MATHEMATICS

S. N. LEVINA

ON THE SOLUTION OF THE WAVE EQUATION ON THE ENTIRE TIME AXIS

(Presented by Academician S. L. Sobolev on 26 X 1956)

The article is devoted to the consideration of the following problem.
Find a solution of the equation

[
u_{tt}-\Delta u-cu=0
\tag{1}
]

under the boundary condition

[
u\big|_{x=0}=f
\tag{2}
]

on the assumption that (x \geqslant 0), while the remaining spatial coordinates and the time variable (t) vary from (-\infty) to (\infty), in the class of functions satisfying the condition:

B. (u=0) at infinity in such a way that, whatever (\eta) may be, there is an (X(\eta)) such that for every point (M) for which (x>X(\eta)), (|u|<\eta).

Solving this problem by classical methods in the case of more than one dimension encounters serious difficulties, connected with the fact that the characteristic surface of the equation, in its intersection with the given time-oriented surface, forms an unbounded domain.

In the one-dimensional case the solution of the posed problem can be obtained from the solution of the Cauchy problem for the equation

[
u_{tt}-u_{xx}=0
\tag{1¹}
]

with the initial conditions

[
u\big|{x=0}=f(t),\qquad u_x\big|=F(t),\qquad -\infty<t<\infty.
\tag{2¹}
]

The solution of this problem is

[
u(t,x)=\frac{f(t+x)+f(t-x)}{2}+\frac{1}{2}\int_{t-x}^{t+x} F(\tau)\,d\tau
\tag{3¹}
]

which, for arbitrary (f) and (F), generally speaking, will not have physical meaning, since the values of the function (u) at the time (t) turn out to depend on the values of the initial functions (f) and (F) at times later than (t).

It is known that, in order to satisfy the requirement of physical consistency, it is sufficient that

[
F(t)=-f'(t),\qquad -\infty<t<\infty.
]

Thus, uniqueness of the solution of the problem for the equation

[
u_{tt}-u_{xx}=0,\qquad x\geqslant0,\qquad -\infty<t<\infty,
]

is ensured by just one initial condition and by the condition that the values of the solution be independent of the values of the initial functions at times following that at which the solution is sought.

The application of methods of operational calculus, based on the two-sided Laplace transform, makes it possible to solve the problem for

cases of two and three dimensions in the presence of the term (cu). Operational calculus makes it possible to establish the equivalence of the independence condition to condition B, which is essentially a radiation condition.

Construct the solution of the equation

[
u_{tt}-u_{xx}-u_{yy}-cu=0
\tag{1^2}
]

under the boundary condition

[
\left.u\right|_{x=0}=f(t,y),\qquad -\infty<y<\infty,\qquad -\infty<t<\infty,
\tag{2^2}
]

and condition B.

Passing to transforms with respect to the variables (t) and (y), and assuming here that

[
u(t,x,y)\ \div\ U(p,q,x)
=
pq\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
e^{-pt-qy}u(t,x,y)\,dy\,dt,
]

we find

[
U(p,q,x)=\varphi(p,q)e^{-\sqrt{p^2-q^2-c}\,x},
]

where (\varphi(p,q)\ \div\ f(t,y)).

Carrying out the inverse passage to originals first with respect to the variable (p), and denoting (f^*(t,q)\ \div\ \varphi(p,q)), we find the function

[
\begin{aligned}
U(p,q,x)\ \div\ u^(t,q,x)
&= f^(t-x,q) \
&\quad + x\int_{-\infty}^{t-x}
\frac{\sqrt{q^2+c}\,I_1!\left[\sqrt{q^2+c}\sqrt{(t-\tau)^2-x^2}\right]}
{\sqrt{(t-\tau)^2-x^2}}\,
f^*(\tau,q)\,d\tau,
\end{aligned}
]

whose original with respect to the variable (q) has the form

[
\begin{aligned}
u(t,x,y) &= f(t-x,y) - \
&\quad - \frac{x}{\pi}\int_{-\infty}^{t-x}
\frac{1}{(t-\tau)^2-x^2}
\int_{y-\sqrt{(t-\tau)^2-x^2}}^{y+\sqrt{(t-\tau)^2-x^2}}
\frac{y-\sigma}
{\sqrt{(t-\tau)^2-x^2-(y-\sigma)^2}}
\
&\qquad\qquad\times
\operatorname{ch}!\left[\sqrt{c}\sqrt{(t-\tau)^2-x^2-(y-\sigma)^2}\right]
f'{\sigma}(\tau,\sigma)\,d\sigma\,d\tau
\
&\quad + \frac{x\sqrt{c}}{\pi}\int}^{t-x
\frac{1}{(t-\tau)^2-x^2}
\int_{y-\sqrt{(t-\tau)^2-x^2}}^{y+\sqrt{(t-\tau)^2-x^2}}
\
&\qquad\qquad\times
\operatorname{sh}!\left[\sqrt{c}\sqrt{(t-\tau)^2-x^2-(y-\sigma)^2}\right]
f(\tau,\sigma)\,d\sigma\,d\tau .
\end{aligned}
\tag{3^2}
]

This is the solution of the problem ((1^2)), ((2^2)).

