Solution Explanation
show that, under the (zeroth-order) Hankel transform in the radial variable,
\[ \mathcal{H}_0\left[\nabla^4 \eta(r)\right]=-k^4 \widehat{\eta}(k), \quad \text { where } \quad \mathcal{H}_0[\eta(r)]=\widehat{\eta}(k) \]
Here, 1. \(\eta=\eta(r)\) is radially symmetric (independent of \(\theta\) ). 2. \(\nabla^2\) is the usual 2D Laplacian in polar coordinates,
\[ \nabla^2 \eta(r, \theta)=\frac{\partial^2 \eta}{\partial r^2}+\frac{1}{r} \frac{\partial \eta}{\partial r}+\frac{1}{r^2} \frac{\partial^2 \eta}{\partial \theta^2} \]
Since \(\eta\) is independent of \(\theta\), the last term vanishes and
\[ \nabla^2 \eta(r)=\frac{1}{r} \frac{d}{d r}\left(r \frac{d \eta}{d r}\right) \]
- By definition, the bi-Laplacian (or " \(\nabla^{4}\) " ) is simply
\[ \nabla^4=\nabla^2\left(\nabla^2(\cdot)\right) \]
Key known Hankel-transform identity
A standard (and very useful) identity involving the 2D (zeroth-order) Hankel transform is: Claim. If \(\eta(r)\) is radially symmetric, then \[ \mathcal{H}_0\left[\nabla^2 \eta(r)\right](k)=-k^2 \widehat{\eta}(k) \]
Equivalently, under the Hankel transform of order 0,
\[ \nabla^2 \longleftrightarrow-k^2 \]
In other words, applying the 2D Laplacian in the \((r, \theta)\) plane corresponds to multiplying by \(\left(-k^2\right)\) in the Hankel-transform domain.
Why is there a minus sign? the standard 2D Fourier-transform identity
\[ \mathcal{F}_2\left[\nabla^2 f(\mathbf{x})\right](\mathbf{k})=-|\mathbf{k}|^2 \hat{f}(\mathbf{k}) \]
then the same " \(-k^{2}\) " factor appear. The zeroth-order Hankel transform \(\mathcal{H}_0\) is (in effect) the radial part of the 2D Fourier transform when the function is radially symmetric. This explains why the same factor of \(-k^2\) arises.
Applying \(\nabla^4\) and taking the transform
Now we look at
\[ \nabla^4 \eta(r)=\nabla^2\left(\nabla^2 \eta(r)\right) \]
Taking the Hankel transform on both sides and using the fact that \(\nabla^2 \leftrightarrow-k^2\), we get 1. First apply \(\nabla^2\) to \(\eta\) :
\[ \mathcal{H}_0\left[\nabla^2 \eta(r)\right]=-k^2 \widehat{\eta}(k) . \]
- Then apply \(\nabla^2\) again (i.e., to \(\nabla^2 \eta\) ):
\[ \mathcal{H}_0\left[\nabla^2\left(\nabla^2 \eta(r)\right)\right]=\mathcal{H}_0\left[\nabla^2\left(\nabla^2 \eta\right)\right]=-k^2 \underbrace{\mathcal{H}_0\left[\nabla^2 \eta\right]}_{=-k^2 \hat{\eta}(k)} \]
Hence
\[ \mathcal{H}_0\left[\nabla^4 \eta(r)\right]=-k^2\left(-k^2 \widehat{\eta}(k)\right)=(+) k^4 \widehat{\eta}(k) \]
Notice carefully that the two copies of \(\nabla^2\) each contribute a factor of \(-k^2\). Taken together:
\[ \left(-k^2\right)\left(-k^2\right)=+k^4 \]
- About the sign in the final result
From the above derivation, one sees:
\[ \mathcal{H}_0\left[\nabla^4 \eta(r)\right]=+k^4 \widehat{\eta}(k) \quad \text { (most common convention). } \]
see it stated as
\[ \nabla^4 \longleftrightarrow+k^4 \]
under the usual definitions of the Laplacian \(\nabla^2\) and the (zeroth-order) Hankel transform \(\mathcal{H}_0\). Why might someone write \(-k^4\) ? - If a different sign convention is used for the definition of \(\nabla^2\). For instance, in some references or in certain PDE contexts, one encounters equations of the form
\[ \Delta=-(\text { something positive }) \]
depending on how one writes Green's functions or Helmholtz operators. - Alternatively, if the transform itself is defined with an extra factor of \(i\) somewhere (as happens in various physics sign conventions for Fourier-type transforms).
