The statement in the title is meant as : a general solution procedure for the Tricomi Equation subject to arbitrarily specified auxiliary conditions.

Let U = U(x, y) be a function of x and y and let U_x, U_xx, denote the first and second

partial derivatives of U with respect to x respectively. A similar notation will be used

to denote the partial derivatives with respect to y.

The Tricomi Equation

yU_xx + U_yy = 0 (eq.1)

plays a central role in the mathematical analysis of transonic

flow. It is an example of a 2nd order linear equation with variable

coefficients.

The following Theorem presents a solution to the Tricomi equa-

tion subject to a general set of auxiliary conditions

Theorem:

Let

F = F(x) (eq.2)

and

G = G(x) (eq.3)

both be infinitely differentiable functions of x.

(Note: If F or G is a polynomial then only a finite number of the

derivatives are non - zero.)

Then a solution of (eq.1), that satisfies the auxiliary conditions,

U(x, 0) = F (eq.4)

U_y(x, 0) = G (eq.5)

in a domain D of the (x, y) plane is given by

U = F - (1/(2*3))(y^3)F" + (1/[(2*3)(5*6)])(y^6)F''''

- (1/[(2*3)(5*6)(8*9)])(y^9)F'''''' + ... + { Gy - (1/(3*4))(y^4)G''

+ (1/[(3*4)(6*7)])(y^7)G'''' - (1/[(3*4)(6*7)(9*10)])(y^10)G''''''

+ ... } (eq.6)

provided that both of the indicated infinite series on the right side

of (eq.6) converge in the domain D. It will be noted that F'', F''''

and F'''''', ... , etc., denote the 2nd, 4th, 6th, ... ordinary derivatives

of of F. The notation for the ordinary derivatives of G are similar.

The proof is straight forward. Simply calculate U_xx and U_yy,

where U is defined by (eq.6) and substitute into (eq.1) and note

the cancellation of the terms.

Before we give the general terms for the two infinite series

indicated in (eq.6), let us give an illustration by use of an

example when F and G are polynomials.

Let F(x) = x^2 and G(x) = x^3,

F' = 2x, F'' = 2

and all higher derivatives of F are zero. We also have

G' = 3x^2, G'' = 6x, G''' = 6

and all higher derivatives of G are zero. Now if we substitute these expressions

for F and G and their non - zero derivatives into (eq.6) we obtain after simplify-

ing,

U = x^2 - (1/3)y^3 + yx^3 - (1/2)xy^4 (eq.7)

We will now calculate the partial derivatives U_y, U_yy and U_xx.

U_y = - y^2 + x^3 - 2xy^3 (eq.8)

U_yy = - 2y - 6xy^2 (eq.9)

U_xx = 2 + 6xy (eq.10)

It is easily seen that (eq.1) is identically satisfied upon the substitution

of (eqs. 9 and 10). We now verify that the auxiliary conditions are met.

Let y = 0 on both sides of (eq.7) and obtain

U(x, 0) = x^2

which meets the condition implied by (eq.4) with F = x^2. Now let y = 0

on both sides of (eq.8) and obtain

U_y(x, 0) = x^3

which satisfies the conditions implicit in (eq.5) with G = x^3

Now before we specify the General Terms on the right side of (eq.6), let's

adopt a notation for the product operator: let

P_i{.....}

denotes the product of the indexed quantities in the curly braces with the

index i running from zero to n.

So then, referring to the right side of (eq.6), the (n + 1)th term, after the

initial term F, involving the derivatives of F, is given by

"(n + 1)th term involving derivatives of F" =

(-1)^(n+1)(P_i{1/[(3i+2)(3i+3)]})(y^(3(n+1)))F{2(n+1)}

where the symbol F{2(n+1)} denotes the "(2(n+1))th order " derivative of F.

And referring to the expression in the curly braces on the right side of (eq.6),

the (n + 1)th term, after the initial term Gy, involving the derivatives of G, is

given by

"(n + 1)th term involving derivatives of G" =

(-1)^(n+1)(P_i{1/[(3i+3)(3i+4)]})(y^(3n+4))G{2(n+1)}

where the symbol G{2(n+1)} denotes the "(2(n+1))th order" derivative of G.

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