In this lecture we derive properties of waves with and without gravity playing a role. In previous lecture here we looked at static settings. But today we will like at gravity waves that move. This will look at using spectral methods and linear perturbation theory to obtain dispersion relations.
When approaching hard PDEs we tend to want to apply the fourier transform to convert a differential equation into an algebraic one which we have tools to solve. To see how this does so consider the Fourier transform: \[F[f(t)] = \int f(t) e^{-iwt}dt\]
Now let us consider the derivative: \[F[\frac{\partial f(t)}{\partial t}] = \int\frac{\partial f(t)}{\partial t}e^{-iwt}dt\] Apply integration by parts \(dv = \frac{\partial f(t)}{\partial t} dt\) and \(u = e^{-iwt}\) we can write \(v = f(t)\)and \(du = -iw e^{-iwt}dt\) \[F[\frac{\partial f(t)}{\partial t}] =f(t)e^{-iwt}|_{-\infty}^\infty - \int -iwf(t) e^{-iwt}dt \] We say that the function vanishes \(f \to 0\) at large ends \[F[\frac{\partial f(t)}{\partial t}] = iw\int f(t) e^{-iwt}dt \] \[F[\frac{\partial f(t)}{\partial t}] = iwF[f(t)] \] This can be repeated for spatial index as well but with a component of \(k\) and a change of sign.
Consider a static medium with no gravity. We can consider perturbed quantities with zeroth and first order terms: \[\rho = \rho_0 + \rho_1 \quad P = P_0 + P_1\] \[\Phi = \Phi_0+ \Phi_1 \quad v = v_0 + v_1\] Remember for our case there is no gravity and so we can obtain our fluid equations. Firstly for mass conservation \[\frac{\partial \rho}{\partial t} + \nabla (\rho v) = 0\] \[\frac{\partial (\rho_0 + \rho_1) }{\partial t} + \nabla ((\rho_0 + \rho_1)(v_0 + v_1)) = 0\] We reconize that initially the fluid is not moving and so \(v_0 = 0\) \[\frac{\partial (\rho_0 + \rho_1) }{\partial t} + \nabla ((\rho_0 + \rho_1)( v_1)) = 0\] \[[\frac{\partial \rho_0 }{\partial t} + \nabla (\rho_0 v_1)] + [\frac{\partial \rho_1 }{\partial t} + \nabla (\rho_1 v_1)]= 0\] By linear perturbation theory we need to group terms of similar order with each other and make them individually satisfy the PDE. This is to match scales. \[\frac{\partial \rho_0 }{\partial t} + \nabla (\rho_0 v_1) =0 \] \[\frac{\partial \rho_1 }{\partial t} + \nabla (\rho_1 v_1)= 0\] But we realise that \(\rho_0\) is just a constant and so the zeroth order is trivially satisfied. We also realise that there are 2 higher order terms multiplied and so those also cancel to zero as they vanish. \[\nabla (\rho_0 v_1) =0 \] \[\frac{\partial \rho_1 }{\partial t} = 0\] Combined to our first order term is \[\boxed{\frac{\partial \rho_1 }{\partial t} + \nabla (\rho_0 v_1)= 0}\] Let us continue but with momentum equation: \[\frac{\partial v}{\partial t} + (v \cdot \nabla v) = \frac{-1}{\rho}\nabla P - \nabla \Phi\] Remember that there is no gravity in this problem. \[\frac{\partial v_1}{\partial t} + (v_1 \cdot \nabla v_1) = \frac{-1}{\rho}\nabla (P_0+P_1)\] Cross out the higher order term. \[\frac{\partial v_1}{\partial t} = \frac{-1}{\rho}\nabla (P_0+P_1)\] We need to match the correct orders \[\frac{\partial v_1}{\partial t} = \frac{-1}{\rho}\nabla P_1\] Now we can compute the time derivative on both sides of the continuity equation \[\frac{\partial }{\partial t}\frac{\partial \rho_1 }{\partial t} + \frac{\partial }{\partial t}\nabla (\rho_0 v_1)= 0\] \[\frac{\partial }{\partial t}\frac{\partial \rho_1 }{\partial t} + \nabla (\rho_0 \frac{\partial }{\partial t}v_1)= 0\] Replace with momentum equation \[\frac{\partial }{\partial t}\frac{\partial \rho_1 }{\partial t} + \nabla (\rho_0 \frac{-1}{\rho}\nabla P_1 )= 0\] \[\frac{\partial^2 \rho_1 }{\partial t^2} = -\nabla^2 P_1\] Recall that this is in a setting where this is an ideal gas and so \(P = P(\rho)\) \[\frac{\partial^2 \rho_1 }{\partial t^2} = -\frac{\partial P}{\partial \rho} \nabla^2 \rho\] We define the sound speed to be \(v_s = \sqrt{\frac{\partial P}{\partial \rho}}\) \[\frac{\partial^2 \rho_1 }{\partial t^2} = -v_s^2\nabla^2 \rho_1\] We can now hit both sides with the fourier transform to obtain \[\boxed{w^2 = -v_s^2 k^2}\] We notice that the phase speed \[v_p = \frac{w}{k} = v_s i\] \[v_g = \frac{dw}{dk} = -v_s^2\] And so they are non despersive. Lastly let us try to get velocity \[\frac{\partial v_1}{\partial t} = \frac{-1}{\rho}\nabla P_1\] We can approach this by rewriting in terms of density \[\frac{\partial v_1}{\partial t} = \frac{-v_s}{\rho}\nabla \rho_1\] We try the fourier transform to obtain \[iwv_1 = \frac{-v_s\rho_1}{\rho_0}ik\] \[\boxed{wv_1 = \frac{-v_s\rho_1}{\rho_0}k}\] We see that none of these depend on \(k\) values explicitly and thus all in phase. \[\] Comment: Dispersion relations are very important in understanding the physics and nature of these complex systems. They help unravel any stabilities or instabilities at long time horizons and how properties that depend on time scales \(w\) relate to length scales \(k\).
