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I asked the question on math.stackexchange.com, but didn't get any reply. So, I asked it again here. Any suggestion or hint is welcome, and thank you for your attention.

Consider the system of first order PDEs

$ \left\{ \begin{eqnarray} \frac{\partial}{\partial t} v_1 + \frac{\partial}{\partial x_1} p_1 + \eta(x_1) v_1 = 0 \\ \frac{\partial}{\partial t} v_2 + \frac{\partial}{\partial x_2} p_2 + \eta(x_2) v_2 = 0 \\ \frac{\partial}{\partial t} p_1 + \frac{\partial}{\partial x_1} v_1 + \eta(x_1) p_1 = f(t) \\ \frac{\partial}{\partial t} p_2 + \frac{\partial}{\partial x_2} v_2 + \eta(x_2) p_2 = 0 \end{eqnarray} \right. $

with initial condition $v_1 = v_2 = p_1 = p_2 = 0$ at $t=0$, where $v_1,v_2,p_1,p_2 : \mathbb{R}^3 \to \mathbb{R} $ are the unknown functions, $\eta, f: \mathbb{R} \to \mathbb{R}$ are given smooth functions.

I would like to consider this system of PDEs with the method of characteristics. I know the method for a single PDE, but I don't know how to find the characteristics of such a system.

Could anyone please give some advice?

Regards.

I could understand that this system is a hyperbolic one.

Take $U=(v_1,v_2,p_1,p_2)'$, then the system can be rewritten as

$\frac{\partial}{\partial t} U + \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 1 ~ & 0 \\ 0 & 0 & 0 & 0 \\ 1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \frac{\partial}{\partial x_1} U + \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 1 \\ 0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) \frac{\partial}{\partial x_2} U + \left( \begin{array}{cccc} ~\eta(x_1) ~ & 0 ~ & 0 ~ & 0 \\ 0 & \eta(x_2) & 0 & 0 \\ 0 ~ & 0 ~ & \eta(x_1) ~ & 0 \\ 0 & 0 & 0 & \eta(x_2) \end{array} \right) U =\left( \begin{array}{cccc} 0 \\ 0 \\ f(t) \\ 0 \end{array} \right) . $

Take $B_1 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 1 ~ & 0 \\ 0 & 0 & 0 & 0 \\ 1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) $, $B_2 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 0 & 0 & 1 \\ 0 ~ & 0 ~ & 0 ~ & 0 \\ 0 & 1 & 0 & 0 \end{array} \right) $, then define a matrix

$B (y_1,y_2)= y_1 B_1 + y_2 B_2 = \left( \begin{array}{cccc} ~0 ~ & 0 ~ & y_1 ~ & 0 \\ 0 & 0 & 0 & y_2 \\ y_1 ~ & 0 ~ & 0 ~ & 0 \\ 0 & y_2 & 0 & 0 \end{array} \right)$.

The eigenvalues of $B$ are $-y_1, y_1, -y_2, y_2$, and the corresponding eigenvectors are $(-1, 0, 1, 0)', (1, 0, 1, 0)', (0, -1, 0, 1)', (0, 1, 0, 1)'$.

Since the eigenvalues of $B$ are distinct and all real and the four eigenvectors are linearly independent, the system of first order PDEs is hyperbolic.

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First of all, your system seems to uncouple quite strongly. The first and third equations only involve the unknowns $v_1$ and $p_1$ and the second and fourth equations only involve $v_2$ and $p_2$, so you are really talking about two independent problems, so you should separate them this way.

Second, if you add the first and third equations and set $z = p_1 + v_1$, this becomes the single PDE $$ \frac{\partial z}{\partial t} + \frac{\partial z}{\partial x_1} + \eta(x_1)z = f(t), $$ so it's an equation for a single unknown, easily solved by the method of characteristics. Similarly, if you subtract the first equation from the third and set $w = p_1 - v_1$, then you get $$ \frac{\partial w}{\partial t} - \frac{\partial w}{\partial x_1} + \eta(x_1)w = f(t), $$ again, an equation for a single unknown.

Similar remarks apply to the second and fourth equations, so you really have four independent first order linear PDE for single unknowns, so the method of characteristics can be applied to each separately.

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  • $\begingroup$ I've never been completely certain of this, but I think the method of characteristics applies only to scalar first order PDEs. So it may be worth underscoring that the only way to apply this method to a system of PDEs is to decouple it into scalar equations, as above. $\endgroup$ – Igor Khavkine Jan 7 '15 at 15:30
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    $\begingroup$ @IgorKhavkine: That's sort of true. It would be more accurate to say that the method of characteristics generalizes to a class of equations that includes the scalar first order PDE as a special case. The key term to look for is 'the method of Darboux', which is a method for searching for higher order Riemann invariants (as they are sometimes called) for higher order PDE or PDE in more unknowns than one. Darboux developed it in the 1870s as a method of integrating a large class of nonlinear PDE. Of course not all PDE are integrable by the method of Darboux, but many interesting ones are. $\endgroup$ – Robert Bryant Jan 7 '15 at 22:02

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