A. P. PRUDNIKOV

ON THE THEORY OF OPERATIONAL CALCULUS

(Presented by Academician A. A. Dorodnitsyn, 10 X 1961)

V. A. Ditkin \((^1)\) constructed an operational calculus for the operator

\[ B=\frac{d}{dt}\,t\,\frac{d}{dt}, \]

generated by Bessel’s equation

\[ y''+\frac{1}{x}y'-\left(1-\frac{\nu^2}{x^2}\right)y=0. \tag{1} \]

In the present note an operational calculus is set forth for the operator

\[ T=\frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}. \]

This operator is closely connected with the equation

\[ y'''+\frac{3}{x}y''+\frac{1-3m^2+3mn-3n^2}{x^2}y' +\left[1+\frac{(m+n)(2m-n)(m-2n)}{x^3}\right]y=0. \tag{2} \]

If in equation (2) \(m=n=0\), then we shall have

\[ y'''+\frac{3}{x}y''+\frac{1}{x^2}y'+y=0. \tag{3} \]

Put \(x=3\sqrt[3]{\lambda t}\). Then the preceding equation takes the form

\[ \frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}y+\lambda y=0. \tag{4} \]

Denote by \(L_T\) the set of all functions \(f(t)\), defined on the half-line \(0\le t<\infty\), Lebesgue integrable on every finite interval \((0,A)\), and satisfying the condition

\[ \int_0^t \frac{dx}{x}\int_0^1 \frac{dy}{y}\int_0^{xy}|f(t)|\,dt<\infty \]

for every \(t_0>0\). Denote by \(M_T\) the set of all functions of the form

\[ F(t)=\int_0^t \frac{dx}{x}\int_0^1 \frac{dy}{y}\int_0^{xy} f(t)\,dt+C, \]

where \(f(t)\) is an arbitrary function from \(L_T\), and \(C\) is an arbitrary constant. In the set \(L_T\) define the product of functions \(f_1(t)\in L_T\) and \(f_2(t)\in L_T\) by the formula

\[ f(t)=\int_0^t d\tau\int_0^1 dx\int_0^1 f_1(xy\tau)\,f_2[(t-\tau)(1-x)(1-y)]\,dy. \]

The function \(f(t)\) belongs to the set \(L_T\). For every function of the set \(M_T\) there exists almost everywhere a third derivative \(F'''(t)\). Define in the set \(M_T\) the operation of multiplication by the formula

\[ F_1(t)*F_2(t)= \frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt} \int_0^t d\tau\int_0^1 dx\int_0^1 F_1(xy\tau)\,F_2[(t-\tau)(1-x)(1-y)]\,dy. \tag{5} \]

If

\[ F_i(t)=\int_0^t \frac{dx}{x}\int_0^1 \frac{dy}{y}\int_0^{xy} f_i(t)\,dt \qquad (i=1,2), \]

\[ f(t)=\int_0^t d\tau\int_0^1 dx\int_0^1 f_1(xy\tau)\,f_2[(t-\tau)(1-x)(1-y)]\,dy, \]

then we shall have

\[ F_1(t)*F_2(t)=\int_0^t \frac{dx}{x}\int_0^1 \frac{dy}{y}\int_0^{xy} f(t)\,dt. \]

Addition in the set \(M_T\) is defined in the natural way. The product defined in (5) again belongs to the set \(M_T\); it is commutative, associative, and has distributivity with respect to addition. Consequently, the set \(M_T\) forms a commutative ring. Moreover, if one of the factors in (5) is a number \(\alpha\), then we have

\[ \alpha f_1(t)=\alpha \frac{d}{dt}t\frac{d}{dt}t\frac{d}{dt} \int_0^t d\tau\int_0^1 dx\int_0^1 f_1[(t-\tau)(1-x)(1-y)]\,dy = \]

\[ =\alpha \frac{d}{dt}t\frac{d}{dt}t\frac{d}{dt} \int_0^t \frac{d\tau}{\tau}\int_0^1 \frac{dx}{x}\int_0^{x\tau} f_1(\xi)\,d\xi = \]

\[ =\alpha \frac{d}{dt}t\frac{d}{dt} \int_0^t \frac{du}{u}\int_0^u f_1(\xi)\,d\xi =\alpha B\frac{1}{B}f_1(t)=\alpha f_1(t), \]

i.e., the product of a number by a function in the ring coincides with the ordinary product of a number by a function. If in (5) both factors are numbers, then the product (5) coincides with the ordinary product of numbers.

