Lemma 2.1
Manasevich-Mawhin [8]
Let Ω be an open bounded set in
\(C^{1}_{T}:=\{x\in C^{1}(\mathbb{R},\mathbb{R}): x(t+T)-x(t)\equiv 0\}\). If
-
(i)
The problem
$$ { } \bigl(\phi\bigl(x'\bigr)\bigr)'=\lambda \tilde{f}\bigl(t,x,x'\bigr),\quad x\in C^{1}_{T}, $$
(2.1)
where
\(\tilde{f}:[0,T]\times\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\)
is assumed to be Carathéodory. For each
\(\lambda\in(0,1)\), problem (2.1) has no solution on
∂Ω.
-
(ii)
The equation
$$F(a):=\frac{1}{T} \int^{T}_{0}\tilde{f}\bigl(t,x,x'\bigr) \,dt=0, $$
has no solution on
\(\partial\Omega\cap\mathbb{R}\).
-
(iii)
The Brouwer degree of
F
$$\deg\{F,\Omega\cap\mathbb{R},0\}\neq0. $$
Then problem (2.1) has at least one periodic solution on Ω̄.
Lemma 2.2
[9]
Suppose that
\(u\in C^{1}_{T}\)
and there exists
\(t_{0}\in[0,T]\)
such that
\(\vert u(t_{0}) \vert < d\). Then
$$\biggl( \int^{T}_{0} \bigl\vert u(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}\leq \biggl(\frac{T}{\pi} \biggr) \biggl( \int^{T}_{0} \bigl\vert u'(t) \bigr\vert ^{2}t \biggr)^{\frac{1}{2}}+dT^{\frac{1}{2}}. $$
Next, we prove that Brillouin electron beam focusing system (1.1) has at least one positive 2π-periodic solution if \(a\in (0,\frac{1}{2} )\). Firstly, we consider the following singular equation:
$$ { } x''(t)+a(t)x(t)=\frac{1}{x(t)}, $$
(2.2)
where \(a(t)\in C(\mathbb{R},[0,+\infty))\) and \(a(t+T)=a(t)\), \(\forall t\in \mathbb{R}\).
Theorem 2.1
Assume that
\(\vert a \vert _{\infty}:=\max_{t\in[0,T]} \vert a(t) \vert <\frac{4\pi^{2}}{T^{2}}\)
holds. Then (2.2) has at least one positive
T-periodic solution.
Proof
Firstly, we consider the following (homotopy) family of (2.2):
$$ { } x''(t)+\lambda \biggl(a(t)x(t)- \frac{1}{x(t)} \biggr)=0,\quad \lambda\in(0,1]. $$
(2.3)
Let \(x(t)\in C^{1}_{T}\) be an arbitrary solution of (2.3). Integrating (2.3) from 0 to T, we get
$$ { } \int^{T}_{0} \biggl(a(t)x(t)-\frac{1}{x(t)} \biggr)\,dt=0 . $$
(2.4)
So, we know that there exist positive constants \(D_{1}< D_{2}\) and \(t_{0}\in(0,T)\) such that
$$ { } D_{1}\leq x(t_{0})\leq D_{2}. $$
(2.5)
Therefore, we have
$$ { } \bigl\vert x(t) \bigr\vert = \biggl\vert x(t_{0})+ \int^{t}_{t_{0}} x'(s)\,ds \biggr\vert \leq D_{2}+ \int^{T}_{0} \bigl\vert x'(s) \bigr\vert \,ds. $$
(2.6)
Let us write \(x(t)=\bar{x}+\tilde{x}(t)\), here \(\tilde{x}(t):=x(t)-\bar{x}\), and \(\bar{x}:=\frac{1}{T}\int^{T}_{0}x(t)\,dt\). Obviously, \(\int^{T}_{0}\tilde{x}(t)\,dt=0\). Now (2.3) for \(\tilde{x}(t)\) is
$$ { } \tilde{x}''(t)+\lambda a(t) \bigl( \bar{x}+\tilde{x}(t)\bigr)=\lambda\frac{1}{x(t)}, $$
