Answer
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Hint:
Here, we are required to state and prove the Euler’s theorem for homogeneous function. Euler’s theorem is used to establish a relationship between the partial derivatives of a function and the product of the function with its degree. Here, we will first write the statement pertaining to the mathematical expression of the Euler’s theorem, and explain it. We will then differentiate the function and manipulate it to obtain the required result.
Complete step by step solution:
Statement –
Euler’s theorem states that if \[f\] is a homogeneous function of degree \[n\] of the variables \[x, y, z\]; then –
\[x\dfrac{{\partial f}}{{\partial x}} + y\dfrac{{\partial f}}{{\partial y}} + z\dfrac{{\partial f}}{{\partial z}} = nf\], where, \[\dfrac{{\partial f}}{{\partial x}}\] is the partial derivative of the function \[f\] w.r.t \[x\], \[\dfrac{{\partial f}}{{\partial y}}\] is the partial derivative of the function \[f\] w.r.t \[y\] and \[\dfrac{{\partial f}}{{\partial z}}\] is the partial derivative of the function \[f\] w.r.t \[z\].
Proof –
Let \[f = u\left[ {x,y,z} \right]\] be a homogenous function of degree \[n\] of the variables \[x,y,z\].
\[f = u\left[ {x,y,z} \right]\]…………\[\left[ 1 \right]\]
Now, we know that –
\[u\left[ {X,Y,Z} \right] = {t^n}u\left[ {x,y,z} \right]\]…………\[\left[ 2 \right]\]
This is because when \[u\] is a function of \[X,Y,Z\], then it becomes a function of \[x,y,z,t\] because \[X,Y,Z\] are a function of \[t\].
This means that –
\[\begin{array}{l}X = x{t^n}\\Y = y{t^n}\\Z = z{t^n}\end{array}\]
Now, we will consider the above equations for degree 1-
\[X = xt\]………\[\left[ 3 \right]\]
\[Y = yt\]……….\[\left[ 4 \right]\]
\[Z = zt\]……….\[\left[ 5 \right]\]
where, \[t\] is any arbitrary parameter.
Now, we will differentiate the equation \[\left[ 3 \right]\] partially w.r.t \[t\].
\[\begin{array}{l}X = xt\\ \Rightarrow \dfrac{\partial }{{\partial t}}X = \dfrac{\partial }{{\partial t}}xt\\ \Rightarrow \dfrac{{\partial X}}{{\partial t}} = x\dfrac{{\partial t}}{{\partial t}}\end{array}\]
\[ \Rightarrow \dfrac{{\partial X}}{{\partial t}} = x\]……….\[\left[ 6 \right]\]
Again, we will differentiate the equation \[\left[ 4 \right]\] partially w.r.t \[t\].
\[\begin{array}{l}Y = yt\\ \Rightarrow \dfrac{\partial }{{\partial t}}Y = \dfrac{\partial }{{\partial t}}yt\end{array}\]
\[ \Rightarrow \dfrac{{\partial Y}}{{\partial t}} = y\]………… \[\left[ 7 \right]\]
We will differentiate the equation \[\left[ 5 \right]\] partially w.r.t \[t\].
\[\begin{array}{l}Z = zt\\ \Rightarrow \dfrac{\partial }{{\partial t}}Z = \dfrac{\partial }{{\partial t}}zt\\ \Rightarrow \dfrac{{\partial Z}}{{\partial t}} = z\dfrac{{\partial t}}{{\partial t}}\end{array}\]
\[ \Rightarrow \dfrac{{\partial Z}}{{\partial t}} = z\]…………\[\left[ 8 \right]\]
Now, we will substitute \[t = 1\] in equation \[\left[ 3 \right]\].
\[\begin{array}{l}X = xt\\ \Rightarrow X = x\left[ 1 \right]\end{array}\]
Multiplying the terms, we get
\[X = x\]…………….\[\left[ 9 \right]\]
We will again substitute \[t = 1\] in equation \[\left[ 4 \right]\].
