# Linearising Wave Equations

We are given the equations

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

and

$$\frac{\partial{h}}{\partial{t}}+\frac{\partial{(hu)}}{\partial{x}} = 0$$

We are then asked

By linearising these equations about a uniform mean flow of speed $u_{0}$ and uniform thickness $h_0$, derive expressions for the phase and group speeds of linear shallow water waves. Are these waves dispersive?

Having attempted $u=u_0 + u'$ and $h=h_0 + h'$ I arrived at

$$\frac{\partial{u'}}{\partial{t}}+u_0\frac{\partial{u'}}{\partial{x}}+g\frac{\partial{h'}}{\partial{x}}=0$$

and

$$\frac{\partial{h'}}{\partial{t}}+u_0\frac{\partial{h'}}{\partial{x}}+h_0\frac{\partial{u'}}{\partial{x}}=0$$

Which I don't see how they come out to be proper wave equations from which I can get the velocities.

• @Winther But $u_{0}$ is a constant (i.e. the mean flow velocity), so by definition $\partial u_{0}/\partial x \equiv 0$? – Thomas Russell Jun 2 '16 at 10:35
• Oh I see. I missed that. Then it's fine! – Winther Jun 2 '16 at 10:35
• Try taking the derivative of one equations wrt $x$, and the other wrt $t$ and combining them to get a wave equation. – David Jun 2 '16 at 23:30
• Try writing your last equations in matrix form. – Semiclassical Jun 4 '16 at 15:27
• The equations can be written $[\partial_t + u_0\partial_x]u' = -g\partial_x h'$ and $[\partial_t + u_0\partial_x]h' = -h_0\partial_x u'$. Applying the operator $[\partial_t + u_0\partial_x]$ to the first equation and using the second equation gives $[\partial_t + u_0\partial_x]^2u' = gh_0\partial_{xx}u'$. Same for $h'$ so this gives that both $u'$ and $h'$ satisfy $(1-gh_0)\partial_{tt}f + 2u_0\partial_{tx}f + u_0^2\partial_{xx}f = 0$ – Winther Jun 4 '16 at 18:11

## 2 Answers

Following a suggestion from Semiclassical on the chat and in comment. We can write the systems in the following form:

\begin{align*}\frac{\partial u^{\prime}}{\partial t} &= -u_{0}\frac{\partial u^{\prime}}{\partial x}-g\frac{\partial h^{\prime}}{\partial x} \\ \frac{\partial h^{\prime}}{\partial t}&= -u_{0}\frac{\partial h^{\prime}}{\partial x} - h_{0}\frac{\partial u^{\prime}}{\partial x}\end{align*}

This can be written in the following matrix form:

$$\begin{pmatrix}\frac{\partial u^{\prime}}{\partial t} \\ \frac{\partial h^{\prime}}{\partial t}\end{pmatrix} = -\begin{pmatrix}u_{0} & g \\ h_{0} & u_{0}\end{pmatrix}\begin{pmatrix}\frac{\partial u^{\prime}}{\partial x} \\ \frac{\partial h^{\prime}}{\partial x}\end{pmatrix}$$

We can diagonalise the matrix to give us:

$$\begin{pmatrix}\frac{\partial u^{\prime}}{\partial t} \\ \frac{\partial h^{\prime}}{\partial t}\end{pmatrix} = \begin{pmatrix}\sqrt{\frac{g}{h_{0}}} & -\sqrt{\frac{g}{h_{0}}} \\ 1 & 1\end{pmatrix}\begin{pmatrix}-\sqrt{gh_{0}}-u_{0} & 0 \\ 0 & \sqrt{gh_{0}}-u_{0}\end{pmatrix}\begin{pmatrix}\sqrt{\frac{g}{h_{0}}} & -\sqrt{\frac{g}{h_{0}}} \\ 1 & 1\end{pmatrix}^{-1}\begin{pmatrix}\frac{\partial u^{\prime}}{\partial x} \\ \frac{\partial h^{\prime}}{\partial x}\end{pmatrix}$$

Premultiplying by the inverse eigenvector matrix:

$$\begin{pmatrix}\sqrt{\frac{h_{0}}{g}} & 1 \\ -\sqrt{\frac{h_{0}}{g}} & 1\end{pmatrix}\begin{pmatrix}\frac{\partial u^{\prime}}{\partial t} \\ \frac{\partial h^{\prime}}{\partial t}\end{pmatrix} = \begin{pmatrix}-\sqrt{gh_{0}}-u_{0} & 0 \\ 0 & \sqrt{gh_{0}}-u_{0}\end{pmatrix}\begin{pmatrix}\sqrt{\frac{h_{0}}{g}} & 1 \\ -\sqrt{\frac{h_{0}}{g}} & 1\end{pmatrix}\begin{pmatrix}\frac{\partial u^{\prime}}{\partial x} \\ \frac{\partial h^{\prime}}{\partial x}\end{pmatrix}$$

