Solving NR corse assignment 2.

Assignments/Assignment_2/Assignment_2.tex

@@ -45,7 +45,7 @@ g_{ab} = \left( \begin{array}{cccc}
 - (1 - 2H) & 2H & 0 & 0 \\
 - (1 - 2H) & 2H & 0 & 0 \\
 2H & (1 + 2H) & 0 & 0 \\
 2H & (1 + 2H) & 0 & 0 \\
 0 & 0 & r^2 & 0 \\
 0 & 0 & r^2 & 0 \\
-00 & 00 & 0 & r^2 \sin^2 \theta
+0 & 0 & 0 & r^2 \sin^2 \theta
 \end{array} \right) \, ,
 \end{array} \right) \, ,
 \end{align*}
 \end{align*}
 and
 and
@@ -54,9 +54,10 @@ g^{ab} = \left( \begin{array}{cccc}
 - (1 + 2H) & 2H & 0 & 0 \\
 - (1 + 2H) & 2H & 0 & 0 \\
 2H & (1 - 2H) & 0 & 0 \\
 2H & (1 - 2H) & 0 & 0 \\
 0 & 0 & r^{-2} & 0 \\
 0 & 0 & r^{-2} & 0 \\
-00 & 00 & 0 & r^{-2} \csc^2 \theta
+0 & 0 & 0 & r^{-2} \csc^2 \theta
 \end{array} \right) \, .
 \end{array} \right) \, .
 \end{align*}
 \end{align*}
+Therefore,~$\sqrt{-g} = r^2 \sin \theta$, which gives~$\sqrt{-g} \, g^{ab} \p_b \phi = r^2 \sin \theta \, [- (1+2H) \p_t \phi + 2H \p_r \phi, 2H \p_t \phi + (1-2H) \p_r \phi, 0, 0]$, and~$ \frac{1}{\sqrt{-g}} \p_a (\sqrt{-g} \, g^{ab} \p_b \phi) = - (1+2H) \p_t^2 \phi + 2H \p_t \p_r \phi + (2/r) [2H \p_t \phi + (1-2H) \p_r \phi] + [(-2H/r) \p_t \phi + 2H \p_r \p_t \phi + (2H/r) \p_r \phi + (1-2H) \p_r^2 \phi] = - (1+2H) \p_t^2 \phi + 4 H \p_t \p_r \phi + (2H/r) \p_t \phi + (2/r)(1-H) \p_r \phi + (1-2H) \p_r^2 \phi$.
 
 
 \item Consider the equation $a u_{xx} + 2b u_{xy} + c _{yy} = f(u, \p u, x,y)$, where $u = u(x,y)$ and the subscripts denote partial derivatives. For what value of $a$, $b$ and $c$ is this equation elliptic, parabolic and elliptic PDEs, give reason: (a). $a = b = c = 1$, (b). $a = c = 1$, $b = 0$, (c) $a = 1$, $b = 0$, $c = -1$. (3 marks) \\
 \item Consider the equation $a u_{xx} + 2b u_{xy} + c _{yy} = f(u, \p u, x,y)$, where $u = u(x,y)$ and the subscripts denote partial derivatives. For what value of $a$, $b$ and $c$ is this equation elliptic, parabolic and elliptic PDEs, give reason: (a). $a = b = c = 1$, (b). $a = c = 1$, $b = 0$, (c) $a = 1$, $b = 0$, $c = -1$. (3 marks) \\