(2.7)
since \(\bar{x}''=0\). Multiplying (2.7) by \(\bar{x}-\tilde{x}(t)\), we have
$$ \bar{x}\tilde{x}''(t)-\tilde{x}(t) \tilde{x}''(t)+\lambda a(t) \bigl(\bar{x}^{2}- \tilde{x}^{2}(t)\bigr)=\lambda \frac{\bar{x}-\tilde{x}(t)}{x(t)}. $$
Integrating this equation over one period and making use of the T-periodicity of \(\tilde{x}(t)\), we get
$$ - \int^{T}_{0}\tilde{x}(t)\tilde{x}''(t) \,dt+\lambda \int^{T}_{0}a(t) \bigl(\bar{x}^{2}- \tilde{x}^{2}(t)\bigr)\,dt=\lambda \int^{T}_{0}\frac{\bar{x}-\tilde{x}(t)}{x(t)}\,dt. $$
So, we have
$$ \int^{T}_{0} \bigl\vert \tilde{x}'(t) \bigr\vert ^{2}\,dt=\lambda \int^{T}_{0}a(t)\tilde{x}^{2}(t)\,dt- \lambda\bar{x}^{2} \int^{T}_{0}a(t)\,dt+\lambda \int^{T}_{0} \frac{\bar{x}-\tilde{x}(t)}{x(t)}\,dt. $$
Since \(a(t)\geq0\), then \(-\bar{x}^{2}\int^{T}_{0}a(t)\,dt\leq0\). So, we have
$$\begin{aligned} \int^{T}_{0} \bigl\vert \tilde{x}'(t) \bigr\vert ^{2}\,dt&\leq\lambda \int^{T}_{0}a(t)\tilde{x}^{2}(t)\,dt+ \lambda \int^{T}_{0}\frac{\bar{x}-\tilde{x}(t)}{x(t)}\,dt \\ &=\lambda \int^{T}_{0}a(t)\tilde{x}^{2}(t)\,dt+ \lambda \int^{T}_{0}\frac{2\bar{x}-x(t)}{x(t)}\,dt \\ &=\lambda \int^{T}_{0}a(t)\tilde{x}^{2}(t)\,dt+ \lambda \int^{T}_{0}\frac{2\bar{x}}{x(t)}\,dt-\lambda T \\ &\leq \int^{T}_{0} \bigl\vert a(t) \bigr\vert \bigl\vert \tilde{x}(t) \bigr\vert ^{2}\,dt+2 \vert \bar{x} \vert \int^{T}_{0}\frac{1}{ \vert x(t) \vert }\,dt. \end{aligned}$$
For any \(\varepsilon>0\), there is \(g_{\varepsilon}^{+}\in L^{2}(0,T)\) and \(g_{\varepsilon}^{+}>0\)
$$ { } \frac{1}{x(t)}\leq \varepsilon x(t)+g^{+}_{\varepsilon}(t) $$
(2.8)
for all \(x(t)>0\) and a.e. \(t\in[0,T]\). So, we have
$$\begin{aligned} \int^{T}_{0} \bigl\vert \tilde{x}'(t) \bigr\vert ^{2} \leq{}& \int^{T}_{0} \bigl\vert a(t) \bigr\vert \bigl\vert \tilde{x}(t) \bigr\vert ^{2}\,dt+2\bar{x} \int_{0}^{T}\bigl(\varepsilon x(t)+g_{\varepsilon}^{+}(t) \bigr)\,dt \\ \leq{}& \int^{T}_{0} \bigl\vert a(t) \bigr\vert \bigl\vert \tilde{x}(t) \bigr\vert ^{2}\,dt+\frac{2}{T} \int^{T}_{0} \bigl\vert x(t) \bigr\vert \,dt \int_{0}^{T}\bigl(\varepsilon x(t)+g_{\varepsilon}^{+}(t) \bigr)\,dt \\ \leq {}&\vert a \vert _{\infty}\int^{T}_{0} \bigl\vert \tilde{x}(t) \bigr\vert ^{2}\,dt+\frac{2\varepsilon}{T} \biggl( \int^{T}_{0} \bigl\vert x(t) \bigr\vert \,dt \biggr)^{2}+ \frac{2}{T} \int^{T}_{0} \bigl\vert x(t) \bigr\vert \,dt \int_{0}^{T}g_{\varepsilon}^{+}(t)\,dt \\ \leq{} &\vert a \vert _{\infty}\int^{T}_{0} \bigl\vert \tilde{x}(t) \bigr\vert ^{2}\,dt+2\varepsilon \int^{T}_{0} \bigl\vert x(t) \bigr\vert ^{2}\,dt\\ &{}+ 2 \biggl( \int^{T}_{0} \bigl\vert x(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \biggl( \int_{0}^{T} \bigl\vert g_{\varepsilon}^{+}(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}, \end{aligned}$$
