Proof of Theorem 1.1
Choose the working space \(E=C(\mathbb{T}^{2})\). Let \(K\subset C(\mathbb{T}^{2})\) be the closed convex cone in \(C(\mathbb{T}^{2})\) defined by (2.10) and \(A:K\to K\) be the completely continuous operator defined by (2.9). Then the positive doubly periodic solution of Equation (1.1) is equivalent to the nontrivial fixed point of A. Let \(0< r< R<+\infty\) and set
$$ \Omega_{1}=\bigl\{ u\in E \mid \Vert u\Vert < r\bigr\} ,\qquad \Omega_{2}=\bigl\{ u\in E \mid \Vert u\Vert < R\bigr\} . $$
(3.1)
We show that the operator A has a fixed point in \(K\cap(\Omega _{2}\setminus \overline{\Omega}_{1})\) when r is small enough and R large enough.
Let \(r\in(0, \delta)\), where δ is the positive constant in condition (F1). We prove that A satisfies the condition of Lemma 2.3 in \(K\cap\partial\Omega_{1}\), namely \(\lambda Au\neq u\) for every \(u\in K\cap\partial\Omega_{1}\) and \(0<\lambda\leq1\). In fact, if there exist \(u_{0}\in K\cap\partial\Omega_{1}\) and \(0<\lambda_{0}\le1\) such that \(\lambda_{0} Au_{0}=u_{0} \), since \(u_{0}=P(\lambda_{0} f(t, x, u_{0}(t-\tau_{1}, x),\ldots,u_{0}(t-\tau_{n},x)))\), by Lemma 2.1 and the definition of P, \(u_{0}\in C(\mathbb{T}^{2})\) satisfies the differential equation
$$ \mathcal{L} u_{0}(t,x)=\lambda_{0} f\bigl(t, x, u_{0}(t-\tau_{1}, x),\ldots ,u_{0}(t- \tau_{n},x)\bigr), \quad \mbox{in }\mathcal{D}'\bigl( \mathbb{T}^{2}\bigr). $$
(3.2)
By definition of the solution in the distribution sense, Equation (3.2) means that \(u_{0}\) satisfies
$$\begin{aligned}& \int_{\mathbb{T}^{2}}u_{0}(t,x) \mathcal{L}\phi(t,x) \, \mathrm{d}t\, \mathrm{d}x \\& \quad =\int_{\mathbb{T}^{2}}\lambda_{0} f\bigl(t, x, u_{0}(t-\tau_{1}, x),\ldots ,u_{0}(t- \tau_{n},x)\bigr) \phi(t,x) \,\mathrm{d}t\,\mathrm{d}x,\quad \forall\phi\in \mathcal{D}\bigl(\mathbb{T}^{2}\bigr). \end{aligned}$$
Choosing \(\phi(t,x)\equiv1\in\mathcal{D}(\mathbb{T}^{2})\), we obtain that
$$ \int_{\mathbb{T}^{2}} a(t, x) u_{0}(t,x)\,\mathrm{d}t\, \mathrm {d}x=\lambda_{0}\int_{\mathbb{T}^{2}} f\bigl(t, x, u_{0}(t-\tau_{1}, x),\ldots ,u_{0}(t- \tau_{n},x)\bigr)\,\mathrm{d}t\,\mathrm{d}x. $$
(3.3)
Since \(u_{0}\in K\cap\partial\Omega_{1}\), by the definitions of K and \(\Omega_{1}\), we have
$$ 0 \le u_{0}(t-\tau_{k},x)\le\|u_{0}\|=r< \delta, \quad t, x\in\mathbb{R}, k=1, \ldots, n. $$
(3.4)
Hence from condition (F1) it follows that
$$f\bigl(t,x,u_{0}(t-\tau_{1},x),\ldots,u_{0}(t- \tau_{n},x)\bigr)\le c_{1}u_{0}(t-\tau _{1},x)+\cdots+c_{n} u_{0}(t-\tau_{n},x) $$
