Commit 95adc8f9 by Shalabh Gautam

assignment 2 added.

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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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\begin{enumerate}
\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 $3 \times 3$ 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*}
\mathbf{A} = \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & c^2 \\
0 & 1 & 0
\end{array} \right) \, , \quad \mathbf{A} = \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{array} \right) \, , \quad \mathbf{A} = \left( \begin{array}{ccc}
a & 0 & 0 \\
0 & 1 & b \\
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) %\\\\
\item Exercise 3.5 of Baumgarte and Shapiro. (3 marks). %\\\\
\end{enumerate}
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