Mathematics

V. N. Karp

On the Existence and Uniqueness of a Periodic Solution of One Nonlinear Problem on Forced Vibrations of a String

(Presented by Academician I. G. Petrovskii, 9 III 1960)

In the present article we prove the existence and uniqueness of a periodic solution of the nonlinear differential equation

[
\frac{\partial^{2}u}{\partial t^{2}}-a^{2}\frac{\partial^{2}u}{\partial x^{2}}
=
\Phi(x,t)+\mu f\left(x,t,u,\frac{\partial u}{\partial t}\right),
\tag{1}
]

satisfying the conditions

[
u(0,t)=u(1,t)=0,
\tag{2}
]

[
u(x,0)=u(x,1),
]

[
\frac{\partial u(x,0)}{\partial t}
=
\frac{\partial u(x,1)}{\partial t},
\tag{3}
]

under the assumption that (\mu) is a small parameter and the constant coefficient (a=p/q), where (p=2k+1) ((k=0,1,2,\ldots)) and ((p,q)=1).

The desired function (u(x,t)) must be periodic in the argument (t), with period equal to one, and continuous together with its partial derivatives of the first two orders in the closed domain

[
\overline{D}=D\left(
\begin{matrix}
0\le x\le 1\
0\le t\le 1
\end{matrix}
\right).
]

Earlier the problem (1), (2), (3) was solved by the author by the method of successive approximations for integral odd values of (a).

To prove the convergence of the successive approximations, the method of wave domains was used, which made it possible to find explicitly the kernel of the integro-differential equation

[
u(x,t)=\frac{2}{\pi a}\int_{0}^{t}d\tau\int_{0}^{1}
\left[
\Phi(\xi,\tau)+\mu f\left(\xi,\tau,u,\frac{\partial u}{\partial \tau}\right)
\right]
K_{1}(x,\xi;t,\tau)\,d\xi+
]

[
+\frac{1}{\pi a}\int_{0}^{1}d\tau\int_{0}^{1}
\left[
\Phi(\xi,\tau)+\mu f\left(\xi,\tau,u,\frac{\partial u}{\partial \tau}\right)
\right]
K_{2}(x,\xi;t,\tau)\,d\xi,
\tag{4}
]

to which the problem (1), (2), (3) is reduced. Essential here was the circumstance that the functions (K_{1}(x,\xi;t,\tau)) and (K_{2}(x,\xi;t,\tau)) are piecewise constant in the domain (\overline{D}) for (a=2k+1).

1. Let (a=\dfrac{2k+1}{2l}). Then

[
K_1(x,\xi;t,\tau)=
\sum_{n=0}^{\infty}
\frac{\sin(2n+1)\pi x\,\sin(2n+1)\pi \xi\,\sin\dfrac{(2n+1)(2k+1)\pi(t-\tau)}{2l}}
{2n+1},
]

[
K_2(x,\xi;t,\tau)=
\sum_{n=0}^{\infty}
\frac{\sin(2n+1)\pi x\,\sin(2n+1)\pi \xi\,\cos\dfrac{(2n+1)(2k+1)\pi(t-\tau+\tfrac12)}{2l}}
{(2n+1)\sin\dfrac{(2n+1)(2k+1)\pi}{4l}} .
]

We have

[
K_1(x,\xi;t,\tau)=\frac14\sum_{m=1}^{4}(-1)^m
\sum_{n=0}^{\infty}\frac{\sin(2n+1)\omega_m}{2n+1},
]

where (\omega_m) is a linear function of (x,\xi,t,\tau), and

[
\sum_{n=0}^{\infty}\frac{\sin(2n+1)\omega_m}{2n+1}
=
\begin{cases}
\pi/4, & 2k\pi<\omega_m<(2k+1)\pi,\
0, & \omega_m=k\pi,\
-\pi/4, & (2k+1)\pi<\omega_m<(2k+2)\pi
\end{cases}
]

[
(m=1,2,3,4;\ k=0,\pm1,\pm2,\ldots).
]

Consequently, the function (K_1(x,\xi;t,\tau)) is piecewise constant in the domain (\overline D). Summing the function (K_2(x,\xi;t,\tau)), we obtain

[
K_2(x,\xi;t,\tau)=
\frac14\sum_{m=1}^{2}\sum_{n=0}^{\infty}
\frac{\cos(2n+1)\omega_m}
{(2n+1)\sin\dfrac{(2n+1)(2k+1)\pi}{4l}}
-
]

