On a Method for Solution of the Ordinary Differential Equations Connected with Huygens’ Equations

A. H. Hovhannisyan, B.-W. Schulze

The problem of describing the linear second order hyperbolic equations, satisfying the Huygens’ principle, is known as the Hadamards’ problem [1], [2], [3]. Ch. Huygens [4] formulated this principle as a geometrical method for constructing the wave front, and it was used for explaining the main properties of light propagation. Its mathematical formulation within the theory of Cauchy problem for hyperbolic partial differential equations has been given by well-known French mathematician J. Hadamard in his 1923 Yale lectures [1]. It has received a good deal of attention and nowadays is a classical problem in mathematical physics, but is has turned out to be very difficult and, at present it is still far from its complete solution.

Let us consider the Cauchy problem for hyperbolic second order differential equations

∑_{i,j=1}^{n} g^{ij}(x)u_{x^{i}x^{j}} + ∑_{i=1}^{n} b^{i}(x)u_{x^{i}} + c(x)u = 0,  (1)

u|_{S} = f, \frac{∂u}{∂ν}|_{S} = g.  (2)

Here S is a space-like manifold of the dimension n − 1 and \frac{∂u}{∂ν} is a derivative with respect to the normal ν to S.

DEFINITION 1. Equation (1) satisfies the Huygens’ principle when the value of the solution of the Cauchy problem (1), (2) for any point P = (x_{0}^{1}, …, x_{0}^{n}) is determined by the Cauchy data given in the intersection of the initial manifold S and the characteristic conoid with vertex at P.

It should be noted that the Huygens’ principle is invariant relative to the following, so-called elemtary transformations:

(a) non-degenerate coordinant transformation;

(b) transformation of the unknown function u → λu, λ(x) ≠ 0;

(c) multiplication of the equation by the function λ(x) ≠ 0. Two equations are called equivalent when one of them is derived from the other by means of elementary

transformations. In our further considerations, these elementary transformations are necessarily taken into account. Papers [5] and [6] contain descriptions of all the Huygens’ equations of the type

L_{n}^{(k)}[u] ≡ u_{tt} − u_{xx} − ∑_{i=1}^{n−2} a^{i}(x − t)u_{y^{i}y^{i}} + a^{k}(x − t)b(y^{k})u = 0,  (3)

k = 1, …, n − 2, where n ≥ 4 is an even number, and a^{i} > 0, i = 1, …, n − 2.

We briefly describe these results assuming k = 1 for definiteness.

We denote

l_{μ} = \frac{∂}{∂y^{1}} − \frac{μ′(y^{1})}{μ(y^{1})}, l_{μ}^{∗} = −\frac{∂}{∂y^{1}} − \frac{μ′(y^{1})}{μ(y^{1})},

where μ(y^{1}) is an arbitrary solution of the equation^{1}

μ′′ − b(y^{1})μ = 0.  (4)

1A.H. Hovhannisyan acknoweldges DAAD for financial support during his stay in Germany.

Then the L_{n}^{(1)} may be written in the form

L_{n}^{(1)} = P + a^{1}(x − t)l_{μ}^{∗}l_{μ},

where

P = \frac{∂^{2}}{∂t^{2}} − \frac{∂^{2}}{∂x^{2}} − ∑_{i=2}^{n−2} a^{i}(x − t)\frac{∂^{2}}{∂(y^{i})^{2}} .

We define the operator

L̃_{n}^{(1)} = P + a^{1}(x − t)l_{μ}l_{μ}^{∗}  (5)

and call it l_{μ}−transform of the operator L_{n}^{(1)}. It is clear that the equality

l_{μ}L_{n}^{(1)} = L̃_{n}^{(1)}l_{μ}  (6)

takes place.

THEOREM 1. If L_{n}^{(1)} satisfies the Huygens’ principle, then L̃_{n+2}^{(1)} also satisfies the Huygens’ principle. We denote the operator L_{n}^{(1)} with b(y^{1}) ≡ 0 as L_{n}^{0}. From [2] it is known that the equation

L_{n}^{0}[u] = 0

satisfies the Huygens’ princple for any even n ≥ 4. Making use of the operator L_{04}, theorem 1 allows to construct Huygens’ operators for n ≥ 6 variables.

There arises a natural question: whether all the Huygens’ operators of the type (3) may be obtained through the described construction basing on L_{4}^{0}. A positive answer to this question is given in [6].

THEOREM 2. Let the operator L_{n}^{(1)} satisfy the Huygens’ princple. Then it may be derived from the operator L_{0n} through l_{μ}−transforms applied no more than \frac{n−4}{2} times. The aim of the present paper is to show how helpful the described algorithm for constructing the Huygens’ equations is and to investigate the equation (4) in detals. Taking account of the equality (6) and equation (4), it follows from theorem 2 that there are operators

l_{μk} = \frac{∂}{∂y^{1}} − \frac{μ_{′}^{k}}{μ_{k}} , k = 0, 1, …, m − 1, m ≤ \frac{n − 4}{2}

such that

l_{μ_{m−1}} ⋯ l_{μ_{0}}L_{0n} = L_{n}^{(1)}l_{μ_{m−1}} ⋯ l_{μ_{0}},  (7)

where

μ_{0} = αy^{1} + β, (α, β = const),
(l_{μ_{k}}l_{μ_{k}}^{∗})μ_{k+1} = 0, k = 0, …, m − 2.  (8)

If m is the minimum number of transformations which transform the operator L_{0n} into L_{n}^{(1)}, then the functions μ_{k} (k = 0, …, m − 1) is uniquely determined with the accuracy up to a constant factor [6].

The following reccurent relation is easily obtained from (8):

μ_{k+1} = \frac{a_{k}}{μ_{k}} + \frac{b_{k}}{μ_{k}} ∫ μ_{k}^{2}(y^{1})dy^{1}.

However, it is not always convenient to use this formula because of the necessity to find the indefinite integral of μ_{k}^{2}. Therefor, further we study properties of the function b(y^{1}) which allow as to determine conditions, easily checked and necessary for equation (3) to acquire Huygens’ properties, and which may indicate us some other method to find functions μ_{k}.

First, we give the known results [7].

THEOREM 3. We assume that the operator has Huygens’ properties for even values n ≥ N, m = \frac{N−4}{2} and {l_{μk}}_{k=0}^{m−1} is a unique sequence of operators determined by the formula (8), so that the equality (7) takes place. Then

(a)  b(y^{1}) = −2\frac{d^{2}}{d(y^{1})^{2}} [ln ∏_{k=0}^{m−1} μ_{k}(y^{1})],

(b)  ∏_{k=0}^{l} μ_{k}(y^{1}) = P_{l}(y^{1})

is a polynomial of degree \frac{1}{2}(l + 1)(l + 2), 0 ≤ l ≥ m − 1. Point (a) is proved by direct computations and point (b) through the following important result.

THEOREM 4. If the sequance of functions {μ_{k}(y^{1})}_{k=0}^{m−1} satisfies the conditions of (8), then each of μ_{k}(y^{1}) is a rational function. Using these results, we shall prove some properties of the function b(y^{1}).

Let the polynomial P_{m−1}(y^{1}) has the form

P_{m−1}(y^{1}) = (y^{1} − a_{1})^{k_{1}} ⋯ (y^{1} − a_{s})^{k_{s}}, (k_{1} + ⋯ + k_{s} = \frac{1}{2}m(m + 1)).

Then from point (a) of theorem 3 we obtain:

b(y^{1}) = 2\frac{∑_{q=1}^{s} ∏_{i≠q}(y^{1} − a_{i})^{2}k_{q}}{∏_{i=1}^{s}(y^{1} − a_{i})^{2}} .

Hence, singularities of the function b(y^{1}) are poles of the second order and (4) is a Funchian equation. We separate one of the poles, e.g. a_{j}, 1 ≤ j ≤ s, and write the function b(y^{1}) in the form

b(y^{1}) = \frac{b_{j}(y^{1})}{(y^{1} − a_{j})^{2},}

where

b_{j}(y^{1}) = 2k_{j} + 2\frac{∑_{q=1}^{s} ∏_{i≠q}(y^{1} − a_{i})^{2}k_{q}}{∏_{i≠j}^{s}(y^{1} − a_{i})^{2}} .  (9)

It immediately follows from this formula that b′(a_{j}) = 0. Let L̃_{n}^{(1)} be the l_{μ_{m}}−transform of the operator L_{n}^{(1)}. Then μ_{m} is the solution of equation (4). According to theorem 4, μ_{m} is a rational fuction. We study its behavior in the neighborhood singular point a_{j} 1 ≤ j ≤ s. With this purpose we write equation (4) as

μ′′_{m} − \frac{b_{j}(y^{1})}{(y^{1} − a_{j})^{2}} = 0.

