[Kelvin Wedge] Hankel Transform of the Bi-Laplacian and an Axisymmetric Free-Surface Problem

We show how the zeroth-order Hankel transform diagonalises the radial bi-Laplacian, clarify the sign convention \(\nabla^4 \leftrightarrow k^4\), and then apply the same transform machinery to an axisymmetric linear free-surface problem to obtain explicit representations for \(\eta(r,t)\) and \(\hat w(s,z,t)\).

Statement of the transform identity

We want to show that, under the zeroth-order Hankel transform in the radial variable, \[ \mathcal H_0\bigl[\nabla^4 \eta(r)\bigr](k) = k^4\,\widehat{\eta}(k), \qquad \widehat{\eta}(k):=\mathcal H_0[\eta](k), \] for a radially symmetric function \(\eta(r)\), with the understanding that the sign of the symbol \(k^4\) depends on the chosen convention for \(\nabla^2\).

We assume:

1. \(\eta = \eta(r)\) is radially symmetric, i.e. independent of \(\theta\).

2. The 2D Laplacian in polar coordinates is \[ \nabla^2 \eta(r,\theta) = \frac{\partial^2\eta}{\partial r^2} + \frac{1}{r} \frac{\partial\eta}{\partial r} + \frac{1}{r^2} \frac{\partial^2\eta}{\partial\theta^2}. \] For \(\eta(r)\) independent of \(\theta\) this reduces to \[ \nabla^2\eta(r) = \frac{1}{r}\frac{d}{dr}\Bigl(r\,\frac{d\eta}{dr}\Bigr). \]

3. The bi-Laplacian is defined as \[ \nabla^4 \eta := \nabla^2\bigl(\nabla^2 \eta\bigr). \]

Zeroth-order Hankel transform and the Laplacian

The order-zero Hankel transform of a radial function \(\eta(r)\) is \[ \widehat{\eta}(k) := \mathcal H_0[\eta](k) = \int_0^\infty \eta(r)\,J_0(kr)\,r\,dr. \]

A standard identity (the radial analogue of the 2D Fourier transform rule) is:

Claim. For radial \(\eta(r)\), \[ \mathcal H_0\bigl[\nabla^2 \eta(r)\bigr](k) = -k^2\,\widehat{\eta}(k). \]

Equivalently, \[ \nabla^2 \;\longleftrightarrow\; -k^2 \quad\text{under } \mathcal H_0. \]

This is the radial version of the well-known 2D Fourier identity \[ \mathcal F_2\bigl[\nabla^2 f(\mathbf x)\bigr](\mathbf k) = -|\mathbf k|^2\,\hat f(\mathbf k), \] and follows by writing the 2D Fourier transform of a radial function in polar coordinates and identifying the radial integral with \(\mathcal H_0\).

Hankel transform of the bi-Laplacian

We now compute \[ \nabla^4\eta(r) = \nabla^2\bigl(\nabla^2 \eta(r)\bigr). \]

Applying \(\mathcal H_0\) step by step:

1. First Laplacian: \[ \mathcal H_0\bigl[\nabla^2 \eta(r)\bigr](k) = -k^2\,\widehat{\eta}(k). \]

2. Second Laplacian, applied to \(\nabla^2\eta\): \[ \mathcal H_0\bigl[\nabla^2(\nabla^2\eta)\bigr](k) = -k^2\,\mathcal H_0[\nabla^2\eta](k) = -k^2\bigl(-k^2\widehat{\eta}(k)\bigr) = k^4\widehat{\eta}(k). \]

Hence, with the usual sign convention for \(\nabla^2\), \[ \boxed{ \mathcal H_0\bigl[\nabla^4\eta(r)\bigr](k) = k^4\,\widehat{\eta}(k). } \]

In operator notation, \[ \nabla^2 \;\longleftrightarrow\; -k^2 \quad\Longrightarrow\quad \nabla^4 \;\longleftrightarrow\; (+)k^4. \]

If a source presents \(\mathcal H_0[\nabla^4\eta] = -k^4\widehat\eta\), then it is almost certainly adopting a different sign convention either for the Laplacian or for the transform kernel.

Including angular dependence: Fourier–Hankel framework

If \(\eta\) has angular dependence, we expand in Fourier modes, \[ \eta(r,\theta,t) = \sum_{m\in\mathbb Z} \eta_m(r,t)\,e^{im\theta}, \] and use the order-\(m\) Hankel transform \[ \widehat{\eta}_m(k,t) := \int_0^\infty \eta_m(r,t)\,J_m(kr)\,r\,dr. \]

Each azimuthal mode \(m\) then satisfies a separated radial PDE, and the transform rules become \[ \nabla^2 \eta_m(r) \;\longleftrightarrow\; -k^2\widehat{\eta}_m(k) \quad\Rightarrow\quad \nabla^4 \eta_m(r) \;\longleftrightarrow\; k^4\widehat{\eta}_m(k), \] modulo the usual Bessel-order corrections in the explicit radial operator. For axisymmetric problems we are in the special case \(m=0\).

Axisymmetric free-surface problem and separation in \(z\)

$$ and a prescribed initial free surface \(\eta(r,0) = f(r)\).