\[\begin{array}{l}Y = yt\\ \Rightarrow Y = y\left[ 1 \right]\end{array}\]
Multiplying the terms, we get
\[Y = y\]…………….\[\left[ {10} \right]\]
We will now substitute \[t = 1\] in equation \[\left[ 5 \right]\].
\[\begin{array}{l}Z = zt\\ \Rightarrow Z = z\left[ 1 \right]\end{array}\]
Multiplying the terms, we get
\[Z = z\]………….\[\left[ {11} \right]\]
We will now differentiate the equation \[\left[ 1 \right]\] partially w.r.t \[x\].
\[\begin{array}{l}f = u\left[ {x,y,z} \right]\\\dfrac{\partial }{{\partial x}}f = \dfrac{\partial }{{\partial x}}u\left[ {x,y,z} \right]\end{array}\]
\[\dfrac{{\partial f}}{{\partial x}} = \dfrac{{\partial u}}{{\partial x}}\]………….\[\left[ {12} \right]\]
Partial differentiation of \[y,z\] w.r.t \[x\] is 0.
Now, we will differentiate the equation \[\left[ 1 \right]\] partially w.r.t \[y\].
\[\begin{array}{l}f = u\left[ {x,y,z} \right]\\\dfrac{\partial }{{\partial y}}f = \dfrac{\partial }{{\partial y}}u\left[ {x,y,z} \right]\end{array}\]
\[\dfrac{{\partial f}}{{\partial y}} = \dfrac{{\partial u}}{{\partial y}}\]………..\[\left[ {13} \right]\]
Partial differentiation of \[x,z\] w.r.t \[y\] is 0.
Again, we will differentiate the equation \[\left[ 1 \right]\] partially w.r.t \[z\].
\[\begin{array}{l}f = u\left[ {x,y,z} \right]\\\dfrac{\partial }{{\partial z}}f = \dfrac{\partial }{{\partial z}}u\left[ {x,y,z} \right]\end{array}\]
\[\dfrac{{\partial f}}{{\partial z}} = \dfrac{{\partial u}}{{\partial z}}\]………..\[\left[ {14} \right]\]
Partial differentiation of \[x,y\] w.r.t \[z\] is 0.
Now, we will differentiate the equation \[\left[ 2 \right]\] partially w.r.t \[t\].
Since, \[u\]is a function of \[X,Y,Z\],and \[X,Y,Z\] are a function of \[x,y,z\], and \[t\], thus apply the chain rule of partial differentiation to differentiate.
\[u\left[ {X,Y,Z} \right] = {t^n}u\left[ {x,y,z} \right]\]
\[\dfrac{{\partial u}}{{\partial x}} \cdot \dfrac{{\partial X}}{{\partial t}} + \dfrac{{\partial u}}{{\partial y}} \cdot \dfrac{{\partial Y}}{{\partial t}} + \dfrac{{\partial u}}{{\partial z}} \cdot \dfrac{{\partial Z}}{{\partial t}} = n{t^{n - 1}}u\left[ {x,y,z} \right]\]………..\[\left[ {15} \right]\]
We will substitute \[x\] for \[\dfrac{{\partial X}}{{\partial t}}\], \[y\] for \[\dfrac{{\partial Y}}{{\partial t}}\], \[z\] for \[\dfrac{{\partial Z}}{{\partial t}}\], \[\dfrac{{\partial f}}{{\partial x}}\] for \[\dfrac{{\partial u}}{{\partial x}}\], \[\dfrac{{\partial f}}{{\partial y}}\] for \[\dfrac{{\partial u}}{{\partial y}}\], \[\dfrac{{\partial f}}{{\partial z}}\] for \[\dfrac{{\partial u}}{{\partial z}}\] in equation \[\left[ {15} \right]\].
