Let's consider process of interaction of two chemical substances which concentration we shall designate u( t ) and w( t ). Let's write the equations chemical kineticsAfter graduating at MSU I considered, that the linear theory of the equations with partial derivatives is already developed. New it is necessary to search in not linearities.
What my surprise when I have come across a new boundary problem in a blast furnace!
Let's consider process of interaction of two chemical substances which concentration we shall designate u( t ) and w( t ). Let's write the equations chemical kinetics
dt u = F( u, w
)
dt w = G( u, w )
about analytical expression for F and G
till now chemists argue.
Now we shall consider a chemical reactor of counterflow type - a component u
goes with a speed v1, and component w goes towards with a speed
v2 (in a blast furnace charge goes from top to down, and gas from below upwards.) We shall receive hyperbolic system of the equations for u( t, x )
and w( t, x )
( ∂t - v1
* ∂x ) u = F( u, w )
( ∂t + v2 * ∂x ) w = G( u, w )
Very few people from chemists (blast-furnace operators) doubts, that if to
maintain constant concentration on an input - output (at x = 0 and x = L), then will be established a stationary mode
- v1 * dx
U = F( U, W )
+ v2 * dx W = G( U, W )
for U( x ) and W( x ).
But we of mathematics. 
The stability of this mode is our interest. Traditional linearization u = U + p, w = W + q
will give us
( ∂t
- v1 * ∂x
) p = ∂U
F( U, W ) * p + ∂W
F( U, W ) * q
( ∂t
+ v2 * ∂x
) q = ∂U
G( U, W ) * p + ∂W
G( U, W ) * q
and boundary problem q( t, 0 ) = 0, p( t, L ) = 0.
And this problem was not considered!
I shall copy the equations in a little less terrible 
kind
( ∂t - v1
* ∂x ) p = A( x ) * p + B( x ) * q
( ∂t + v2 * ∂x ) q = C( x ) * p + D(
x ) * q
Is here and degenerated case: B( x ) = 0, C( x ) = 0 - The system breaks up to two independent equations of 1-st order. The solution since some moment - identical zero.
I investigated a problem on a computer for various v1, v2, A( x ), B( x ), C( x ), D( x ). Everywhere I observed an output of the solution on function of type
exp( μ * t ) * [ R( x ) * Cos( ν * t ) + S( x ) * Sin( ν * t ) ]
- it is simply real part of exp[ ( μ + i * ν ) t ] * [ R( x ) - i * S( x ) ].
Let's consider some examples. System
( ∂t - ∂x ) p =
- q
( ∂t + ∂x ) q = p
It can be turned in one equation of 2-nd order
( ∂tt
- ∂xx
) p
= - p
And the boundary problem for it is put rather strange ( ∂t - ∂x ) p( t,
0 ) =0, p( t, L ) = 0.
Probably, this is the reason, that classics have bypassed this problem with the attention.
Typical example of evolution of the solution (an axis t - from below upwards):
red lines - p( t, x ) = 0,
blue lines - q( t, x ) = 0
Believing
p = exp( μ * t ) * [ Rp( x ) * Cos( ν * t ) + Sp( x ) * Sin( ν * t ) ]
q = exp( μ * t ) * [ Rq( x ) * Cos( ν * t ) + Sq( x ) * Sin( ν * t ) ]
let's receive system of the ordinary differential equations of 4-th order with four boundary conditions
Why linear systems with constant factors are good? - there is a formula for the solution. We shall designate a vector as Y( x ), a matrix as M, matrix of final transformation as K. Let's receive
Y( L ) = exp( M * L ) * Y( 0 ) = K * Y( 0 )
What should be a matrix K, to provide boundary conditions? - With a zero minor in the left top corner!
And a spectral problem: to find such μ and ν, that this minor was zero (by numerical search "as it is possible closer to zero"). And results of numerical search
Three eigen values are visible. I shall note, that there is no only real eigen value. In a vicinity of 1-st eigen value a matrix K ( μ ~ -0.49, ν ~ 1.69 ) turns out about such
| -0.0025 | 0.0023 | 0.0027 | -0.9946 |
| -0.0023 | -0.0025 | 0.9946 | 0.0027 |
| 0.0027 | -0.9946 | -3.3617 | -0.9814 |
| 0.9946 | 0.0027 | 0.9814 | -3.3617 |
And other system
( ∂t - ∂x ) p
= q
( ∂t + ∂x ) q = p
results in a spectrum, where 1-st eigen value is real ( μ ~ +0.32, ν ~ 0.0 )
Very similar on a discrete spectrum. The hypothesis is natural:
There is an infinite discrete spectrum and system of eigen functions is complete.
There is a natural question: how the coordinated set of Rp( x ) and Rq( x ) is possible to approach two any functions p( 0, x ) and q( 0, x ) ? - The matter is that here to one eigen value there correspond linearly independent two sets of functions: that mentioned above, and
p = exp( μ * t ) * [ -Sp( x ) * Cos( ν * t ) + Rp( x ) * Sin( ν * t ) ]
q = exp( μ * t ) * [ -Sq( x ) * Cos( ν * t ) + Rq( x ) * Sin( ν * t ) ]
- these are simply imaginary parts of exp[ ( μ + i * ν ) t ] * [ R( x ) - i * S( x ) ].
So, there is also a second coordinated set Sp( x )
è
Sq( x ).
Prof. A.A.Shkalikov has proved completeness of system of eigen functions for the similar equations, but the proof is not published yet.