Solution μ_{m} is found in the form of

μ_{m} = (y^{1} − a_{j})^{ρ}[1 + ∑_{k=1}^{∞} μ_{m}^{k}(y^{1} − a_{j})^{k}]

where ρ is the root of the defining equation

ρ^{2} − ρ − b_{j}(a_{j}) = 0.  (10)

Hence

ρ_{1,2} = \frac{1}{2} ± \sqrt{\frac{1}{4} + b_{j}(a_{j})}.

As μ_{s} is a rational function, then ρ_{1} and ρ_{2} must be integers

\frac{1}{2} ± \frac{\sqrt{1 + 4b_{j}(a_{j})}}{2} = q_{j} + 1, q_{j} = 0, ±1, ….

Hence

b_{j}(a_{j}) = q_{j}(q_{j} + 1).  (11)

As the negative values q_{j} do not suggest new values for b_{j}(a_{j}), it may be assumed that q_{j} = 0, 1, ….

We obtain from formula (9) that

b_{j}(a_{j}) = 2k_{j}.

Therefore, multiplicity of the root a_{j} of the polynomial P_{m−1} may assume the following values:

k_{j} = \frac{q_{j}(q_{j} + 1)}{2}, j = 1, 2, …, s.

As the order of the polynomial P_{m−1} is \frac{1}{2}m(m + 1) and m ≤ \frac{n−4}{2} we obtain

∑_{j=1}^{s} q_{j}(q_{j} + 1) ≤ \frac{n − 4}{2} · \frac{n − 2}{2}.

If s = 1, then

q ≤ \frac{n − 4}{2}

which is in agreement with the known result [2].

Roots of equation(10) are

ρ_{1} = q_{j} + 1, ρ_{2} = −q_{j}.

Therefore, signularity of the rational function μ_{m} in the point a_{j} may have only the form (y^{1} − a_{j})^{−q_{j}}. Since μ_{m} is the solution of a linear equation, its singularities may appear only in the points where singularities of the equation coefficients are. Therefore the function μ_{m} may be found as

μ_{m} = \frac{T_{r}(y^{1})}{(y^{1} − a_{1})^{q_{1}} ⋯ (y^{1} − a_{s})^{q_{s}}} ,  (12)

where T_{r}(y^{1}) is a polynomial of the order r. Acording to theorem 3 the polynomial

P_{m} = ∏_{k=0}^{m} μ_{k} = P_{m−1}μ_{m}

has the order \frac{1}{2}(m − 1)(m + 2), and the order of P_{m−1} is \frac{1}{2}m(m + 1), accordingly.

Therefore,

r = m + 1 + ∑_{k=1}^{s} q_{k}.

Writing T_{r}(y^{1}) with indefinite coefficients and substituting μ_{m} into the equation, it is possible to determine these coefficients, and hence, to find μ_{m}.

The facts, proved above, may be presented as the following result.

THEOREM 5. Let equation (3) satisfy the Huygens’ principle and let b(y^{1}) ≠ 0. Then

(a) singularities if the function b(y^{1}) a_{1}, …, a_{s} are poles of the second order, being s ≤ \frac{1}{2}m(m+1) ≤ \frac{(n−2)(n−4)}{8}.

(b) if the function b(y^{1}) is presented in the neighbourhood of a singular point a_{j} 1 ≤ j ≤ s as

b(y^{1}) = \frac{b_{j}(y^{1})}{(y^{1} − a_{j})^{2}} ,

then b_{j}(a_{j}) may assume only one of the following values

b_{j}(a_{j}) = q_{j}(q_{j} + 1), q_{j} = 0, 1, …,

where

∑_{j=1}^{s} q_{j}(q_{j} + 1) ≤ \frac{(n − 4)(n − 2)}{4}

and b′_{j}(a_{j}) = 0.

(c) solution of equation (4) has the form of (12). Let us consider the Huygens’ equation [5], [6]

L_{8}[u] ≡ u_{tt} − u_{xx} − ∑ a^{i}(x − t)u_{y^{i}y^{i}} + a^{1}(x − t)\frac{6y^{1}[(y^{1})^{3} + 2]}{[(y^{1})^{3} − 1]^{2}}u = 0.