[
-\frac14\sum_{m=3}^{4}\sum_{n=0}^{\infty}
\frac{\cos(2n+1)\omega_m}
{(2n+1)\sin\dfrac{(2n+1)(2k+1)\pi}{4l}},
]

where

[
\omega_1=\pi\left[x-\xi-\frac{2k+1}{2l}\left(t-\tau+\frac12\right)\right],
\quad
\omega_2=\pi\left[x-\xi+\frac{2k+1}{2l}\left(t-\tau+\frac12\right)\right],
]

[
\omega_3=\pi\left[x+\xi+\frac{2k+1}{2l}\left(t-\tau+\frac12\right)\right],
\quad
\omega_4=\pi\left[x+\xi-\frac{2k+1}{2l}\left(t-\tau+\frac12\right)\right].
]

Denoting

[
A_\lambda=
\frac{1}{\sin\dfrac{(2\lambda+1)(2k+1)\pi}{4l}}
]

and putting (\lambda'-\lambda=2l), we obtain (A_\lambda=A_{\lambda'}). Then

[
\sum_{n=0}^{\infty}
\frac{\cos(2n+1)\omega}
{(2n+1)\sin\dfrac{(2n+1)(2k+1)\pi}{4l}}
=
\sum_{\lambda=0}^{2l-1}
A_\lambda\,\operatorname{Re}\int_{0}^{z}\frac{z^{2\lambda}}{1+z^{4l}}\,dz,
\tag{5}
]

where (z=e^{\omega i}).

The function (\dfrac{z^{2\lambda}}{1+z^{4l}}) has singular points

[
z_q=e^{\pi i\frac{2q+1}{4l}}
\qquad(q=0,\ldots,2l-1).
]

According to Cauchy’s theorem we obtain

[
\operatorname{Re}\int_{0}^{z}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
=
\operatorname{Re}\left[
\int_{0}^{1}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
+
\int_{L_0}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
+
\int_{l_0}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
+
\int_{L_1}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
+\right.
]

[
\left.
+
\int_{l_1}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
+\cdots+
\int_{L_p}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
+
\int_{l_p}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
+
\int_{L'}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
\right],
]

where (L_0, L_1, \ldots, L_p) are arcs of the upper semicircle of unit radius with center at the point (z=0), from the point (z=1) to the nearest singular point (z_0) of the function (\dfrac{z^{2\lambda}}{1+z^{4l}}), then from the singular point (z_0) to the next (z_1), and so on; (l_0,l_1,\ldots,l_p) are semicircles of radius (\varepsilon) with centers at these singular points; (L') is the arc of the upper semicircle from the singular point (z_p) to an arbitrary point (z) (Fig. 1).

Next we find

[
\operatorname{Re}\int_{l_q}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
=
]

[

\operatorname{Re}\frac{z_q^{2\lambda-4l+1}}{4l}\int_{l_q}\frac{dz}{z-z_q}

]

[

-\frac{\pi}{4l}\sin\frac{(2q+1)(2\lambda+1)\pi}{4l},
]

where (l_q) is a semicircle of radius (\varepsilon), traversed clockwise ((q=0,\ldots,p)).

Fig. 1

It is easy to establish that

[
\sum_{\lambda=0}^{2l-1} A_\lambda \operatorname{Re}\int_{L'}\frac{z^{2\lambda}}{1+z^{4l}}\,dz=0,\qquad
\sum_{\lambda=0}^{2l-1} A_\lambda \operatorname{Re}\int_{L_s}\frac{z^{2\lambda}}{1+z^{4l}}\,dz=0
\quad (s=0,\ldots,p);
]

[
\sum_{\lambda=0}^{2l-1}A_\lambda\int_{0}^{1}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
=
\frac14\sum_{\lambda=0}^{2l-1}A_\lambda\int_{-\infty}^{\infty}\frac{z^{2\lambda}}{1+z^{4l}}\,dz.
\tag{6}
]

Further, by a well-known theorem of residue theory we obtain

[
\int_{-\infty}^{\infty}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
=
2\pi i\sum_{q=0}^{2l-1}\operatorname{Res}\frac{z^{2\lambda}}{1+z^{4l}}
]

with respect to the points (z_q) lying in the upper half-plane of (z). Then

[
\operatorname{Re}\int_{-\infty}^{\infty}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
=
\operatorname{Re}\left[
2\pi i\sum_{q=0}^{2l-1}
\left(\frac{z_q^{2\lambda}}{4lz_q^{4l-1}}\right)_{z_q=\exp\left[\pi i\,\frac{2q+1}{4l}\right]}
\right]
=
]

[

\frac{\pi}{2l}\sum_{q=0}^{2l-1}\sin\frac{(2\lambda+1)(2q+1)\pi}{4l}.
\tag{7}
]

Taking (6) and (7) into account, we find

[
\sum_{\lambda=0}^{2l-1} A_\lambda \operatorname{Re}\int_{0}^{1}\frac{z^{2\lambda}}{1+z^{4l}}\,dz
=
\frac{\pi}{8l}\sum_{\lambda=0}^{2l-1}A_\lambda
\sum_{q=0}^{2l-1}
\sin\frac{(2\lambda+1)(2q+1)\pi}{4l}.
]

Taking (5) into account, we finally obtain

[
\sum_{n=0}^{\infty}
\frac{\cos (2n+1)\omega}
{(2n+1)\sin \dfrac{(2n+1)(2k+1)\pi}{4l}}
=
]

[

\frac{\pi}{8l}
\sum_{\lambda=0}^{2l-1}
\frac{1}
{\sin \dfrac{(2\lambda+1)(2k+1)\pi}{4l}}
\sum_{q=0}^{2l-1}
\sin \frac{(2\lambda+1)(2q+1)\pi}{4l}
-
]

[

\frac{\pi}{4l}
\sum_{\lambda=0}^{2l-1}
\frac{1}
{\sin \dfrac{(2\lambda+1)(2k+1)\pi}{4l}}
\sum_{q=0}^{p}
\sin \frac{(2\lambda+1)(2q+1)\pi}{4l},
]

if (\omega_p<\omega<\omega_{p+1}). For (0<\omega<\omega_0) the second term is absent. Thus it has been proved that the function (K_2(x,\xi;t,\tau)) is piecewise constant in the domain (\overline{D}).

II. Let

[
a=\frac{2k+1}{2l+1}.
]

Then the functions (K_1(x,\xi;t,\tau)) and (K_2(x,\xi;t,\tau)) in the integro-differential equation (4) are replaced by the functions (K_3(x,\xi;t,\tau)) and (K_4(x,\xi;t,\tau)), respectively. These functions are also piecewise constant in the domain (\overline{D}), and for the function (K_3(x,\xi;t,\tau)) this is proved in the same way as for the function (K_1(x,\xi;t,\tau)).

Summation of the trigonometric series (K_4) gives

[
K_4(x,\xi;t,\tau)
=
\frac{1}{4}
\sum_{m=1}^{2}
\sum_{n=0}^{\infty}
\frac{\cos (2n+1)\omega_m}
{(2n+1)\sin \dfrac{(2n+1)(2k+1)\pi}{2(2l+1)}}
-
]

[

\frac{1}{4}
\sum_{m=3}^{4}
\sum_{n=0}^{\infty}
\frac{\cos (2n+1)\omega_m}
{(2n+1)\sin \dfrac{(2n+1)(2k+1)\pi}{2(2l+1)}} .
]

With the aid of transformations analogous to those carried out earlier, we obtain

[
\sum_{n=0}^{\infty}
\frac{\cos (2n+1)\omega}
{(2n+1)\sin \dfrac{(2n+1)(2k+1)\pi}{2(2l+1)}}
=
]

[

\frac{\pi}{4(2l+1)}
\sum_{\lambda=0}^{2l}
\frac{1}
{\sin \dfrac{(2\lambda+1)(2k+1)\pi}{2(2l+1)}}
\sum_{q=0}^{2l}
\sin \frac{(2\lambda+1)(2q+1)\pi}{2(2l+1)}
-
]

[

\frac{\pi}{2(2l+1)}
\sum_{\lambda=0}^{2l}
\frac{1}
{\sin \dfrac{(2\lambda+1)(2k+1)\pi}{2(2l+1)}}
\sum_{q=0}^{p}
\sin \frac{(2\lambda+1)(2q+1)\pi}{2(2l+1)},
]

if (\omega_p<\omega<\omega_{p+1}). For (0<\omega<\omega_0) the second term is absent.

It follows from this that the function (K_4(x,\xi;t,\tau)) is piecewise constant in the domain (\overline{D}).

In the proof of existence and uniqueness of the solution of problem (1), (2), (3), carried out by the author earlier for the case of integral odd values of the number (a), the only essential point was that the kernel of the integro-differential equation (4) is piecewise constant. Consequently, this proof remains valid also for

[
a=\frac{2k+1}{2l}
\quad\text{and}\quad
a=\frac{2k+1}{2l+1}.
]

Odessa Polytechnic Institute

Received
7 III 1960