for every \(t, x\in\mathbb{R}\). By this inequality and (3.3), using the periodicity of \(u_{0}\) and (1.9), we have
$$\begin{aligned} \int_{\mathbb{T}^{2}} a(t, x) u_{0}(t,x)\,\mathrm{d}t\, \mathrm{d}x =&\lambda_{0}\int_{\mathbb{T}^{2}} f\bigl(t, x, u_{0}(t-\tau_{1}, x),\ldots ,u_{0}(t- \tau_{n},x)\bigr)\,\mathrm{d}t\,\mathrm{d}x \\ \le& \int_{\mathbb{T}^{2}} f\bigl(t, x, u_{0}(t- \tau_{1}, x),\ldots ,u_{0}(t-\tau_{n},x)\bigr)\, \mathrm{d}t\,\mathrm{d}x \\ \le& \int_{\mathbb{T}^{2}}\bigl(c_{1}u_{0}(t- \tau_{1},x)+\cdots+c_{n} u_{0}(t-\tau _{n},x)\bigr)\,\mathrm{d}t\,\mathrm{d}x \\ =& (c_{1}+\cdots+c_{n})\int_{\mathbb{T}^{2}}u_{0}(t,x) \,\mathrm{d}t\, \mathrm{d}x. \end{aligned}$$
Consequently, we obtain that
$$\begin{aligned} \underline{a}\int_{\mathbb{T}^{2}}u_{0}(t,x) \, \mathrm{d}t\,\mathrm {d}x & \le \int_{\mathbb{T}^{2}} a(t, x) u_{0}(t,x) \,\mathrm{d}t\, \mathrm{d}x \\ & \le (c_{1}+\cdots+c_{n})\int_{\mathbb{T}^{2}}u_{0}(t,x) \,\mathrm {d}t\,\mathrm{d}x. \end{aligned}$$
(3.5)
By the definition of cone K, \(\int_{\mathbb{T}^{2}}u_{0}(t,x) \, \mathrm{d}t\,\mathrm{d}x \ge\sigma\|u_{0}\|\cdot4\pi^{2}> 0\). From (3.5) it follows that \(\underline{a}\le c_{1}+\cdots+c_{n}\), which contradicts the assumption in condition (F1). Hence A satisfies the condition of Lemma 2.3 in \(K\cap{\partial\Omega}_{1}\). By Lemma 2.3 we have
$$ i (A, K\cap{\Omega}_{1}, K)=1. $$
(3.6)
On the other hand, choose \(R>\max \{H/\sigma, \delta \}\), where H is the positive constant in condition (F2), and let \(e(t,x)\equiv1\). Clearly, \(e\in K\setminus\{\theta\}\). We show that A satisfies the condition of Lemma 2.4 in \(K\cap\partial\Omega_{2}\), namely \(u-Au\ne\mu e \) for every \(u\in K\cap\partial\Omega_{2}\) and \(\mu\ge0\). In fact, if there exist \(u_{1}\in K\cap\partial\Omega_{2}\) and \(\mu _{1}\ge0\) such that \(u_{1}-Au_{1}=\mu_{1} e\), since \(u_{1}-\mu_{1} e=Au_{1}=P(f(t, x, u_{1}(t-\tau_{1}, x),\ldots,u_{1}(t-\tau_{n},x)))\), by the definition of P and Lemma 2.1, \(u_{1}-\mu_{1} e \in C(\mathbb{T}^{2})\) satisfies the differential equation
$$ \mathcal{L} (u_{1}-\mu_{1} e)=f\bigl(t,x,u_{1}(t- \tau_{1}),\ldots,u_{1}(t-\tau _{n})\bigr),\quad \mbox{in }\mathcal{D}'\bigl(\mathbb{T}^{2}\bigr). $$
(3.7)
In the definition of the solution of Equation (3.7) in the distribution sense, choosing \(\phi(t,x)\equiv1\in\mathcal{D}(\mathbb{T}^{2})\), we have
