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Circulated on 18/02/2022
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{\bf\Large NR Topical Course Spring 2022 Assignment 2}\\
Please submit your solutions in the LATEXed pdf form to be considered. \\(Due on 04/03/2022)
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\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.~$\blacksquare$
\textit{Ex 6.2 Proof.} Substituting $\phi= g(x + ct)+ h(x - ct)$, we get $u =-2c h'(x - ct)$ and $v =2c g'(x + ct)$. This solution also gives $\p_t \phi= c (g' - h')=(u + v)/2$ and $\p_x \phi= g' + h' =(v - u)/(2c)$. This gives, $k =-(u + v)/2$ and $l =(v - u)/(2c)$. Substituting these in the equations, we get
\begin{align*}
&\p_t \phi = (u + v)/2 \\
& - \p_t (u + v)/2 + c \p_x (v - u)/2 = - \rho\\
&\p_t (v - u)/2 - c \p_x (u + v)/2 = 0 \, ,
\end{align*}
which on solving gives
\begin{align*}
&\p_t \phi = (u + v)/2 \\
&\p_t u + c \p_x u = \rho\\
&\p_t v - c \p_x v = \rho\, .
\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 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 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: