Existence and uniqueness and first order approximation of solutions to atmospheric Ekman flows

Abstract

In this paper, we study the classical problem of the wind in the steady atmospheric Ekman layer with classical boundary conditions and the eddy viscosity is an arbitrary height-dependent function with a finite limit value. We present existence and uniqueness and smooth results to justify computing first order approximation of solutions. Using a different argument that in previous works, we construct the Green’s function to derive the solution by a perturbation approach.

1 Introduction

The atmospheric boundary layer has three parts [1, 2], i.e., the lamina sublayer, surface (Prandtl) layer and the Ekman layer. The Ekman layer covers 90% of the atmospheric boundary layer and it is driven by a three-way balance among frictional effects, pressure gradient and the influence of the coriolis force [1, 3, 4]. In general, textbooks on geophysical fluid dynamics and dynamic meteorology contain the derivation for the Ekman layer with a constant eddy viscosity k [5, 6]. However k usually varies with the height. As a result it is necessary to find the explicit solution of Ekman flows with a non-constant eddy viscosity, but unfortunately explicit solutions are scare in the literature and are restricted to linear [7, 8] or quadratic and cubic poly-nominal [9]. For arbitrary k(z) or k(z, t), we often have to rely on approximation and numerical simulation; the authors in [10,11,12,13] apply the Wentzel, Kramers and Brillouin’s method to get the approximation solution.

Constantin and Johnson [14] studied Ekman flows with variable eddy viscosity k(z) and they derived the explicit solution through an unclosed form and verified the existence of the solution by transformation and the iterative technique. The authors in [15] studied the horizontal wind drift currents which spiral and decay with depth and they obtained the solution by a perturbation approach. In this paper, we adopt the linearization approach to establish existence and uniqueness results and we consider the smoothness of solutions, which justify computing first order approximation of solutions. Also, based on [15], we regard the eddy viscosity k(z) as perturbations of the asymptotic reference value and we perform the variable change and get a linear, non-homogeneous second order differential equation and then we show the existence of a solution using the Green’s function.

2 Model description

The Ekman layer is governed by the following equations, see [1, 2]

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{Du}{Dt}=-\frac{1}{\rho }\frac{dP}{dx}+fv-\frac{\partial \overline{u'w'}}{\partial z},\\ \frac{Dv}{Dt}=-\frac{1}{\rho }\frac{dP}{dy}-fu-\frac{\partial \overline{v'w'}}{\partial z}, \end{array}\right. } \end{aligned}$$

where u, v and w are the components of the wind in the x, y and z directions respectively, P is the atmospheric pressure, $\rho$ is the reference density, $f=2\Omega sin\theta$ is the Coriolis parameter at the fixed latitude $\theta$, $\Omega \approx 7.29\times 10^{-5}$ is the angular speed of the roattion of the earth in the northern Hemisphere, and $\theta \in (0, \pi /2]$ is the angle of latitude in right-handed rotating spherical cooridates, t is time and k is the eddy diffusivity for momentum.

Assuming a steady state we get $\frac{Du}{Dt}=0, \frac{Dv}{Dt}=0.$ From the geostrophic balance, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{\rho }\frac{dP}{dx}=fv_{g},\\ \frac{1}{\rho }=-fu_{g}. \end{array}\right. } \end{aligned}$$

From the Flux-Gradient theory, we get

$$\begin{aligned} {\left\{ \begin{array}{ll} \overline{u'w'}=-k(\frac{\partial u}{\partial z}),\\ \overline{v'w'}=-k(\frac{\partial v}{\partial z}), \end{array}\right. } \end{aligned}$$

where k is the eddy viscosity coefficient. Then we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} f(v-v_{g})=-\frac{\partial }{\partial z}(k\frac{\partial u}{\partial z}),\\ f(u-u_{g})=\frac{\partial }{\partial z}(k\frac{\partial v}{\partial z}), \end{array}\right. } \end{aligned}$$

(1)

where $u_{g}$ and $v_{g}$ are the corresponding constant geostrophic wind components. We use the traditional boundary conditions for (1) as

$$\begin{aligned}&\displaystyle u=0,~~~v=0~~~\text{ at } ~~z=0,\end{aligned}$$

(2)

$$\begin{aligned}&\displaystyle u\rightarrow u_{g},~~~v\rightarrow v_{g}~~~\text{ for } ~~z\rightarrow \infty . \end{aligned}$$

(3)

Let $\Phi =(u-u_{g})+i(v-v_{g})$, and from (1), we will get

$$\begin{aligned} k\frac{\partial ^{2}\Phi }{\partial z^{2}}+\frac{\partial k}{\partial z}\frac{\partial \Phi }{\partial z}-i\cdot f\Phi =0. \end{aligned}$$

(4)