where \(\vert a \vert _{\infty}=\max_{t\in[0,T]} \vert a(t) \vert \). Since \(D_{1}\leq x(t_{0})\leq D_{2}\), by Lemma 2.2, we have
$$ { } \biggl( \int^{T}_{0} \bigl\vert x(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}\leq \biggl(\frac{T}{\pi} \biggr) \biggl( \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}+D_{2}\sqrt{T}. $$
(2.9)
By applications of Wirtinger’s inequality (in [10] Lemma 2.4) and (2.9), we have
$$\begin{aligned} \int^{T}_{0} \bigl\vert \tilde{x}'(t) \bigr\vert ^{2}\leq {}& \vert a \vert _{\infty}\biggl( \frac{T}{2\pi} \biggr)^{2} \int^{T}_{0} \bigl\vert \tilde{x}'(t) \bigr\vert ^{2}\,dt+2\varepsilon \biggl( \biggl(\frac{T}{\pi} \biggr) \biggl( \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}+D_{2}\sqrt{T} \biggr)^{2} \\ &{}+ 2 \biggl( \biggl(\frac{T}{\pi} \biggr) \biggl( \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}+D_{2}\sqrt{T} \biggr) \bigl\Vert g_{\varepsilon}^{+} \bigr\Vert _{2} \\ ={}& \vert a \vert _{\infty}\biggl(\frac{T}{2\pi} \biggr)^{2} \int^{T}_{0} \bigl\vert \tilde{x}'(t) \bigr\vert ^{2}\,dt+2\varepsilon \biggl(\frac{T}{\pi} \biggr)^{2} \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt\\ &{}+ 2 \bigl(2D_{2}\sqrt{T}\varepsilon+ \bigl\Vert g_{\varepsilon}^{+} \bigr\Vert _{2} \bigr) \biggl(\frac{T}{\pi} \biggr) \biggl( \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}}\\ &{} +2D_{2}^{2}T \varepsilon+2D_{2}\sqrt{T} \bigl\Vert g_{\varepsilon}^{+} \bigr\Vert _{2}, \end{aligned}$$
where \(\Vert g_{\varepsilon}^{+} \Vert _{2}= (\int^{T}_{0} \vert g_{\varepsilon}^{+}(t) \vert ^{2}\,dt )^{\frac{1}{2}}\). Since \(\tilde{x}'(t)=x'(t)\), then we have
$$\begin{aligned} \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\leq{} & \biggl( \vert a \vert _{\infty}\biggl( \frac{T}{2\pi} \biggr)^{2}+2\varepsilon \biggl(\frac{T}{\pi} \biggr)^{2} \biggr) \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt\\ &{}+ 2 \bigl(2D_{2}\sqrt{T}\varepsilon+ \bigl\Vert g_{\varepsilon}^{+} \bigr\Vert _{2} \bigr) \biggl( \frac{T}{\pi} \biggr) \biggl( \int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \\ &{}+2D_{2}^{2}T\varepsilon+2D_{2}\sqrt{T} \bigl\Vert g_{\varepsilon}^{+} \bigr\Vert _{2}. \end{aligned}$$
From \(\vert a \vert _{\infty}<\frac{4\pi^{2}}{T^{2}}\) for \(\varepsilon>0\) sufficiently small, there exists a positive constant \(M_{1}'\) such that