Thus, the equation \[\left[ {15} \right]\] becomes –
\[\dfrac{{\partial u}}{{\partial x}} \cdot \dfrac{{\partial X}}{{\partial t}} + \dfrac{{\partial u}}{{\partial y}} \cdot \dfrac{{\partial Y}}{{\partial t}} + \dfrac{{\partial u}}{{\partial z}} \cdot \dfrac{{\partial Z}}{{\partial t}} = n{t^{n - 1}}u\left[ {x,y,z} \right]\]
Using the result of equations \[\left[ 9 \right]\], \[\left[ {10} \right]\], and \[\left[ {11} \right]\], we can write above equation as
\[\dfrac{{\partial f}}{{\partial x}} \cdot x + \dfrac{{\partial f}}{{\partial y}} \cdot y + \dfrac{{\partial f}}{{\partial z}} \cdot z = n{t^{n - 1}}u\left[ {x,y,z} \right]\]……….\[\left[ {16} \right]\]
We will again substitute \[t = 1\] and \[f\] for \[u\left[ {x,y,z} \right]\] in the equation \[\left[ {16} \right]\], we get
\[\begin{array}{l}\dfrac{{\partial f}}{{\partial x}} \cdot x + \dfrac{{\partial f}}{{\partial y}} \cdot y + \dfrac{{\partial f}}{{\partial z}} \cdot z = n{\left[ 1 \right]^{n - 1}}f\\\dfrac{{\partial f}}{{\partial x}} \cdot x + \dfrac{{\partial f}}{{\partial y}} \cdot y + \dfrac{{\partial f}}{{\partial z}} \cdot z = nf\end{array}\]
Hence, the Euler’s equation is finally proved.
Note:
A homogeneous function of degree \[n\] of the variables \[x,y,z\] can be defined as a function in which all terms have degree \[n\]. For example, consider the following function –
\[f\left[ {x,y,z} \right] = A{x^3} + B{y^3} + C{z^3} + Dx{y^2} + Ex{z^2} + Gy{x^2} + Hy{z^2} + Iz{x^2} + Jz{y^2} + Kxyz\]
We can see that the degree of all the terms in the above function is three. Since, the degree is constant for all the terms, hence, the above function is a homogeneous function of degree 3.
Here, we are required to state and prove the Euler’s theorem for homogeneous function. Euler’s theorem is used to establish a relationship between the partial derivatives of a function and the product of the function with its degree. Here, we will first write the statement pertaining to the mathematical expression of the Euler’s theorem, and explain it. We will then differentiate the function and manipulate it to obtain the required result.
Complete step by step solution:
Statement –
Euler’s theorem states that if \[f\] is a homogeneous function of degree \[n\] of the variables \[x, y, z\]; then –
\[x\dfrac{{\partial f}}{{\partial x}} + y\dfrac{{\partial f}}{{\partial y}} + z\dfrac{{\partial f}}{{\partial z}} = nf\], where, \[\dfrac{{\partial f}}{{\partial x}}\] is the partial derivative of the function \[f\] w.r.t \[x\], \[\dfrac{{\partial f}}{{\partial y}}\] is the partial derivative of the function \[f\] w.r.t \[y\] and \[\dfrac{{\partial f}}{{\partial z}}\] is the partial derivative of the function \[f\] w.r.t \[z\].
Proof –
Let \[f = u\left[ {x,y,z} \right]\] be a homogenous function of degree \[n\] of the variables \[x,y,z\].
\[f = u\left[ {x,y,z} \right]\]…………\[\left[ 1 \right]\]
Now, we know that –
\[u\left[ {X,Y,Z} \right] = {t^n}u\left[ {x,y,z} \right]\]…………\[\left[ 2 \right]\]
This is because when \[u\] is a function of \[X,Y,Z\], then it becomes a function of \[x,y,z,t\] because \[X,Y,Z\] are a function of \[t\].
This means that –
\[\begin{array}{l}X = x{t^n}\\Y = y{t^n}\\Z = z{t^n}\end{array}\]
Now, we will consider the above equations for degree 1-
\[X = xt\]………\[\left[ 3 \right]\]
\[Y = yt\]……….\[\left[ 4 \right]\]
\[Z = zt\]……….\[\left[ 5 \right]\]
where, \[t\] is any arbitrary parameter.