If $k=$ constant, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} u(z)=u_{g}-e^{-\gamma z}[u_{g}\cos (\gamma z)+v_{g}\sin (\gamma z)],\\ v(z)=v_{g}+e^{-\gamma z}[u_{g}\sin (\gamma z)-v_{g}\cos (\gamma z)], \end{array}\right. } \end{aligned}$$

(5)

where $\gamma =\sqrt{\frac{f}{2k}}.$ However, the eddy viscosity k always varies with height [10], so (4) will become

$$\begin{aligned} {\left\{ \begin{array}{ll} k'(z)\frac{dv}{dz}+k(z)\frac{d^{2}v}{dz^{2}}=f(u-u_{g}),\\ -k'(z)\frac{du}{dz}-k(z)\frac{d^{2}u}{dz^{2}}=f(v-v_{g}). \end{array}\right. } \end{aligned}$$

(6)

Here, we regard the physically relevant eddy viscosity k(z) as perturbations of the asymptotic reference value $k_{*}=\lim \limits _{z\rightarrow \infty }k(z)>0,$ so

$$\begin{aligned} k(z)=k_{*}+\varepsilon k_{1}(z),\quad \text{ at } ~~z\ge 0, \end{aligned}$$

(7)

where $\varepsilon \ll 1$, and the asymptotic rate of convergence is faster than quadratic [14], that is, there exist constants $a, b, c>0$ such that

$$\begin{aligned} |k_{1}(z)|\le \frac{a}{1+b|z|^{2+c}},\quad ~z\ge 0. \end{aligned}$$

3 Main results

3.1 Existence and uniqueness

Let

$$\begin{aligned} \begin{aligned} x=k(z)\frac{\partial u}{\partial z},\\ y=k(z)\frac{\partial v}{\partial z}, \end{aligned} \end{aligned}$$

so (1) is replaced with

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\partial u}{\partial z}=\hat{k}(z)x,\\ \frac{\partial v}{\partial z}=\hat{k}(z)y,\\ \frac{\partial x}{\partial z}=-f(v-v_{g}),\\ \frac{\partial y}{\partial z}=f(u-u_{g}) \end{array}\right. } \end{aligned}$$

(8)

for $\hat{k}(z)=\frac{1}{k(z)}$. The affine system (8) has a unique equilibrium

$$\begin{aligned} p=(u_g,v_g,0,0) \end{aligned}$$

(9)

with the linearization

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\partial u}{\partial z}=\hat{k}(z)x,\\ \frac{\partial v}{\partial z}=\hat{k}(z)y,\\ \frac{\partial x}{\partial z}=-fv,\\ \frac{\partial y}{\partial z}=fu \end{array}\right. } \end{aligned}$$

(10)

When $\varepsilon =0$, $k(z)=k_*$ is a constant function, and then the matrix

$$\begin{aligned} A=\begin{pmatrix} 0 &{}\quad 0 &{}\quad \hat{k}_* &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \hat{k}_*\\ 0 &{}\quad - f &{}\quad 0 &{}\quad 0 \\ f &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix} \end{aligned}$$

has eigenvalues

$$\begin{aligned} \lambda _1=-\sqrt{\frac{f\hat{k}_*}{2}}(1+i),\quad \lambda _2=-\sqrt{\frac{f\hat{k}_*}{2}}(1-i),\quad \lambda _3=\sqrt{\frac{f\hat{k}_*}{2}}(1-i),\lambda _4=\sqrt{\frac{f\hat{k}_*}{2}}(1+i) \end{aligned}$$

with corresponding eigenvectors

$$\begin{aligned} \begin{aligned} \left( -\frac{(1+i) \sqrt{\hat{k}_*}}{\sqrt{2} \sqrt{f}},-\frac{(1-i) \sqrt{\hat{k}_*}}{\sqrt{2} \sqrt{f}},i,1\right) \\ \left( -\frac{(1-i) \sqrt{\hat{k}_*}}{\sqrt{2} \sqrt{f}},-\frac{(1+i) \sqrt{\hat{k}_*}}{\sqrt{2} \sqrt{f}},-i,1\right) \\ \left( \frac{(1-i) \sqrt{\hat{k}_*}}{\sqrt{2} \sqrt{f}},\frac{(1+i) \sqrt{\hat{k}_*}}{\sqrt{2} \sqrt{f}},-i,1\right) \\ \left( \frac{(1+i) \sqrt{\hat{k}_*}}{\sqrt{2} \sqrt{f}},\frac{(1-i) \sqrt{\hat{k}_*}}{\sqrt{2} \sqrt{f}},i,1\right) . \end{aligned} \end{aligned}$$

