Modified differential transform method for solving linear and nonlinear pantograph type of differential and Volterra integro-differential equations with proportional delays

Abstract

In this study, a hybrid technique for improving the differential transform method (DTM), namely the modified differential transform method (MDTM) expressed as a combination of the differential transform method, Laplace transforms, and the Padé approximant (LPDTM) is employed for the first time to ascertain exact solutions of linear and nonlinear pantograph type of differential and Volterra integro-differential equations (DEs and VIDEs) with proportional delays. The advantage of this method is its simple and trusty procedure, it solves the equations straightforward and directly without requiring large computational work, perturbations or linearization, and enlarges the domain of convergence, and leads to the exact solution. Also, to validate the reliability and efficiency of the method, some examples and numerical results are provided.

1 Introduction

Mathematical modeling of various phenomena in science and engineering such as biological population management, chemistry, physics, physiological and pharmaceutical kinetics and chemical kinetics, medicine, infectious diseases, economy, nonlinear dynamical system, communication networks, number theory, electrodynamics, the navigational control of ships and aircraft and control problems and electronic systems leads to one of the most important kinds of delay differential equations (DDEs), namely pantograph equation [1–7]. The term pantograph was first used by Ockendon and Tayler in [8] which modeled and redesigned the collection system for an electric locomotive.

At this point, it is usually difficult to solve these kinds of DDEs analytically. Therefore, in the literature, there are some valuable efforts that focus on finding the analytical and numerical methods for solving the pantograph type of DEs and VIDEs with proportional delays. (see, e.g., [9–21] and the references therein). The differential transform method (DTM) is an analytical-numerical technique introduced for the first time by Zhou [22] to study the electrical circuits. Like any subject in mathematics, DTM has grown over a period of time. In 1999, Chen and Ho [23] developed this method for partial differential equations with two independent variables. In 2004, Ayaz [24] extended the two-dimensional DTM into the three-dimensional DTM and used it to solve linear and nonlinear partial differential equations. In 2005, Arikoglu and Ozkol [25] used the differential transform method for integro-differential equations. With the advent of fractional calculus, the differential transform method was also modified to solve derivative and integral problems of any order. In 2007, Arikoglu and Ozkol [26] proposed a numerical-analytical method similar to the DTM, called the fractional differential transform method (FDTM), which they used to solve fractional differential equations. In 2008, Odibat and Momani [27] introduced the generalized differential transform method (GDTM) based on the differential transform method and generalized Taylor’s formula and the Caputo fractional derivative and used it to solve fractional partial differential equations. Also in the same year, Momani and Erturk [28] learned to correct and improve the accuracy of the solution of convergent series obtained by the differential transform method, they introduced the modified differential transform method (MDTM), and Chang [29] used the DTM for one-dimensional nonlinear functions. In 2009, Keskin [30, 31] introduced the reduced form of the DTM as the reduced differential transform method (RDTM), which Keskin and Oturanc [32, 33] used to solve partial differential equations and fractional differential equations. Many authors, during recent years, have used this method for solving various types of equations. For example, differential-algebraic equations [34–36], Volterra integral equation [37–39], integro-differential equations [40–43] and fractional differential equations [44–47] are solved using this method. This suggested technique is highly efficient and powerful in obtaining the exact solutions and approximate solutions of mathematical modeling of many problems, gives the solution in the form of rapidly convergent successive approximations, and is capable of handling linear and nonlinear equations in a similar manner. Moreover the comparison of our method with other analytical methods available in the literature show that although the results of these methods are the same, RDTM is a lot easier, more convenient, and reliable than them [48–55].

In this paper, we present the application of the modified differential transform method (MDTM) as a hybrid approach, for improving DTM’s truncated series solutions in convergence rate combining DTM, Laplace transforms, and Padé approximant. The solutions series obtained by the differential transform method, even if they contain a large number of terms, may converge in a limited area. Therefore, the domain of convergence of the truncated power series expands by the Laplace–Padé differential transform method (LPDTM) and often leads to the exact solution. To improve the solution of convergent series obtained by the DTM, we apply Laplace transform to it, and then by forming its Padé approximant, the transformed series convert into a meromorphic function. Finally, to obtain the analytical solution, we take the inverse Laplace transform from the function obtained of the Padé approximant. Therefore, in the light of the above-mentioned method, we will study the exact solution of linear and nonlinear pantograph type of DEs and VIDEs with proportional delays

$$\begin{aligned} & F \bigl(t,u(\alpha _{0}t),u^{\prime }( \alpha _{1}t),\ldots , u^{(n)}( \alpha _{n}t) \bigr)=0, \\ \end{aligned}$$