Now, we will differentiate the equation \[\left[ 3 \right]\] partially w.r.t \[t\].
\[\begin{array}{l}X = xt\\ \Rightarrow \dfrac{\partial }{{\partial t}}X = \dfrac{\partial }{{\partial t}}xt\\ \Rightarrow \dfrac{{\partial X}}{{\partial t}} = x\dfrac{{\partial t}}{{\partial t}}\end{array}\]
\[ \Rightarrow \dfrac{{\partial X}}{{\partial t}} = x\]……….\[\left[ 6 \right]\]
Again, we will differentiate the equation \[\left[ 4 \right]\] partially w.r.t \[t\].
\[\begin{array}{l}Y = yt\\ \Rightarrow \dfrac{\partial }{{\partial t}}Y = \dfrac{\partial }{{\partial t}}yt\end{array}\]
\[ \Rightarrow \dfrac{{\partial Y}}{{\partial t}} = y\]………… \[\left[ 7 \right]\]
We will differentiate the equation \[\left[ 5 \right]\] partially w.r.t \[t\].
\[\begin{array}{l}Z = zt\\ \Rightarrow \dfrac{\partial }{{\partial t}}Z = \dfrac{\partial }{{\partial t}}zt\\ \Rightarrow \dfrac{{\partial Z}}{{\partial t}} = z\dfrac{{\partial t}}{{\partial t}}\end{array}\]
\[ \Rightarrow \dfrac{{\partial Z}}{{\partial t}} = z\]…………\[\left[ 8 \right]\]
Now, we will substitute \[t = 1\] in equation \[\left[ 3 \right]\].
\[\begin{array}{l}X = xt\\ \Rightarrow X = x\left[ 1 \right]\end{array}\]
Multiplying the terms, we get
\[X = x\]…………….\[\left[ 9 \right]\]
We will again substitute \[t = 1\] in equation \[\left[ 4 \right]\].
\[\begin{array}{l}Y = yt\\ \Rightarrow Y = y\left[ 1 \right]\end{array}\]
Multiplying the terms, we get
\[Y = y\]…………….\[\left[ {10} \right]\]
We will now substitute \[t = 1\] in equation \[\left[ 5 \right]\].
\[\begin{array}{l}Z = zt\\ \Rightarrow Z = z\left[ 1 \right]\end{array}\]
Multiplying the terms, we get
\[Z = z\]………….\[\left[ {11} \right]\]
We will now differentiate the equation \[\left[ 1 \right]\] partially w.r.t \[x\].
\[\begin{array}{l}f = u\left[ {x,y,z} \right]\\\dfrac{\partial }{{\partial x}}f = \dfrac{\partial }{{\partial x}}u\left[ {x,y,z} \right]\end{array}\]
\[\dfrac{{\partial f}}{{\partial x}} = \dfrac{{\partial u}}{{\partial x}}\]………….\[\left[ {12} \right]\]
Partial differentiation of \[y,z\] w.r.t \[x\] is 0.
Now, we will differentiate the equation \[\left[ 1 \right]\] partially w.r.t \[y\].
\[\begin{array}{l}f = u\left[ {x,y,z} \right]\\\dfrac{\partial }{{\partial y}}f = \dfrac{\partial }{{\partial y}}u\left[ {x,y,z} \right]\end{array}\]
\[\dfrac{{\partial f}}{{\partial y}} = \dfrac{{\partial u}}{{\partial y}}\]………..\[\left[ {13} \right]\]
Partial differentiation of \[x,z\] w.r.t \[y\] is 0.
Again, we will differentiate the equation \[\left[ 1 \right]\] partially w.r.t \[z\].
\[\begin{array}{l}f = u\left[ {x,y,z} \right]\\\dfrac{\partial }{{\partial z}}f = \dfrac{\partial }{{\partial z}}u\left[ {x,y,z} \right]\end{array}\]
\[\dfrac{{\partial f}}{{\partial z}} = \dfrac{{\partial u}}{{\partial z}}\]………..\[\left[ {14} \right]\]
Partial differentiation of \[x,y\] w.r.t \[z\] is 0.