Thus the linear system (10) has a stable space

$$\begin{aligned} X_s=\left[ \begin{array}{c} \left( -\frac{\sqrt{\hat{k}}}{\sqrt{2} \sqrt{f}},-\frac{\sqrt{\hat{k}}}{\sqrt{2} \sqrt{f}},0,1\right) \\ \left( -\frac{\sqrt{\hat{k}}}{\sqrt{2} \sqrt{f}},\frac{\sqrt{\hat{k}}}{\sqrt{2} \sqrt{f}},1,0\right) \end{array}\right] . \end{aligned}$$

Introduce a linear subspace

$$\begin{aligned} W=\{(0,0,x,y)\mid x,y\in \mathbb {R}\}, \end{aligned}$$

(11)

and condition (2) means $(u(0),v(0),x(0),y(0))\in W$. Note

$$\begin{aligned} \mathbb {R}^n=X_s\oplus W. \end{aligned}$$

First we study a general affine system

$$\begin{aligned} q'(z)=(A+\varepsilon B(\varepsilon ,z))(q(z)-p) \end{aligned}$$

(12)

for $A\in L(\mathbb {R}^n)$, $B:(-\delta ,\delta )\times \mathbb {R}_+\rightarrow L(\mathbb {R}^n)$, $\delta >0$ and $p\in \mathbb {R}^n$, where $\mathbb {R}_+=[0,\infty )$ and $L(\mathbb {R}^n)$ is the space of linear maps on $\mathbb {R}^n$. We suppose

• (A1) A is hyperbolic, i.e., the real parts of eigenvalues of A are nonzero.

• (A2) $B(\varepsilon ,z)$ has continuous partial derivatives $\partial _{\varepsilon }^iB(\varepsilon ,z)$ with

  $$\begin{aligned} \sup _{(\varepsilon ,z)\in (-\delta ,\delta )\times \mathbb {R}_+}\Vert \partial _{\varepsilon }^iB(\varepsilon ,z)\Vert <\infty \end{aligned}$$

  for $i=0,1,\cdots ,r$.

Recall that (A1) is equivalent to the existence of constants $K>0$, $\alpha >0$ and a splitting $\mathbb {R}^n=X_s\oplus X_u$ such that $A(X_s)=X_s$, $A(X_u)=X_u$ along with $\Vert e^{A_sz}\Vert \le Ke^{-\alpha z}$ and $\Vert e^{-A_uz}\Vert \le Ke^{-\alpha z}$ for $z\in \mathbb {R}_+$, where $A_s=A/X_s$ and $A_u=A/X_u$. We consider a projection $P:\mathbb {R}^n\rightarrow \mathbb {R}^n$ with $\ker P=X_u$ and ${\text {im}\,}P=X_s$.

Let $W\subset \mathbb {R}^n$ be a linear subspace such that $\mathbb {R}^n=X_s\oplus W$, i.e., W is transversal to $X_s$. We look for solutions of (12) satisfying

$$\begin{aligned} q(0)\in W,\quad \lim _{z\rightarrow \infty }q(z)=p. \end{aligned}$$

(13)

Let $Q:\mathbb {R}^n\rightarrow \mathbb {R}^n$ be a projection with $\ker Q=W$ and ${\text {im}\,}Q=X_s$.

Proposition 3.1

Assume (A1) and (A2). Set

$$\begin{aligned} B_\infty =\sup _{(\varepsilon ,z)\in (-\delta ,\delta )\times \mathbb {R}_+}\Vert B(\varepsilon ,z)\Vert . \end{aligned}$$

For any $\varepsilon \in (-\delta ,\delta )$ satisfying,

$$\begin{aligned} |\varepsilon |<\frac{\alpha }{4KB_\infty (\Vert P\Vert +(1+K\Vert Q\Vert )\Vert I-P\Vert )}, \end{aligned}$$

(14)

there is a unique solution $q(\varepsilon ,z)$ of (12) satisfying (13). Moreover, $q(\varepsilon ,z)$ is $C^r$-smooth in $\varepsilon$.

Proof

Following [16], any bounded solution of (12) is given by

$$\begin{aligned} q(z)= & {} p+e^{Az}q_s+\varepsilon \int _0^ze^{A(z-r)}PB(\varepsilon ,r)(q(r)-p)dr\nonumber \\&-\varepsilon \int _z^\infty e^{A(r-z)}(I-P)B(\varepsilon ,r)(q(r)-p)dr \end{aligned}$$

(15)

for $q_s\in X_s$. We consider a Banach space $C_b(\mathbb {R}_+,\mathbb {R}^n)$ of all bounded and continuous functions $q:\mathbb {R}_+\rightarrow \mathbb {R}^n$ endowed with a norm $\Vert q\Vert =\sup _{z\in \mathbb {R}_+}|q(z)|$. A solution of (15) is a fixed point of the map

$$\begin{aligned} H(\varepsilon ,q_s,q)(z)= & {} p+e^{Az}q_s+\varepsilon \int _0^ze^{A(z-r)}PB(\varepsilon ,r)(q(r)-p)dr\nonumber \\&-\varepsilon \int _z^\infty e^{A(r-z)}(I-P)B(\varepsilon ,r)(q(r)-p)dr. \end{aligned}$$