Now let us explore the case where there IS gravity. Let us assume that we are at hydrostatic equilibrium. \[\frac{-1}{\rho_0}\nabla P_0 = \nabla \Phi\] \[\frac{-1}{\rho_0}\nabla P_0 = gz\] We integrate since pressure is only depth (\(z\)) dependent we can just obtain: where \(P_s\) is the surface. \[ P_0 = P_s - \rho gz\] We can then continue with our first order equations \[\nabla v_1 = 0 \tag{1st order mass eq.}\] \[\frac{\partial v_1}{\partial t} = - \frac{-1}{\rho_0}\nabla P_1 \tag{1st order mom. eq}\] We apply the same trick and take the derivative on both sides of the mass equation: \[\nabla \frac{\partial}{\partial t}v_1 = 0 \] \[\nabla \frac{\partial v_1}{\partial t} = - \frac{-1}{\rho_0}\nabla^2 P_1 \] This time we have the following: \[- \frac{-1}{\rho_0}\nabla^2 P_1 = 0\] \[\nabla^2 P_1 = 0\] We can then look at this and guess that this has some wave property: \[P_1 \propto e^{i(kx - wt))} f(z)\] We can try and solve for \(f(z)\) \[i^2k^2 e^{i(kx - wt))} f(z) + e^{i(kx - wt))} \frac{d^2}{dz^2}f(z) = 0\] \[i^2k^2 f(z) +\frac{d^2}{dz^2}f(z) = 0\] We can then rearrange the \[\boxed{\frac{d^2}{dz^2}f(z) = -k^2 f(z) \tag{harmonic ocillator}}\] This is can be solved with: \[f(z) \propto e^{\pm kz}\] We can be given boundary conditions, such as at the bottom of the wave \(-H\) the pressure change must drop to zero. \[\frac{\partial P_1}{\partial z}|_H = 0 \Rightarrow \frac{\partial f}{\partial z}|_H = 0\] Since we know that the only way to satisfy this is if we have the sum of the solutions: \[f(z) = A (e^{k(z+H)} - e^{-k(z+H)}) = A \cosh[k(z+H)]\] This means pressure is of the following form: \[P_1 = A e^{i(kx - wt))} \cosh[k(z+H)]\] We can then plug this back into momentum equation \[\frac{\partial v_1}{\partial t} = - \frac{-1}{\rho_0}\nabla P_1\] \[\frac{\partial v_{1x}}{\partial t} = - \frac{-1}{\rho_0} \frac{\partial }{\partial x}A e^{i(kx - wt))} \cosh[k(z+H)]\] integrating out we get \[\boxed{\partial v_{1x} = - \frac{k}{w\rho_0}A e^{i(kx - wt))} \cosh[k(z+H)]}\] We can repeat and plug in to the z values \[\frac{\partial v_{1z}}{\partial t} = - \frac{-1}{\rho_0} \frac{\partial }{\partial z}A e^{i(kx - wt))} \cosh[k(z+H)]\] \[\boxed{\frac{\partial v_{1z}}{\partial t} = - \frac{-k}{iw\rho_0} A e^{i(kx - wt))} \sinh[k(z+H)]}\] If we only focused on the vertical displacement velocity \[v_z(z= 0) =- \frac{-k}{iw\rho_0} A e^{i(kx - wt))} \sinh[kH] \] \[\xi = \int v_z(z=0) dt = \frac{k}{w^2\rho_0}Ae^{i(kx-wt)}\sinh(kH)\] Finally to get the dispersion relation we notice we can get the LAGRANGIAN perturbation of the pressure and satisfy the boundary condition that \[P_{1,L} = P_1 + \xi \cdot \nabla P_0 =0\] Plugging things in we get \[A e^{i(kx - wt))} \cosh[k(z+H)] + \frac{k}{w^2\rho_0}Ae^{i(kx-wt)}\sinh(kH) \nabla P_0 =0\] Use the momentum zeroth order equation \[A e^{i(kx - wt))} \cosh[k(z+H)] + \frac{k}{w^2\rho_0}Ae^{i(kx-wt)}\sinh(kH) \rho g =0\] Simplifying gives us \[\boxed{w^2 = kg \tanh(kH)}\] We see that this does depend on k and thus is dispersive and it has two limits one where it is \(w = \sqrt{kg}\) for shallow water and one where it is \(w = k\sqrt{gH}\) for deep water