Quiz Chapter 10.02: Parabolic 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

B^{2} - 4AC

<

0

B^{2}-4AC>0

B^{2}-4AC=0

B^{2}-4AC \neq 0

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

x > \left( \dfrac{1}{12} \right)^{\frac{1}{3}}

x

<

\left( \dfrac{1}{12} \right)^{\frac{1}{3}}

for all values of

x

x = \left( \dfrac{1}{12} \right)^{\frac{1}{3}}

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

40.7134^{\circ}

C

40.6882^{\circ}

C

40.7033^{\circ}

C

40.6956^{\circ}

C

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

40.7134^{\circ}

C

40.6882^{\circ}

C

40.7033^{\circ}

C

40.6956^{\circ}

C

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

40.7134^{\circ}

C

40.6882^{\circ}

C

40.7033^{\circ}

C

40.6956^{\circ}

C

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 $0$ )

41.6478^{\circ}

C

38.435^{\circ}

C

39.9983^{\circ}

C

37.5798^{\circ}

C