Now H is linear in q and $C^r$-smooth in $\varepsilon$. For any $q_1,q_2\in C_b(\mathbb {R}_+,\mathbb {R}^n)$, we have

$$\begin{aligned} \begin{aligned}&|H(\varepsilon ,q_s,q_1)(z)-H(\varepsilon ,q_s,q_2)(z)|\\&\quad \le |\varepsilon |\int _0^z|e^{A(z-r)}PB(\varepsilon ,r)(q_1(r)-q_2(r))|dr\\&\qquad +\,|\varepsilon |\int _z^\infty |e^{A(z-r)}(I-P)B(\varepsilon ,r)(q_1(r)-q_2(r))|dr\\&\quad \le |\varepsilon |K\Vert P\Vert B_\infty \int _0^ze^{-\alpha (z-r)}dr\Vert q_1-q_2\Vert \\&\qquad +\,|\varepsilon |K\Vert I-P\Vert B_\infty \int _z^\infty e^{\alpha (z-r)}dr\Vert q_1-q_2\Vert \\&\quad \le |\varepsilon |KB_\infty \alpha ^{-1}(\Vert P\Vert +\Vert I-P\Vert )\Vert q_1-q_2\Vert . \end{aligned} \end{aligned}$$

Condition (14) guarantees the contraction of $H(\varepsilon ,q_s,q)$ in q, so the Banach fixed point theorem ensures a unique solution $\hat{q}(\varepsilon ,q_s,z)$ of (15). Next, for any $q_{s1},q_{s2}\in X_s$, we have

$$\begin{aligned} |H(\varepsilon ,q_{s1},q)(z)-H(\varepsilon ,q_{s2},q)(z)\le |e^{Az}(q_{s1}-q_{s2})|\le K|q_{s1}-q_{s2}|. \end{aligned}$$

Consequently, $\hat{q}(\varepsilon ,q_s,z)$ is globally Lipschitz in $q_s$ with a constant

$$\begin{aligned} l_1=\frac{K}{1-|\varepsilon |KB_\infty \alpha ^{-1}(\Vert P\Vert +\Vert I-P\Vert )} \end{aligned}$$

and $C^r$-smooth in $\varepsilon$. To finish the proof, we need to solve

$$\begin{aligned} Q\hat{q}(\varepsilon ,q_s,0)=0. \end{aligned}$$

(16)

We set

$$\begin{aligned} \hat{q}(\varepsilon ,q_s,0)=p+q_s-\varepsilon \tilde{q}(\varepsilon ,q_s). \end{aligned}$$

From (15), for any $q_{s1},q_{s2}\in X_s$, we have

$$\begin{aligned} \begin{aligned} |\tilde{q}(\varepsilon ,q_{s1})-\tilde{q}(\varepsilon ,q_{s2})|&\le \int _z^\infty |e^{A(z-r)}(I-P)B(\varepsilon ,r)(\hat{q}(\varepsilon ,q_{s1},r)-\hat{q}(\varepsilon ,q_{s2},r))|dr\\&\le KB_\infty \alpha ^{-1}\Vert I-P\Vert l_1\Vert q_1-q_2\Vert . \end{aligned} \end{aligned}$$

Clearly (16) is equivalent to

$$\begin{aligned} \varepsilon Q\tilde{q}(\varepsilon ,q_s)-Qp=q_s. \end{aligned}$$

(17)

Assumption (14) ensures that the map

$$\begin{aligned} q_s\rightarrow \varepsilon Q\tilde{q}(\varepsilon ,q_s)-Qp \end{aligned}$$

is a contraction. Thus (17) has a unique solutions $q_s(\varepsilon )$. Next, by using [16, Proposition 1, p. 34], we see that (12) is dichotomous for any $\varepsilon$ satisfying (14). Consequently, $q(\varepsilon ,z)=\hat{q}(\varepsilon ,q_s(\varepsilon ,z)$ is the desired unique solution of (12) satisfying (13). The proof is finished. $\square$

Now we apply Proposition 3.1 to (8). We already verified (A1). To show (A2), we note that

$$\begin{aligned} B(\varepsilon ,z)=\begin{pmatrix} 0 &{}\quad 0 &{}\quad \hat{k}_2(\varepsilon ,z) &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \hat{k}_2(\varepsilon ,z)\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix} \end{aligned}$$

for

$$\begin{aligned} \hat{k}_2(\varepsilon ,z)=\frac{k_1(z)}{k_*(k_*+\varepsilon k_1(z))}. \end{aligned}$$

Thus supposing that $k_1(z)\ge 0$ is continuous with $\sup _{z\in \mathbb {R}_+}|k_1(z)|<\infty$, condition (A2) holds. Consequently, we have the following result.

