Formation of Partial Differential Equations (Part-2)

Formation of PDE by Elimination of Arbitrary Function

To form a partial differential equation (PDE) by eliminating an arbitrary function from a given equation, follow these steps:

Method-2

Algorithm (Method of Elimination of arbitrary function)

• Step 1: Differentiate partially the given static equation.
• Step 2: Eliminate the arbitrary function(s) and its derivables using the expression for partial differentials.

Example 1: Find the PDE of the equation \(z = f(x, y)\). where \(f(x, y)\) is an arbitrary function of \(x\) and \(y\).

Solution:

Step 1: Differentiate with Respect to \(x\) and \(y\)

Differentiate the given equation with respect to \(x\) and \(y\) to form new equations involving the partial derivatives.

Differentiate with respect to \(x\):

\[ \frac{\partial z}{\partial x} = \frac{\partial f(x, y)}{\partial x} \]

Let \(p = \frac{\partial z}{\partial x}\):

\[ p = f_x(x, y) \]

Differentiate with respect to \(y\):

\[ \frac{\partial z}{\partial y} = \frac{\partial f(x, y)}{\partial y} \]

Let \(q = \frac{\partial z}{\partial y}\):

\[ q = f_y(x, y) \]

Step 2: Form the PDE

Divide the equations for \(p\) and \(q\):

\[ \frac{p}{q} = \frac{f_x(x, y)}{f_y(x, y)} \]

Since \(f(x, y)\) is arbitrary, \(f_x(x, y)\) and \(f_y(x, y)\) are just partial derivatives of the function. The relationship simplifies to:

\[ \frac{p}{q} = 1 \]

So, the PDE is:

\[ p = q \]

Example 2: Form a PDE from the general dependancy relationship, \(z=yf\left(\frac{y}{x}\right)\) by eliminating the arbitrary function.

Solution:

Given static equation is \[z=yf\left(\frac{y}{x}\right) \tag{1}\]

Differentiating (Equation 1) partially with respect to \(x\), we get:

\[\dfrac{\partial z}{\partial x}=yf'\left(\frac{y}{x}\right)\left(-\dfrac{y}{x^2}\right)\]

or \[f'\left(\frac{y}{x}\right)=-\frac{x^2}{y^2}p \tag{2}\]

Differentiating (Equation 1) partially with respect to \(x\), we get:

\[\dfrac{\partial z}{\partial y}=yf'\left(\frac{y}{x}\right)\left(\dfrac{1}{x}\right)+f\left(\frac{y}{x}\right) \tag{3}\]

Using (Equation 1), and (Equation 2) in (Equation 3), we get:

\[\begin{align*} q&=\dfrac{y}{x}\times \left(\dfrac{x^2}{y^2}\right)p+\dfrac{z}{y}\\ &=-\frac{x}{y}p+\frac{z}{y}\\ \implies z&=px+qy \end{align*}\]

Example 3: Form a PDE from the equation \(z=f(x^2-y^2)\), where \(f\) is an arbitrary function.

Solution:

Given \[z=f(x^2-y^2) \tag{4}\]

Differentiating (Equation 4) partially with respect to \(x\) and \(y\), we get:

\[\begin{align} p&=f'(x^2-y^2)\times 2x \\ q&=f'(x^2-y^2)\times -2y \end{align}\]

\[\therefore \dfrac{p}{q}=-\frac{x}{y}\implies py+qx=0\]

Additional Problems:

Problem:

Form the partial differential equation by eliminating the arbitrary function from \(f(x+y+z,x^2+y^2+z^2)=0\).

Solution:

Let \(x+y+z=u\), \(x^2+y^2+z^2=v\), then \(f(u,v)=0\)

Differentiating with respect to \(x\) and \(y\), we get:

\[\begin{eqnarray*} \frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}&=&0\\ \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}&=&0 \end{eqnarray*}\]

or \(\dfrac{\partial f}{\partial u}(1+p)+\dfrac{\partial f}{\partial v}(2x+2zp)=0\)………….(1)

\(\dfrac{\partial f}{\partial u}\left(\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}q\right)+\dfrac{\partial f}{\partial v}\left(\dfrac{\partial v}{\partial y}+\dfrac{\partial v}{\partial z}q\right)=0\)

or \(\dfrac{\partial f}{\partial u}(1+q)+\dfrac{\partial f}{\partial v}(2y+2zq)=0\)………….(2)

Eliminating \(\dfrac{\partial f}{\partial u}\) and \(\dfrac{\partial f}{\partial v}\) from (1) and (2), we get:

\[\begin{align*} \begin{vmatrix} ~~1+p &2x+2zp~\\ ~~1+q &2y+2zq~ \end{vmatrix}&=0\\ (1+p)(2y+2zq)-(1+q)(2x+2zp)&=0\\ \text{ Or}\quad (y-z)p+(z-x)q&=x-y \end{align*}\]

, which is a partial differential equation of the first order.