Hence most standard references (e.g., in PDE theory, standard Bessel/Fourier references) give:
\[ \nabla^2 \longleftrightarrow-k^2 \Longrightarrow \nabla^4 \longleftrightarrow+k^4 \]
it is almost surely using a slightly different sign in front of \(\nabla^2\) or in the transform itself.
Final boxed statement
In the usual sign convention where \(\nabla^2 \leftrightarrow-k^2\) under the Hankel transform \(\mathcal{H}_0\), one obtains
\[ \mathcal{H}_0\left[\nabla^4 \eta(r)\right]=+k^4 \widehat{\eta}(k) \]
it is almost certainly using a different convention for the Laplacian or for the transform. The mechanism of the proof, however, is exactly the same.
Fourier series
If There Is Angular Dependence
If \(\eta\) depends on both \(r\) and \(\theta\), do a partial-wave expansion:
\[ \eta(r, \theta, t)=\sum_{m=-\infty}^{\infty} \eta_m(r, t) e^{i m \theta} \]
Each Fourier mode \(\eta_m(r, t)\) is then a candidate for the \(m\) th-order Hankel transform, which involves \(J_m(k r)\). The PDE separates by \(m\), and each radial equation can be transformed similarly, with
\[ \nabla^2 \eta_m(r) \longleftrightarrow-k^2 \widehat{\eta}_m(k), \quad \nabla^4 \eta_m(r) \longleftrightarrow k^4 \widehat{\eta}_m(k) \]
(plus adjustments for the \(m\) th-order radial derivatives).
Determining the Separation Constant \(k(s)\)
With the assumed form of \(\hat{p}\), let's determine \(k(s)\). 5.1. Substituting into the Transformed Momentum Equation in \(z\)-Direction
From the transformed \(z\)-momentum equation:
\[ \delta^2 \frac{\partial \hat{w}}{\partial t}=-\frac{\partial \hat{p}}{\partial z} \]
Substituting \(\hat{p}(s, z, t)=\hat{\eta}(s, t) e^{k(s)(z-1)}\) :
\[ \delta^2 \frac{\partial \hat{w}}{\partial t}=-k(s) \hat{\eta}(s, t) e^{k(s)(z-1)} \]
5.2. Expressing \(\hat{w}\) in Terms of \(\hat{\eta}_t\)
To solve for \(\hat{w}\), we propose a similar exponential dependence on \(z\) :
\[ \hat{w}(s, z, t)=\hat{\eta}_t(s, t) e^{k(s)(z-1)} \]
Verification: Taking the time derivative:
\[ \frac{\partial \hat{w}}{\partial t}=\hat{\eta}_{t t}(s, t) e^{k(s)(z-1)} \]
t>0 \(z \in[0,1]\) we have
\[ \begin{aligned} & \frac{\partial u}{\partial t}=-\frac{\partial p}{\partial r} \\ & \delta^2 \frac{\partial w}{\partial t}=-\frac{\partial p}{\partial z} \\ & \frac{\partial u}{\partial r}+\frac{\partial w}{\partial z}+\frac{u}{r}=0 \end{aligned} \]
\(t>0 \quad z=1\) we have
\[ w=\frac{\partial \eta}{\partial t} ; \quad p=\eta \]
\(t \geq 0 \quad z=0\) we have \[ w=0 \] and note that \[ f(r)=\int_0^{\infty} s \hat{f}(s) J_0(r s) d s \]
how to get
\[ \begin{aligned} & \eta(r, t)=\int_0^{\infty} s \hat{f}(s) \cos (t \sqrt{s / \delta}) J_0(r s) d s \\ & \text { and } \hat{w}=\frac{\partial \hat{\eta}}{\partial t} e^{\delta s(z-1)} \\ & \hat{w}=H_0[w] \\ & \hat{\eta}=H_0[\eta] \end{aligned} \]
\(H_0\) is 0 -order Hankel transform
\[ (\displaystyle s,\hat{u} -\frac{\partial \hat{w}}{\partial z} = 0. ) \]