(1.1)

$$\begin{aligned} & G \bigg(t,u(\alpha _{0}t),u^{\prime }(\alpha _{1}t),\ldots , u^{(n)}( \alpha _{n}t), \int _{0}^{lt} K \bigl(t,\xi ,u(\beta _{0}\xi ),u^{\prime }( \beta _{1}\xi ),\ldots , u^{(m)}(\beta _{m}\xi )\,\mathrm{d}\xi \bigg) \\ &\quad =0, \\ & 0\leq t\leq T,\qquad \alpha _{i}, \beta _{j}, l \in (0,1),\quad i=0,1, \ldots ,n, j=0,1,\ldots ,m, m< n, \end{aligned}$$

(1.2)

where u is the unknown function and the functions F, G, and K are analytic in the domain of interest.

The rest of this study is presented in the following sections: In Sect. 2, The main idea behind the Padé approximant is introduced. In Sect. 3, we simply introduce the modified differential transform method (MDTM) as a combined form of the DTM with Laplace transforms, and the Padé approximant. In Sect. 4, we prove several important Theorems. In Sect. 5, we apply the MDTM to obtain the exact solutions for linear and nonlinear pantograph type of DEs and VIDEs with proportional delays. Finally, we offer some summaries and a conclusion in Sect. 6.

2 Padé approximant

The best approximation of a function with a rational function of a certain order is the Padé approximant. Under this technique, the approximant’s power series agrees with the power series of the function it is approximating.

Let $u(t)$ be an analytical function that corresponds to the Maclaurin series:

$$ u(t)=\sum_{k=0}^{\infty } c_{k} t^{k},\quad 0\leq t\leq T. $$

(2.1)

Then the Padé approximant to $u(t)$ is a rational fraction which we denote as $[\frac{m}{n} ]$ and is defined by [56, 57]

$$ \biggl[\frac{m}{n} \biggr] = \frac{a_{0}+a_{1} t+a_{2} t^{2}+\cdots +a_{m} t^{m}}{b_{0}+b_{1} t+b_{2} t^{2}+\cdots +b_{n} t^{n}}, $$

(2.2)

there are $m+1$ independent numerator coefficients and n independent denominator coefficients, making $m+n+1$ unknown coefficients in all. We have

$$ u(t)- \biggl[\frac{m}{n} \biggr](t)=O \bigl(t^{m+n+1} \bigr). $$

(2.3)

From Eq. (2.3), we have

$$ u(t)\sum_{j=0}^{n} b_{j} t^{j}-\sum_{i=0}^{m} a_{i} t^{i}=O \bigl(t^{m+n+1} \bigr). $$

(2.4)

From Eq. (2.4), we get the following algebraic linear systems:

$$\begin{aligned} &c_{m}b_{1}+\cdots +c_{m-n+1}b_{n}=-c_{m+1}, \\ &c_{m+1}b_{1}+\cdots +c_{m-n+2}b_{n}=-c_{m+2}, \\ &\vdots \\ &c_{m+n-1}b_{1}+\cdots +c_{m}b_{n}=-c_{m+n}, \end{aligned}$$

(2.5)

and

$$\begin{aligned} &a_{0}=c_{0}, \\ &a_{1}=c_{1}+b_{1} c_{0}, \\ &a_{2}=c_{2}+b_{1} c_{1}+b_{2} c_{0}, \\ &a_{3}=c_{3}+b_{1}c_{2}+b_{2} c_{1} +b_{3} c_{0}, \\ & \vdots \\ &a_{n}= c_{n}+ \sum_{k=1}^{n} b_{k} c_{n-k}. \end{aligned}$$

(2.6)

We determine first all the denominator coefficients $b_{j}$, $1\leq j\leq n$, from Eq. (2.5). Then, we calculate the numerator coefficients $a_{i}$, $0\leq i \leq m$, from Eq. (2.6).

Remark 2.1

For a fixed value of $m+n+1$, error Eq. (2.3) is the smallest when the numerator has one degree higher than the denominator of (2.2) or when the numerator and denominator have the same degree.

The advantage of Padé approximant is that it often gives a better approximation than the truncated series solutions from the Taylor series. This is because sometimes the Taylor series may not be convergent, and the Padé approximant has the potential to expand the domain of convergence of solutions or includes finding exact solutions.

3 Summary of the method

We present some important definitions and mathematical preliminaries operations of the modified differential transform method which can help to gain more understanding of the method stated in this section.

Definition 3.1

The differential transform function of $u(t)$ can be written in the following form:

$$ \mathcal{U}(k)=\frac{1}{k!} \biggl[ \frac{\mathrm{d}^{k}}{\mathrm{d} t^{k}}u(t) \biggr]_{t=t_{0}}, $$

(3.1)

where $u(t)$ represents the main original analytic and differentiated continuously function with regard to time t, in the domain of interest, and $\mathcal{U}(k)$ describes the transformed function.