Theorem 3.2

Assume $k_1(z)\ge 0$ is continuous with $\sup _{z\in \mathbb {R}_+}|k_1(z)|<\infty$. Then for any $\varepsilon$ small, there is a unique solution of (8) satisfying (13) with p and W given by (9) and (11), respectively. This solution is $C^\infty$ smooth in $\varepsilon$.

Since k(z) is continuous, here we have a solution u(z), v(z) of (12) such that u(z), v(z), $\frac{\partial u(z)}{\partial z}$, $\frac{\partial v(z)}{\partial z}$, $\frac{\partial }{\partial z}(k(z)\frac{\partial u(z)}{\partial z})$ and $\frac{\partial }{\partial z}(k(z)\frac{\partial u(z)}{\partial z})$ exist and are continuous on $\mathbb {R}_+$.

3.2 First order approximation

We use Theorem 3.2 to consider first order approximations of solutions. Consider the following second order boundary-value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} w''(s)-\frac{i\cdot f}{k_{*}}w(s)=0,\quad ~s \in (0, t) \cup (t, \infty ),~~k_{*}>0,\\ w(0)=0,~~ w'(t_{-})-w'(t_{+})=1,~~ w(s)\rightarrow 0 ~{as} ~~s\rightarrow \infty . \end{array}\right. } \end{aligned}$$

(18)

Lemma 3.3

If $w(\cdot ):=G(\cdot ,t),t\in (0,\infty )$ (called the Green’s function) solves (18), then

$$\begin{aligned} w(s)= {\left\{ \begin{array}{ll} a_{1}x_{1}(s)+a_{2}x_{2}(s),\quad ~s \in (0, t) ,\\ b x_{2}(s),~~~~ ~~~~~s \in (t, \infty ), \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} x_{1}(s)=&e^{(1+i)\lambda s}, ~ x_{2}(s)=e^{-(1+i)\lambda s},~~\lambda =\sqrt{f/2k_{*}},\\ a_{1}=&\frac{1}{2(1+i)\lambda }e^{-(1+i)\lambda t},~~a_{2}=-\frac{1}{2(1+i)\lambda }e^{-(1+i)\lambda t}, \end{aligned}$$

and

$$\begin{aligned} b=-\frac{1}{2(1+i)\lambda }e^{-(1+i)\lambda t}+\frac{1}{2(1+i)\lambda }e^{(1+i)\lambda t}. ~~~~ \end{aligned}$$

Proof

Note $x_{1}$ and $x_{2}$ are two linearly independent solutions of $w''(s)-\frac{i\cdot f}{k_{*}}w(s)=0$. Suppose

$$\begin{aligned} w(s)= {\left\{ \begin{array}{ll} a_{1}x_{1}(s)+a_{2}x_{2}(s),~~~ ~s \in (0, t) ,\\ b_{1}x_{1}(s)+b_{2}x_{2}(s),~~ ~s \in (t, \infty ), \end{array}\right. } \end{aligned}$$

is a solution of (18). Then using the boundary conditions one obtains

$$\begin{aligned} b_{1}=0, ~~~~a_{1}+a_{2}=0, \end{aligned}$$

and

$$\begin{aligned} a_{1}(1+i)\lambda e^{(1+i)\lambda t}-a_{2}(1+i)\lambda e^{-(1+i)\lambda t}=-b_{2}(1+i)\lambda e^{-(1+i)\lambda t}+1. \end{aligned}$$

Using the continuity of w at $s=t$, we find

$$\begin{aligned} a_{1}e^{(1+i)\lambda t}+a_{2}e^{-(1+i)\lambda t}=b_{2}e^{-(1+i)\lambda t}. \end{aligned}$$

From the above we get the value of $a_{1},~a_{2},~b_{1}$ and $b_{2}$, and therefore we obtain

$$\begin{aligned} w(s)= {\left\{ \begin{array}{ll} \frac{1}{2(1+i\lambda )}[e^{(1+i)\lambda (s-t)}-e^{-(1+i)\lambda (s+t)}],~~~ ~s \in (0, t) ,\\ \frac{1}{2(1+i\lambda )}[e^{(1+i)\lambda t}-e^{-(1+i)\lambda t}]e^{-(1+i)\lambda s},~~ ~s \in (t, \infty ). \end{array}\right. } \end{aligned}$$

The proof is complete. $\square$

Let

$$\begin{aligned} s=s(z)=k_{*}\int _{0}^{z}\frac{1}{k(t)}dt. \end{aligned}$$

(19)