$$\int^{T}_{0} \bigl\vert x'(t) \bigr\vert ^{2}\,dt\leq M_{1}'. $$
From (2.6) and by applying Hölder’s inequality, we have
$$ { } \vert x \vert _{\infty}\leq D_{2}+ \int^{T}_{0} \bigl\vert x'(s) \bigr\vert \,ds\leq D_{2}+ \sqrt{T} \biggl( \int^{T}_{0} \bigl\vert x'(s) \bigr\vert ^{2}\,ds \biggr)^{\frac{1}{2}}\leq D_{2}+ \sqrt{T}M_{1}^{\prime\frac{1}{2}}=M_{1}. $$
(2.10)
On the other hand, from \(x(0)=x(T)\), we know that there is a point \(t_{1}\in[0,T]\) such that \(x'(t_{1})=0\), and then \(\vert x'(t) \vert = \vert x'(t_{1})+\int^{t}_{t_{1}} x''(s)\,ds \vert \leq \int^{T}_{0} \vert x''(s) \vert \,ds\). From (2.3) and (2.8), we have
$$\begin{aligned} \bigl\vert x' \bigr\vert _{\infty}&\leq \int^{T}_{0} \bigl\vert x''(t) \bigr\vert \,dt \\ &\leq\lambda \int^{T}_{0} \bigl\vert a(t) \bigr\vert \bigl\vert x(t) \bigr\vert \,dt+\lambda \int^{T}_{0}\frac{1}{x(t)}\,dt \\ &\leq\lambda \vert a \vert _{\infty}M_{1}T+\lambda \int^{T}_{0}\bigl(\varepsilon x(t)+g_{\varepsilon}^{+}(t) \bigr)\,dt \\ &\leq\lambda \vert a \vert _{\infty}M_{1}T+\lambda \varepsilon M_{1}T+\lambda\sqrt{T} \biggl( \int^{T}_{0} \bigl\vert g_{\varepsilon}^{+}(t) \bigr\vert ^{2}\,dt \biggr)^{\frac{1}{2}} \\ &\leq\lambda \vert a \vert _{\infty}M_{1}T+\lambda \varepsilon M_{1}T+\lambda\sqrt{T} \bigl\Vert g_{\varepsilon}^{+} \bigr\Vert _{2}:=\lambda M_{2}, \end{aligned}$$
i.e.,
$$ \bigl\vert x' \bigr\vert _{\infty}\leq \lambda M_{2}. $$
(2.11)
Multiplying (2.3) by \(x'(t)\), we get
$$ { } x''(t)x'(t)+\lambda a(t)x(t)x'(t)=\lambda\frac{x'(t)}{x(t)}. $$
(2.12)
Let \(\tau\in[0,T]\) be as in (2.5). For any \(\tau\leq t\leq T\), we integrate (2.12) on \([\tau,t]\) and get
$$ { } \begin{aligned} \lambda \int^{x(t)}_{x(\tau)}\frac{1}{u}\,du&=\lambda \int^{t}_{\tau}\frac{x'(s)}{x(t)}\,ds \\ &=\frac{1}{2}x'(t)^{2}-\frac{1}{2}x'( \tau)^{2}+\lambda \int^{t}_{\tau}a(s)x(s)x'(s)\,ds. \end{aligned} $$
(2.13)
By (2.11) we have
$$\begin{aligned} &x'(t)^{2}\leq \lambda^{2}M_{2}^{2}, \\ & \biggl\vert \int^{t}_{\tau}a(s)x(s)x'(s)\,ds \biggr\vert \leq \lambda \vert a \vert _{\infty}M_{1}M_{2}T. \end{aligned}$$
With these inequalities we can derive from (2.13) that
$$ { } \biggl\vert \int^{x(t)}_{x(\tau)}\frac{1}{u}\,du \biggr\vert \leq M_{2}^{2}+ \vert a \vert _{\infty}M_{1}M_{2}T. $$
(2.14)
So, we know that there exists \(M_{3}>0\) such that
$$ { } x(t)\geq M_{3},\quad \forall t\in[\tau,T], $$
(2.15)
since \(\lim_{x\to 0^{+}}\int^{x}_{1}\frac{1}{u}\,du=+\infty\). The case \(t\in[0,\tau]\) can be treated similarly.
