Solving NR corse assignment 2.

Assignments/Assignment_2/Assignment_2.tex

@@ -57,7 +57,34 @@ g^{ab} = \left( \begin{array}{cccc}
 0 & 0 & 0 & r^{-2} \csc^2 \theta
 \end{array} \right) \, .
 \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$.
+Therefore,~$\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$, where we used~$H = M/r$. Substituting~$\phi = \Phi/r$, we get
+\begin{align*}
+\nabla_a \nabla^a \phi \equiv & \frac{1}{\sqrt{-g}} \p_a (\sqrt{-g} \, g^{ab} \p_b \phi) \\
+= & - \frac{(1+2H)}{r} \p_t^2 \Phi - \frac{4H \p_t \Phi}{r^2} + \frac{4 H \p_t \p_r \Phi}{r} + \frac{2H \p_t \Phi}{r^2} - \frac{2 (1-H) \Phi}{r^3} + \frac{2(1-H) \p_r \Phi}{r^2} \\
+& + (1-2H) (\frac{2 \Phi}{r^3} - \frac{\p_r \Phi}{r^2} - \frac{\p_r \Phi}{r^2} + \frac{\p_r^2 \Phi}{r}) \\
+= & - \frac{(1+2H)}{r} \p_t^2 \Phi - \frac{2H \p_t \Phi}{r^2} + \frac{4 H \p_t \p_r \Phi}{r} - \frac{2 H \Phi}{r^3} + \frac{2H \p_r \Phi}{r^2} + \frac{(1-2H) \p_r^2 \Phi}{r} \, .
+\end{align*}
+Therefore,~$\nabla_a \nabla^a \phi = 0$ gives
+\begin{align*}
+- (1+2H) \p_t^2 \Phi + 4H \p_r \p_t \Phi + (1-2H) \p_r^2 \Phi - \frac{2H}{r} \p_t \Phi + \frac{2H}{r} \p_r \Phi - \frac{2 H}{r^2} \Phi = 0 \, .
+\end{align*}
+Taking~$A \equiv - (1+2H)$,~$B \equiv 4H$ and~$C \equiv (1-2H)$, we get~$AC - B^2 = - (1+2H) (1-2H) - (4H)^2 = - 1 < 0$ for all~$r$. Therefore, this equation iss hyperbolic both inside and outside the horizon at~$2H = 1$, or~$r = 2M$.
+
+Let,~$\p_t \Phi \equiv - \Pi$ and~$\p_r \Phi \equiv \Psi$. This gives the following system of eqations
+\begin{align*}
+& \p_t \Phi = - \Pi \, , \\
+& \p_t \Psi + \p_r \Pi = 0 \, , \\
+& \p_t \Pi + \frac{4H}{1+2H} \p_r \Pi - \frac{1-2H}{1+2H} \p_r \Psi = \frac{2H}{r} (\Pi + \Psi) - \frac{2H}{r^2} \Phi \, .
+\end{align*}
+This gives
+\begin{align*}
+\mathbf{A} = \left( \begin{array}{ccc}
+0 & 0 & 0 \\
+0 & 0 & 1 \\
+0 & - \frac{1-2H}{1+2H} & \frac{4H}{1+2H}
+\end{array} \right)
+\end{align*}
+Eigenvalues of this matrix are~$0$,~$\frac{4H \pm \sqrt{16 H^2 - 4(1-4H^2)}}{2(1 + 2H)}$, or~$0$,~$-1$ and~$\frac{1-2H}{1 + 2H}$.
 
 \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) \\