# Quiz Chapter 10.02: Parabolic Partial Differential Equations

 MULTIPLE CHOICE TEST PARABOLIC PARTIAL DIFFERENTIAL EQUATIONS PARTIAL DIFFERENTIAL EQUATIONS

1. In a general second order linear partial differential equation with two independent variables,

• $A \dfrac{\partial^{2} u}{\partial x^{2}} + B \dfrac{\partial^{2} u}{\partial x \partial y} + C \dfrac{\partial^{2} u}{\partial y^{2}} + D = 0$

Where $A, \, B, \, C$ are functions of $x$ and $y$, and $D$ is a function of $x, \, y, \, \dfrac{\partial u}{\partial x}, \, \dfrac{\partial u}{\partial y}$, then the PDE is parabolic if

2. The region in which the following equation

• $x^{3} \dfrac{\partial^{2} u}{\partial x^{2}} + 27 \dfrac{\partial^{2} u}{\partial y^{2}} + 3 \dfrac{\partial^{2} u}{\partial x \partial y} + 5u = 0$

acts as a parabolic equation is

3. The partial differential equation of the temperature in a long thin rod is given by

• $\dfrac{\partial T}{\partial t} = \alpha \dfrac{\partial^{2} T}{\partial x^{2}}$

If $\alpha = 0.8$ cm2/s, the initial temperature of the rod is $40^{\circ}$C, and the rod is divided into three equal segments, the temperature at nod $1$ (using $\triangle t = 0.1$s) by using an explicit solution at $t=0.2$s is

4. The partial differential equation of the temperature in a long thin rod is given by

• $\dfrac{\partial T}{\partial t} = \alpha \dfrac{\partial^{2} T}{\partial x^{2}}$

If $\alpha = 0.8$ cm2/s, the initial temperature of the rod is $40^{\circ}$C, and the rod is divided into three equal segments, the temperature at nod $1$ (using $\triangle t = 0.1$s) by using an implicit solution at $t=0.2$s is

5. The partial differential equation of the temperature in a long thin rod is given by

• $\dfrac{\partial T}{\partial t} = \alpha \dfrac{\partial^{2} T}{\partial x^{2}}$

If $\alpha = 0.8$ cm2/s, the initial temperature of the rod is $40^{\circ}$C, and the rod is divided into three equal segments, the temperature at nod $1$ (using $\triangle t = 0.1$s) by using an Crank-Nicolson solution at $t=0.2$s is

6. The partial differential equation of the temperature in a long thin rod is given by

• $\dfrac{\partial T}{\partial t} = \alpha \dfrac{\partial^{2} T}{\partial x^{2}}$

If $\alpha = 0.8$ cm2/s, the initial temperature of the rod is $40^{\circ}$C, and the rod is divided into three equal segments, the temperature at nod $1$ (using $\triangle t = 0.1$s) by using an explicit solution at $t=0.2$s is

(For node $0, \, k \dfrac{\partial T}{\partial x} = h (T_{a} - T_{0})$) where $k=9$ W/(m °C), $h=20$ W/m2, $T_{a}=25^{\circ}$C and $T_{0}=$(the temperature at node $0$)