It can be proved that the ring \(M_T\) has no zero divisors. Consequently, the ring \(M_T\) can be extended to a field of quotients. We shall denote the extended ring by \(\mathfrak{M}_T\). The elements of the set \(\mathfrak{M}_T\) will be called operators. For the operator \(\frac{1}{t}\) we introduce the notation \(\frac{1}{t}=T\). Then for the inverse operator \(T^{-1}=\frac{1}{T}\) we shall have \(\frac{1}{T}=t\). Therefore

\[ \frac{1}{T}f(t)=\frac{d}{dt}t\frac{d}{dt}t\frac{d}{dt} \int_0^t d\tau\int_0^1 dx\int_0^1 xy\tau f[(t-\tau)(1-x)(1-y)]\,dy, \]

or

\[ \frac{1}{T}f(t)=\int_0^t \frac{d\tau}{\tau}\int_0^1 \frac{dx}{x}\int_0^{x\tau} f(\xi)\,d\xi . \tag{6} \]

If \(F(t)\in M_T\) and \(F(0)=0\), then

\[ TF(t)=\frac{d}{dt}t\frac{d}{dt}t\frac{d}{dt}F(t) =t^2F'''(t)+3tF''(t)+F'(t). \tag{7} \]

From (6) it follows that

\[ \frac{1}{T^n}f(t)=\frac{1}{(n!)^3}\int_0^t (t-\tau)^n\,d\tau \int_0^1 (1-x)^n\,dx\int_0^1 f(xy\tau)(1-y)^n\,dy. \tag{8} \]

In the particular case we have

\[ \frac{1}{T^n}=\frac{t^n}{(n!)^3}. \tag{9} \]

One of the linearly independent solutions of equation (4), namely

\[ J_{0,0}^{(2)}(3\sqrt[3]{\lambda t})={}_0F_2(1,1;-\lambda t), \]

for \(t=0\) takes the value \(J_{0,0}^{(2)}(0)=1\). Therefore from (4), (6), (7) it follows that

\[ T\left[J_{0,0}^{(2)}(-3\sqrt[3]{\lambda t})-J_{0,0}^{(2)}(0)\right] =\frac{d}{dt}t\frac{d}{dt}t\frac{d}{dt}J_{0,0}^{(2)}(-3\sqrt[3]{\lambda t}) \]

or

\[ T\left[J_{0,0}^{(2)}\left(-3\sqrt[3]{\lambda t}\right)-1\right] =\lambda J_{0,0}^{(2)}\left(-3\sqrt[3]{\lambda t}\right). \]

Hence

\[ \frac{T}{T+\lambda}=J_{0,0}^{(2)}\left(3\sqrt[3]{\lambda t}\right), \qquad \frac{T}{T-\lambda}=I_{0,0}^{(2)}\left(3\sqrt[3]{\lambda t}\right), \tag{10} \]

where

\[ I_{0,0}^{(2)}(x)={}_0F_2\left[1,1;\left(\frac{x}{3}\right)^3\right]. \]

With the aid of formulas (10) it is easy to obtain

\[ \frac{T^2}{T^2-\lambda^2} =\frac{1}{2}\left[ I_{0,0}^{(2)}\left(3\sqrt[3]{\lambda t}\right) +J_{0,0}^{(2)}\left(3\sqrt[3]{\lambda t}\right) \right]; \tag{11} \]

\[ \frac{\lambda T}{T^2-\lambda^2} =\frac{1}{2}\left[ I_{0,0}^{(2)}\left(3\sqrt[3]{\lambda t}\right) -J_{0,0}^{(2)}\left(3\sqrt[3]{\lambda t}\right) \right]; \tag{12} \]

\[ \frac{T^2}{T^2+\omega^2} =\operatorname{ber}_{0,0}^{(2)}\left(3\sqrt[3]{\omega t}\right), \qquad \frac{\omega T}{T^2+\omega^2} =\operatorname{bei}_{0,0}^{(2)}\left(3\sqrt[3]{\omega t}\right), \tag{13} \]

where

\[ \operatorname{ber}_{0,0}^{(2)}(x) =\sum_{k=0}^{\infty}\frac{(-1)^k(x/3)^{6k}}{[(2k)!]^3}, \qquad \operatorname{bei}_{0,0}^{(2)}(x) =\sum_{k=0}^{\infty}\frac{(-1)^k(x/3)^{3(2k+1)}}{[(2k+1)!]^3}. \]

Applying to the field of operators \(\mathfrak{M}_T\) the known operational methods \((^2,^3)\), one can substantially extend the table of values of the operators (10), (11), (12), (13). For example, differentiating (10) with respect to the parameter \(\lambda\) and then putting \(\lambda=1\) and \(\lambda=-1\), we obtain

\[ \frac{T}{(T+1)^{n+1}} =\frac{1}{n!}\,t^{n/3}J_{n,n}^{(2)}\left(3\sqrt[3]{t}\right); \tag{14} \]

\[ \frac{T}{(T-1)^{n+1}} =\frac{1}{n!}\,t^{n/3}I_{n,n}^{(2)}\left(3\sqrt[3]{t}\right), \tag{15} \]

where

\[ J_{n,n}^{(2)}(x) =\frac{(x/3)^{2n}}{\Gamma^2(n+1)} \,{}_0F_2\left[n+1,n+1;-\left(\frac{x}{3}\right)^3\right], \]

\[ I_{n,n}^{(2)}(x) =\frac{(x/3)^{2n}}{\Gamma^2(n+1)} \,{}_0F_2\left[n+1,n+1;\left(\frac{x}{3}\right)^3\right]. \]

