SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS BY THE OPERATOR METHOD IN GENERALIZED FUNCTIONS

V. P. TIMOSHENKO

(Presented by the scientific seminar of the Computing Laboratory of TPI)

In works [3, 4], operational calculus was constructed on the basis of the concept of a generalized function, and an apparatus was developed for solving differential equations with constant coefficients. In the present work, the operator method in generalized functions is extended to ordinary differential equations having both variable and constant coefficients.

1. The set \(K\) of all real functions \(\varphi(x)\), each of which has continuous derivatives of all orders and is finite, i.e., vanishes outside some finite region (depending on \(\varphi\)), is called the basic space of basic finite functions.

A sequence of basic functions \(\varphi_1(x), \varphi_2(x), \ldots, \varphi_n(x), \ldots\) is called convergent to zero in the space \(K\) if these functions vanish outside one and the same bounded region and converge uniformly to zero, as do their derivatives of any order.

Every linear continuous functional defined on the basic space \(K\) is called a generalized function (see [1], p. 13).

The functional that assigns to each function \(\varphi(x)\) its value at the point \(x_0 = 0\) will be called the delta function and denoted by \(\delta(x)\); thus,

\[ (\delta(x), \varphi(x)) = \varphi(0). \]

A generalized function \(f\) is said to be equal to zero in a neighborhood \(\varepsilon\) of the point \(x_0\) if, for every basic function \(\varphi(x)\) different from zero only within this neighborhood \(\varepsilon\), the equality \((f,\varphi)=0\) holds.

If, however, the generalized function \(f\) is not equal to zero in any neighborhood of the point \(x_0\), then \(x_0\) is called essential for the functional \(f\).

The set of all essential points of a generalized function is called its support.

The functional \(f'\), defined for each generalized function \(f\) on any basic function \(\varphi\) by the equality \((f',\varphi)=-(f,\varphi')\), where \(\varphi'(x)\) is the derivative of the function \(\varphi(x)\), is called the generalized derivative of \(f\).

Let us denote by \(s\) the operator of differentiation of a generalized function, i.e., we have \(f' = sf\).

Introduce operators of the following form:

\[ \begin{aligned} T(s,x)f &=\bigl([p_0 g_0(x)S^{n-1}+p_1 g(x)S^{n-2}+\cdots+p_{n-1}g_{n-1}(x)]f,\varphi\bigr) \\ &=\Bigl(f,(-1)^{n-1}\frac{d^{\,n-1}}{dx^{n-1}}[p_0g_0(x)\varphi(x)] \\ &\qquad\quad+(-1)^{n-2}\frac{d^{\,n-2}}{dx^{n-2}}[p_1g_1(x)\varphi(x)] +\cdots+p_{n-1}(x)\varphi(x)\Bigr), \end{aligned} \]

\[ S_{(s)}^n=S^n+\alpha_0 a_0S^{n-1}+a_1a_1S^{n-2}+\cdots+\alpha_{n-1}a_{n-1}, \]

where \(\varphi(x)\) is an arbitrary basic function from \(K\); \(a_i=\mathrm{const}\); \(p_i=0,1\); \(\alpha_i=0,1\); \(g_i(x)\) are infinitely differentiable functions; \(i=0,1,2,\ldots,n-1\).

Let us note that for the operator \(T(s,x)\) the operations of differentiation and multiplication by a function are not commutative.

The convolution \(f_1*f_2\) of two generalized functions \(f_1\) and \(f_2\) is called the generalized function defined on any basic function \(\varphi\) by the equality

\[ (f_1*f_2,\varphi)=(f_2(y),(f_1(x),\varphi(x+y))). \]

In [3] it is shown that the set \(K'_+\) of all generalized functions whose supports lie on the half-axis \([0,\infty)\) forms a ring without zero divisors with the usual operation of addition of generalized functions and the multiplication operation—the convolution of generalized functions.

1. Suppose we have the differential equation

\[ y^{(n)}(x)-g_0(x)y^{(n-1)}(x)-\cdots-g_{n-1}(x)y(x)=f(x),\quad x\geqslant 0 \tag{1} \]

with the following Cauchy initial data:

\[ y^{(k)}(0)=0,\quad k=0,1,2,\ldots,n-2 \]

\[ y^{(n-1)}(0)=1. \tag{2} \]

Such a specification of the initial conditions does not restrict the generality of the Cauchy problem for equation (1), since any equation of order \(n\) satisfying arbitrary initial data

\[ y^{(k)}(0)=y_0^{(k)},\quad k=0,1,2,\ldots,n-1, \]

can, by means of a certain transformation, be reduced to a new equation satisfying the initial conditions (2).

