= Solution
Keep the prescribed outer slip <velocity> fixed when perturbing the <unsteady Prandtl equation>. Cancelling the base equation and retaining terms linear in the disturbance gives the <linearized unsteady Prandtl equation>
$$
\tilde u_t+U\tilde u_x+V\tilde u_y+U_x\tilde u+U_y\tilde v
=\tilde u_{yy},\qquad
\tilde u_x+\tilde v_y=0.
$$
The <no-slip boundary condition> supplies $\tilde u=\tilde v=0$ at the wall; the fixed outer <velocity> supplies $\tilde u\to0$ at infinity. There is no prescribed zero normal-velocity disturbance at infinity: a finite displacement-related value is permitted.
With the frozen parallel profile $U=U(y)$, $V=0$, the coefficients are invariant under translations in $x$ and $t$. <Normal modes> therefore separate those variables. For the mode convention used here the equations are
$$
i\alpha(U-c)u+U'v=u'',\qquad i\alpha u+v'=0.
$$
Eliminate $u=-v'/(i\alpha)$ to obtain the <Prandtl normal-mode equation>
$$
\boxed{v'''=i\alpha[(U-c)v'-U'v]},\qquad
v(0)=v'(0)=0,\quad v'(\infty)=0,
$$
with a bounded far-field $v$. Requiring $v(\infty)=0$ would incorrectly eliminate the proposed outer profile. For real $U$ and real $\alpha$, complex conjugation of the equation and its <boundary conditions> replaces $(\alpha,c,v)$ by $(-\alpha,c^*,v^*)$. This proves the stated <eigenvalue> symmetry. The paired modes have the same temporal <growth rate> $\alpha\operatorname{Im}c$, so choosing positive <wavenumber> loses no real physical disturbance.
Here the prescribed parallel profile is a local frozen-coefficient model. A nontrivial arbitrary $U(y)$ with $V=0$ and constant outer slip is not generally an exact steady solution of the unforced <unsteady Prandtl equation>; its base equation would require $U''=0$. The following stability calculation uses the simplifying local model stipulated for the mode analysis.
Set $c_0=U_c=U(y_c)$. Away from the critical point the dominant equation is $(U-c_0)v_0'-U'v_0=0$, whose solutions are multiples of $U-c_0$. Choosing the coefficient to be zero below $y_c$ and one above it satisfies the wall and tangential far-field conditions. The choice $c_0=U_c$ makes $v_0$ continuous across the joining point; $U'(y_c)=0$ also makes its first <derivative> continuous. Its second <derivative> jumps. Thus it is a valid leading outer solution away from $y_c$, but viscosity must smooth the join in a <critical layer in a shear flow>.
Near the join, $U-c_0\sim(y-y_c)^2/2$. If the layer width is $\ell$, matching gives $v=O(\ell^2)$. The advection side of the mode equation scales as $\alpha\ell^3$ and $v'''$ scales as $\ell^{-1}$. Balance gives $\ell^4\sim\alpha^{-1}=\varepsilon/k$. Equivalently, the <eigenvalue> correction $\varepsilon^{1/2}c_1$ balances $(y-y_c)^2$, so the <quarter-power critical layer> has
$$
\boxed{p=\frac14,\qquad q=\frac12},\qquad
y-y_c=\varepsilon^{1/4}Y,\quad v=\varepsilon^{1/2}w(Y)+\cdots.
$$
Substitution, retaining the leading terms and cancelling their common factor, gives
$$
\boxed{kYw-k\left(\frac{Y^2}{2}-c_1\right)w_Y=iw_{YYY}}.
$$
At order $\varepsilon^{1/2}$ outside the layer,
$$
(U-c_0)v_1'-U'v_1=c_1v_0'.
$$
A solution consistent with the chosen lower branch and the wall conditions is $v_1=0$ below the join. Above it a particular solution is $v_1=-c_1$; a multiple of $U-c_0$ represents an arbitrary amplitude renormalization and may be set to zero. Matching consequently requires
$$
\boxed{w\to0\quad(Y\to-\infty),\qquad
w-\left(\frac{Y^2}{2}-c_1\right)\to0\quad(Y\to+\infty)}.
$$
The corresponding <derivatives> match as $w_Y\sim Y$, $w_{YY}\to1$ on the positive side and as zero on the negative side. The matching statements require that growing homogeneous corrections be absent.
To symmetrize these different end conditions and remove $k$ and $i$, choose
$$
a=(ik)^{1/4}=k^{1/4}e^{i\pi/8},\qquad
z=aY,\quad C=a^2c_1,\quad
W=2a^2w-\left(\frac{z^2}{2}-C\right).
$$
The polynomial $A(z)=z^2/2-C$ is itself a solution of the transformed homogeneous equation. Direct substitution, using $a^4=ik$, gives
$$
\boxed{W_{zzz}-\left(\frac{z^2}{2}-C\right)W_z+zW=0},
\qquad W\mp\left(\frac{z^2}{2}-C\right)\to0\quad(z\to\pm\infty).
$$
The <derivative> is third order, as required by the original mode equation. Initially the two ends are along the rotated rays $z=aY$ with real $Y$. Continuing them to the real $z$ axis is compatible with the asymptotic end conditions: the decreasing homogeneous correction behaves exponentially as $\exp[-z^2/(2\sqrt2)]$ and still decreases throughout the rotation from angle $\pi/8$ to zero. Thus the supplied real-axis spectral normalization uses the same recessive conditions; a complex coordinate change should not be mistaken for a real stretching alone.
For each supplied real <eigenvalue> $C_n=(4n+7)/\sqrt2$, the phase-speed correction is
$$
c_1=k^{-1/2}e^{-i\pi/4}C_n,\qquad
\operatorname{Im}c=-\frac{C_n}{\sqrt{2k}}\varepsilon^{1/2}+o(\varepsilon^{1/2}).
$$
The mode factor has temporal magnitude $e^{\alpha\operatorname{Im}c\,t}$. Therefore
$$
\boxed{\gamma_n=\alpha\operatorname{Im}c
=-\frac{4n+7}{2}\sqrt\alpha+o(\sqrt\alpha)}.
$$
Among the listed indices, $n=-2$ gives positive growth $\gamma\sim\sqrt\alpha/2$; $n=-1,1,2,\ldots$ give decay. Hence the frozen non-monotone flow has <high-wavenumber instability of a non-monotone Prandtl layer>, with arbitrarily rapid linear growth as the positive <wavenumber> increases. The normal-mode calculation demonstrates <linear instability>; it is not by itself a claim about the final nonlinear state.
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