Problem:

Form the partial differential equation by eliminating the arbitrary function from \(f(xy+z^2,x+y+z)=0\).

Solution:

Let \(xy+z^2=u\), \(x+y+z=v\), then \(f(u,v)=0\).

Differentiating with respect to \(x\) and \(y\), we get:

\[\begin{eqnarray*} \frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}&=&0\\ \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}&=&0 \end{eqnarray*}\]

or \(\dfrac{\partial f}{\partial u}(y+2zp)+\dfrac{\partial f}{\partial v}(1+p)=0\)………….(1)

\(\dfrac{\partial f}{\partial u}\left(\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}q\right)+\dfrac{\partial f}{\partial v}\left(\dfrac{\partial v}{\partial y}+\dfrac{\partial v}{\partial z}q\right)=0\)

or \(\dfrac{\partial f}{\partial u}(x+2zq)+\dfrac{\partial f}{\partial v}(1+q)=0\)………….(2)

Eliminating \(\dfrac{\partial f}{\partial u}\) and \(\dfrac{\partial f}{\partial v}\) from (1) and (2), we get:

\[\begin{align*} \begin{vmatrix} ~~y+2zp &1+p~\\ ~~x+2zq &1+q~ \end{vmatrix}&=0\\ (y+2zp)(1+q)-(x+2zq)(1+p)&=0\\ \text{ or } \,(2z-x)p+(y-2z)q&=x-y \end{align*}\] which is a partial differential equation of the first order.

Problem:

Form the partial differential equation by eliminating the arbitrary function from \(f(x^2+y^2+z^2,z^2-2xy)=0\).

Solution:

Let \(x^2+y^2+z^2=u\), \(z^2-2xy=v\), then \(f(u,v)=0\).

Differentiating with respect to \(x\) and \(y\), we get:

\[\begin{eqnarray*} \frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}&=&0\\ \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}&=&0 \end{eqnarray*}\]

or \(\dfrac{\partial f}{\partial u}(2x+2zp)+\dfrac{\partial f}{\partial v}(2zp-2y)=0\)………….(1)

\(\dfrac{\partial f}{\partial u}\left(\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}q\right)+\dfrac{\partial f}{\partial v}\left(\dfrac{\partial v}{\partial y}+\dfrac{\partial v}{\partial z}q\right)=0\)

or \(\dfrac{\partial f}{\partial u}(2y+2zq)+\dfrac{\partial f}{\partial v}(2zq-2x)=0\)………….(2)

Eliminating \(\dfrac{\partial f}{\partial u}\) and \(\dfrac{\partial f}{\partial v}\) from (1) and (2), we get:

\[\begin{align*} \begin{vmatrix} ~~2x+2zp &2zp-2y\\ ~~2y+2zq &2zq-2x \end{vmatrix}&=0\\ (2x+2zp)(2zq-2x)-(2y+2zq)(2zp-2y)&=0\\ \text{or} \,(y+x)zp+(y+x)zq&=y^2-x^2\\ \implies\, (p-q)z&=y-x \end{align*}\] which is a partial differential equation of the first order.

Problem: Form the partial differential equation by eliminating the arbitrary function from \(f(x^2+y^2,x^2-z^2)=0\)

Solution:

Let \(x^2+y^2=u\), \(x^2-z^2=v\), then \(f(u,v)=0\).

Differentiating with respect to \(x\) and \(y\), we get:

\[\begin{eqnarray*} \frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}&=&0\\ \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}&=&0 \end{eqnarray*}\]

or \(\dfrac{\partial f}{\partial u}(2x)+\dfrac{\partial f}{\partial v}(2x-2zp)=0\)………….(1)

\(\dfrac{\partial f}{\partial u}\left(\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}q\right)+\dfrac{\partial f}{\partial v}\left(\dfrac{\partial v}{\partial y}+\dfrac{\partial v}{\partial z}q\right)=0\)

or \(\dfrac{\partial f}{\partial u}(2y)+\dfrac{\partial f}{\partial v}(-2zq)=0\)………….(2)

Eliminating \(\dfrac{\partial f}{\partial u}\) and \(\dfrac{\partial f}{\partial v}\) from (1) and (2), we get:

\[\begin{align*} \begin{vmatrix} ~~2x &2x-2zp\\ ~~2y &-2zq \end{vmatrix}=0\\ (2x)(-2zq)-(2y)(2x-2zp)&=0\\ \implies yp-xq&=\dfrac{xy}{z}, \end{align*}\]

which is a partial differential equation of the first order.

