@@ -19,7 +19,7 @@ Please submit your solutions in the LATEXed pdf form to be considered. \\(Due on
%\vspace{1.0em}
\begin{enumerate}
\item Exercises 6.1 and 6.2 of Baumgarte and Shapiro. (3 marks). %\\\\
\item Exercises 6.1 and 6.2 of Baumgarte and Shapiro. (3 marks). \\
\textit{Ex 6.1 Proof.} The equation can also be written as $(c^{-1}\p_t +\p_r)(c^{-1}\p_t -\p_r)\phi=0$. As $(c^{-1}\p_t +\p_r) f(x - ct)=(c^{-1}\p_t +\p_r) g(x + ct)=0$, we get the result.~$\blacksquare$
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@@ -37,9 +37,30 @@ which on solving gives
\end{align*}
Therefore, with $\mathbf{u}=(\phi, u, v)$, the velocity matrix $\mathbf{A}=$ diag$(0, c, -c)$.~$\blacksquare$
\item Exercise 6.3 of Baumgarte and Shapiro. (6 marks). %\\\\
\item Exercise 6.3 of Baumgarte and Shapiro. (6 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) %\\\\
\textit{Proof.} Here, $l_a =(1,1,0,0)$ and $l^a =(-1,1,0,0)$. Taking $x^a =(t,r,\theta,\phi)$, we get $\eta_{ab}=$ diag$(-1,1,r^2,r^2\sin^2\theta)$, and the Schwarzschild metric in Kerr-Schild coordinates takes form $ds^2=(\eta_{ab}+2 H l_a l_b) dx^a dx^b =-(1-2H) dt^2+4 H dt dr +(1+2H) dr^2+ r^2(d \theta^2+\sin^2\theta d \phi^2)$. This gives
\begin{align*}
g_{ab} = \left( \begin{array}{cccc}
- (1 - 2H) & 2H & 0 & 0 \\
2H & (1 + 2H) & 0 & 0 \\
0 & 0 & r^2 & 0 \\
00 & 00 & 0 & r^2 \sin^2 \theta
\end{array}\right) \, ,
\end{align*}
and
\begin{align*}
g^{ab} = \left( \begin{array}{cccc}
- (1 + 2H) & 2H & 0 & 0 \\
2H & (1 - 2H) & 0 & 0 \\
0 & 0 & r^{-2}& 0 \\
00 & 00 & 0 & r^{-2}\csc^2 \theta
\end{array}\right) \, .
\end{align*}
\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) \\
\textit{Proof.} Since (a). $ac - b^2=0$, the system is parabolic. (b). $ac - b^2 > 0$, the system is elliptic. (c) $ac - b^2 < 0$, the system is hyperbolic.~$\blacksquare$
\item Consider the hyperbolic differential equation $\p_t \mathbf{u}+\mathbf{A}\p_x =\mathbf{S}$, where $\mathbf{u}=(u_1, u_2, u_3)^T$ is a state vector, $A$ is a $3\times3$ velocity matrix and $\mathbf{S}=(s_1, s_2, s_3)^T$ is the source vector which depends at most on $\mathbf{u}$ and the coordinates, but not on $\p\mathbf{U}$. For which of the following values of $\mathbf{A}$ is the equation is weakly, strongly and symmetric hyperbolic, give reasons:
\begin{align*}
...
...
@@ -57,7 +78,9 @@ a & 0 & 0 \\
0 & b & -1
\end{array}\right) \, ,
\end{align*}
where $a$, $b$ and $c$ are all real numbers? (See Sec. 11.1 of Baumgarte and Shapiro.) (3 marks) %\\\\
where $a$, $b$ and $c$ are all real numbers? (See Sec. 11.1 of Baumgarte and Shapiro.) (3 marks) \\
\textit{Proof.} The first matrix is diagonalizable with all real eigenvalues but not symmetric, the second one is not diagonalizable but has real eigenvalues, and the third one is symmetric, and hence diagonalizable, and also has real eigenvalues. Therefore, the first value of the velocity matrix~$\mathbf{A}$, the equation is strongly hyperbolic. For the second one, the equation is weakly hyperbolic. And for the third one, the equation is symmetric hyperbolic.~$\blacksquare$
\item Exercise 3.5 of Baumgarte and Shapiro. (3 marks). %\\\\