Definition 3.2

The differential inverse transform of $\mathcal{U}(k)$ is determined as

$$ u(t)=\sum_{k=0}^{\infty } \mathcal{U}(k) (t-t_{0})^{k}. $$

(3.2)

Then, consolidating Eqs. (3.2) and (3.1) yields

$$ u(t)=\sum_{k=0}^{\infty } \frac{1}{k!} \biggl[ \frac{\mathrm{d}^{k}}{\mathrm{d} t^{k}}u(t) \biggr]_{t=t_{0}}(t-t_{0})^{k}. $$

(3.3)

By the help of the upper definitions, to illustrate the basic idea of the DTM, consider the following form of nonlinear ordinary differential equations:

$$ \frac{\mathrm{d}u(t)}{\mathrm{d}t}=f \bigl(u(t),t \bigr),\quad t\geq 0, $$

(3.4)

with the following initial condition:

$$ u(0)=c, $$

(3.5)

where $f(u(t),t)$ indicates a nonlinear smooth function.

After applying the DTM definition on both sides of Eq. (3.4), we can write the following iteration formula:

$$ (k+1)\mathcal{U}(k+1)=\mathcal{F} \bigl(\mathcal{U}(0), \ldots , \mathcal{U}(k),k \bigr),\quad k\geq 0, $$

(3.6)

where $\mathcal{F}(\mathcal{U}(0), \ldots ,\mathcal{U}(k),k)$ is the differential transform functions of $f(u(t),t)$.

Implementing the aforesaid method to the initial condition (3.5), we have

$$ \mathcal{U}(0)=c. $$

(3.7)

To discover the remaining iteration, we plug Eq. (3.7) into Eq. (3.6) and by simple reiterative calculation, we get the subsequent $\mathcal{U}(k)$ values. Then, by using the inverse transformation of the set of values $\{\mathcal{U}(k)\}^{n}_{k=0}$, the approximation solution can be written as follows:

$$ \tilde{u}_{n}(t)=\sum _{k=0}^{n}\mathcal{U}(k) (t-t_{0})^{k}. $$

(3.8)

Thus, the exact solution of the considered equation can be gained by

$$ u(t)=\lim_{n\rightarrow \infty }\tilde{u}_{n}(t). $$

(3.9)

The solutions series derived from the DTM, even if they contain a large number of terms, may converge in a limited area. To improve the solution obtained in convergent series form using the DTM, and enlarge the domain of convergence of solutions we apply the Laplace–Padé method by following steps.

• Step 1: Applying Laplace transformation with respect to t for the obtained power series (3.8).

• Step 2: Substituting s by $\frac{1}{t}$ in the resulting equation.

• Step 3: Creating the Padé approximant of order $[\frac{m}{n} ]$ for the transformed series and convert it into a meromorphic function.

Remark 3.3

m and n are arbitrarily chosen, but they should be of smaller values than the order of the power series. In this step, to obtain better convergence and accuracy, the Padé approximant enlarges the domain of the truncated series solution.

• Step 4: Substituting t by $\frac{1}{s}$.

• Step 5: Applying the inverse Laplace transformation with respect to s for obtaining the exact or approximate solution.

Table 1 contains the basic mathematical operations carried out by DTM.

Table 1 The fundamental operations of RDTM

4 Main results

The principal main of this section is to study the fundamental Theorem of this paper.

Theorem 4.1

If $w(t) =u(\alpha t)$, then $\mathcal{W}(k)=\alpha ^{k} \mathcal{U}(k)$ where $\alpha \in (0,1)$.

Proof

From Eq. (3.1), we get

$$\begin{aligned} \mathcal{W}(k) &=\frac{1}{k!} \biggl[ \frac{d^{k}}{dt^{k}} w(t) \biggr]_{t=t_{0}} =\frac{1}{k!} \biggl[ \frac{d^{k}}{dt^{k}} u( \alpha t) \biggr]_{t=t_{0}} \\ &= \frac{1}{k!} \biggl[\alpha ^{k} \frac{d^{k}}{d\hat{t}^{k}} u(\hat{t}) \biggr]_{\hat{t}=t_{0}}= \frac{1}{k!} \alpha ^{k} k! \mathcal{U}(k), \end{aligned}$$

where $\hat{t}=\alpha t$, then

$$ \mathcal{W}(k)=\alpha ^{k} \mathcal{U}(k). $$

 □

Theorem 4.2

If $w(t) =u(\alpha _{1}t) v(\alpha _{2}t)$, then, for $\alpha _{1}, \alpha _{2}\in (0,1)$, we have

$$ \mathcal{W}(k)=\sum_{r=0}^{k} \alpha _{1}^{r} \alpha _{2}^{k-r} \mathcal{U}(r) \mathcal{V}(k-r),\quad k\geq 0. $$

