Question

The one dimensional heat conduction partial differential equation $\frac{\partial T}{\partial t}=\frac{{\partial }^{2}T}{\partial x^2}$ is

• A. Parabolic
• B. Hyperbolic
• C. Elliptic
• D. Mixed

Answer 1

The correct option is A Parabolic

The general form of $2nd$ order partial differential equation is:
$A\frac{{\partial }^{2}U}{\partial x^2}+B\frac{{\partial }^{2}U}{\partial x\partial y}+C\frac{{\partial }^{2}U}{\partial y^2}+f(x,y,u,\frac{\partial U}{\partial x},\frac{\partial U}{\partial y})=0$ .... (i)
$\frac{\partial T}{\partial t}=\frac{{\partial }^{2}T}{\partial x^2}\Rightarrow 1.\frac{{\partial }^{2}T}{\partial x^2}+0.\frac{{\partial }^{2}T}{\partial x\partial Y}+0.\frac{{\partial }^{2}T}{\partial Y^2}+\frac{\partial T}{\partial t}=0$
On comparision, $A=1,B=0,C=0$
$B^2-4AC=0-4\left(1\right)\left(0\right)=0$
$\therefore$ Given equatin is parabolic.