2.8. Hyperbolic first order systems in 1D


## 2.8. Hyperbolic first order systems with one spatial variable

### Definition

We consider system $$E U_t + AU_x+ BU=F \label{eq-2.8.1}$$ where $E,A,B$ are $n\times n$-matrices, $U$ is unknown $n$-vector (column) and $F$ is known $n$-vector (column).

### Completely separable systems

Assume that $E$ and $A$ are constant matrices, $E$ is non-degenerate, $E^{-1}A$ has real eigenvalues $\lambda_1,\ldots, \lambda_n$ and is diagonalisable: $E^{-1}A= Q^{-1}\Lambda Q$ with $\Lambda =\diag(\lambda_1,\ldots, \lambda_n)$ (diagonal matrix with $\lambda_1,\ldots, \lambda_n$ on the diagonal.

Then substituting $U=QV$ (or $V=Q^{-1}U$) we have \begin{equation*} QV_t + E^{-1}AQV_x+ E^{-1}BQV=E^{-1}F \end{equation*} or $$V_t + \Lambda V_x+ Q^{-1}E^{-1}BQV=Q^{-1}E^{-1}F. \label{eq-2.8.2}$$ In particular if $Q^{-1}E^{-1}BQ$ is also a diagonal matrix: $Q^{-1}E^{-1}BQ=\diag(\alpha_1,\ldots,\alpha_n)$ (which is the case provided $B=0$) we have $n$ separate equations $$V_j,_t +\lambda_j V_j,_x +\alpha_j V_j = f_j \label{eq-2.8.3}$$ and we can apply the theory of Section 2.1.

Definition 1. Lines $x-\lambda_j t= \const$ are characteristics, $V_j$ are called Riemannian invariants. If $\alpha_j=0$, $f_j=0$ these Riemannian invariants are constant along characteristics.

### IVP (Cauchy problem)

Consider Cauchy problem: $U|_t=0= G(x)$, $x\in \mathbb{R}$.

Proposition 1.

1. Let $E=I$, $A=\Lambda$ (already diagonalized) and $B=0$. Then $U_j$ at point $P)$ is defined by $G_j(P_j)$ and $F_j$ on a segment of characteristics connecting $P$ and $P_j$ where $P_j$ is an intersection of $x-\lambda_j t=\const$ passing through $P$ and $\{t=0\}$; ($j=1,\ldots,n$).
2. Let $B=0$. Then $U$ at point $P$ is defined by $G(P_j)$ and by $F$ on segments of characteristics connecting $P$ and $P_j$; ($j=1,\ldots,n$).

Proof. The first statement is obvious and the second follows from it. Note that transform by $Q$ messes up components of $U$, and $F$, and $G$.

### IBVP

Consider now equations (\ref{eq-2.8.3}) in domain $\Omega=\{t>0, x> \mu t\}$ assuming that the lateral boundary $\Gamma=\{x=\mu t, t>0\}$ is not a characteristic, i.e. $\mu$ is not one of the numbers $\lambda_1,\ldots,\lambda_n$. Then (renumbering Riemannian invariants if necessary) we have $$\lambda_1\le \ldots \le \lambda_m < \mu <\lambda_{m+1}\le \ldots \le \lambda_n. \label{eq-2.8.4}$$ Then with equation in $\Omega$ and initial data on $\{t=0, x>0\}$ we can find $V_1,\ldots,V_m$ everywhere in $\Omega$ and thus on $\Gamma$; we call $V_1,\ldots,V_m$ incoming Riemannian invariants. On the other hand, we define this way $V_{m+1},\ldots, V_n$ only as $x\ge \lambda_{m+1},\ldots, \lambda_n$ respectively and therefore not on $\Gamma$; we call them outgoing Riemannian invariants.

To define outgoing Riemannian invariants $V_{m+1},\ldots, V_n$ on $\Gamma$ and thus in the rest of $\Omega$ we need boundary condition $CU|_\Gamma =H$ where $C$ is $(n-m)\times n$-matrix and $H=H(t)$ is $(n-m)$-vector.

Indeed we need as many equations as outgoing Riemannian invariants. However it is not sufficient. We need also to assume that the following non-degeneracy assumption is fulfilled: $(n-m)\times (n-m)$-matrix $C'$ obtained from $C$ by selecting last $(n-m)$ columns (corresponding to outgoing Riemannian invariants) is non-degenerate.

### Compatibility condition

The solution is continuous if and only if compatibility condition $CG(0)=H(0)$. Think why. If this condition fails $U$ is discontinuous (has jumps) along characteristics going into $\Omega$ from the corner point $(0,0)$.

But even if solution is continuous it is not necessarily continuously differentiable (one needs more compatibility conditions for this); even more compatibility conditions for $U$ be twice continuously differentiable etc.

Problem 1.

a. Prove compatibility condition $CG(0)=H(0)$ for continuity of solution $U$; b. Derive a compatibility condition for continuity of the first derivatives $U_t,U_x$ of the solution; c. Derive a compatibility condition for continuity of the second derivatives $U_{tt},U_{tx}, U_{xx}$ of the solution. \end{problem}

### General case

What happens if $Q^{-1}E^{-1}BQ$ is not a diagonal matrix? More generally consider $E=E(x,t)$, $A=A(x,t)$, $B=B(x,t)$. Then assuming that $E^{-1}A$ is smooth diagonalisable (t.m. one can select $Q=Q(x,t)$ a smooth matrix; this is the case provided $\lambda_1,\ldots,\lambda_n$ are real and distinct at every point) we have $$V_t + \Lambda V + Q^{-1}(Q_t+AQ_x +CQ)V=Q^{-1}F; \label{eq-2.8.5}$$ so while main part of system broke into separate equations, they are entangled through lower order terms.

Then main conclusions are the same:

#### IVP

For Cauchy problem consider point $P$ and a triangle $\Delta(P)$ formed by two characteristics: the leftmost and the rightmost going back in time and by initial line. This triangle (curvilinear if $\lambda_j$ are not constant) is the domain of dependence of $P$.

#### IBVP

Again we need to assume that $\Gamma$ (which could be a smooth curve rather than the straight ray) is non-characteristic at each point, then the numbers of incoming and outgoing Riemannian invariants remain constant along $\Gamma$ and we need impose as many boundary condition as there are outgoing Riemannian invariants and we need to impose non-degeneracy condition in each point of $\Gamma$.