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Again I need help. This time it's exercise 18.1 from LeVeque Finite Volume Methods.

Consider the system $q_t+Aq_x+Bq_y=0$ with $A=\begin{pmatrix}3&1\\1&3\end{pmatrix}$ and $B=\begin{pmatrix}0&2\\2&0\end{pmatrix}$.
Show that these matrices are simultaneously diagonalizible and determine the general solution to this system with arbitrary initial data. In particular sketch how the solution evolves in the $x-y$ plane with data
$$\begin{align*} &q^1(x,y,0)=\left\{\begin{array}{ll}1&\text{ if } x^2+y^2\leq 1\\0&\text{ otherwise }\end{array}\right.\\ &q^2(x,y,0)\equiv 0 \end{align*}$$

So far I've got $A=R\Lambda^xR^{-1}$ and $B=R\Lambda^yR^{-1}$ with

$$\begin{align*} &R&&=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\\ &R^{-1}&&=\frac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\\ &\Lambda^x&&=\begin{pmatrix}4&0\\0&2\end{pmatrix}\\ &\Lambda^y&&=\begin{pmatrix}2&0\\0&-2\end{pmatrix} \end{align*}$$

but now I don't know how to move on. Any tips are appreciated :)

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1 Answer 1

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Setting $p = R^{-1}q$, we have $$ p_t + \Lambda_x p_x + \Lambda_y p_y = 0 \, . $$ Since the $\Lambda^\alpha$ matrices are diagonal, this system is decoupled. Applying the method of characteristics componentwise yields $$ p(x,y,t) = R^{-1} q(x,y,t) = \begin{bmatrix} F_1(\lambda^x_1 x -t, \lambda^y_1 x - \lambda^x_1 y) \\ F_2(\lambda^x_2 x -t, \lambda^y_2 x - \lambda^x_2 y) \end{bmatrix} $$ where $F_1$, $F_2$ are two arbitrary functions, and $\lambda_i^\alpha$ denotes the diagonal entries of $\Lambda_i^\alpha$ (see this post, where the notations correspond to $\Phi = F_i$, $a=1$, $b=\lambda_i^x$, and $c=\lambda_i^y$). Now, it remains to apply the intial condition to determine the arbitrary functions: $$ q(x,y,0) = R\, p(x,y,0) = \begin{bmatrix} F_1(4 x, 2 x - 4 y) + F_2(2 x, -2 (x + y)) \\ F_1(4 x, 2 x - 4 y) - F_2(2 x, -2 (x + y)) \end{bmatrix} . $$ From the initial condition $q_1(x,y,0) = \Bbb I_{x^2 + y^2 < 1}$ and $q_2(x,y,0) = 0$ in OP, we deduce that \begin{aligned} \tfrac12\Bbb I_{x^2 + y^2 < 1} &= F_2(2x,-2(x+y)) \\ & = F_2(X,Y) = \tfrac12\Bbb I_{X^2 + (Y+X)^2 < 4} \\ \tfrac12\Bbb I_{x^2 + y^2 < 1} &= F_1(4x,2 x - 4 y) \\ & = F_1(\xi,\eta) = \tfrac12\Bbb I_{\xi^2 + (\eta-\xi/2)^2 < 16} \end{aligned} where $\Bbb I$ is the indicator function. Finally, $$ q(x,y,t) = \frac12 \begin{bmatrix} \Bbb I_{(x-t/4)^2 + (y-t/8)^2 < 1} + \Bbb I_{(x-t/2)^2 + (y+t/2)^2 < 1} \\ \Bbb I_{(x-t/4)^2 + (y-t/8)^2 < 1} - \Bbb I_{(x-t/2)^2 + (y+t/2)^2 < 1} \end{bmatrix} $$ with the above expressions.

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  • $\begingroup$ ok, so now I have $F_1(4x-t,2x-4y)$ and $F_2(2x-t,-2x-2y)$. But still I have no initial data and, to be honest, no idea how to continue :( $\endgroup$
    – jigga
    Jan 25, 2021 at 13:27
  • $\begingroup$ Wow! Just wow! Never ever I would come to this! I hoped for something nice of a function as a solution. One final question, why the $\frac{1}{2}$ in front of the indicator function? $\endgroup$
    – jigga
    Jan 25, 2021 at 17:52
  • $\begingroup$ @jigga The boundary condition $q_2=0$ gives $F_1(4x,\dots) = F_2(2x,\dots)$. The other boundary condition then gives $$F_1(4x,\dots) + F_2(2x,\dots) = 2F_2(2x,\dots) = \Bbb I . $$ Division by two adds a factor $\frac12$ in front of the indicator function. $\endgroup$
    – EditPiAf
    Jan 26, 2021 at 6:50

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