Then

$$\begin{aligned} \frac{ds}{dz}=\frac{k_{*}}{k(z)}. \end{aligned}$$

Let

$$\begin{aligned} k_{2}(s)=\frac{f}{k_{*}^{2}}k_{1}(z), ~~~\Psi (s)=U(s)+iV(s), \end{aligned}$$

(20)

where

$$\begin{aligned} U(s)=u(z)-u_{g}, ~~V(s)=v(z)-v_{g}. \end{aligned}$$

(21)

Set

$$\begin{aligned} \Psi (s)=\Psi _{0}(s)+\epsilon \varphi (s)+o(\epsilon ),~~s\ge 0, \end{aligned}$$

(22)

where $\Psi _{0}(s)$ is the classical Ekman solution

$$\begin{aligned} \Psi _{0}(s)=-e^{-(1+i)\lambda s}(u_{g}+iv_{g}). \end{aligned}$$

Theorem 3.4

The function defined by (22) is the solution of (6) with (2) and (3) if

$$\begin{aligned} \begin{aligned} \varphi (s)=&\frac{v_{g}-iu_{g}}{2(1+i)\lambda }\int _{0}^{s}k_{2}(t)e^{-(1+i)\lambda (s+t)}(e^{(1+i)\lambda t}-e^{-(1+i)\lambda t})dt\\&+\frac{v_{g}-iu_{g}}{2(1+i)\lambda }\int _{s}^{\infty }k_{2}(t)(e^{(1+i)\lambda (s-t)}-e^{-(1+i)\lambda (s+t)})e^{-(1+i)\lambda t}dt. \end{aligned} \end{aligned}$$

(23)

Proof

Suppose for simplicity that k(z) is $C^2$-smooth. From the definition of $\Psi (s)$, we have

$$\begin{aligned} \Psi '(s)=\frac{k(z)}{k_{*}}\frac{dU}{dz}+i\frac{k(z)}{k_{*}}\frac{dV}{dz} \end{aligned}$$

(24)

and

$$\begin{aligned} \Psi ''(s)=\frac{k'(z)k(z)}{k_{*}^{2}}\frac{dV}{dz}+\frac{k^{2}(z)}{k_{*}^{2}}\frac{d^{2}U}{dz^{2}}+i(\frac{k'(z)k(z)}{k_{*}^{2}}\frac{dV}{dz}+\frac{k^{2}(z)}{k_{*}^{2}}\frac{d^{2}U}{dz^{2}}). \end{aligned}$$

(25)

Using (21), (6) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} k'(z)\frac{dV}{dz}+k(z)\frac{d^{2}V}{dz^{2}}=fU,\\ k'(z)\frac{dU}{dz}+k(z)\frac{d^{2}U}{dz^{2}}=-fV. \end{array}\right. } \end{aligned}$$

From (24) and (25) we obtain

$$\begin{aligned} \Psi ''(s)=i\frac{k(z)}{k_{*}^{2}}f\Psi (s), \end{aligned}$$

(26)

and the corresponding boundary conditions are

$$\begin{aligned} \Psi (s)=&-(u_{g}+iv_{g}),~~~\text{ at } ~~s=0,\\ \Psi (s)\rightarrow&0,~~~\text{ as } ~~s\rightarrow \infty . \end{aligned}$$

Inserting (22) and (7) into (26) one obtains

$$\begin{aligned} \Psi ''_{0}(s)+\epsilon \varphi ''(s)=i\left( \frac{f}{k_{*}}+\epsilon k_{2}(s)\right) (\Psi _{0}(s)+\epsilon \varphi (s)+o(\epsilon )). \end{aligned}$$

From the definition of $\Psi _{0}(s)$, we have

$$\begin{aligned} \Psi ''_{0}(s)=\frac{if}{k_{*}}\Psi _{0}(s). \end{aligned}$$

Dividing by $\epsilon$ and letting $\epsilon \rightarrow 0$, this yields a second order differential equation for $\varphi$, namely

$$\begin{aligned} \varphi ''(s)-\frac{if}{k_{*}}\varphi (s)=ik_{2}(s)\Psi _{0}(s), \end{aligned}$$

(27)

The boundary conditions are transformed to

$$\begin{aligned} \varphi (0)=0,~~~~\varphi \rightarrow 0~~~\text{ as } ~~s\rightarrow \infty . \end{aligned}$$

(28)

From Lemma 3.3, we know that $G(\cdot ,\cdot )$ is the Green’s function of (27) and (28), and the solution can be expressed as the convolution

$$\begin{aligned} \varphi (s)=i\int _{0}^{\infty }k_{2}(t)\Psi _{0}(t)G(s,t)dt ~~~ ~s\ge 0. \end{aligned}$$

