{\partial f}{\partial y} \cdot y=n f$

Hence proved.

Limitations of Euler's Theorem

• Euler’s Theorem cannot be applied to non-homogeneous differential equations. It is only applicable to homogeneous differential equations.

• Euler’s Theorem is very complex to understand and needs knowledge of ordinary and partial differential equations.

Application of Euler’s Theorem

• Euler’s theorem has wide application in electronic devices which work on the AC principle.

• Euler’s formula is used by scientists to perform various calculations and research.

Solved Examples

1. If $u(x, y)=\dfrac{x^{2}+y^{2}}{\sqrt{x+y}}$, prove that $x \dfrac{\partial u}{\partial x}+y \dfrac{\partial u}{\partial y}=\dfrac{3}{2} u$.

Ans:

Given $u(x, y)=\dfrac{x^{2}+y^{2}}{\sqrt{x+y}}$

We can say that

$\Rightarrow u(\lambda x, \lambda y)=\dfrac{\lambda^{2} x^{2}+\lambda^{2} y^{2}}{\sqrt{\lambda x+\lambda y}} \\$

$\Rightarrow u(\lambda x, \lambda y)=\dfrac{\lambda^{2}\left(x^{2}+y^{2}\right)}{\lambda^{1 / 2} \sqrt{x+y}} \\$

$\Rightarrow u(\lambda x, \lambda y)=\dfrac{\lambda^{3 / 2}\left(x^{2}+y^{2}\right)}{\sqrt{x+y}} u$

is a homogeneous function of degree $\dfrac{3}{2}$

By Euler's Theorem,

$x \dfrac{\partial u}{\partial x}+y \dfrac{\partial u}{\partial y}=\dfrac{3}{2} u$

2. If $v(x, y)=\log \left(\dfrac{x^{2}+y^{2}}{x+y}\right)$, prove that $x \dfrac{\partial v}{\partial x}+y \dfrac{\partial u}{\partial y}=1$

Ans:

Given that

$v(x, y)=\log \left(\dfrac{x^{2}+y^{2}}{x+y}\right)$

Change into an exponential function.

Let $e^{v}=\dfrac{x^{2}+y^{2}}{x+y}=f(x, y)$

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

$\Rightarrow f(x, y)=\dfrac{\lambda^{2}\left(x^{2}+y^{2}\right)}{\lambda(x+y)} f$ is a homogeneous function of degree 1.

By Euler's Theorem,

$x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=1 \times f$

$\Rightarrow f x \dfrac{\partial}{\partial x} e^{v}+y \dfrac{\partial}{\partial y} e^{v}=e^{v}$ exists.

$\Rightarrow x^{v} \dfrac{\partial f}{\partial x}+y e^{v} \dfrac{\partial f}{\partial y}=e^{v} \\$

$\Rightarrow x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=\dfrac{e^{v}}{e^{v}}=1$

Hence Proved.

3. If $u=\tan ^{-1}(x+y)$, then $x \dfrac{\partial u}{\partial x}+y \dfrac{\partial u}{\partial y}=$

Ans:

Given that

$\tan u=x+y \tan u$ is a homogeneous of degree $=1$

$\sum_{l} \dfrac{x d \tan u}{d x}=n \tan u \sec ^{2} u \\$

$\Rightarrow \dfrac{x d u}{d x}=\tan u \\$

$\Rightarrow \sum_{a} \dfrac{x d u}{d x}=\dfrac{1}{2} \sin 2 u \\$

$\Rightarrow x \dfrac{\partial u}{\partial x}+y \dfrac{\partial u}{\partial y}=\dfrac{1}{2} \sin 2 u$

Important Formulas to Remember

The function $f(x, y)$ is said to be a homogeneous function if

$f(\lambda x, \lambda y)=\lambda^{n} f(x, y), \text { for any non zero constant } \lambda .$

Here, $n$ is the degree of homogeneous function $f$.

If $f$ is a homogeneous function of degree $n$ of variables $x$ and $y$, then

$x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=n f$

Conclusion

Euler’s Theorem has a wide range of applications in daily life and it is a fundamental tool of algebra. In this article, we have discussed Euler's theorem and its proof along with its applications of it in daily life.