We shall assume the coefficients \(g_0(x),g_1(x),\ldots,g_{n-1}(x)\) to be infinitely differentiable functions, and some of them may be constants, while the right-hand side \(f(x)\) is a continuous function on the whole axis \([0,\infty)\). We shall seek the solution of equation (1) on the set \(K'_+\).

In operator form, equation (1), taking into account conditions (2), is written as follows:

\[ \bigl(S_{(s)}^n-T(s,x)\bigr)y=f+\hat{\delta},\quad x\geqslant 0. \]

If there exists a unique generalized function \(y\in K'_+\) satisfying equation (1) and conditions (2), then we shall call \(y\) the solution of this equation and denote it by

\[ y=\frac{f+\hat{\delta}}{S_{(s)}^n-T}\equiv \frac{f}{S_{(s)}^n-T}+\frac{\hat{\delta}}{S_{(s)}^n-T}. \]

Performing formally the division of the function \(f\), and then of the \(\delta\)-function, by the operator \(S^n(s)-T(s,x)\), we obtain the solution in the form of the following series:

\[ \begin{aligned} y={}&\frac{1}{S^n(s)}+\frac{1}{S^n(s)}* \left(T(s,x)\left[\frac{1}{S^n(s)}\right]\right) +\frac{1}{S^n(s)}* \left(T(s,x)\left[\frac{1}{S^n(s)}*\right.\right.\\ &\left.\left.*\left(T(s,x)\left[\frac{1}{S^n(s)}\right]\right)\right]\right) +\cdots \frac{f}{S^n(s)} +\frac{1}{S^n(s)}* \left(T(s,x)\left[\frac{1}{S^n(s)}\right]\right)+\\ &+\frac{1}{S^n(s)}* \left[T(s,x)\left(\frac{1}{S^n(s)}* \left(T(s,x)\left[\frac{1}{S^n(s)}\right]\right)\right)\right]+\cdots, \end{aligned} \]

where \(1\equiv\delta(x)\).

Or, briefly:

\[ y=\sum_{\kappa=0}^{\infty} \left(\frac{1}{S^n(s)}*T(s,x)\right)^\kappa \frac{f}{S^n(s)} + \sum_{\kappa=0}^{\infty} \left(\frac{1}{S^n(s)}*T(s,x)\right) \frac{1}{S^n(s)} . \tag{3} \]

We shall show the uniform convergence of the series (3) on any finite interval \([0,a]\subset[0,\infty)\), and also verify that it satisfies equation (1) under the initial conditions (2).

For simplicity of the argument, equation (1) will be considered homogeneous. In this case its solution takes the form

\[ y=\sum_{\kappa=0}^{\infty} \left(\frac{1}{S^n(s)}*T(s,x)\right) \frac{1}{S^n(s)} . \tag{3'} \]

Let us note that \(S^n(s)=P_n(s)\), that is, \(S^n(s)\) is a polynomial of degree \(n\). Then one may write

\[ \frac{1}{S^n(s)} = \frac{1}{(S-s_1)(S-s_2)\ldots(S-s_n)}, \]

where \(s_1,s_2,\ldots,s_n\) are the roots of the polynomial \(S^n(s)\).

In view of the fact that in the space \(K_+\) the operation of multiplication is defined as the convolution of generalized functions, we have

\[ \frac{1}{S^n(s)} = \frac{1}{S-s_1}* \frac{1}{S-s_2}*\cdots* \frac{1}{S-s_n}. \]

The last expression is realized in functions of the argument \(x\) in the following way (see [4]):

\[ \frac{1}{S^n(s)} = \begin{cases} l^{s_1x}*l^{s_2x}*\cdots*l^{s_nx}, & x\ge 0,\\ 0, & x<0, \end{cases} \]

or

\[ \frac{1}{S^n(s)} = \begin{cases} \displaystyle \int_0^x\int_0^t\int_0^{t_1}\cdots\int_0^{t_{n-2}} l^{s_1t_{n-2}}l^{s_2(t_{n-3}-t_{n-2})}\cdots l^{s_n(x-t)}\,dt_{n-2}\cdots dt_1\,dt =:\psi(x), & x\ge0,\\[2ex] 0, & x<0. \end{cases} \]

