Jtdylktuy

I know that for a system to be hyperbolic it must have 2 real distinct eigenvalues $lambda$ where $det(B-lambda A)=0$. My system of equations are:
begin{aligned}
(pu)_x + (pv)_y &= 0 \
p(uu_x + vu_y) + c(p)^2p_x &= 0 \
p(uv_x +vv_y)+c(p)^2p_y &= 0
end{aligned}

I need to show that the system is hyperbolic if $u^2 + v^2 > c^2$. I am stuck on how to formulate the matrices $A,B$. Could someone please help?

In order for the system to be hyperbolic, it needs to have $n$ distinct real eigenvalues if you are working with $n$ dependent variables. In this case, $n=3$.

The first thing to do is to write it in the form

$$mathbf Afrac{partial mathbf u}{partial x} + mathbf Bfrac{partial mathbf u}{partial y} = mathbf c$$

where $mathbf u = (p,u,v)^T$ is the vector of dependent variables.

All you need to do is expand out the derivatives in the given equations and write them out neatly:

begin{matrix}
up_x & + & pu_x & + & & & vp_y & + & & & pv_y & = & 0 \
c^2p_x & + & puu_x & + & & & & & pvu_y & & & = & 0 \
& & & & puv_x & + & c^2p_y & + & & & pvv_y & = & 0
end{matrix}

Notice how I have arranged each equation in the order $p_x$, $u_x$, $v_x$, $p_y$, $u_y$, $v_y$. We can now write it in the form as previously mentioned:

$$
begin{pmatrix}
u & p & 0 \
c^2 & pu & 0 \
0 & 0 & pu
end{pmatrix}frac{partial}{partial x} begin{pmatrix} p \ u \ vend{pmatrix} + begin{pmatrix}
v & 0 & p \
0 & pv & 0 \
c^2 & 0 & pv
end{pmatrix}frac{partial}{partial y} begin{pmatrix} p \ u \ vend{pmatrix} = 0
$$

There you have your $mathbf A$ and $mathbf B$, and the next step should be straightforward.

You then compute $text{det}(mathbf B- lambda mathbf A)$:

begin{align}
text{det}(mathbf B- lambda mathbf A) = & text{det} begin{pmatrix} v - lambda u & - lambda p & p \ -lambda c^2 & pv-lambda pu & 0 \ c^2 & 0 & pv - lambda pu end{pmatrix} \
= & (v-lambda u)(pv-lambda pu)(pv-lambda pu)-p(pv-lambda pu)c^2-(pv-lambda pu)(-lambda p)(-lambda c^2) \
= & p^2(v-lambda u)big[(v-lambda u)^2-c^2-lambda ^2 c^2big] \
= & p^2(v-lambda u)big[(u^2-c^2)lambda^2-2uvlambda+v^2-c^2big]
end{align}

Set this to $0$ and solve for $lambda$:

begin{align}
& text{det}(mathbf B- lambda mathbf A)=0 \
implies & p^2(v-lambda u)big[(u^2-c^2)lambda^2-2uvlambda+v^2-c^2big]=0 \
implies & lambda = frac vu, frac{uvpm sqrt{u^2v^2-(u^2-c^2)(v^2-c^2)}}{u^2-c^2} \
implies & lambda = frac vu, frac{uvpm csqrt{u^2+v^2-c^2}}{u^2-c^2}
end{align}

In order for the system to by hyperbolic, these three roots must be real and distinct.

$frac vu neq frac{uvpm csqrt{u^2+v^2-c^2}}{u^2-c^2}$ because $lambda = frac vu$ does not satisfy $big[(u^2-c^2)lambda^2-2uvlambda+v^2-c^2big]=0$, whereas $frac{uv + csqrt{u^2+v^2-c^2}}{u^2-c^2} neq frac{uv - csqrt{u^2+v^2-c^2}}{u^2-c^2}$ as long as $u^2+v^2-c^2 neq 0$ (i.e. $u^2+v^2 neq c^2$).

Finally, these roots are real if the expression in the square root is non-negative, i.e. if $u^2+v^2 geq c^2$.

Hence, the system is hyperbolic if $u^2+v^2>c^2$.