Therefore the solution to (27) and (28) is given by (23). $\square$

4 An example

Constantin and Johnson [14] considered the case of an eddy viscosity which is constant above a certain height, with a non-constant below, that is

$$\begin{aligned} k(z)= {\left\{ \begin{array}{ll} a+bz,~~0\le z\le z_{0},\\ a+bz_{0},~~~z>z_{0}, \end{array}\right. } \end{aligned}$$

(29)

where $a>0,~z_{0}>0$, $b\in R$. Motivated by [14], we change (29) to

$$\begin{aligned} k(z)= {\left\{ \begin{array}{ll} a+bz,~~0\le z\le z_{0},\\ k_{*}+e^{-z},~~~z>z_{0}, \end{array}\right. } \end{aligned}$$

here $a>0,~z_{0}>0$, and we assume $a+bz_{0}=k_{*}+e^{-z_{0}}$ for continuity. Here, $\varepsilon =1$. From (19), we get

$$\begin{aligned} s= {\left\{ \begin{array}{ll} \frac{k_{*}}{b}\ln \frac{a+bz}{a},\quad 0\le s\le s_{0},\\ s_{0}+\ln \frac{k_{0}e^{z}+1}{k_{0}e^{z_{0}}+1},\quad s>s_{0}, \end{array}\right. } \end{aligned}$$

with $s_{0}=\frac{k_{*}}{b}\ln \frac{k_{*}}{a}>0$. Note that

$$\begin{aligned} k_{3}(s)=\frac{f}{k_{*}^{2}}k(z)= {\left\{ \begin{array}{ll} \frac{af}{k_{*}^{2}}e^{\frac{bs}{k_{*}}},~~0\le s\le s_{0},\\ \frac{f}{k_{*}}+\frac{f}{k_{*}}\frac{1}{(k_{*}e^{z_{0}}+1)e^{s-s_{0}}-1},~~s>s_{0}, \end{array}\right. } \end{aligned}$$

(30)

with $z_{0}=\frac{a}{b}(e^{\frac{bs_{0}}{k^{*}}}-1).$

For the solution in the lower part of the layer $s\in [0, s_{0}]$, we set

$$\begin{aligned} \Psi (s)=\Phi (x),~~~ \text{ with }~~ x=e^{\frac{bs}{2k_{*}}}. \end{aligned}$$

Then (26) will be transformed into the Bessel equation

$$\begin{aligned} x^{2}\frac{d^{2}\Phi }{dx^{2}}+x\frac{d\Phi }{dx}-\frac{4iaf}{b^{2}}x^{2}\Phi =0. \end{aligned}$$

Let $\beta = \sqrt{-\frac{4iaf}{b^{2}}}=\frac{2\sqrt{af}}{|b|}e^{-\frac{\pi }{4}i}$, and the general solution of above Bessel equation can be expressed as

$$\begin{aligned} \Phi (x)=c_{1}J_{0}(\beta x)+c_{2}Y_{0}(\beta x), \end{aligned}$$

where $J_{0}(x)$ and $Y_{0}(x)$ are the first and second kind of the Bessel functions with order 0 respectively. In the upper part of the layer, $s\in (s_{0}, \infty )$, we obtain the solution $\varphi (s)$ by (23).

Also motivated by [14] and (7) with

$$\begin{aligned} k_1(z)= {\left\{ \begin{array}{ll} e^{-z_0}-bz_0+bz,~~0\le z\le z_{0},\\ e^{-z},~~~z>z_{0}, \end{array}\right. } \end{aligned}$$

we could change (29) to

$$\begin{aligned} k(z)= {\left\{ \begin{array}{ll} k_*+\varepsilon (e^{-z_0}-bz_0+bz),~~0\le z\le z_{0},\\ k_{*}+\varepsilon e^{-z},~~~z>z_{0}, \end{array}\right. } \end{aligned}$$

with $k_{\star }>0, b>0$ and $z_0>0$. It follows from (19) that

$$\begin{aligned} s= {\left\{ \begin{array}{ll} \frac{k_{*}}{b\varepsilon }\ln \frac{k_{*}+e^{-z_{0}}\varepsilon -\varepsilon bz_{0}+\varepsilon bz}{a},~~0\le s\le s_{0},\\ s_{0}+\ln \frac{k_{*}e^{z}+\varepsilon }{k_{*}e^{z_{0}}+\varepsilon },~~s>s_{0}, \end{array}\right. } \end{aligned}$$

with $s_{0}=\frac{k_{*}}{b\varepsilon }\ln \frac{k_{*}+e^{-z_{0}}\varepsilon }{k_{*}+e^{-z_{0}}\varepsilon -b\varepsilon z_{0}}>0$. From (20), we see that

$$\begin{aligned} k_{2}(s)= {\left\{ \begin{array}{ll} \frac{fe^{\frac{b\varepsilon s}{k_{*}}}(e^{-z_{0}}-bz_{0}+\frac{k_{*}}{\varepsilon })}{k_{*}^{2}}-\frac{f}{\varepsilon k_{*}},~~0\le s\le s_{0},\\ \frac{f}{k_{*}(k_{*}e^{z_{0}}+\varepsilon )e^{s-s_{0}}-\varepsilon },~~s>s_{0}, \end{array}\right. } \end{aligned}$$

(31)

then we obtain the solution $\varphi (s)$ by (23) and (31).