The convergence of the improper integrals entering into this formula is ensured by the following assumptions on the function (f(t,y)):

[
\begin{aligned}
\text{A.}\qquad
|f(t,y)| &< C'e^{-\varepsilon |y|}e^{\sqrt{c}\,t}, \qquad t>0;\
|f(t,y)| &< Ce^{-\varepsilon |y|}e^{(\sqrt{c}+\alpha)t};\
|f(t,y)| &< C_1e^{(\sqrt{c}+\alpha)t},\qquad \varepsilon>0,\quad \alpha>0,\quad t\to-\infty .
\end{aligned}
]

The function (u(t,x,y)) satisfies all the conditions of the posed problem. Indeed:

1) The validity of condition ((2^2)) is obviously ensured.

2) The function (u) satisfies condition B. Indeed, (f(t-x,y)\to0) as (x\to\infty) by condition A. The integral

[
\frac{1}{c}\int_{0}^{\infty}\rho\,\frac{x\,d\rho}{\sqrt{\rho^2+x^2}}
\int_{-\rho}^{\rho}
\frac{\lambda}{\sqrt{\rho^2-\lambda^2}}\,
\operatorname{ch}!\left[\sqrt{c}\sqrt{\rho^2-\lambda^2}\right]\,
f'_{\sigma}!\left(t-\sqrt{\rho^2+x^2},\,y-\lambda\right)\,d\lambda
]

(we have put (y-\sigma=\lambda,\ \sqrt{(t-\sigma)^2-x^2}=\rho)) for sufficiently large (x) in absolute value is less than

[
\frac{1}{\sqrt c}\int_0^\infty \frac{x\,d\rho}{\sqrt{\rho^2+x^2}}\,
C_1 e^{(\sqrt c+\alpha)\left(t-\sqrt{\rho^2+x^2}\right)}
\operatorname{sh}(\sqrt c\,\rho)=
]

[
=\frac{C_1}{\sqrt c}\int_0^\infty
\frac{x\,d\rho}{\rho\sqrt{\rho^2+x^2}}\,
\operatorname{sh}(\sqrt c\,\rho)\,
e^{(\sqrt c+\alpha)\left(t-\sqrt{\rho^2+x^2}\right)}
<
]

[
<C\int_0^\infty \frac{d\rho}{\rho}\,
\operatorname{sh}(\sqrt c\,\rho)\,
e^{(\sqrt c+\alpha)\left(t-\sqrt{\rho^2+x^2}\right)},
]

whence it is clear that as (x\to\infty) it tends to zero. Further,

[
\left|
\int_0^\infty
\frac{x\,d\rho}{\rho\sqrt{\rho^2+x^2}}
\int_{-\rho}^{\rho}
\operatorname{sh}!\left[\sqrt c\,\sqrt{\rho^2-\lambda^2}\right]
f\left(t-\sqrt{\rho^2+x^2},\,y-\lambda\right)\,d\lambda
\right|
\le
]

[
\le
\int_0^\infty
\frac{x\,d\rho}{\rho\sqrt{\rho^2+x^2}}
\left|
\int_{-\rho}^{\rho}
\operatorname{sh}!\left[\sqrt c\,\sqrt{\rho^2-\lambda^2}\right]
f\left(t-\sqrt{\rho^2+x^2},\,y-\lambda\right)\,d\lambda
\right|
<
]

[
<
\int_0^\infty d\rho\, e^{\sqrt c\rho}\,C'e^{(\sqrt c+\alpha)\left(t-\sqrt{\rho^2+x^2}\right)}
\to 0
\quad \text{as } x\to\infty .
]

3) Under the condition of sufficient differentiability of the function (f(t,y)), by direct verification one can establish that the function ((3^2)) satisfies the differential equation ((1^2)).

Remark. If, along with condition A, one requires only the existence of (f'_y(t,y)) and the fulfillment of the Lipschitz condition for (f'_y(t,y)) with respect to (y), then one can assert the existence of a unique generalized* solution in the same class of functions.

For the three-dimensional case (without dispersion), the solution of the problem has been obtained by us in the form

[
u(t,x,y)=f(t-x,y,z)-
]

[
-\frac{x}{2\pi}\int_{-\infty}^{t-x}
\frac{d\tau}{(t-\tau)^2-x^2}
\int_{y-\sqrt{(t-\tau)^2-x^2}}^{\,y+\sqrt{(t-\tau)^2-x^2}}
d\sigma\,
\left{f'{\sigma}(\tau,\sigma,z+R)+f'(\tau,\sigma,z-R)\right}
\frac{y-\sigma}{R}
+
]

[
+\frac{x}{2\pi}\int_{-\infty}^{t-x}
\frac{d\tau}{(t-\tau)^2-x^2}
\int_{y-\sqrt{(t-\tau)^2-x^2}}^{\,y+\sqrt{(t-\tau)^2-x^2}}
\left{f'{\zeta}(\tau,\sigma,z+R)-f'(\tau,\sigma,z-R)\right}d\sigma,
]

where

[
R=\sqrt{(t-\tau)^2-x^2-(y-\sigma)^2}.
]

The conditions A here have the form:

[
|f(t,y,z)|0;
]

[
|f(t,y,z)|<Ce^{\alpha t}e^{-\varepsilon|y|-\eta|z|};
]

[
|f'_y(t,y,z)|<C_1e^{\alpha t};
]

[
|f'_z(t,y,z)|0,\ t\to-\infty .
]

Tula State
Pedagogical Institute

Received
16 IV 1956

* By a generalized solution we mean the original (u(t,x,y)) of the solution (U(p,q,x)) of the representing equation.