### 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 2*π*-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 2*π*-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 2*π*-periodic weak solution (in the distribution sense).