(1)&&(3) It is possible to construct analytical solutions for even and odd modes, using the boundary conditions at the center of the channel and on one of the walls. These solutions depend on an arbitrary amplitude (A, B). Using these solutions on the second wall, we find the dispersion equation

```
solodd = DSolve[{-I [Omega]*(v''[y] - [Beta]^2 v[y]) -
1/Rey (v''''[y] - 2 [Beta]^2 v''[y] + [Beta]^4 v[y]) == 0,
v[1] == 0, v[0] == 0, v'[1] == 0, v'[0] == A}, v[y], y]
soleven =
DSolve[{-I [Omega]*(v''[y] - [Beta]^2 v[y]) -
1/Rey (v''''[y] - 2 [Beta]^2 v''[y] + [Beta]^4 v[y]) == 0,
v[1] == 0, v[0] == B, v'[1] == 0, v'[0] == 0}, v[y], y]
```

(2) The dependence of the dispersion equation on the amplitude appears in the second approximation, whereas the Orr-Sommerfeld equation is the first (linear) approximation.

The complete solution is a linear combination of even and odd solutions, we have

```
V=First[v[y] /. soleven]+First[v[y] /. solodd]; V1=D[V,y];
```

We calculate the values of these functions on the wall

```
eq = {V, V1} /. y -> -1
```

By the conditions of the problem `eq==0`

, from this we find a linear system of equations for A,B, the determinant of which must be equal to zero. This is the dispersion equation.

Using V, we can try to solve equation

```
sol = DSolve[{-I*[Omega] [Eta][y] -
1/Rey*([Eta]''[
y] - [Beta]^2*[Eta][
y]) == -I [Beta] (-2 y) V, [Eta][-1] == 0, [Eta][1] ==
0}, [Eta][y], y]
```

But the solution turns out to be so cumbersome that it can not be surveyed.

I derived the dispersion equation in two ways and got the same result.The first method is shown above. The second way is shown below.It is possible to simplify the derivation of the dispersion equation. To do this, we find a solution that satisfies the boundary conditions on one wall:

```
sol1 = DSolve[{-I [Omega]*(v''[y] - [Beta]^2 v[y]) -
1/Rey (v''''[y] - 2 [Beta]^2 v''[y] + [Beta]^4 v[y]) == 0,
v[1] == 0, v'[1] == 0}, v[y], y];
V = ExpToTrig[First[v[y] /. sol1]];
```

Then on the other wall we have a system of two equations:

```
eq = {V, D[V, y]} /. y -> -1
m = CoefficientArrays[eq, Array[C, 2]] // Last
FullSimplify[Det[m] /. {[Omega] -> -I ([Beta]^2 - [Mu]^2)/Rey}]
```

The answer looks like neither in the book, nor at the fact that bbgodfrey

```
2 [Beta] [Mu] -
2 [Beta] [Mu] Cosh[2 [Beta]] Cosh[
2 [Mu]] + ([Beta]^2 + [Mu]^2) Sinh[2 [Beta]] Sinh[2 [Mu]] = 0
```

However, if we compare the solutions of this equation and the system of equations `{-[Mu] Tanh[[Beta]] + [Beta] Tanh[[Mu]] ==`

using

0, -[Mu] Coth[[Beta]] + [Beta] Coth[[Mu]] == 0}`ContourPlot[]`

, then we see that their solutions are identical $beta =pm mu$. This is a trivial solution $omega =0$. We can indicate other non-trivial solutions, for example

```
sol = DSolve[{-I [Omega]*(v''[y] - [Beta]^2 v[y]) -
1/Rey (v''''[y] - 2 [Beta]^2 v''[y] + [Beta]^4 v[y]) == 0,
v[1] == 0, v[-1] == 0}, v[y], y];
eq = {(D[V, y] /. y -> -1), (D[V, y] /. y -> 1)};
m = CoefficientArrays[eq, Array[C, 2]] // Last;
FullSimplify[Det[m]]
V = ExpToTrig[First[v[y] /. sol]]
```