Having in mind (2.5), (2.10), (2.11) and (2.15), we define
$$ { } \Omega=\bigl\{ x\in C_{T}^{1}: E_{1}< x(t)< E_{2} \mbox{ and } \bigl\vert x'(t) \bigr\vert < E_{3}\ \forall t\in \mathbb{R}\bigr\} , $$
(2.16)
where \(0< E_{1}<\min\{M_{3},D_{1}\}\), \(E_{2}>\max\{M_{1},D_{2}\} \) and \(E_{3}>M_{2}\). Then condition (i) of Lemma 2.1 is satisfied. For a constant \(x \in\ker L\), \(x>0\), we have
$$\bar{g}(x):=\frac{1}{T} \int^{T}_{0} \biggl(a(t)x(t)-\frac{1}{x(t)} \biggr)\,dt. $$
Obviously, \(\bar{g}(x)<0\) for all \(x\in(0,E_{1})\), \(\bar{g}(x)>0\) for all \(x>E_{2}\), so condition (ii) of Lemma 2.1 holds. Set
$$H(x,\mu)=\mu x+(1-\mu)\frac{1}{T} \int^{T}_{0} \biggl(a(t)x(t)-\frac{1}{x(t)} \biggr)\,dt, $$
we have \(xH(x,\mu)>0\). Thus \(H(x,\mu)\) is a homotopic transformation and
$$\begin{aligned} \deg\{F,\Omega\cap \mathbb{R},0\}&=\deg \biggl\{ \frac{1}{T} \int^{T}_{0} \biggl(a(t)x(t)-\frac{1}{x(t)} \biggr)\,dt,\Omega\cap\mathbb{R},0 \biggr\} \\ &=\deg\{x,\Omega\cap \mathbb{R},0\}\neq0. \end{aligned}$$
Thus assumption (iii) of Lemma 2.1 is also verified. Therefore (2.2) has at least one positive T-periodic solution. □
Next, we apply Theorem 2.1 to Brillouin electron beam focusing system (1.1). Equation (1.1) is of the form (2.2) with \(a(t)=a(1+\cos2t)\).
Theorem 2.2
If
\(a\in (0,\frac{1}{2} )\), then (1.1) has at least one positive 2π-periodic solution.
Proof
If \(a<\frac{1}{2}\), then
$$\vert a \vert _{\infty}=2a< 1=\frac{4\pi^{2}}{T^{2}}, $$
i.e., \(\vert a \vert _{\infty}<\frac{4\pi^{2}}{T^{2}}\) holds. Theorem 2.1 implies that (1.1) has at least one 2π-periodic positive solution. □
Finally, we present an example to illustrate our result.
Example 2.1
Consider the second order differential equation with singularity:
$$ { } x''(t)+(1+\cos t)=\frac{1}{x}. $$
(2.17)
It is clear that \(T=\pi\), \(a(t)=1+\cos t\). Obviously,
$$\vert a \vert _{\infty}=\max_{t\in[0,T]} \vert 1+\cos t \vert =2< 4=\frac{4\pi^{2}}{\pi^{2}}. $$
Therefore, (2.17) has at least one π-periodic solution by application of Theorem 2.1.