From (13) we find

\[ T\left(\frac{\omega T}{T^2+\omega^2}\right) =\frac{\omega T^2}{T^2+\omega^2} =\omega\,\operatorname{ber}_{0,0}^{(2)}\left(3\sqrt[3]{\omega t}\right), \]

\[ T\left(\frac{T^2}{T^2+\omega^2}-1\right) =-\frac{\omega^2T}{T^2+\omega^2} =-\omega\left(\frac{\omega T}{T^2+\omega^2}\right) =-\omega\,\operatorname{bei}_{0,0}^{(2)}\left(3\sqrt[3]{\omega t}\right). \]

Hence it follows that

\[ \frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}\, \operatorname{bei}_{0,0}^{(2)}\left(3\sqrt[3]{\omega t}\right) =\omega\,\operatorname{ber}_{0,0}^{(2)}\left(3\sqrt[3]{\omega t}\right), \]

\[ \frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}\, \operatorname{ber}_{0,0}^{(2)}\left(3\sqrt[3]{\omega t}\right) =-\omega\,\operatorname{bei}_{0,0}^{(2)}\left(3\sqrt[3]{\omega t}\right). \]

Multiplying equality (15) by \(\mu^n\) and summing with respect to \(n\) from \(0\) to \(\infty\), we find

\[ \sum_{n=0}^{\infty}\frac{T\mu^n}{(T-1)^{n+1}}=\frac{T}{T-1-\mu}, \]

whence

\[ \sum_{n=0}^{\infty}\frac{\mu^n}{n!}\,t^{n/3} I_{n,n}^{(2)}\left(3\sqrt[3]{t}\right) = I_{0,0}^{(2)}\left(3\sqrt[3]{(1+\mu)t}\right). \]

The operational calculus for the operator \(T=\dfrac{d}{dt}\,t\,\dfrac{d}{dt}\,t\,\dfrac{d}{dt}\) can be constructed starting from the corresponding integral transform. The analogue of the Laplace transform here will be the integral transform

\[ f^*(T)=2\int_{0}^{\infty} f(t)\,M(Tt)\,dt, \]

where

\[ M(t)=2\int_{0}^{\infty} e^{-4t/u^2}K_0(u)\,\frac{du}{u} \]

is the solution of equation (4) for \(\lambda=1\). The operational calculus for the operator

\[ T=\frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt} \]

can also be applied to the solution of differential equations. The solutions of differential equations of the form

\[ L\left(\frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}\right)x(t)=f(t), \]

where \(L(\lambda)=\lambda^n+a_1\lambda^{n-1}+\cdots+a_n\) is a polynomial with constant coefficients \(a_i\), are found most simply. Replacement of the operator \(\dfrac{d}{dt}\,t\,\dfrac{d}{dt}\,t\,\dfrac{d}{dt}\) by the operator \(T\) is carried out according to the formula

\[ \left(\frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}\right)^n x(t) = T^n x(t)-T x_{n-1}-T^2 x_{n-2}-\cdots-T^n x_0, \]

where

\[ x_k=T^k x(t)\big|_{t=0}\qquad (k=0,1,2,\ldots,n-1). \]

For example, let us find the solution of the equation

\[ \left(\frac{d}{dt}\,t\,\frac{d}{dt}\,t\,\frac{d}{dt}\right)^2 x(t)+x(t)=0 \]

under the condition

\[ x(0)=1,\qquad T x(t)\big|_{t=0} = \left[t^2x'''(t)+3t x''(t)+x'(t)\right]_{t=0} =1. \]

After replacing the operator \(\dfrac{d}{dt}\,t\,\dfrac{d}{dt}\,t\,\dfrac{d}{dt}\) by the operator \(T\), we shall have

\[ T^2x(t)-T-T^2+x(t)=0, \]

whence

\[ x(t)=\frac{T^2}{T^2+1}+\frac{T}{T^2+1} = \operatorname{ber}_{0,0}^{(2)}\left(3\sqrt[3]{t}\right) + \operatorname{bei}_{0,0}^{(2)}\left(3\sqrt[3]{t}\right). \]

Computing Center
Academy of Sciences of the USSR

Received
5 X 1961

CITED LITERATURE

1. V. A. Ditkin, DAN, 116, No. 1 (1957).
2. Ya. Mikusinski, Operational Calculus, IL, 1956.
3. V. A. Ditkin, A. P. Prudnikov, Integral Transforms and Operational Calculus, 1961.
4. R. Delerue, Sur le calcul symbolique à n variables, Thèse, Montpellier, 1951.