Proof

From Eq. (3.1), we get

$$\begin{array}{rl}\mathcal{W}(k)& =\frac{1}{k!}{[\frac{d^k}{d{t}^k}w(t)]}_{t=t_0}\\ & =\frac{1}{k!}{\left[\frac{d^k}{d{t}^k}(u({\alpha }_{1}t)v({\alpha }_{2}t))\right]}_{t=t_0}\\ & =\frac{1}{k!}{[\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\alpha }_1^r\frac{d^r}{d{\hat{t}}^r}u(\hat{t}){\alpha }_2^{k-r}\frac{d^{k-r}}{d{\tilde{t}}^{k-r}}v(\tilde{t})]}_{\hat{t},\tilde{t}=t_0}\\ & =\frac{1}{k!}\sum _{r=0}^k\frac{k!}{r!(k-r)!}{\alpha }_1^{r}r!\mathcal{U}(r){\alpha }_2^{k-r}(k-r)!\mathcal{V}(k-r),\end{array}$$

where $\hat{t}=\alpha _{1} t$ and $\tilde{t}=\alpha _{2} t$, then

$$ \mathcal{W}(k)=\sum_{r=0}^{k} \alpha _{1}^{r} \alpha _{2}^{k-r} \mathcal{U}(r) \mathcal{V}(k-r). $$

 □

Theorem 4.3

If $w(t)=\frac{d^{r}}{dt^{r}} u(\alpha t)$, then $\mathcal{W}(k)=\alpha ^{k+r} \frac{(k+r) !}{k !} \mathcal{U}(k+r)$.

Proof

From Eq. (3.1), we have

$$\begin{aligned} \mathcal{W}(k) &=\frac{1}{k!} \biggl[\frac{d^{k}}{dt^{k}} w(t) \biggr]_{t=t_{0}}= \frac{1}{k!} \biggl[\frac{d^{k+r}}{dt^{k+r}} u( \alpha t) \biggr]_{t=t_{0}}= \frac{1}{k!} \biggl[\alpha ^{k+r} \frac{d^{k+r}}{d\hat{t}^{k+r}} u( \hat{t}) \biggr]_{\hat{t}=t_{0}} \\ &=\frac{1}{k!} \alpha ^{k+r}(k+r) ! \mathcal{U}(k+r), \end{aligned}$$

therefore

$$ \mathcal{W}(k)=\alpha ^{k+r} \frac{(k+r) !}{k !} \mathcal{U}(k+r). $$

 □

Theorem 4.4

If $w(t) =\frac{d^{m}}{dt^{m}} u(\alpha _{1}t) \frac{d^{n}}{dt^{n}} v( \alpha _{2}t)$, then

$$ \mathcal{W}(k)=\sum_{r=0}^{k} \alpha _{1}^{r+m} \alpha _{2}^{k-r+n} \frac{(r+m)!(k-r+n)!}{r! (k-r)!} \mathcal{U}(r+m) \mathcal{V}(k-r+n),\quad k\geq 0. $$

Proof

From Eq. (3.1), we have

$$\begin{array}{rl}\mathcal{W}(k)& =\frac{1}{k!}{[\frac{d^k}{d{t}^k}w(t)]}_{t=t_0}=\frac{1}{k!}{\left[\frac{d^k}{d{t}^k}(\frac{d^m}{d{t}^m}u({\alpha }_{1}t)\frac{d^n}{d{t}^n}v({\alpha }_{2}t))\right]}_{t=t_0}\\ & =\frac{1}{k!}{\left[\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right)\frac{d^r}{d{t}^r}(\frac{d^m}{d{t}^m}u({\alpha }_{1}t))\frac{d^{k-r}}{d{t}^{k-r}}(\frac{d^n}{d{t}^n}v({\alpha }_{2}t))\right]}_{t=t_0}\\ & =\frac{1}{k!}{[\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\alpha }_1^{r+m}\frac{d^{r+m}}{d{\hat{t}}^{r+m}}u(\hat{t}){\alpha }_2^{k-r+n}\frac{d^{k-r+n}}{d{\tilde{t}}^{k-r+n}}v(\tilde{t})]}_{\hat{t},\tilde{t}=t_0}\\ & =\frac{1}{k!}\sum _{r=0}^k\frac{k!}{r!(k-r)!}{\alpha }_1^{r+m}(r+m)!\mathcal{U}(r+m){\alpha }_2^{k-r+n}(k-r+n)!\mathcal{V}(k-r+n),\end{array}$$

where $\hat{t}=\alpha _{1} t$ and $\tilde{t}=\alpha _{2} t$, then

$$ \mathcal{W}(k)=\sum_{r=0}^{k} \alpha _{1}^{r+m} \alpha _{2}^{k-r+n} \frac{(r+m)!(k-r+n)!}{r! (k-r)!} \mathcal{U}(r+m) \mathcal{V}(k-r+n). $$