The function

b(y^{1}) = \frac{6y^{1}[(y^{1})^{3} + 2]}{[(y^{1})^{3} − 1]^{2}}

has the singularities in the points a_{1} = 1, a_{2} = −\frac{1}{2} + i\frac{\sqrt{3}}{2}, a_{3} = −\frac{1}{2} − i\frac{\sqrt{3}}{2}. For the b_{j}(y^{1}), i = 1, 2, 3 we have

b_{1}(1) = 1 · 2, q_{1} = 1,

b_{1}( − \frac{1}{2} + i\frac{\sqrt{3}}{2}) = 1 · 2, q_{2} = 1,

b_{1}( − \frac{1}{2} − i\frac{\sqrt{3}}{2}) = 1 · 2, q_{2} = 1.

Operator L_{8} is obtained from the operator L_{8}^{0}

L_{8}^{0} = \frac{∂^{2}}{∂t^{2}} − \frac{∂^{2}}{∂x^{2}} − ∑ a^{i}(x − t)\frac{∂^{2}}{∂y^{i2}}  (13)

with the operators l_{μ_{0}}, l_{μ}, where μ_{0} = y^{1} and μ_{1} is a solution of the equation

μ′′ − \frac{2}{y^{i2}} μ = 0

and has the form

μ_{1} = \frac{(y^{1})^{3} − 1}{y^{1}}.

Then

P_{0} = μ_{0} = y^{1},
P_{1} = μ_{0}μ_{1} = (y^{1})^{3} − 1.

The solution of equation

μ′′ − \frac{6y^{1}[(y^{1})^{3} + 2]}{[(y^{1})^{3} − 1]^{2}}μ = 0  (14)

according to the (12) can be find in the form

μ_{2} = \frac{T_{2}(y^{1})}{(y^{1})^{3} − 1} = \frac{∑_{i=1}^{6} α_{i}(y^{1})^{i}}{(y^{1})^{3} − 1} .

Substituding this function in (14) we obtain

30α_{6} − 36α_{6} + 6α_{6} = 0,
20α_{5} − 30α_{5} + 6α_{5} = 0,
12α_{4} − 24α_{4} + 6α_{4} = 0,
6α_{3} − 30α_{6} − 18α_{3} + 6α_{3} = 0,
2α_{2} − 20α_{5} − 12α_{2} + 6α_{2} = 0,
−12α_{4} − 6α_{1} + 6α_{1} = 0,
−6α_{3} + α_{0} = 0,
−2α_{2} = 0.

From here we have

α_{5} = α_{4} = α_{2} = 0,
α_{0} = α_{3} = −5α_{6}

and α_{1}, α_{6} are any numbers. If we take α_{6} = 0, α_{1} = 1, then

μ_{2} = \frac{y^{1}}{(y^{1})^{3} − 1}.

If α_{6} = −1, α_{1} = 0

μ_{2} = \frac{5 + 5(y^{1})^{3} − (y^{1})^{6}}{(y^{1})^{3} − 1} .

Therefore, the common solution of the equation (14) has the form

μ_{2} = C_{1} \frac{y^{1}}{(y^{1})^{3} − 1} + C_{2} \frac{5 + 5(y^{1})^{3} − (y^{1})^{6}}{(y^{1})^{3} − 1} .  (15)

For C_{1} = 1, C_{2} = 0 we have

P_{2} = y^{1}, b(y^{1}) = \frac{2}{(y^{1})^{2}}

and we obtain the known Huygens’ equation

u_{tt} − u_{xx} − ∑_{i=1}^{8} a^{i}(x − t)u_{y^{i}y^{i}} + a^{1}(x − t)\frac{2}{(y^{1})^{2}} u = 0.

If C_{1} = 0, C_{2} = 1, then

P_{2} = 5 + 5(y^{1})^{3} − (y^{1})^{6}

u_{tt} − u_{xx} − ∑_{i=1}^{8} a^{i}(x − t)u_{y^{i}y^{i}} + a^{1}(x − t)\frac{2[6(y^{1})^{10} + 225(y^{1})^{4} − 150y^{1}]}{[(y^{1})^{6} − 5(y^{1})^{3} − 5]^{2}}u = 0.