Now, we will differentiate the equation \[\left[ 2 \right]\] partially w.r.t \[t\].
Since, \[u\]is a function of \[X,Y,Z\],and \[X,Y,Z\] are a function of \[x,y,z\], and \[t\], thus apply the chain rule of partial differentiation to differentiate.
\[u\left[ {X,Y,Z} \right] = {t^n}u\left[ {x,y,z} \right]\]
\[\dfrac{{\partial u}}{{\partial x}} \cdot \dfrac{{\partial X}}{{\partial t}} + \dfrac{{\partial u}}{{\partial y}} \cdot \dfrac{{\partial Y}}{{\partial t}} + \dfrac{{\partial u}}{{\partial z}} \cdot \dfrac{{\partial Z}}{{\partial t}} = n{t^{n - 1}}u\left[ {x,y,z} \right]\]………..\[\left[ {15} \right]\]
We will substitute \[x\] for \[\dfrac{{\partial X}}{{\partial t}}\], \[y\] for \[\dfrac{{\partial Y}}{{\partial t}}\], \[z\] for \[\dfrac{{\partial Z}}{{\partial t}}\], \[\dfrac{{\partial f}}{{\partial x}}\] for \[\dfrac{{\partial u}}{{\partial x}}\], \[\dfrac{{\partial f}}{{\partial y}}\] for \[\dfrac{{\partial u}}{{\partial y}}\], \[\dfrac{{\partial f}}{{\partial z}}\] for \[\dfrac{{\partial u}}{{\partial z}}\] in equation \[\left[ {15} \right]\].
Thus, the equation \[\left[ {15} \right]\] becomes –
\[\dfrac{{\partial u}}{{\partial x}} \cdot \dfrac{{\partial X}}{{\partial t}} + \dfrac{{\partial u}}{{\partial y}} \cdot \dfrac{{\partial Y}}{{\partial t}} + \dfrac{{\partial u}}{{\partial z}} \cdot \dfrac{{\partial Z}}{{\partial t}} = n{t^{n - 1}}u\left[ {x,y,z} \right]\]
Using the result of equations \[\left[ 9 \right]\], \[\left[ {10} \right]\], and \[\left[ {11} \right]\], we can write above equation as
\[\dfrac{{\partial f}}{{\partial x}} \cdot x + \dfrac{{\partial f}}{{\partial y}} \cdot y + \dfrac{{\partial f}}{{\partial z}} \cdot z = n{t^{n - 1}}u\left[ {x,y,z} \right]\]……….\[\left[ {16} \right]\]
We will again substitute \[t = 1\] and \[f\] for \[u\left[ {x,y,z} \right]\] in the equation \[\left[ {16} \right]\], we get
\[\begin{array}{l}\dfrac{{\partial f}}{{\partial x}} \cdot x + \dfrac{{\partial f}}{{\partial y}} \cdot y + \dfrac{{\partial f}}{{\partial z}} \cdot z = n{\left[ 1 \right]^{n - 1}}f\\\dfrac{{\partial f}}{{\partial x}} \cdot x + \dfrac{{\partial f}}{{\partial y}} \cdot y + \dfrac{{\partial f}}{{\partial z}} \cdot z = nf\end{array}\]
Hence, the Euler’s equation is finally proved.
Note:
A homogeneous function of degree \[n\] of the variables \[x,y,z\] can be defined as a function in which all terms have degree \[n\]. For example, consider the following function –
\[f\left[ {x,y,z} \right] = A{x^3} + B{y^3} + C{z^3} + Dx{y^2} + Ex{z^2} + Gy{x^2} + Hy{z^2} + Iz{x^2} + Jz{y^2} + Kxyz\]
We can see that the degree of all the terms in the above function is three. Since, the degree is constant for all the terms, hence, the above function is a homogeneous function of degree 3.
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