On the Equations of Nonstationary Transonic Gas Flows

where variable z is considered as the time-variable . Three-dimensional generalization of this equation is defined by the equation

where variable z here is considered as the time-variable. The solutions of these equations and a corresponding bibliography have been considered recently in [1] In this article we apply the method of solution of the p.d.e.’s described first in [2] and developed then in [3], [4].

This method allow us to construct particular solutions of the partial nonlinear differential equation

with the help of transformation of the function and variables. Essence of method consists in a following presentation of the functions and variables

where variable t is considered as parameter.

Remark that conditions of the type

are fulfilled at the such type of presentation.

In result instead of equation ( 3) one get the relation between the new variables u(x, t, z) and v(x, t, z) and their partial derivatives

(5)

This relation coincides with initial p.d.e at the condition v(x, t, z, s) = t and takes more general form after presentation of the functions u, v in form u(x, t, z, , s) = F (ω, ω t …) and v(x, t, z, s) = Φ(ω, ω t …) with some function ω(x, t, z, s) .

Example.

The equation of Riemann wave

after (u, v)-transformation takes the form

The substitution here of the expressions

give us the linear equation

where F1 (tx) is arbitrary function. Choice of the function F1 (tx) and elimination of the parameter t from the relations

lead to the function f (x, y) satisfying the Riemann wave equation.

The equation (1) after applying (u, v)-transformation with conditions

takes the form

Its solution of the form

lead to the equation on the function A(t, z)

where F2 (t), B(t), F1 (z) are arbitrary functions.

In result we find that the function

is the solution of the equation (6).

After the choice of the functions F2 (t), B(t) and elimination of the parameter t from the relations

one gets the solution of the equation ( 1). Let us consider some examples.

In the case

we find the relations

Elimination of the parameter t from these relations in the case F1 (z) = 0 lead to the solution f (x, y, z) of the equation (1) satisfying the algebraic equation 248832 (f (x, y, z)) 5 z 2 -248832 x 2 (f (x, y, z)) 4 z + -1451520 z 2 xy -221184 y 3 z (f (x, y, z)) 3 + + -216000 z 4 x + 475200 z 3 y 2 + 1327104 yzx 3 + 221184 y 3 x 2 (f (x, y, z)) 2 + + 90000 yz 5 + 995328 y 4 zx + 614400 z 3 x 3 + 1290240 y 2 z 2 x 2 f (x, y, z)–518400 y 3 z 3 x -393216 x 5 z 2 -1179648 x 4 y 2 z + 3125 z 7 -373248 y 5 z 2 –192000 z 4 x 2 y -884736 y 4 x 3 = 0.

In a three dimensional case the equation of Nonstationary Transonic Gas Flow takes the form

Recall that the variable z in this equation play the role of a time-variable. After application of u, v-transformation of the form u(x, t, z, s) = t ∂ ∂t ω(x, t, z, s) -ω(x, t, z, s), v(x, t, z, s) = ∂ ∂t ω(x, t, z, s)

we find from (7) the equation

From here in the case ω(x, t, z, s) = A(t, s) + k (x + z) t one gets the Monge-Ampere equation on the function A(t, s)

It is possible to show that the equation ( 9) can be integrated with the help of corresponding (u, v)-transformation.

Its solutions are dependent from solutions of the linear Laplace equation.

In fact, the equation

takes the form of linear Laplace equation

and its particular solutions after elimination of parameter t give us the solutions of the Monge-Ampere equation ( 9). The substitution of another form ω(x, t, z, s) = A(t, z) + (s + x) t into the equation ( 8) lead to the equation on the function A(t, z)

having the general solution

where F2 (t), F1 (z) are arbitrary functions. The choice of the functions F2 (t), F1 (z) allow us to construct solutions of initial equation. For example in the case

elimination of the parameter t from the relations

give us the solution of the equation (7) satisfying the algebraic equation 16 (f (x, y, z, s))

This equation has solution of the form

where function B(t) is arbitrary and the function A(x, z) satisfies the equation

The solutions of the equation ( 11) can be obtained with the help of (u, v)-transformation and a simplest of them has the form

Using the expression ω(x, t, z) = A(x, z)t + B(t) with a given function A(x, z) and arbitrary function B(t) the solution of the equation (10) can be constructed.

As example, in the case B(t) = ln(t) the elimination of the parameter t from the relations ) + e -y+1 x -1 e -y+1 -1 .

In the case B(t) = te t by analogy way we find the solution

The solution of the equation (11)

A(x, z) = 2/9 x 3 z 2 lead to the function ω(x, t, z) = 2/9 x 3 t z 2 + B(t) where B(t) is arbitrary function.

In the case B(t) = ln(t) we find f (x, y, z) = 2/9 x 3 z 2 + e -y+1 -1 .

In the case B(t) = t