After carrying out the corresponding operations, the series (3′) takes the form

\[ y=\psi(x)+\sum_{\kappa=1}^{\infty}\int_0^x \psi(x-t)F_\kappa(t)\,dt, \tag{4} \]

where

\[ F_k=\int_0^x F_{k-1}(t)\sum_{\alpha=0}^{n-1}p_{n-1-\alpha}g_{n-1-\alpha}(t)\psi^{(\alpha)}(x-t)\,dt, \]

\[ F_1=p_0g_0(x)\psi^{(n-1)}(x)+p_1g_1(x)\psi^{(n-2)}(x)+\cdots+p_{n-1}g_{n-1}(x)\psi(x) \]

and

\[ p_i=0,1\quad \text{for } i=0,1,2,\ldots,n-1 \quad (p_i=0,\ \text{if } g_i(x)=\mathrm{const}). \]

Let us estimate the series (4). From the condition of infinite differentiability of the coefficients \(g_0(x), g_1(x),\ldots,g_{n-1}(x)\) of equation (1) there follows their boundedness on any finite interval \([0,a]\subset[0,\infty)\). The function \(\psi(x)\), by virtue of its definition, will also be an infinitely differentiable function and bounded on any finite interval together with all its derivatives. Taking this into account, set

\[ M=\max_{x\in[0,a]}\{|p_0g_0(x)|,\ |p_1g_1(x)|,\ldots,|p_{n-1}g_{n-1}(x)|\} \]

and

\[ A=p_0A_0+p_1A_1+\cdots+p_{n-1}A_{n-1},\qquad A_i=\max_{x\in[0,a]}|\psi^{(i)}(x)|,\quad i=0,1,2,\ldots,n-1. \]

We have

\[ |F_1(x)|\le \sum_{\alpha=0}^{n-1}|p_{n-1-\alpha}g_{n-1-\alpha}(x)|\,|\psi^{(\alpha)}(x)| \le M\sum_{\alpha=0}^{n-1}p_\alpha|\psi^{(\alpha)}(x)| \le MA. \]

\[ |F_2(x)|\le \int_0^x |F_1(t)|\sum_{\alpha=0}^{n-1}|p_{n-1-\alpha}g_{n-1-\alpha}(t)|\,|\psi^{(\alpha)}(x-t)|\,dt \le \]

\[ \le M^2A^2\int_0^x dt = M^2A^2\frac{x}{1!}. \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ |F_k(x)|\le \int_0^x |F_{k-1}(t)|\sum_{\alpha=0}^{n-1}p_{n-1-\alpha}g_{n-1-\alpha}(t)\,|\psi^{(\alpha)}(x-t)|\,dt \le \]

\[ \le M^kA^k\int_0^x\frac{t^{k-2}}{(k-2)!}\,dt = M^kA^k\frac{x^{k-1}}{(k-1)!}. \]

The validity of the last equality for arbitrary \(k\), \(k=2,3,\ldots\), is easily proved by the method of mathematical induction. The general estimate (3) will have the form:

\[ \left|\psi(x)+\sum_{k=1}^{\infty}\int_0^x \psi(x-t)F_k(t)\,dt\right| \le |\psi(x)|+ \sum_{k=1}^{\infty}\int_0^x |\psi(x-t)|\,|F_k(t)|\,dt \le \]

\[ \le A_0+\sum_{k=1}^{\infty}A_0\int_0^x |F_k(t)|\,dt \le A_0+\sum_{k=1}^{\infty}A_0M^kA^k\int_0^x\frac{t^{k-1}}{(k-1)!}\,dt = A_0+ \]

\[ +\sum_{k=0}^{\infty}A_0M^kA^k\frac{x^k}{k!} \le \sum_{k=0}^{\infty}A_0M^kA^k\frac{a^k}{k!}. \]

Consequently, the series \((3')\) converges in the space of generalized functions and converges uniformly on every finite interval \([0,a]\subset[0,\infty)\). Moreover, it is obvious that the series \((3')\) admits termwise differentiation any number of times in the domain under consideration. By direct substitution we verify that the series \((3')\) satisfies equation (1) (for \(f(x)\equiv 0\)), as well as the initial conditions (2), i.e., it is the desired solution.

