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Question

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 correct answer is

Non-linear equation of order 2

Partial Differential Equation Classification

Understanding the nature of a given partial differential equation (PDE) is fundamental in mathematics. PDEs are classified based on several characteristics, primarily their linearity and order. Let's analyze the provided equation:

\[\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\]

Equation Linearity Analysis

To determine if a partial differential equation is linear or non-linear, we examine how the dependent variable (in this case, \(u\)) and its derivatives appear in the equation. A PDE is considered linear if:

  • The dependent variable and all its derivatives appear only to the first power.
  • There are no products of the dependent variable with itself or with its derivatives.
  • There are no products of the derivatives with each other.
  • There are no transcendental functions (like \(\sin(u)\), \(e^u\), \(\log(u)\)) of the dependent variable or its derivatives.

Conversely, if any of these conditions are not met, the PDE is non-linear.

Let's look at the terms in our given partial differential equation:

  • The term \(\frac{{\partial u}}{{\partial t}}\) is linear.
  • The term \(u\frac{{\partial u}}{{\partial x}}\) involves the product of the dependent variable \(u\) and its derivative \(\frac{{\partial u}}{{\partial x}}\). This multiplication makes the term, and thus the entire partial differential equation, non-linear.
  • The term \(\frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is linear.

Since the term \(u\frac{{\partial u}}{{\partial x}}\) contains a product of the dependent variable \(u\) and its derivative \(\frac{{\partial u}}{{\partial x}}\), the given partial differential equation is a Non-linear equation.

Equation Order Determination

The order of a partial differential equation is determined by the highest order of the partial derivatives present in the equation. Let's identify the order of each derivative term in the given partial differential equation:

  • The term \(\frac{{\partial u}}{{\partial t}}\) is a first-order partial derivative with respect to \(t\). Its order is 1.
  • The term \(u\frac{{\partial u}}{{\partial x}}\) involves a first-order partial derivative \(\frac{{\partial u}}{{\partial x}}\) with respect to \(x\). Its order is 1.
  • The term \(\frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is a second-order partial derivative with respect to \(x\). Its order is 2.

Comparing the orders of all derivatives, the highest order derivative present in the partial differential equation is \(\frac{{{\partial ^2}u}}{{\partial {x^2}}}\), which is of second order. Therefore, the order of the given partial differential equation is 2.

Equation Classification Summary

Based on our analysis:

  • The partial differential equation contains a non-linear term \(u\frac{{\partial u}}{{\partial x}}\), making it Non-linear.
  • The highest order derivative in the partial differential equation is \(\frac{{{\partial ^2}u}}{{\partial {x^2}}}\), making its order 2.

Thus, the partial differential equation \(\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is a Non-linear equation of order 2.

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Important Questions from Order and Degree of a Differential Equation

  1. 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?

  2. 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:

  3. In the following partial differential equation, θ is a function of t and z, and D and K are functions of θ

    \(D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} + \frac{{\delta K\left( \theta \right)}}{{\delta z}} - \frac{{\delta \theta }}{{\delta t}} = 0\)

    The above equation is
  4. The solution of the equation \({\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0{\rm{}}\) passing through the point (1,1) is

  5. The order and degree of the differential equation

    \(\frac{{{d^3}y}}{{d{x^3}}} + 4\sqrt{\left[{{{{\left( {\frac{{dy}}{{dx}}} \right)}^3} + {y^2}}}\right]}= 0\;\)

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