$$ \int_{\mathbb{T}^{2}} a(t, x) \bigl(u_{1}(t,x)- \mu_{1}\bigr)\,\mathrm{d}t\, \mathrm{d}x=\int_{\mathbb{T}^{2}} f \bigl(t, x, u_{1}(t-\tau_{1}, x),\ldots ,u_{1}(t- \tau_{n},x)\bigr)\,\mathrm{d}t\,\mathrm{d}x. $$
(3.8)
Since \(u_{1}\in K\cap\partial\Omega_{2}\), by the definition of K, we have
$$ u_{1}(t-\tau_{k},x)\ge\sigma\|u_{1}\|=\sigma R>H, \quad t, x\in\mathbb{R}, k=1,\ldots,n . $$
(3.9)
From this and condition (F2), it follows that
$$f\bigl(t, x, u_{1}(t-\tau_{1}),\ldots, u_{1}(t- \tau_{n})\bigr)\ge d_{1} u_{1}(t-\tau_{1}, x)+\cdots+d_{n}u_{n}(t-\tau_{n}, x) $$
for every \(t, x\in\mathbb{R}\). By this inequality and (3.8), using the periodicity of \(u_{1}(t,x)\) and (1.9), we have
$$\begin{aligned} \int_{\mathbb{T}^{2}} a(t, x) u_{1}(t,x)\,\mathrm{d}t\, \mathrm{d}x \ge& \int_{\mathbb{T}^{2}} a(t, x) \bigl(u_{1}(t,x)- \mu_{1}\bigr)\,\mathrm{d}t\, \mathrm{d}x \\ \ge& \int_{\mathbb{T}^{2}} f\bigl(t, x, u_{1}(t- \tau_{1}, x),\ldots ,u_{1}(t-\tau_{n},x)\bigr)\, \mathrm{d}t\,\mathrm{d}x \\ \ge& \int_{\mathbb{T}^{2}}\bigl(d_{1} u_{1}(t- \tau_{1},x)+\cdots+ d_{n} u_{1}(t- \tau_{n},x)\bigr) \,\mathrm{d}t\,\mathrm{d}x \\ =& (d_{1}+\cdots+d_{n})\int_{\mathbb{T}^{2}} u_{1}(t,x) \,\mathrm{d}t\, \mathrm{d}x. \end{aligned}$$
Hence, we get that
$$\begin{aligned} \overline{a}\int_{\mathbb{T}^{2}} u_{1}(t,x) \, \mathrm{d}t\,\mathrm {d}x & \ge \int_{\mathbb{T}^{2}} a(t,x) u_{1}(t,x ) \,\mathrm{d}t\, \mathrm{d}x \\ & \ge (d_{1}+\cdots+d_{n})\int_{\mathbb{T}^{2}}u_{1}(t,x) \,\mathrm {d}t\,\mathrm{d}x. \end{aligned}$$
(3.10)
Since \(\int_{\mathbb{T}^{2}} u_{1}(t,x)\,\mathrm{d}t\,\mathrm {d}x\ge\sigma\|u_{1}\|\cdot4\pi^{2}> 0\), from this inequality it follows that \(\overline{a}\ge d_{1}+\cdots+d_{n}\), which contradicts the assumption in condition (F2). This means that A satisfies the condition of Lemma 2.4 in \(K\cap \partial\Omega_{2}\). By Lemma 2.4,
$$ i (A, K\cap{\Omega}_{2}, K)=0. $$
(3.11)
Now, using the additivity of fixed point index in cone K, by (3.6) and (3.11) we have that
$$i \bigl(A, K\cap(\Omega_{2}\setminus\overline{\Omega}_{1}), K\bigr)= i (A, K\cap{\Omega}_{2}, K)-i (A, K\cap{\Omega}_{1}, K)=-1. $$
Hence, A has a fixed-point in \(K\cap({\Omega}_{2}\setminus\overline {\Omega}_{1})\), which is a positive doubly periodic solution of Equation (1.1). □
Proof of Theorem 1.2
Let \(\Omega_{1}, \Omega_{2}\subset C(\mathbb {T}^{2})\) be defined by (3.1). We prove that the operator A defined by (2.9) has a fixed point in \(K\cap(\Omega_{2}\setminus \overline{\Omega}_{1})\) when r is small enough and R large enough.