In the case C_{1} ≠ 0, C_{2} ≠ 0, it is follow that
P_{2} = y^{1} + γ[5 + 5(y^{1})^{3} − (y^{1})^{6}]

and the following Huygens’ equation

u_{tt} − u_{xx} − ∑_{i=1}^{8} a^{i}(x − t)u_{y^{i}y^{i}} + a^{1}(x − t)\frac{2 + 36γ(y^{1})^{5} + 6γ^{2}y^{1}[2(y^{1})^{9} + 75(y^{1})^{3} − 50]}{\{y^{1} + γ[5 + 5(y^{1})^{3} − (y^{1})^{6}]\}^{2}}u = 0.  (16)

The solution of the equation (14) can be find in the form

μ_{2} = \frac{(y^{1} − 1)^{2}T_{3}(t^{1})}{(y^{1})^{3} + y^{1} + 1}

too, where

T_{3}(y^{1}) = ∑_{i=0}^{3} α_{i}(y^{1})^{i}.

Substituding this function in the equation () gives

μ_{2} = \frac{(y^{1} − 1)^{2}[(y^{1})^{3} + 3(y^{1})^{2} + 6^{1} + 5]}{(y^{1})^{2} + y^{1} + 1}.  (17)

Using this solution it is possible to find the common solution of the equation (14). The function (17) can be find from (15) if we take C_{1} = 9, C_{2} = −1 (γ = −\frac{1}{9}). In this case the equation (16) can be written in the form

u_{tt} − u_{xx} − ∑_{i=1}^{8} a^{i}(x − t)u_{y^{i}y^{i}} + a^{1}(x − t)[\frac{6}{(y^{1} − 1)^{2}} + \frac{6[(y^{1})^{4} + 4(y^{1})^{3} − 6(y^{1})^{2} + 2y^{1} + 2]}{[(y^{1})^{3} + 3(y^{1})^{2} + 6(y^{1})^{1} + 5]^{2}}]u = 0.

To constuct the new Huygens’ equation, we have to know the factorization of the denominator of the function b(y^{1}). However using the fact, that P_{3} = P_{2}μ_{3} is the polynomial of degree 10, it is possible to assume, that the equation

μ′′_{3} − \frac{2 + 36γ(y^{1})^{5} + 6γ^{2}y^{1}[2(y^{1})^{9} + 75(y^{1})^{3} − 50]}{\{y^{1} + γ[5 + 5(y^{1})^{3} − (y^{1})^{6}]\}^{2}}μ_{3} = 0  (18)

has the solution of the form

μ_{3} = \frac{T_{10}(y^{1})}{y^{1} + γ[5 + 5(y^{1})^{3} − (y^{1})^{6}]}.

Easy to see that substitution of the function

μ = \frac{T_{r}(y^{1})}{Q_{s}(y^{1})},

in the equation

μ′′ − \frac{R}{Q_{s}^{2}} = 0

lead to expresion

Q_{s}^{2}T′′_{r} − 2Q_{s}Q′_{s}T′_{r} + [2(Q′_{s})^{2} − Q_{s}Q′′_{s} − R]T_{r} = 0.  (19)

For μ_{3} we have

T_{r} = T_{10} = ∑_{i=0}^{10} α_{i}(y^{1})^{i},

Q_{s} = Q_{6} = y^{1} + γ[5 + 5(y^{1})^{3} − (y^{1})^{6}],
R = 2 + 36γ(y^{1})^{5} + 6γ^{2}y^{1}[2(y^{1})^{9} + 75(y^{1})^{3} − 50].

Substituding these functions in (19) we can find the common solution of the equation (18)

μ_{3} = C_{1}μ_{3}^{(1)} + C_{2}μ_{3}^{(2)},

where

μ_{3}^{(1)} = \frac{3γ^{2}(y^{1})^{10} − 45γ^{2}(y^{1})^{7} − 21γ(y^{1})^{5} − 105γ(y^{1})^{2} − 525γ^{2}y^{1} − 7}{y^{1} + γ[5 + 5(y^{1})^{3} − (y^{1})^{6}]}

and

μ_{3}^{(2)} = \frac{(y^{1})^{3} − 1}{y^{1} + γ[5 + 5(y^{1})^{3} − (y^{1})^{6}]}.

Now let us consider the following Huygens’ equation

L_{8}[u] ≡ u_{tt} − u_{xx} − ∑_{i=1}^{6} a^{i}(x − t)u_{y^{i}y^{i}} + \frac{6a^{1}(x − t)}{(y^{1})^{2}}u = 0  (20)

Operator L_{8} can be obtained from L_{8}^{0} (13) by the transformations with operators l_{μ_{0}} and l_{μ_{0}}, μ_{0} = y^{1}, μ_{1} = (y^{1})^{2}. To construct the next Huygens’ equation we have to solve

μ′′_{2} − \frac{6}{(y^{1})^{2}} μ_{2} = 0.