 □

Theorem 4.5

If $w(t)=\int _{0}^{lt}u(\alpha \xi ) \,\mathrm{d}\xi$, then $\mathcal{W}(k)=\frac{1}{k} l^{k} \alpha ^{k-1} \mathcal{U}(k-1)$.

Proof

$$\begin{aligned} \mathcal{W}(k) =\frac{1}{k!} \biggl[\frac{d^{k}}{dt^{k}} w(t) \biggr]_{t=t_{0}}= \frac{1}{k!} \biggl[l \frac{d^{k-1}}{dt^{k-1}} u(l \alpha t) \biggr]_{t=t_{0}}= \frac{1}{k!} \bigl[l (k-1)! (l\alpha )^{k-1}\mathcal{U}(k-1) \bigr], \end{aligned}$$

therefore

$$ \mathcal{W}(k)=\frac{1}{k} l^{k} \alpha ^{k-1} \mathcal{U}(k-1). $$

 □

Theorem 4.6

If $w(t)=\int _{0}^{lt}u(\alpha _{1}\xi ) v(\alpha _{2}\xi )\,\mathrm{d} \xi$, then

$$ \mathcal{W}(k)=\frac{1}{k}\sum_{r=0}^{k-1} l^{k} \alpha _{1}^{r} \alpha _{2}^{k-r-1} \mathcal{U}(r) \mathcal{V}(k-r-1),\quad k\geq 1. $$

Proof

$$\begin{array}{rl}\mathcal{W}(k)& =\frac{1}{k!}{[\frac{d^k}{d{t}^k}w(t)]}_{t=t_0}\\ & =\frac{1}{k!}{\left[l\frac{d^{k-1}}{d{t}^{k-1}}(u(l{\alpha }_{1}t)v(l{\alpha }_{2}t))\right]}_{t=t_0}\\ & =\frac{1}{k!}{[l\sum _{r=0}^{k-1}\left(\begin{array}{c}k-1\\ r\end{array}\right){(l{\alpha }_1)}^r\frac{d^r}{d{\hat{t}}^r}u(\hat{t}){(l{\alpha }_2)}^{k-r-1}\frac{d^{k-r-1}}{d{\tilde{t}}^{k-r-1}}v(\tilde{t})]}_{\hat{t},\tilde{t}=t_0}\\ & =\frac{1}{k!}l\sum _{r=0}^{k-1}\frac{(k-1)!}{r!(k-r-1)!}{l}^r{\alpha }_1^{r}r!\mathcal{U}(r)l^{k-r-1}{\alpha }_2^{k-r-1}(k-r-1)!\mathcal{V}(k-r-1),\end{array}$$

where $\hat{t}=l\alpha _{1} t$ and $\tilde{t}=l\alpha _{2} t$, then

$$ \mathcal{W}(k)=\frac{1}{k}\sum_{r=0}^{k-1} l^{k} \alpha _{1}^{r} \alpha _{2}^{k-r-1} \mathcal{U}(r) \mathcal{V}(k-r-1). $$

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Theorem 4.7

If $w(t)=v(\beta t)\int _{0}^{lt}u_{1}(\alpha _{1}\xi ) u_{2}(\alpha _{2} \xi )\,\mathrm{d}\xi$, then

$$ \mathcal{W}(k)=\sum_{r=0}^{k-1}\sum _{s=0}^{k-r-1}\frac{1}{k-r} l^{k-r} \beta ^{r} \alpha _{1}^{s} \alpha _{2}^{k-r-s-1} \mathcal{V}(r) \mathcal{U}_{1}(s) \mathcal{U}_{2}(k-r-s-1),\quad k\geq 1. $$