Let \(r\in(0, \delta)\), where δ is the positive constant in condition (F3), and choose \(e(t,x)\equiv1\). We prove that A satisfies the condition of Lemma 2.4 in \(K\cap\partial\Omega_{1}\), namely \(u-Au\ne\mu e\) for every \(u\in K\cap\partial\Omega_{1}\) and \(\mu\ge0\). In fact, if there exist \(u_{0}\in K\cap\partial\Omega_{1}\) and \(\mu _{0}\ge0\) such that \(u_{0}-Au_{0}=\mu_{0} e\), since \(u_{0}-\mu_{0} e=Au_{0}=P(f(t, x, u_{0}(t-\tau_{1}, x),\ldots,u_{0}(t-\tau _{n},x)))\), by the definition of P and Lemma 2.1, \(u_{0}-\mu_{0} e\in C(\mathbb{T}^{2})\) satisfies the differential equation
$$ \mathcal{L} (u_{0}-\mu_{0} e)=f\bigl(t,x,u_{0}(t- \tau_{1}),\ldots,u_{0}(t-\tau _{n})\bigr),\quad \mbox{in }\mathcal{D}'\bigl(\mathbb{T}^{2}\bigr). $$
(3.12)
In the definition of the solution of Equation (3.12) in the distribution sense, choosing \(\phi(t,x)\equiv1\in\mathcal {D}(\mathbb{T}^{2})\), we get that
$$ \int_{\mathbb{T}^{2}} a(t, x) \bigl(u_{0}(t,x)- \mu_{0}\bigr)\,\mathrm{d}t\, \mathrm{d}x=\int_{\mathbb{T}^{2}} f\bigl(t, x, u_{0}(t-\tau_{1}, x),\ldots ,u_{0}(t-\tau_{n},x)\bigr)\,\mathrm{d}t\, \mathrm{d}x. $$
(3.13)
Since \(u_{0}\in K\cap\partial\Omega_{1}\), by the definitions of K and \(\Omega_{1}\), \(u_{0}\) satisfies (3.4). From (3.4) and condition (F3) we see that
$$f\bigl(t,x, u_{0}(t-\tau_{1},x),\ldots,u_{0}(t- \tau_{n},x)\bigr)\ge d_{1} u_{0}(t-\tau _{1},x)+\cdots+d_{n}u_{0}(t-\tau_{n},x) $$
for every \(t, x\in\mathbb{R}\). By this inequality and (3.13), we have
$$\begin{aligned} \int_{\mathbb{T}^{2}} a(t, x) u_{0}(t,x)\,\mathrm{d}t\, \mathrm{d}x \ge& \int_{\mathbb{T}^{2}} a(t, x) \bigl(u_{0}(t,x)- \mu_{0}\bigr)\,\mathrm{d}t\, \mathrm{d}x \\ \ge& \int_{\mathbb{T}^{2}} f\bigl(t, x, u_{0}(t- \tau_{1}, x),\ldots ,u_{0}(t-\tau_{n},x)\bigr)\, \mathrm{d}t\,\mathrm{d}x \\ \ge& \int_{\mathbb{T}^{2}}\bigl(d_{1} u_{0}(t- \tau_{1},x)+\cdots+ d_{n} u_{0}(t- \tau_{n},x)\bigr) \,\mathrm{d}t\,\mathrm{d}x \\ =& (d_{1}+\cdots+d_{n})\int_{\mathbb{T}^{2}} u_{0}(t,x) \,\mathrm{d}t\, \mathrm{d}x. \end{aligned}$$
From this it follows that
$$\begin{aligned} \begin{aligned}[b] \overline{a}\int_{\mathbb{T}^{2}} u_{0}(t,x) \, \mathrm{d}t\,\mathrm {d}x & \ge \int_{\mathbb{T}^{2}} a(t,x) u_{0}(t,x ) \,\mathrm{d}t\, \mathrm{d}x \\ & \ge (d_{1}+\cdots+d_{n})\int_{\mathbb{T}^{2}}u_{0}(t,x) \,\mathrm {d}t\,\mathrm{d}x. \end{aligned} \end{aligned}$$
(3.14)