Next, we change (29) to

$$\begin{aligned} k(z)= {\left\{ \begin{array}{ll} 5e^{-z}-2,~~0\le z\le z_{0},\\ 5e^{-z_{0}}-2,~~~z>z_{0}, \end{array}\right. } \end{aligned}$$

where $z_{0}$ satisfies $k_{*}=5\cdot e^{-z_{0}}-2>0.$ From (19), we have

$$\begin{aligned} s= {\left\{ \begin{array}{ll} -\frac{k_{*}}{2}\ln \frac{a+bz}{a},~~0\le s\le s_{0},\\ s_{0}+\ln \frac{k_{0}e^{z}+1}{k_{0}e^{z_{0}}+1},~~s>s_{0}, \end{array}\right. } \end{aligned}$$

with $s_{0}=-\frac{k_{*}}{2}\ln \frac{5-2e^{z_{0}}}{3}>0$. Then we have

$$\begin{aligned} \begin{aligned} k_{3}(s)&=\frac{f}{k_{*}^{2}}k(z)=\frac{f}{k_{*}^{2}}\left( 5\cdot \frac{2}{5-3 e^{-\frac{2}{k_{*}}s}}-2\right) =\frac{2f}{k_{*}^{2}}\frac{\frac{3}{5}e^{-\frac{2}{k_{*}}s}}{1-\frac{3}{5}e^{-\frac{2}{k_{*}}s}}, \end{aligned}~~~~0<s\le s_{0}. \end{aligned}$$

Therefore

$$\begin{aligned} k_{3}(s)= {\left\{ \begin{array}{ll} \frac{2f}{k_{*}^{2}}\frac{\frac{3}{5}e^{-\frac{2}{k_{*}}s}}{1-\frac{3}{5}e^{-\frac{2}{k_{*}}s}},~~0\le s\le s_{0},\\ \frac{f}{5e^{-z_{0}}-2},~~s>s_{0}. \end{array}\right. } \end{aligned}$$

(32)

For the solution in the upper part of the layer, $s>s_{0}$, we have

$$\begin{aligned} \Psi ''(s)=i\frac{f}{5e^{-z_{0}}-2}\Psi (s)=\frac{(1+i)^{2}}{2}\frac{f}{5e^{-z_{0}}-2}\Psi (s), \end{aligned}$$

so the general solution can be expressed by

$$\begin{aligned} \Psi (s)=c_{1}e^{(1+i)\sqrt{\frac{f}{2(5e^{-z_{0}}-2)}}s}+c_{2}e^{-(1+i)\sqrt{\frac{f}{2(5e^{-z_{0}}-2)}}s}. \end{aligned}$$

For the solution in the lower part of the layer $s\in [0,s_{0}]$, we set

$$\begin{aligned} \zeta =\frac{3}{5}e^{-\frac{2}{k_{*}}s},~~~\Psi (s)=\Phi (\zeta ). \end{aligned}$$

From (32), $k_{3}(s)$ can be expressed by

$$\begin{aligned} k_{3}(s)=\frac{2f}{k_{*}^{2}}\frac{\zeta }{1-\zeta }. \end{aligned}$$

Now (26) can be written as

$$\begin{aligned} \frac{36}{25k_{*}^{2}}e^{-\frac{4s}{k_{*}}}\frac{d^{2}\Phi (\zeta )}{d\zeta ^{2}}+\frac{12}{5k_{*}^{2}}e^{-\frac{2s}{k_{*}}}\frac{d\Phi (\zeta )}{d\zeta } =i\frac{2f}{k_{*}^{2}}\frac{\zeta }{1-\zeta }\Phi (\zeta ), \end{aligned}$$

and then we get the hypergeometric equation

$$\begin{aligned} \frac{d^{2}\Phi (\zeta )}{d\zeta ^{2}}+\frac{1}{\zeta }\frac{d\Phi (\zeta )}{d\zeta } =if\frac{25}{18}\frac{\zeta }{1-\zeta }\Phi (\zeta ). \end{aligned}$$

(33)

If we let

$$\begin{aligned} \beta =\frac{5(1+i)}{6}\sqrt{f},~~~~\gamma =-\beta , \end{aligned}$$

then the solution of (33) is

$$\begin{aligned} \Phi (\zeta )=c\digamma (\beta ,\gamma ,1,\zeta ), \end{aligned}$$

where c is an arbitrary complex constant and $\digamma$ is the Gauss hypergeometric function.