Proof

Let $y(t)=\int _{0}^{lt}u_{1}(\alpha _{1}\xi ) u_{2}(\alpha _{2}\xi ) \,\mathrm{d}\xi$. From Theorem 4.6 we have

$$\frac{d^k}{d{t}^k}w(t)=\frac{d^k}{d{t}^k}(v(\beta t)y(t))=\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\beta }^r\frac{d^r}{d{\overline{t}}^r}v(\overline{t})\frac{d^{k-r}}{d{t}^{k-r}}y(t),$$

where $\bar{t}=\beta t$ and from Theorem 4.5 we have

$$\begin{array}{rl}\frac{d^{k-r}}{d{t}^{k-r}}y(t)& =l\frac{d^{k-r-1}}{d{t}^{k-r-1}}(u_1(l{\alpha }_{1}t)u_2(l{\alpha }_{2}t))\\ & =l\sum _{s=0}^{k-r-1}\left(\begin{array}{c}k-r-1\\ s\end{array}\right){(l{\alpha }_1)}^s\frac{d^s}{d{\hat{t}}^s}{u}_1(\hat{t}){(l{\alpha }_2)}^{k-r-s-1}\frac{d^{k-r-s-1}}{d{\tilde{t}}^{k-r-s-1}}{u}_2(\tilde{t}),\end{array}$$

where $\hat{t}=l\alpha _{1} t$ and $\tilde{t}=l\alpha _{2} t$, then

$$\begin{array}{rl}\mathcal{W}(k)=& \frac{1}{k!}{[\frac{d^k}{d{t}^k}w(t)]}_{t=t_0}\\ =& \sum _{r=0}^k\sum _{s=0}^{k-r-1}\left(\begin{array}{c}k\\ r\end{array}\right)\left(\begin{array}{c}k-r-1\\ s\end{array}\right)l^{k-r}{\beta }^r{\alpha }_1^s{\alpha }_2^{k-r-s-1}\\ & \times r!\mathcal{V}(r)s!{\mathcal{U}}_1(s)(k-r-s-1)!{\mathcal{U}}_2(k-r-s-1),\end{array}$$

therefore

$$ \mathcal{W}(k)=\sum_{r=0}^{k-1}\sum _{s=0}^{k-r-1}\frac{1}{k-r} l^{k-r} \beta ^{r} \alpha _{1}^{s} \alpha _{2}^{k-r-s-1} \mathcal{V}(r) \mathcal{U}_{1}(s) \mathcal{U}_{2}(k-r-s-1). $$

 □

Theorem 4.8

If $w(t)=\int _{0}^{lt} \frac{d^{m}}{dt^{m}} u(\alpha _{1}\xi ) \frac{d^{n}}{dt^{n}} v(\alpha _{2}\xi )\,\mathrm{d}\xi$, then for $k\geq 1$

$$ \mathcal{W}(k)=\frac{1}{k}\sum_{r=0}^{k-1} \frac{(r+m)! (k-r+n-1)!}{r! (k-r-1)!} l^{k+m+n} \alpha _{1}^{r+m} \alpha _{2}^{k-r+n-1} \mathcal{U}(r+m) \mathcal{V}(k-r+n-1). $$

Proof

$$\begin{array}{rl}\mathcal{W}(k)=& \frac{1}{k!}{[\frac{d^k}{d{t}^k}w(t)]}_{t=t_0}\\ =& \frac{1}{k!}{\left[l\frac{d^{k-1}}{d{t}^{k-1}}(\frac{d^m}{d{t}^m}u(l{\alpha }_{1}t)\frac{d^n}{d{t}^n}v(l{\alpha }_{2}t))\right]}_{t=t_0}\\ =& \frac{1}{k!}{[l\sum _{r=0}^{k-1}\left(\begin{array}{c}k-1\\ r\end{array}\right){(l{\alpha }_1)}^{r+m}\frac{d^{r+m}}{d{\hat{t}}^{r+m}}u(\hat{t}){(l{\alpha }_2)}^{k-r+n-1}\frac{d^{k-r+n-1}}{d{\tilde{t}}^{k-r+n-1}}v(\tilde{t})]}_{\hat{t},\tilde{t}=t_0}\\ =& \frac{1}{k!}l\sum _{r=0}^{k-1}\frac{(k-1)!}{r!(k-r-1)!}{(l{\alpha }_1)}^{r+m}(r+m)!\mathcal{U}(r+m)\\ & \times {(l{\alpha }_2)}^{k-r+n-1}(k-r+n-1)!\mathcal{V}(k-r+n-1),\end{array}$$

where $\hat{t}=l\alpha _{1} t$ and $\tilde{t}=l\alpha _{2} t$, then

$$ \mathcal{W}(k)=\frac{1}{k}\sum_{r=0}^{k-1} \frac{(r+m)! (k-r+n-1)!}{r! (k-r-1)!} l^{k+m+n} \alpha _{1}^{r+m} \alpha _{2}^{k-r+n-1} \mathcal{U}(r+m) \mathcal{V}(k-r+n-1). $$