Since \(\int_{\mathbb{T}^{2}} u_{0}(t,x) \,\mathrm{d}t\,\mathrm {d}x\ge\sigma\|u_{0}\|\cdot4\pi^{2}> 0\), from inequality (3.14) it follows that \(\overline{a}\ge d_{1}+\cdots+d_{n}\), which contradicts the assumption in (F3). Hence A satisfies the condition of Lemma 2.4 in \(K\cap\partial\Omega_{1}\). By Lemma 2.4 we have
$$ i (A, K\cap{\Omega}_{1}, K)=0. $$
(3.15)
Choosing \(R>\max \{H/\sigma, \delta \}\), we show that A satisfies the condition of Lemma 2.3 in \(K\cap\partial\Omega_{2}\), namely \(\lambda Au\ne u\) for every \(u\in K\cap\partial\Omega_{2}\) and \(0<\lambda\le1\). In fact, if there exist \(u_{1}\in K\cap\partial\Omega_{2}\) and \(0<\lambda_{1}\le1\) such that \(\lambda_{1} Au_{1}=u_{1}\), since \(u_{1}=P(\lambda_{1} f(t, x, u_{1}(t-\tau_{1}, x),\ldots,u_{1}(t-\tau _{n},x)))\), by the definition of P and Lemma 2.1, \(u_{1}\in C(\mathbb{T}^{2})\) satisfies the differential equation
$$ \mathcal{L} u_{1} (t,x)=\lambda_{1} f\bigl(t, x, u_{1}(t-\tau_{1}, x),\ldots ,u_{1}(t- \tau_{n},x)\bigr),\quad \mbox{in }\mathcal{D}'\bigl( \mathbb{T}^{2}\bigr). $$
(3.16)
In the definition of the solution of Equation (3.16) in the distribution sense, choosing \(\phi(t,x)\equiv1\in\mathcal {D}(\mathbb{T}^{2})\), we have
$$ \int_{\mathbb{T}^{2}} a(t, x) u_{1}(t,x)\,\mathrm{d}t\,\mathrm {d}x=\lambda_{1}\int_{\mathbb{T}^{2}} f\bigl(t, x, u_{1}(t-\tau_{1}, x),\ldots ,u_{1}(t- \tau_{n},x)\bigr)\,\mathrm{d}t\,\mathrm{d}x. $$
(3.17)
Since \(u_{1}\in K\cap\partial\Omega_{2}\), by the definition of K, \(u_{1}\) satisfies (3.9). From (3.9) and condition (F4) it follows that
$$f\bigl(t,x,u_{1}(t-\tau_{1},x),\ldots,u_{1}(t- \tau_{n},x)\bigr)\le c_{1} u_{1}(t-\tau _{1},x)+\cdots+c_{n}u_{1}(t-\tau_{n},x) $$
for every \(t, x\in\mathbb{R}\). By this inequality and (3.17), we have
$$\begin{aligned} \int_{\mathbb{T}^{2}} a(t, x) u_{1}(t,x)\,\mathrm{d}t\, \mathrm{d}x =&\lambda_{1}\int_{\mathbb{T}^{2}} f\bigl(t, x, u_{1}(t-\tau_{1}, x),\ldots, u_{1}(t- \tau_{n},x)\bigr)\,\mathrm{d}t\,\mathrm{d}x \\ \le&\int_{\mathbb{T}^{2}} f\bigl(t, x, u_{1}(t- \tau_{1}, x),\ldots, u_{1}(t-\tau_{n},x)\bigr)\, \mathrm{d}t\,\mathrm{d}x \\ \le& \int_{\mathbb{T}^{2}}\bigl(c_{1}u_{1}(t- \tau_{1},x)+\cdots+c_{n} u_{1}(t-\tau _{n},x)\bigr)\,\mathrm{d}t\,\mathrm{d}x \\ =& (c_{1}+\cdots+c_{n})\int_{\mathbb{T}^{2}}u_{1}(t,x) \,\mathrm{d}t\, \mathrm{d}x. \end{aligned}$$
This means that
