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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?

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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

$$\mathbf A\frac{\partial \mathbf u}{\partial x} + \mathbf B\frac{\partial \mathbf u}{\partial y} = \mathbf c$$

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:

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

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^2\big] \\ = & p^2(v-\lambda u)\big[(u^2-c^2)\lambda^2-2uv\lambda+v^2-c^2\big] \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^2-2uv\lambda+v^2-c^2\big]=0 \\ \implies & \lambda = \frac vu, \frac{uv\pm \sqrt{u^2v^2-(u^2-c^2)(v^2-c^2)}}{u^2-c^2} \\ \implies & \lambda = \frac vu, \frac{uv\pm c\sqrt{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.

$\frac vu \neq \frac{uv\pm c\sqrt{u^2+v^2-c^2}}{u^2-c^2}$ because $\lambda = \frac vu$ does not satisfy $\big[(u^2-c^2)\lambda^2-2uv\lambda+v^2-c^2\big]=0$, whereas $\frac{uv + c\sqrt{u^2+v^2-c^2}}{u^2-c^2} \neq \frac{uv - c\sqrt{u^2+v^2-c^2}}{u^2-c^2}$ 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 $u^2+v^2>c^2$.

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  • $\begingroup$ hi when solving this, i get $v-\lambda u=c$ where am i going wrong? $\endgroup$ Dec 12, 2018 at 16:20

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