New results for Brillouin electron beam focusing system

Abstract

An experimental conjecture on the existence of positive periodic solutions for the Brillouin electron beam focusing system $x''+a(1+\cos 2t)x=\frac{1}{x}$ for $a\in(0,\frac{1}{2})$ is proved by applications of the Manasevich-Mawhin theorem.

1 Introduction

In this paper, we consider the 2π-periodic boundary value problem for the equation

$$ x''+a(1+\cos 2t)x=\frac{1}{x}, $$

(1.1)

where $a>0$ is constant.

The equations arise in the study of electronics and govern the motion of a magnetically focused axially symmetric electron beam under the influence of a Brillouin flow [1]. When the negative pole in a traveling-wave tube is shielded completely by a magnetic field screen, the electron beam focusing system can be described by (1.1). Besides, from a mathematical point of view, equation (1.1) is a singular perturbation of the Mathieu equation.

Motivated by the results of laboratory experiments experts realized in [1], it was conjectured that (1.1) should have a positive periodic solutions if $a\in(0,\frac{1}{4})$ [2]. In the last fifty years, many mathematicians have given birth to extensive literature about this topic (see [3–7]). Although numerical studies back up the experimental conjecture, an analytical proof of the existence of periodic solutions of (1.1) for $a\in(0,\frac{1}{4})$ is still lacking.

The first analytic work on periodic solution of (1.1) was obtained by Ding [3]. Ding proved that (1.1) had at least one positive periodic solution if $a\in(0,\frac{1}{16})$. Afterwards, Ye and Wang [4] obtained that (1.1) had at least one positive periodic solution if $a\in(0,0.1442)$. In [5], Zhang investigated a kind of singular Liénard equation, and by applications of his theory, they extended the existence result of (1.1) to $a\in(0,0.1532)$.

However, in the above works, authors were not able to prove or disprove the result which was conjectured in [1]. In this paper, we will show that (1.1) has at least one positive 2π-periodic solution when the parameter $a\in(0,\frac{1}{2})$ other than $(0,\frac{1}{4})$.

2 Brillouin electron beam focusing system

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.