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The question is as followed:

"Consider the initial value problem (IVP) for the linearised shallow water equations

(1) $\frac{\partial h}{\partial t} + H_0 \frac{\partial u}{\partial x} = 0$ ;

(2) $\frac{\partial u}{\partial t} + g \frac{\partial h}{\partial x} = 0$

With the initial conditions $h(x,0)=h_0$ ; $u(x,0)=0$ ; for $-\infty<x<\infty$

Specify the function $h_0(x)$ and find the corresponding solution to the IVP. "

Now I have attempted this to the best of my ability, but this topic was one which we were not formally taught due to the current circumstances meaning that my knowledge is entirely from a set of typed notes which only show how the linearised equations are generated and not how they are solved.

Firstly I realised that differentiating equation (2) w.r.t. $t$ and then substituting the expression for $\frac{\partial h}{\partial t}$ gives and expression in the form of the wave equation: $u_{tt}-c^2u_{xx}=0$ where $c^2=gH_0$

The same can be done to equation (1) w.r.t. $x$ to give the wave equation: $h_{tt}-c^2h_{xx}=0$ where $c^2=gH_0$.

I have found examples on how to solve the wave equation assuming the solution is in the form $u(x,t)=p(x-ct)+q(x+ct)$ when given initial conditions $u(x,0)=u_0(x)$ and $u'(x,0)=a_0(x)$ but I cannot understand how to do it with only $u(x,0)$ and $h(x,0)$ provided?

Any tips would be greatly appreciated.

Edit: I have since found that you can deduce that $h_t(x,0)=0$ given that $u(x,0)=0$. Can anyone confirm this?

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1 Answer 1

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The problem with differentiating the PDEs is that you don't know a priori how differentiable the functions are i.e whether the functions are $C^{1}, C^{2}$ etc. If you assume the solutions are regular enough, then there is no problem (and if this is for an undergraduate class, this is probably assumed). However, you can avoid these regularity issues by using the method of characteristics. Your problem in matrix form is

\begin{align} \begin{pmatrix} h \\ u \end{pmatrix}_{t} &= \begin{pmatrix} 0 & - H_{0} \\ - g & 0 \end{pmatrix} \begin{pmatrix} h \\ u \end{pmatrix}_{x} \\ \implies \vec{h}_{t} &= M \vec{h}_{x} \end{align}

Computing the eigenvalues and eigenvectors of $M$ allows us to diagonalise $M$ as $M = P D P^{-1}$, where

\begin{align} P &= \begin{pmatrix} \sqrt{H_{0}/g} & -\sqrt{H_{0}/g} \\ 1 & 1 \end{pmatrix} \\\\ \implies P^{-1} &= \frac{1}{2 \sqrt{H_{0}/g}} \begin{pmatrix} 1 & \sqrt{H_{0}/g} \\ -1 & \sqrt{H_{0}/g} \end{pmatrix} \\\\ D &= \begin{pmatrix} \sqrt{g H_{0}} & 0 \\ 0 & -\sqrt{g H_{0}} \end{pmatrix} \end{align}

and hence your equation set becomes

$$\vec{h}_{t} = P D P^{-1} \vec{h}_{x}$$

Multiplying both sides by $P^{-1}$ and setting $P^{-1} \vec{h} = \vec{v} = (v_{1}, v_{2})^{T}$ yields the system

$$\vec{v}_{t} = D \vec{v}_{x}$$

which is directly solvable

\begin{align} v_{1} &= f_{1}(x + \sqrt{g H_{0}} t) \\ v_{2} &= f_{2}(x - \sqrt{g H_{0}} t) \end{align}

with $f_{1}, f_{2}$ arbitrary differentiable functions. Inverting the relationship $P^{-1} \vec{h} = \vec{v} \implies \vec{h} = P \vec{v}$ yields the general solution to the original problem

\begin{align} h &= \sqrt{H_{0}/g} v_{1} - \sqrt{H_{0}/g} v_{2} \\ u &= v_{1} + v_{2} \end{align}

after which the initial conditions can be applied.

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