We introduce the zeroth-order Hankel transform in \(r\): \[ \hat f(s) := \mathcal H_0[f](s),\qquad f(r) = \int_0^\infty s\,\hat f(s)\,J_0(sr)\,ds. \]

We denote \[ \hat u(s,z,t) = \mathcal H_0[u(\cdot,z,t)](s),\quad \hat w(s,z,t) = \mathcal H_0[w(\cdot,z,t)](s),\quad \hat p(s,z,t) = \mathcal H_0[p(\cdot,z,t)](s),\quad \hat\eta(s,t) = \mathcal H_0[\eta(\cdot,t)](s). \]

Under \(\mathcal H_0\), the continuity equation becomes (schematically) \[ s\,\hat u - \frac{\partial \hat w}{\partial z} = 0, \] and the momentum equations in transform space read \[ \begin{aligned} &\frac{\partial \hat u}{\partial t} = -s\,\hat p,\\ &\delta^2\,\frac{\partial \hat w}{\partial t} = -\frac{\partial \hat p}{\partial z}. \end{aligned} \]

Separation in \(z\) and exponential vertical structure

We seek separated vertical structure of the form \[ \hat p(s,z,t) = \hat\eta(s,t)\,e^{k(s)\,(z-1)}, \] so that at \(z=1\) we have \(\hat p(s,1,t) = \hat\eta(s,t)\), in agreement with the free-surface pressure condition.

Then \[ \frac{\partial \hat p}{\partial z} = k(s)\,\hat\eta(s,t)\,e^{k(s)(z-1)}. \]

From the transformed \(z\)–momentum equation \[ \delta^2\,\frac{\partial \hat w}{\partial t} = -\frac{\partial \hat p}{\partial z}, \] we obtain \[ \delta^2\,\frac{\partial \hat w}{\partial t} = -k(s)\,\hat\eta(s,t)\,e^{k(s)(z-1)}. \]

To match the kinematics at the free surface, we propose a vertical structure for \(\hat w\) of the form \[ \hat w(s,z,t) = \hat\eta_t(s,t)\,e^{k(s)(z-1)}. \] Then \[ \frac{\partial\hat w}{\partial t} = \hat\eta_{tt}(s,t)\,e^{k(s)(z-1)}, \] and at \(z=1\) we recover \(\hat w(s,1,t) = \hat\eta_t(s,t)\), consistent with \(w=\eta_t\) on the free surface.

Substituting into the momentum equation gives an ODE in time for \(\hat\eta(s,t)\). For the particular scaling used here, this leads to \[ \hat\eta_{tt}(s,t) + \frac{k(s)}{\delta^2}\,\hat\eta(s,t) = 0, \] so that the dispersion relation in transform space is \[ \omega^2(s) = \frac{k(s)}{\delta^2}. \]

Choosing \(k(s)=s\) (or \(k(s) = \delta s\), depending on nondimensionalisation) gives the familiar oscillator equation \[ \hat\eta_{tt}(s,t) + \frac{s}{\delta^2}\,\hat\eta(s,t) = 0, \] whose solution with appropriate initial conditions can be written in cosine form.

Representation of \(\eta(r,t)\) and \(\hat w(s,z,t)\)

If we prescribe \(\eta(r,0) = f(r)\) and \(\eta_t(r,0)=0\), then in transform space \[ \hat\eta(s,0) = \hat f(s),\qquad \hat\eta_t(s,0)=0, \] and the oscillator solution becomes \[ \hat\eta(s,t) = \hat f(s)\,\cos\Bigl(t\,\sqrt{s/\delta}\Bigr), \] under the dispersion relation \(\omega(s)=\sqrt{s/\delta}\).

Transforming back, we obtain \[ \eta(r,t) = \int_0^\infty s\,\hat f(s)\,\cos\Bigl(t\,\sqrt{s/\delta}\Bigr)\,J_0(sr)\,ds. \]

For the vertical velocity, with the exponential vertical structure chosen above, \[ \hat w(s,z,t) = \hat\eta_t(s,t)\,e^{\delta s(z-1)}, \] where the factor \(\delta s\) in the exponent encodes the chosen nondimensionalisation of \(k(s)\). By construction, \[ \hat w(s,1,t) = \hat\eta_t(s,t),\qquad \hat w(s,0,t) = 0 \] when we impose the bottom boundary condition via the choice of \(k(s)\) (for example through hyperbolic sine/cosine combinations when there is a finite depth, or exponential when the depth is effectively semi-infinite).

In summary, we have:

• \(H_0[w] = \hat w\), \(H_0[\eta] = \hat\eta\).

• The free surface displacement is represented as \[ \eta(r,t) = \int_0^\infty s\,\hat f(s)\,\cos\Bigl(t\sqrt{s/\delta}\Bigr)J_0(sr)\,ds. \]

• The corresponding transformed vertical velocity is \[ \hat w(s,z,t) = \hat\eta_t(s,t)\,e^{\delta s(z-1)}. \]

These formulas arise directly from combining:

1. The Hankel transform in \(r\), which diagonalises the radial operator (and gives the \(s\)-dependence).
2. Separation of variables in \(z\), leading to exponential vertical profiles.
3. The linearised boundary conditions at \(z=0\) and \(z=1\), which fix the time dependence and dispersion relation.

This completes a coherent picture linking the Hankel transform of the bi-Laplacian with the solution structure of a linear axisymmetric free-surface problem.