Now consider the case where equation (1) is nonhomogeneous. The solution in this case is the series (3), which can be written in the form

\[ y=\psi(x)+\sum_{k=1}^{\infty}\int_{0}^{x}\psi(x-t)F_k(t)\,dt +\omega(x)+\sum_{k=1}^{\infty}\int_{0}^{x}\omega(x-t)G_k(t)\,dt, \]

where

\[ \psi(x)\ \text{and}\ F_k(x),\quad k=1,2,\ldots \]

are the functions defined by us above;

\[ \omega(x)=\int_{0}^{x}\int_{0}^{t}\int_{0}^{t_1}\cdots\int_{0}^{t_{n-2}} f(t_n)\,l_{s_1}(t_{n-1}-t_n)\cdots l_{s_n}(x-t)\, dt_{n-1}\cdots dt_1\,dt, \]

and \(G_1(x)\) and \(G_k(x)\), \(k=2,3,\ldots\), have the form:

\[ G_1(x)=p_0g_0(x)\omega^{(n-1)}(x)+p_1g_1(x)\omega^{(n-2)}(x)+\cdots+p_{n-1}g_{n-1}(x)\omega(x), \]

\[ G_k(x)=\int_{0}^{x}\omega(x-t)G_{k-1}(t)\,dt, \]

where

\[ p_i=0,1\quad (i=0,1,2,\ldots). \]

The convergence of the series

\[ \omega(x)+\sum_{k=1}^{\infty}\int_{0}^{x}\omega(x-t)G_k(t)\,dt \]

is established in exactly the same way as we did this for the series \((3')\). In this case the following inequality holds:

\[ \left|\omega(x)+\sum_{k=1}^{\infty}\int_{0}^{x}\omega(x-t)G_k(t)\,dt\right| \leq B_0\sum_{k=0}^{\infty}M^kB^k\frac{a^k}{k!}, \]

where

\[ M=\max_{x\in[0,a]}\{|p_0g_0(x)|,\ |p_1g_1(x)|,\ldots,\ |p_{n-1}g_{n-1}(x)|\} \]

and

\[ B=p_0B_0+p_1B_1+\cdots+p_{n-1}B_{n-1}; \qquad B_i=\max_{x\in[0,a]}|\omega^{(i)}(x)|\quad (i=0,1,\ldots,n-1). \]

Thus, we have verified that the series (3) is a solution of the nonhomogeneous differential equation (1), satisfying condition (2).

Remark. It is not difficult to see that all our arguments concerning equation (1) will remain valid if its coefficients \(g_0(x), g_1(x),\ldots,g_{n-1}(x)\) are assumed to be merely continuous functions and are not required to be infinitely differentiable.

3. Example. Consider the equation from [5]:

\[ y''-4xy'+(4x^2-2)y=0 \tag{5} \]

under the following initial conditions

\[ y(0)=0,\quad y'(0)=1. \tag{6} \]

Equation (5), taking into account conditions (6), is written in operator form as follows:

\[ \left[S^2 - T(s,x)\right]y = \delta(x), \]

where

\[ T(s,x)=4xS-(4x-2). \]

The solution in this case will have the form:

\[ y=\frac{1}{S^2}+\frac{1}{S^2}*\left[T(s,x)\left(\frac{1}{S^2}\right)\right] +\frac{1}{S^2}*\left[T(s,x)\left(\frac{1}{S^2}*\left[T(s,x)\left(\frac{1}{S^2}\right)\right]\right)\right]+\ldots \tag{7} \]

Carrying out in (7), term by term, the corresponding operations, we obtain the sequence of approximate solutions:

\[ y_0=x+R_0, \qquad R_0=0, \]

\[ y_1=x+\frac{x^3}{1!}+R_1, \qquad R_1=\frac{1}{5}x^5, \]

\[ y_2=x+\frac{x^3}{1!}+\frac{x^5}{2!}+R_2, \qquad R_2=\frac{1}{7}x^7+\frac{1}{90}x^9, \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ y_m=\sum_{k=0}^{m}\frac{x^{2k+1}}{k!}+R_m =x\sum_{k=0}^{m}\frac{x^{2k}}{k!}+R_m, \]

where \(R_m\) is a polynomial containing \(m\) terms with all odd powers from \(2m+3\) to \(4m+1\).

Passing in the last equality to the limit as \(m\to\infty\) and denoting \(R_\infty=\lim\limits_{m\to\infty}R_m\), we have:

\[ \lim_{m\to\infty}y_m=xe^{x^2}+R_\infty. \]

Next we note that \(xe^{x^2}\) satisfies equation (1) under conditions (2). Consequently, \(R_\infty\), by virtue of the uniqueness theorem, is identically equal to zero.

Thus, the analytic solution of equation (1) under conditions (2) has the form

\[ y=xe^{x^2}. \]

REFERENCES

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3. R. M. Malakhovskaya. Uchenye zapiski TGU, No. 36, pp. 13–32, 1960.
4. R. M. Malakhovskaya. Trudy TGU, vol. 163, pp. 28–43, 1963.
5. E. Kamke. Handbook of Ordinary Differential Equations, IL, 1961.