 □

Theorem 4.9

If $w(t)=\frac{d^{\lambda }}{dt^{\lambda }} v(\beta t)\int _{0}^{lt} \frac{d^{m}}{dt^{m}} u_{1}(\alpha _{1}\xi ) \frac{d^{n}}{dt^{n}} u_{2}( \alpha _{2}\xi )\,\mathrm{d}\xi$, then for $k\geq 1$

$$\begin{aligned} \mathcal{W}(k)={}&\sum_{r=0}^{k-1}\sum _{s=0}^{k-r-1}\frac{1}{k-r} \frac{(r+\lambda )! (s+m)! (k-r-s+n-1)!}{r! s! (k-r-s-1)!} \\ &{}\times l^{k-r+m+n} \beta ^{r+\lambda } \alpha _{1}^{s+m} \alpha _{2}^{k-r-s+n-1} \mathcal{V}(r+\lambda ) \mathcal{U}_{1}(s+m) \mathcal{U}_{2}(k-r-s+n-1). \end{aligned}$$

Proof

Let $y(t)=\int _{0}^{lt} \frac{d^{m}}{dt^{m}}u_{1}(\alpha _{1}\xi ) \frac{d^{n}}{dt^{n}} u_{2}(\alpha _{2}\xi )\,\mathrm{d}\xi$. From Theorem 4.8 we have

$$\frac{d^k}{d{t}^k}w(t)=\frac{d^k}{d{t}^k}(\frac{d^{\lambda }v(\beta t)}{d{t}^{\lambda }}y(t))=\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\beta }^{r+\lambda }\frac{d^{r+\lambda }}{d{\overline{t}}^{r+\lambda }}v(\overline{t})\frac{d^{k-r}}{d{t}^{k-r}}y(t),$$

where $\bar{t}=\beta t$ and from Theorem 4.5 we have

$$\begin{array}{rl}\frac{d^{k-r}}{d{t}^{k-r}}y(t)=& l\frac{d^{k-r-1}}{d{t}^{k-r-1}}[\frac{d^m}{d{t}^m}{u}_1(l{\alpha }_{1}t)\frac{d^n}{d{t}^n}{u}_2(l{\alpha }_{2}t)]\\ =& l\sum _{s=0}^{k-r-1}\left(\begin{array}{c}k-r-1\\ s\end{array}\right){(l{\alpha }_1)}^{s+m}\frac{d^{s+m}}{d{\hat{t}}^{s+m}}{u}_1(\hat{t})\\ & \times {(l{\alpha }_2)}^{k-r-s+n-1}\frac{d^{k-r-s+n-1}}{d{\tilde{t}}^{k-r-s+n-1}}{u}_2(\tilde{t}),\end{array}$$

where $\hat{t}=l\alpha _{1} t$ and $\tilde{t}=l\alpha _{2} t$, then

$$\begin{array}{rl}\mathcal{W}(k)=& \frac{1}{k!}{[\frac{d^k}{d{t}^k}w(t)]}_{t=t_0}\\ =& \sum _{r=0}^k\sum _{s=0}^{k-r-1}\left(\begin{array}{c}k\\ r\end{array}\right)\left(\begin{array}{c}k-r-1\\ s\end{array}\right)l^{k-r+m+n}{\beta }^{r+\lambda }{\alpha }_1^{s+m}{\alpha }_2^{k-r-s+n-1}\\ & \times (r+\lambda )!\mathcal{V}(r+\lambda )(s+m)!{\mathcal{U}}_1(s+m)(k-r-s+n-1)!{\mathcal{U}}_2(k-r-s+n-1),\end{array}$$

therefore

$$\begin{aligned} \mathcal{W}(k)={}&\sum_{r=0}^{k-1}\sum _{s=0}^{k-r-1}\frac{1}{k-r} \frac{(r+\lambda )! (s+m)! (k-r-s+n-1)!}{r! s! (k-r-s-1)!} \\ &{}\times l^{k-r+m+n} \beta ^{r+\lambda } \alpha _{1}^{s+m} \alpha _{2}^{k-r-s+n-1} \mathcal{V}(r+\lambda ) \mathcal{U}_{1}(s+m) \mathcal{U}_{2}(k-r-s+n-1). \end{aligned}$$

 □

5 Applications

The principal aim of this section is to apply the modified differential transform method to a class of linear and nonlinear pantograph type of DEs and VIDEs with proportional delays. We present the following examples to illustrate the accuracy of the presented method and compare the new method with the previous result.