Then

μ_{2} = \frac{(y^{1})^{5} + γ}{(y^{1})^{2}}

and

P_{2} = μ_{0}μ_{1}μ_{2} = y^{1}[(y^{1})^{5} + γ].

The coresponding Huygens’ equation has the form

u_{tt} − u_{xx} − ∑_{i=1}^{6} a^{i}(x − t)u_{y^{i}y^{i}} + a^{1}(x − t)\frac{12(y^{1})^{10} − 36γ(y^{1})^{5} + 2γ^{2}}{(y^{1})^{2}[(y^{1})^{5} + γ]^{2}} u = 0.  (21)

Note that in spite of the fact that in the equation (20) q = 2, nevertheless in (21) q_{i} = 1, i = 1, …, 6. Therefor the solution of the equation

μ′′ − \frac{12(y^{1})^{1}0 − 36γ(y^{1})^{5} + 2γ^{2}}{(y^{1})^{2}[(y^{1})^{5} + γ]^{2}} μ_{3} = 0  (22)

can be find in the form

μ_{3} = \frac{T_{10}}{y^{1}[(y^{1})^{5} + γ]}.

Substuding of this function in (22) gives

μ_{3}^{(1)} = \frac{3(y^{1})^{10} + 21γ(y^{1})^{5} − 7γ^{2}}{y^{1}[(y^{1})^{5} + γ]}

and the Huygens’ equation

u_{tt} − u_{xx} − ∑_{i=1}^{10} a^{i}(x − t)u_{y^{i}y^{i}} + a^{1}(x − t)\frac{180(y^{1})^{18} − 1260γ(y^{1})^{13} + 8190γ^{2}(y^{1})^{8} − 5880γ^{3}(y^{1})^{3}}{[3(y^{1})^{10} + 21γ(y^{1})^{5} − 7γ^{2}]^{2}} u = 0.

The second linear independent solution is

μ_{3}^{(2)} = \frac{(y^{1})^{2}}{[(y^{1})^{5} + γ]}

and the corresponding Huygens’ equation

u_{tt} − u_{xx} − ∑_{i=1}^{10} a^{i}(x − t)u_{y^{i}y^{i}} + a^{1}(x − t)\frac{6}{(y^{1})^{2}} u = 0.

All these examples of the odinary differential equations of the form (4) conectes with Huygens’ principle can be usefull for the investigating of the following problem.

Let us write the equation (4) with the coefficient (a) in theorem 3 in the form

μ′′ + ∑_{i=1}^{s} \frac{2k_{i}}{(y^{1} − a_{i})^{2}} μ = 0,  (23)

where a_{i} (i = 1, …, s) are the root of the polynomial P_{m−1}(y^{1}) in formula (b) and k_{i} are there multiplicities. As it was shown before for an equations, conected with Huygens’ equations 2k_{i} = q_{i} (q_{i} + 1) with some positive integers q_{i}. The equations of the form (23) are known as Fuchs equations [9]. The arrising problem is to discribe the a_{i} and k_{i}, i = 1, …, s such that the equation (23) will be conected with Huygens’ principle, in particular all the solutions of (23) are rational functions. Below we give a necessary and suficient condition on a_{i} and k_{i}, i = 1, …, n for which the solutions of the equation (23) do not consist the logarithmic singularities.

In the neighborhood of a singular point a_{r} (r = 1, …, s) an equation (23) can be written in the form

(y^{1} − a_{r})^{2}μ′′ − [q_{r}(q_{r} + 1) + ∑_{{k=1 \atop k≠r}}^{s} q_{k}(q_{k} + 1)\frac{(y^{1} − a_{r})^{2}}{(y^{1} − a_{k})^{2}} ]μ = 0.

