In the following partial differential equation, θ is a function of t and z, and D and K are functions of θ \(D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} + \frac{{\delta K\left( \theta \right)}}{{\delta z}} - \frac{{\delta \theta }}{{\delta t}} = 0\)
a second order non-linear equation
Partial Differential Equation Analysis
A partial differential equation (PDE) is an equation that involves an unknown function of multiple independent variables and its partial derivatives with respect to those variables. To classify a given PDE, we typically determine its order and linearity.
The given partial differential equation is:
\(D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} + \frac{{\delta K\left( \theta \right)}}{{\delta z}} - \frac{{\delta \theta }}{{\delta t}} = 0\)
Here, \( \theta \) is the dependent variable, which is a function of independent variables \( t \) and \( z \). Also, \( D \) and \( K \) are functions of \( \theta \).
Order of the Partial Differential Equation
The order of a partial differential equation is determined by the highest order partial derivative present in the equation.
- The term \( \frac{{\delta ^2}\theta }}{{\delta {z^2}}} \) involves a second-order partial derivative of \( \theta \) with respect to \( z \).
- The term \( \frac{{\delta K\left( \theta \right)}}{{\delta z}} \) involves a first-order partial derivative with respect to \( z \). If we apply the chain rule, it becomes \( \frac{dK}{d\theta} \frac{\delta\theta}{\delta z} \), which still contains a first-order derivative of \( \theta \).
- The term \( \frac{{\delta \theta }}{{\delta t}} \) involves a first-order partial derivative of \( \theta \) with respect to \( t \).
Comparing these, the highest order derivative in the given partial differential equation is \( \frac{{{\delta ^2}\theta }}{{\delta {z^2}}} \), which is a second-order derivative. Therefore, the partial differential equation is a second-order equation.
Linearity of the Partial Differential Equation
A partial differential equation is considered linear if the dependent variable and all its partial derivatives appear only in the first power, and there are no products of the dependent variable with its derivatives, nor are there any transcendental functions of the dependent variable or its derivatives. Additionally, the coefficients of the dependent variable and its derivatives must only be functions of the independent variables, not the dependent variable itself.
Let's examine the terms in the given partial differential equation:
- Term 1: \( D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} \)
- Here, \( D\left( \theta \right) \) is the coefficient of the second-order derivative \( \frac{{{\delta ^2}\theta }}{{\delta {z^2}}} \). Since \( D \) is explicitly stated as a function of \( \theta \) (the dependent variable), this term violates the condition for linearity. If a coefficient depends on the dependent variable, the equation is non-linear.
- Term 2: \( \frac{{\delta K\left( \theta \right)}}{{\delta z}} \)
- Using the chain rule, this term can be written as \( \frac{dK}{d\theta} \frac{\delta\theta}{\delta z} \). Since \( K \) is a function of \( \theta \), \( \frac{dK}{d\theta} \) will also be a function of \( \theta \). This means we have a coefficient \( \frac{dK}{d\theta} \) that depends on the dependent variable \( \theta \), multiplying the derivative \( \frac{\delta\theta}{\delta z} \). This also indicates non-linearity. For example, if \( K(\theta) = \theta^2 \), then \( \frac{\delta K(\theta)}{\delta z} = \frac{\delta (\theta^2)}{\delta z} = 2\theta \frac{\delta\theta}{\delta z} \). The presence of the product \( \theta \frac{\delta\theta}{\delta z} \) makes the equation non-linear.
- Term 3: \( - \frac{{\delta \theta }}{{\delta t}} \)
- This term involves a linear derivative with a constant coefficient (-1). This term alone does not introduce non-linearity.
Due to the presence of \( D\left( \theta \right) \) and \( K\left( \theta \right) \) as functions of the dependent variable \( \theta \), and how they interact with the derivatives, the partial differential equation is classified as non-linear.
Equation Classification
Based on our analysis:
- The highest derivative is second order, making it a second-order equation.
- The coefficients depend on the dependent variable \( \theta \), and terms like \( \frac{{\delta K\left( \theta \right)}}{{\delta z}} \) can introduce products of \( \theta \) and its derivatives, making it a non-linear equation.
Therefore, the given partial differential equation is a second order non-linear equation.
Important Questions from Order and Degree of a Differential Equation
Consider the following in respect of the differential equation:
\(\frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\)
1. The degree of the differential equation is 1.
2. The order of the differential equation is 2.
Which of the above statements is/are correct?
The partial differential equation \(\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is a
The degree of the differential equation \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\frac{{dy}}{{dx}}} \right)^2} + \sin x\left( {\frac{{dy}}{{dx}}} \right) + y = 0\) is:
The solution of the equation \({\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0{\rm{}}\) passing through the point (1,1) is
The order and degree of the differential equation
\(\frac{{{d^3}y}}{{d{x^3}}} + 4\sqrt{\left[{{{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}}}\right]}= 0\;\)
