Classifying linear first-order PDE system (elliptic, hyperbolic, or parabolic)

Question

2. Consider the constants $$\begin{aligned} & (\text i.)\; a_1 = b_1 = a_2 = b_2 = 1 \\ & (\text {ii}.)\; a_1 = b_2 = 1, \quad b_1 = 0, \quad a_2 = -1 \\ & (\text {iii}.)\; a_1 = b_1 = b_2 = 1, \quad a_2 = -1 \end{aligned} \tag{3}$$ classify the following system for the constants given above $$\begin{aligned} & a_1 u_x + a_2 v_y = g_1 \\ & b_1 v_x + b_2 u_y = g_2 \end{aligned} \tag{4}$$ where $g_1$ and $g_2$ are +ive constants.

I have already put the system in the form $$ Aq_x + Bq_y = C, $$ with $$ A = \begin{bmatrix} a_1 & 0\\ 0 &b_1 \end{bmatrix},\quad B= \begin{bmatrix} 0 & a_2\\ b_2 & 0 \end{bmatrix},\quad q= \begin{pmatrix} u\\ v \end{pmatrix}. $$ But then, I do not know how to continue from here on to find out the eigenvalues due to the term $$ C = \begin{pmatrix} g_1\\ g_2 \end{pmatrix} $$ that is confusing me. If $C$ were $0$, then I'd know how to proceed, but unfortunately this is not the case.

Any help would be appreciated.

Answer 1

Let us differentiate the first equation w.r.t. $y$ and the second one w.r.t. $x$: \begin{aligned} a_1 u_{xy} + a_2 v_{yy} &= 0 \\ b_1 v_{xx} + b_2 u_{yx} &= 0 \end{aligned} Using the equality of mixed derivatives for $u$ and the fact that $a_1 = b_2 = 1 \neq 0$, we obtain the equations $$ u_{xy} = u_{yx} = -a_2/a_1 v_{yy} = -b_1/b_2 v_{xx} $$ Since $a_2\neq 0$, we are left with the second-order linear PDE $A v_{xx} + 2 B v_{xy} + Cv_{yy} = 0$ satisfied by $v$, with $A = \frac{a_1 b_1}{a_2 b_2}$, $B = 0$ and $C = -1$.

• The case (i) is hyperbolic since $B^2 - A C > 0$.
• The case (ii) is parabolic since $B^2 - A C = 0$.
• The case (iii) is elliptic since $B^2 - A C < 0$.

An alternative derivation would consist in the diagonalization of the matrices $$ \mathcal A = \begin{bmatrix} a_1 & 0 \\ 0 & b_1 \end{bmatrix} \qquad\text{and}\qquad \mathcal B = \begin{bmatrix} 0 & a_2 \\ b_2 & 0 \end{bmatrix} $$ as $\mathcal A = P D_{\mathcal A} P^{-1}$ and $\mathcal B = P D_{\mathcal B} P^{-1}$, where $D_{\mathcal A}$, $D_{\mathcal B}$ are diagonal matrices. For instance, (i) leads to $$ P = \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} \qquad\text{and}\qquad P^{-1} = \begin{bmatrix} -1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix} $$ so that $D_{\mathcal A} = \text{diag}(1,1)$ and $D_{\mathcal B} = \text{diag}(-1,1)$. Hence, the first-order system $\mathcal A q_x + \mathcal B q_y = g$ with $q = (u,v)^\top$ and $g = (g_1, g_2)^\top$ rewrites as $p_x + D_{\mathcal B} p_y = f$ with $p = P^{-1} q$ and $f = P^{-1} g$. Since $D_{\mathcal B}$ has real eigenvalues, this system is a first-order hyperbolic system.