If we write

(\frac{y^{1} − a_{r}}{y^{1} − a_{k}} )^{2} = ∑_{n=2}^{∞}(n − 1)\frac{(y^{1} − a_{r})^{n}}{(a_{k} − a_{r})^{n}}

and look for the solution in the form

μ(y^{1}) = (y^{1} − a_{r})^{q} ∑_{m=0}^{∞} α_{mr}(y^{1} − a_{r})^{m}

we obtain

∑_{m=0}^{∞} α_{mr}(m + q)(m + q − 1)(y^{1} − a_{r})^{m} − [q_{r}(q_{r} + 1) + ∑_{{k=1 \atop k≠r}}^{s} q_{k}(q_{k} + 1) ∑_{n=2}^{∞} \frac{(n − 1)}{(a_{k} − a_{r})^{n}} (y^{1} − a_{r})^{n}]+

+ ∑_{m=0}^{∞} α(y^{1} − a_{r})^{m} = 0.  (24)

For m = 0 we have

q_{1} = −q_{r}, q_{r} = q_{r} + 1

and α_{0r} an arbitrary number.

Let us consider q = −q_{r}, when the solution has the singularity. In this case the equation (24) has the form

∑_{m=0}^{∞} α_{mr}(m−q_{r})(m−q_{r} −1)(y^{1} −a_{r})^{m} −[q_{r}(q_{r} +1)+ ∑_{{k=1 \atop k≠r}}^{s} q_{k}(q_{k} +1) ∑_{n=2}^{∞} \frac{(n − 1)(y^{1} − a_{r})^{n}}{(a_{k} − a_{r})^{n}} ] ∑_{m=0}^{∞} α_{mr}(y^{1} −a_{r})^{m} = 0.

For m = 1 we obtain α_{1r} = 0 and for m = p ≥ 2

α_{pr}[p(p − 2q_{r} − 1)] = ∑_{{k=1 \atop k≠r}}^{s} q_{k}(q_{k} + 1) ∑_{l=0}^{p−2} \frac{(p − l − 1)}{(a_{k} − a_{r})^{p−l}} , r = 1, …, s.  (25)

From (25) we obtain THEOREM 6. The solutions of the equation (23) have not a logarithmic singularities if and only if

∑_{{k=1 \atop k≠r}}^{s} q_{k}(q_{k} + 1) ∑_{l=0}^{2q_{r}+1} α_{lr} \frac{(2q_{r} − l)}{(a_{k} − a_{r})^{2q_{r}+1−l}} = 0.  (26)

If s = 1 the equation (23) is none other tahn Euler equation and has the common solution

μ = C_{1}(y^{1} − a_{1})^{q_{1}+1} + C_{2}(y^{1} − a_{1})^{−q_{1}} (q_{1} > 0 is integer)

which does not contain the logarithmic singularity for each value of a_{1}.

For s = 2 it follows from (25) and (26), that the common solution of the equation (23) has the logarithmic singularities for any values of a_{1} and a_{2}.

If s > 2, as is shown in for example (14), there are some distribution of the roots a_{i}, for which the common solution of (23) has no the logarithmic singularities.

R E F E R E N C E S 1. J. HADAMARD, Lectures on Cauchy’s Problem in Linear Partial Differential Equations, New Haven, Yale

University Press, 1923.

2. N. H. IBRAGIMOV, Transformation Groups in Mathematical Physics, ”Nauka”, Moscow, 1983 (in Russian).

3. P. GÜNTHER, Huygens’ Principle and Hyperbolic Equations, Acad. Press, Boston, 1988.

4. CH. HUYGENS’, Traite’ de la Lumiere. Van de Aa, Leyden, 1690. (English translation: Treaties of Light, 1960, reprint by Dover Publications, N.-Y., 1962).

5. A. O. OGANESIAN, “Generalization of Lagnese-Stellmacher’s method on a class of equations satisfying Huygens’ principle” [in Russian], Izv. Akad. Nauk Armenii, Matematika [English translation: Journal of Contemporary Math. Anal. (Armenian Academy of Sciences)], vol. 25, no. 5, pp. 462 — 473, 1990.

6. N. H. IBRAGIMOV, A. O. OGANESIAN, “Hierarchy of Huygens’ equations in the spaces with a nontrivial conformal group” [in Russian], Uspekhi Math. Nauk, 46, no. 3, pp. 111 – 146, 1991.

7. J. E. LAGNESE, The Structure of a Class of Huygens’ Operators. J. Math. Mech., 18 (1969) No. 12, p. 1195 – 1201.

8. WEISS, M. TABOR, G. CARNEVALE, The Painleve Property for Partial Differential Equations. J. Math. Phys., 24 (1983), p. 522 – 526.

9. V. V. GOLUBEV, Lectures on Analitical Theory of Differential Equations (in Russian), M., 1941.