This leads to two decoupled partial differential equations:

\begin{align}\frac{\partial}{\partial t}\left(\sqrt{\frac{h_{0}}{g}}u^{\prime} + h^{\prime}\right)&=-\left(\sqrt{gh_{0}}+u_{0}\right)\frac{\partial}{\partial x}\left(\sqrt{\frac{h_{0}}{g}}u^{\prime} + h^{\prime}\right) \\ \frac{\partial}{\partial t}\left(h^{\prime}-\sqrt{\frac{h_{0}}{g}}u^{\prime}\right) &= \left(\sqrt{gh_{0}}-u_{0}\right)\frac{\partial}{\partial x}\left(h^{\prime}-\sqrt{\frac{h_{0}}{g}}u^{\prime}\right)\end{align}

We now can assume the two ansätze expressions: $$\sqrt{\frac{h_{0}}{g}}u^{\prime} + h^{\prime} = Ae^{i(k_{1}x - \omega_{1} t)},\quad h^{\prime}-\sqrt{\frac{h_{0}}{g}}u^{\prime} = Be^{i(k_{2}x - \omega_{2} t)}$$

From this and our two derived equations we find:

$$-\omega_{1} = -(\sqrt{gh_{0}}+u_{0})k_{1} \implies \frac{\omega_{1}}{k_{1}} = \frac{\partial \omega_{1}}{\partial k_{1}} = \sqrt{gh_{0}}+u_{0}$$

And:

$$-\omega_{2} = (\sqrt{gh_{0}}-u_{0})k_{2} \implies \frac{\omega_{2}}{k_{2}} = \frac{\partial \omega_{2}}{\partial k_{2}} = u_{0} - \sqrt{gh_{0}}$$

I'm not sure if this is correct, as this talks about the velocity of propagation of the normal modes of vibration, so I would appreciate it if anyone has any comments as to the validity of this approach.

Differentiating each equation with respect to $t$ and $x$ we have \begin{align} \frac{\partial^2 u'}{\partial t^2} + u_0\frac{\partial^2 u'}{\partial t \partial x} + g \frac{\partial^2 h'}{\partial t \partial x} = 0 \label{1}\tag{1}\\ \frac{\partial^2 u'}{\partial t \partial x} + u_0\frac{\partial^2 u'}{\partial x^2} + g \frac{\partial^2 h'}{\partial x^2} = 0 \label{2}\tag{2}\\ \frac{\partial^2 h'}{\partial t^2} + u_0\frac{\partial^2 h'}{\partial t \partial x} + h_0 \frac{\partial^2 u'}{\partial t \partial x} = 0 \label{3}\tag{3}\\ \frac{\partial^2 h'}{\partial t \partial x} + u_0\frac{\partial^2 h'}{\partial x^2} + h_0 \frac{\partial^2 u'}{\partial x^2} = 0 \label{4}\tag{4} \end{align} To find the wave equation for $u'$, rearrange \eqref{4} to make $\frac{\partial^2 h'}{\partial t \partial x}$ the subject, and substitute into \eqref{1}: $$\frac{\partial^2 u'}{\partial t^2} + u_0\frac{\partial^2 u'}{\partial t \partial x} - g \left[u_0\frac{\partial^2 h'}{\partial x^2} + h_0 \frac{\partial^2 u'}{\partial x^2}\right] = 0$$ Then rearrange \eqref{2} to make $\frac{\partial^2 h'}{\partial x^2}$ the subject and substitute this in as well. The resulting equation is \begin{align} 0 &= \frac{\partial^2 u'}{\partial t^2} + u_0\frac{\partial^2 u'}{\partial t \partial x} - g h_0 \frac{\partial^2 u'}{\partial x^2} + u_0\left[\frac{\partial^2 u'}{\partial t \partial x} + u_0\frac{\partial^2 u'}{\partial x^2}\right]\\ &=\frac{\partial^2 u'}{\partial t^2} + 2u_0 \frac{\partial^2 u'}{\partial t \partial x} + (u_0^2 - gh_0) \frac{\partial^2 u'}{\partial x^2} \end{align} This is an inhomogeneous wave equation: $$\frac{\partial^2 u'}{\partial t^2} - (gh_0 - u_0^2) \frac{\partial^2 u'}{\partial x^2} = -2u_0 \frac{\partial^2 u'}{\partial t \partial x},$$ note that it can be 'factored' into the form $$\left[\frac{\partial}{\partial t} + (u_0 + \sqrt{gh_0})\frac{\partial}{\partial x}\right]\left[\frac{\partial}{\partial t} + (u_0 - \sqrt{gh_0})\frac{\partial}{\partial x}\right] u' = 0$$ A similar form can be found for $h'$ by substituting \eqref{2} and then \eqref{4} into \eqref{3}.

Using the wavelike ansatz $$u' = Ae^{-i(\omega t - kx)}$$ we find the dispersion relation $$-\omega^2 + 2u_0 \omega k - k^2(u_0^2 - gh_0) = 0$$ or (solving as a quadratic in $\omega$) $$\omega = \left(u_0 \pm \sqrt{gh_0}\right)k$$