Here the dispersion equation has the form (with `Rey [Omega] -> [CapitalOmega]`

)

```
deq= -2 [Beta] Sqrt[[Beta]^2 -
I [CapitalOmega]] + [Beta] Sqrt[[Beta]^2 - I [CapitalOmega]]
Cosh[2 [Beta] -
2 Sqrt[[Beta]^2 - I [CapitalOmega]]] + [Beta] Sqrt[[Beta]^2 -
I [CapitalOmega]]
Cosh[2 ([Beta] +
Sqrt[[Beta]^2 - I [CapitalOmega]])] + (-2 [Beta]^2 +
I [CapitalOmega]) Sinh[2 [Beta]] Sinh[2 Sqrt[[Beta]^2 - I [CapitalOmega]]]
ContourPlot[
Re[deq] == 0, { [Beta], -10, 10}, {[CapitalOmega], -10, 10}]
```

Using ContourPlot, we find the roots of equation

Dispersion relations for this question can be derived in a straightforward manner, as I described in an earlier comment above. Solve the two ODEs in turn but without boundary conditions.

```
s = DSolveValue[{-I ω*(v''[y] - β^2 v[y]) - 1/Rey (v''''[y] - 2 β^2 v''[y]
+ β^4 v[y]) == 0} /. ω -> -I (β^2 - μ^2)/Rey, v[y], y]
(* E^(-y μ) C[1] + E^(y μ) C[2] + E^(-y β) C[3] + E^(y β) C[4] *)
s1 = DSolveValue[(-I*ω η[y] - 1/Rey*(η''[y] - β^2*η[y]) ==
-I β (-2 y) s) /. ω -> -I (β^2 - μ^2)/Rey, η[y], y] // Simplify
(* (1/(4 (β - μ)^2 μ^3 (β + μ)^2))E^(-y (β + 3 μ)) (8 I E^(3 y μ)
Rey β μ^3 (-2 β - y β^2 + y μ^2) C[3] - 8 I E^(2 y β + 3 y μ)
Rey β μ^3 (-2 β + y β^2 - y μ^2) C[4] + E^(y (β + 4 μ)) (β^2 - μ^2)^2
(-I Rey β (1 - 2 y μ + 2 y^2 μ^2) C[2] + 4 μ^3 C[5]) + E^(y (β + 2 μ))
(β^2 - μ^2)^2 (I Rey β (1 + 2 y μ + 2 y^2 μ^2) C[1] + 4 μ^3 C[6])) *)
```

where `ω`

has been replaced by `-I (β^2 - μ^2)/Rey`

, consistent with the substitution shown in the question. Now, applying the six boundary conditions yields six linear, homogeneous equations for the six `C[_]`

, which must all equal zero, unless the determinant of the coefficient matrix vanishes. The determinant, therefore, is the dispersion relation.

```
{s /. y -> 1, s /. y -> -1, D[s, y] /. y -> 1, D[s, y] /. y -> -1,
s1 /. y -> -1, s1 /. y -> 1};
Det[CoefficientArrays[%, Array[C, 6]] // Last] // Factor;
[email protected][-% // ExpToTrig]/64
(* Cosh[μ] Sinh[μ] (μ Cosh[μ] Sinh[β] - β Cosh[β] Sinh[μ])
(β Cosh[μ] Sinh[β] - μ Cosh[β] Sinh[μ]) *)
```

The third and fourth factors, which also can be obtained by solving the first ODE only, can be rewritten as

```
Thread[Simplify[-%[[3]]/(Cosh[μ] Cosh[β]), %[[4]]/(Sinh[μ] Sinh[β])}] == 0]
(* -μ Tanh[β] + β Tanh[μ] == 0, -μ Coth[β] + β Coth[μ] == 0} *)
```

which are similar to, but not identical with, the odd and even mode dispersion results cited in the question. Note that the only solutions of interest to these two equations are pure imaginary.

The first two factors of the determinant arise from the second ODE.

```
Thread[List @@ %%[[1 ;; 2]] == 0]
(* {Cosh[μ] == 0, Sinh[μ] == 0} *)
```

Here, too, roots are pure imaginary, given by `I (n-1/2) Pi`

and `I n Pi`

, respectively. The first differs from the corresponding expression in the question by a factor of `I`

; the second is not given in the question.

Even though these solutions differ from those cited in the question, I am inclined to believe them (except, perhaps, for `Sinh[μ] == 0`

, which may be spurious), because the computation is so straightforward.