Question

Find the nature of the one-dimensional heat equation.

(a) Circular

(b) Elliptic

(c) Hyperbolic

(d) Parabolic

This question was posed to me during an online exam.

The doubt is from Classification of PDE in section Mathematical Behaviour of Partial Differential Equations of Computational Fluid Dynamics

Answer 1

Correct answer is (d) Parabolic

To explain: The one-dimensional heat equation is

\(\frac{\partial T}{\partial t}=α\frac{\partial^2 T}{\partial x^2}\)

The general equation is in this form.

A\(\frac{\partial^2 \Phi}{\partial x^2}+B\frac{\partial ^2 \Phi}{\partial x\partial y}+C\frac{\partial^2 \Phi}{\partial y^2}+D\frac{\partial \Phi}{\partial x}+E\frac{\partial\Phi}{\partial y}+F\Phi +G=0\)

Comparing \(α\frac{\partial^2 T}{\partial x^2}-\frac{\partial T}{\partial t}=0\) with the above equation, (let ‘y’ be ‘t’).

A=α

B=0

C=0

To find the type,

d=B^2-4AC

d=0

As d is zero, the one-dimensional heat equation is parabolic.