Quiz Chapter 10.03: Elliptic 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 elliptic 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 finite difference approximation of $\dfrac{\partial^{2} u}{\partial x^{2}}$ in the elliptic equation

- - $\dfrac{\partial^{2} u}{\partial x^{2}} + \dfrac{\partial^{2} u}{\partial y^{2}} = 0$

at $(x, \, y)$ can be approximated as

\dfrac{\partial^{2} u}{\partial x^{2}} \cong \dfrac{u(x + \Delta x, \, y) - 2u(x, \, y) + u(x - \Delta x, \, y)}{(\Delta x)^{2}}

\dfrac{\partial^{2} u}{\partial x^{2}} \cong \dfrac{u(x + \Delta x, \, y) - u(x, \, y) + u(x - \Delta x, \, y)}{(\Delta x)^{2}}

\dfrac{\partial^{2} u}{\partial x^{2}} \cong \dfrac{u(x, \, y + \Delta y, \, y) - 2u(x, \, y) + u(x, \, y - \Delta y, \, y)}{(\Delta x)^{2}}

\dfrac{\partial^{2} u}{\partial x^{2}} \cong \dfrac{u(x + \Delta x, \, y) - u(x - \Delta x, \, y)}{2 \Delta x}

4. Find the temperature at the interior node given in the following figure using the direct method

45.19^{\circ}

48.64^{\circ}

50.00^{\circ}

56.79^{\circ}

5. Find the temperature at the interior node given in the following figure

Using the Lieberman method and relaxation factor of $1.2$ , the temperature at $x=3, \, y=6$ estimated after $2$ iterations is (use the temperature of interior nodes as $50^{\circ}$ C for the initial guess)

52.36^{\circ}

53.57^{\circ}

56.20^{\circ}

58.64^{\circ}

6. Find the steady-state temperature at the interior node as given in the following figure

53.57^{\circ}

66.40^{\circ}

68.20^{\circ}

69.59^{\circ}