Problem: Form PDE using, \(f(x^2+y^2,z-xy)=0\)

Solution:

Given, \[\begin{eqnarray} f(x^2+y^2,z-xy)&=&0\\ or\,\,f(u,v)&=&0 \end{eqnarray}\] where \(u=x^2-y^2\) and \(v=z-xy\).

Differentiating given equation partially with respect to \(x\) and \(y\), we get:

\[\begin{eqnarray*} \frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}&=&0\\ \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}&=&0 \end{eqnarray*}\]

\[\begin{eqnarray*} \frac{\partial f}{\partial u}2x+\frac{\partial f}{\partial v}.(p-y)&=&0\\ \frac{\partial f}{\partial u}2y+\frac{\partial f}{\partial v}.(q-x)&=&0 \end{eqnarray*}\]

A non-trivial solution of the above equation is:

\[\begin{equation*} \begin{array}{|cc|} 2x&p-y\\ 2y&q-x \end{array}=0 \end{equation*}\]

On expanding the determinant, we get \[xq-yp=x^2-y^2\]

Practice Problems

1. Form the partial differential equation by eliminating arbitrary constants from \(z=ax+by+a^2+b^2\).
2. Form the partial differential equation by eliminating arbitrary constants from \((x-a)^2+(y-b)^2+z^2=1\).
3. Form the partial differential equation by eliminating arbitrary constants from \(z=xy+y\sqrt{x^2-a^2}+b\).
4. Form the partial differential equation by eliminating arbitrary constants from \(z=a(x+y)+b(x-y)+abt+c\).
5. Form the partial differential equation by eliminating arbitrary constants from \(z=ax^3+by^3\).
6. Form the partial differential equation by eliminating the arbitrary function from \(z=f(x^2+y^2)\).
7. Form the partial differential equation by eliminating the arbitrary function from \(z=f(y/x)\).
8. Form the partial differential equation by eliminating the arbitrary function from \(xyz=f(x+y+z)\).
9. Form the partial differential equation by eliminating the arbitrary functions from \(z=f(x)g(y)\).
10. Form the partial differential equation by eliminating the arbitrary functions from \(z=yf(x)+xg(y)\).
11. Form the partial differential equation by eliminating the arbitrary functions from \(z=f(x+ay)+g(x-ay)\).
12. Form the partial differential equation by eliminating the arbitrary functions from \(z=f(x+4t)+g(x-4t)\).
13. Form the partial differential equation by eliminating the arbitrary functions from \(z=f(y+2x)+g(y-3x)\).
14. Form the partial differential equation by eliminating the arbitrary function from \(f(z/x^2,x-y)=0\).
15. Form the partial differential equation by eliminating the arbitrary function from \(f(x+y,x^2+y^2+z^2)=0\).

Answers:

1. \(z=px+qy+p^2+q^2\)
2. \(z^2(p^2+q^2+1)=1\)
3. \(px+qy=pq\)
4. \(4\dfrac{\partial z}{\partial t}=\left(\dfrac{\partial z}{\partial x}\right)^2-\left(\dfrac{\partial z}{\partial y}\right)^2\)
5. \(xp+yq=3z\)
6. \(py-qx=0\)
7. \(px+qy=0\)
8. \(x(y-z)p+y(z-x)q=z(x-y)\)
9. \(z\dfrac{\partial^2z}{\partial x\partial y}-\dfrac{\partial z}{\partial x}.\dfrac{\partial z}{\partial y}=0\)
10. \(xy\dfrac{\partial^2z}{\partial x\partial y}=x\dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}-z\)
11. \(\dfrac{\partial^2z}{\partial y^2}=a^2\dfrac{\partial^2z}{\partial x^2}\)
12. \(16\dfrac{\partial^2z}{\partial x^2}-\dfrac{\partial^2z}{\partial t^2}=0\)
13. \(\dfrac{\partial^2z}{\partial x^2}+\dfrac{\partial^2z}{\partial x\partial y}-6\dfrac{\partial^2z}{\partial y^2}=0\)
14. \(2z=x(p+q)\)
15. \(z(p-q)=y-x\)