$$\begin{aligned} \underline{a}\int_{\mathbb{T}^{2}} u_{1}(t,x) \, \mathrm{d}t\,\mathrm {d}x & \le \int_{\mathbb{T}^{2}} a(t,x) u_{1}(t,x ) \,\mathrm{d}t\, \mathrm{d}x \\ & \le (c_{1}+\cdots+c_{n})\int_{\mathbb{T}^{2}}u_{1}(t,x) \,\mathrm {d}t\,\mathrm{d}x. \end{aligned}$$
(3.18)
Since \(\int_{\mathbb{T}^{2}} u_{1}(t,x) \,\mathrm{d}t\,\mathrm {d}x\ge\sigma\|u_{0}\|\cdot4\pi^{2}> 0\), from inequality (3.18) it follows that \(\underline{a}\le c_{1}+\cdots+c_{n}\), which contradicts the assumption in condition (F4). Hence A satisfies the condition of Lemma 2.3 in \(K\cap{\partial \Omega}_{1}\). By Lemma 2.3 we have
$$ i (A, K\cap{\Omega}_{2}, K)=1. $$
(3.19)
Now, from (3.15) and (3.19) it follows that
$$i \bigl(A, K\cap(\Omega_{2}\setminus\overline{\Omega}_{1}), K\bigr)= i (A, K\cap{\Omega}_{2}, K)-i (A, K\cap{\Omega}_{1}, K)=1. $$
Hence A has a fixed point in \(K\cap({\Omega}_{2}\setminus\overline {\Omega}_{1})\), which is a positive doubly periodic solution of Equation (1.1). □
Example 3.1
Consider the existence of doubly 2π-periodic solution of the following telegraph equation with time delay:
$$ u_{tt}-u_{xx}+2 u_{t}+ u=b_{1}(t,x) u^{2}(t,x)+b_{2}(t,x) u^{2}(t-\pi,x), $$
(3.20)
where \(b_{i}\in C(\mathbb{T}^{2})\) and \(b_{i}(t,x)>0\) for \((t,x)\in\mathbb {T}^{2}\), \(i=1, 2\). Corresponding to the general equation (1.1),
$$\begin{aligned}& c=2,\qquad a(t,x)\equiv1,\qquad n=2,\qquad \tau_{1}=0, \qquad \tau_{2}=\pi, \\& f(t,x,\eta_{1},\eta_{2})=b_{1}(t,x) \eta_{1}^{2}+b_{2}(t,x) \eta_{2}^{2}. \end{aligned}$$
(3.21)
We easily see that assumptions (H1) and (H2) hold. From (3.21) we can directly verify that f satisfies conditions (F1) and (F2). By Theorem 1.1, Equation (3.20) has at least one positive doubly 2π-periodic weak solution (in the distribution sense).
Example 3.2
Consider the following nonlinear telegraph equation with time delays
$$ u_{tt}-u_{xx}+4 u_{t}+\bigl(2+\sin(t+x)\bigr) u=u^{2/3}(t-\pi/2,x)+u^{1/3}(t-\pi ,x). $$
(3.22)
Corresponding to Equation (1.1),
$$\begin{aligned}& c=4,\qquad a(t,x)=2+\sin(t+x),\qquad n=2,\qquad \tau_{1}=\pi/2, \qquad \tau_{2}=\pi, \\& f(t, x, \eta_{1}, \eta_{2})=\eta_{1}^{2/3}+\eta_{2}^{1/3}. \end{aligned}$$
For these, we can directly verify that the conditions of Theorem 1.2 are satisfied. By Theorem 1.2, Equation (3.22) has a positive doubly 2π-periodic weak solution (in the distribution sense).