# Linearize non-linear PDE with BCs to hyperbolic problem: How does linearization affect BCs?

I am working with the Shallow Water equations that is a system of non-linear PDEs that simulate water waves propagation on some domain, in my cases the $$x$$-axis. I have here a GIF showing the results (is it possible to insert the GIF into the post? Because I have not figured out how). I will try to keep it short and what I am about to do is linearize the system below and then solve the linearized hyperbolic system analytically (my problem is how I should deal with the boundary conditions for the characteristic variables I obtain). So we have

$$\begin{cases} h_{t} + (hv)_{x} = 0, \\[2mm] (hv)_{t} + (hv^{2} + \frac{1}{2}gh^{2})_{x} = 0, \\[2mm] \text{on} \quad (x,t)\in [0,L]\times[0,\infty) \end{cases} \quad \iff \begin{pmatrix}h \\ hv\end{pmatrix}_{t} + \begin{pmatrix}hv \\ hv^{2} + \frac{1}{2}gh^{2}\end{pmatrix}_{x} = u_{t} + f_{x}(u) = 0.$$

The initial and boundary conditions are given by

$$\begin{cases} \text{Initial conditions:} \quad h(x,0) = H + \varepsilon e^{-(x-L/2)^{2}/w^{2}} \quad \text{and} \quad v(x,0) = 0, \\[2mm] \text{Boundary conditions:} \quad v(0) = 0 \quad \text{and} \quad v(L) = 0. \end{cases}$$

Now if a linearize this system around some constant state, say $$u_{0} = (h_{0},h_{0}v_{0})$$, then I get a hyperbolic system. If we denote

$$q = u - u_{0}$$

the perturbation from this constant state, then we may linearize the system by the following approach:

$$f(u) = \begin{pmatrix}hv \\ hv^{2} + \frac{1}{2}gh^{2}\end{pmatrix} = \left.\begin{cases} u_{1} &= h \\ u_{2} &= hv \end{cases}\right\} = \begin{pmatrix}u_{2} \\ \frac{u_{2}^{2}}{u_{1}} + \frac{1}{2}gu_{1}^{2}\end{pmatrix} \quad \implies f'(u) = \begin{pmatrix}0 & 1\\ - \frac{u_{2}^{2}}{u_{1}^{2}} + gu_{1} & 2\frac{u_{2}}{u_{1}}\end{pmatrix}.$$

If we evaluate $$f'(u)$$ in $$u_{0}$$ we obtain the linearized system matrix $$A$$,

$$A = f'(u)\vert_{u=u_{0}} = \left.\begin{pmatrix}0 & 1\\ - \frac{u_{2}^{2}}{u_{1}^{2}} + gu_{1} & 2\frac{u_{2}}{u_{1}}\end{pmatrix}\right\vert_{u=u_{0}} = \begin{pmatrix}0 & 1\\ - v_{0}^{2} + gh_{0} & 2v_{0}\end{pmatrix}.$$

Furthermore if we choose $$v_{0}=0$$ we obtain the linearized system

$$q_{t} + \begin{pmatrix}0 & 1\\ gh_{0} & 0\end{pmatrix}q_{x} = 0$$

which can be seen to be hyperbolic. Now to my question: It is possible to transform this system and decouple it into the so called characteristic equations (these are two partial differential equations modeling advection). But what will the boundary conditions be for each of these equations? I have read in Leveque $$\S$$ 7.3 that these equations each have one boundary condition, but what are the domains of definition for each characteristic variable?Is it not still that $$x\in[0,L]$$? For me it seems that one bounary condition would not require each equation to stay in the interior.

Let me show in mathematical terms what I mean: After linearization we may transform the system by diagnoalizing $$A$$. We obtain

$$A = \begin{pmatrix}\xi_{1} & \xi_{2}\end{pmatrix}\begin{pmatrix}\lambda_{1} & 0 \\ 0 & \lambda_{2}\end{pmatrix}\begin{pmatrix}\xi_{1} & \xi_{2}\end{pmatrix}^{-1} \quad \text{where} \quad \lambda_{1} = -\sqrt{gh_{0}},\; \xi_{1} = \begin{pmatrix}-1 \\ \sqrt{gh_{0}}\end{pmatrix},\; \lambda_{2} = \sqrt{gh_{0}},\; \xi_{2} = \begin{pmatrix}1 \\ \sqrt{gh_{0}}\end{pmatrix}.$$

Now if we denote $$R = \begin{pmatrix}\xi_{1} & \xi_{2}\end{pmatrix}$$ and set $$z = R^{-1}q$$ we obtain the decoupled system

$$z_{t} + \begin{pmatrix}-\sqrt{gh_{0}} & 0 \\ 0 & \sqrt{gh_{0}}\end{pmatrix}z_{x} = z_{t} + \Lambda z_{x} = 0.$$

The initial conditions for this transformed system become

$$\begin{cases} \text{Initial conditions:} \quad z(x,0) = R^{-1}u(x,0) = R^{-1}\begin{pmatrix}h(x,0)\\ h(x,0)v(x,0)\end{pmatrix} = \frac{1}{2}\begin{pmatrix}-1\\ 1\end{pmatrix}h(x,0) \end{cases}.$$

But what happens to the boundary conditions? Are they given by

$$\begin{cases} \text{Boundary condition } x=0:\quad z(0,t) = R^{-1}u(0,t) = R^{-1}\begin{pmatrix}h(0,t)\\ h(0,t)v(0,t)\end{pmatrix} = \frac{1}{2}\begin{pmatrix}-1\\ 1\end{pmatrix}h(0,t), \\[2mm] \text{Boundary condition } x=L:\quad z(L,t) = R^{-1}u(L,t) = R^{-1}\begin{pmatrix}h(L,t)\\ h(L,t)v(L,t)\end{pmatrix} = \frac{1}{2}\begin{pmatrix}-1\\ 1\end{pmatrix}h(L,t)\;? \end{cases}$$

Because there is no boundary value prescribed for $$h(0,t)$$ or $$h(L,t)$$. So how do I go about solving this analytically from here, how do I treat reflecting boundary conditions like this on a finite domain for a system of hyberbolic equations?

\begin{align} z_{2}(0,t) = -z_{1}(0,t) \quad \text{and} \quad z_{1}(L,t) = -z_{2}(L,t), \end{align}
• Yes, I have read that section. But it only deals with one boundary while I have two. I have solved the linearized system analytically, but my solution shows that the two waves just travel through the boundaries. I have to impose the boundary conditions somehow, but it seems I do not need them to find an analytical solution. Perhaps my solution is just defined on some interval $t\in[0,T]$? Maybe, although this doesn't sound right, I should use some finite volume method (Lax-Wendroff scheme?) for the linearized system? I will edit the plot above. – SimpleProgrammer Mar 23 at 13:38