Example 5.1

As the first example, we consider the following linear pantograph equation:

$$\begin{aligned} u^{\prime }(t)=-u(t)+\frac{1}{10}u \biggl( \frac{t}{5} \biggr)-\frac{1}{10} e^{- \frac{t}{5}}, \quad 0\leq t \leq 1, \end{aligned}$$

(5.1)

subject to the initial condition

$$\begin{aligned} u(0)=1. \end{aligned}$$

(5.2)

By using Table 1 and Theorem 4.1, Eq. (5.1) transforms to the following recurrence relations:

$$\begin{aligned} (k+1)\mathcal{U}(k+1)=-\mathcal{U}(k)+\frac{1}{10} \biggl(\frac{1}{5} \biggr)^{k} \mathcal{U}(k)- \frac{1}{10}\frac{(\frac{-1}{5})^{k}}{k!}, \end{aligned}$$

(5.3)

and from the initial condition (5.2), we write

$$ \mathcal{U}(0)=1. $$

(5.4)

Substituting Eq. (5.4) in Eq. (5.3) recursively we derive the following results:

$$\begin{aligned} \begin{aligned}&\mathcal{U}(1)=-1,\qquad \mathcal{U}(2)=\frac{1}{2!},\qquad \mathcal{U}(3)= \frac{-1}{3!}, \\ & \mathcal{U}(4)=\frac{1}{4!},\qquad \mathcal{U}(5)= \frac{-1}{5!},\qquad \ldots . \end{aligned} \end{aligned}$$

(5.5)

Therefore, from (3.2) we have

$$\begin{aligned} u(t)=\sum_{k=0}^{\infty } \mathcal{U}(k)t^{k}=1-t+\frac{t^{2}}{2!}- \frac{t^{3}}{3!}+ \frac{t^{4}}{4!}-\frac{t^{5}}{5!}+\cdots . \end{aligned}$$

(5.6)

To improve Eq. (5.1), we implement the (LPDTM) for the third-order approximation solution

$$\begin{aligned} \tilde{u}_{3}(t)=\sum_{k=0}^{3} \mathcal{U}(k)t^{k}=1-t+ \frac{t^{2}}{2!}-\frac{t^{3}}{3!}. \end{aligned}$$

(5.7)

Applying the Laplace transformation with respect to t for $u(t)$ yields

$$ \mathcal{L} \bigl[u(t) \bigr]=\frac{1}{s}-\frac{1}{s^{2}}+ \frac{1}{s^{3}}- \frac{1}{s^{4}}. $$

(5.8)

Let $s=\frac{1}{t}$, we have

$$ \mathcal{L} \bigl[u(t) \bigr]=t-t^{2}+t^{3}-t^{4}. $$

(5.9)

For $m\geq 1$, $n\geq 1$, and $m+n\leq 4$, all of the $[\frac{m}{n} ](t)$-Padé approximants of Eq. (5.9) yield

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{t}{1+t}. $$

(5.10)

In this step, by writing $\frac{1}{s}$ instead of t in Eq. (5.10) we obtain

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{1}{s+1}. $$

(5.11)

Finally, using the inverse Laplace transform on the Padé approximants (5.11), we arrive at the improved solution that corresponds to the exact solution $u(t)=e^{-t}$.

Example 5.2

As the second example, we consider the following linear pantograph equation:

$$\begin{aligned} u^{\prime \prime }(t)=4 e^{-\frac{t}{2}} \sin \biggl( \frac{t}{2} \biggr) u \biggl( \frac{t}{2} \biggr),\quad 0\leq t\leq 1, \end{aligned}$$

(5.12)

subject to the initial conditions

$$\begin{aligned} u(0)=1, \qquad u^{\prime }(0)=-1. \end{aligned}$$

(5.13)

By using Table 1, Theorems 4.1 and 4.2, Eq. (5.12) transforms to the following recurrence relations:

$$\begin{aligned} &(k+1) (k+2)\mathcal{U}(k+2) \\ &\quad =4 \sum _{r_{2}=0}^{k} \sum_{r_{1}=0}^{r_{2}} \biggl(\frac{-1}{2} \biggr)^{r_{1}} \biggl( \frac{1}{2} \biggr)^{k-r_{1}} \frac{1}{r_{1}!(r_{2}-r_{1})!}\sin \biggl({ \frac{(r_{2}-r_{1})\pi }{2}} \biggr) \mathcal{U}(k-r_{2}), \end{aligned}$$

(5.14)

and from the initial conditions (5.13), we write

$$ \mathcal{U}(0)=1,\qquad \mathcal{U}(1)=-1. $$

(5.15)

Consequently, we find

$$\begin{aligned} \mathcal{U}(2)=0,\qquad \mathcal{U}(3)=\frac{1}{3},\qquad \mathcal{U}(4)= \frac{-1}{6}, \qquad \mathcal{U}(5)=\frac{1}{30}, \qquad \ldots . \end{aligned}$$

(5.16)

Therefore, from (3.2) we have

$$\begin{aligned} u(t)=\sum_{k=0}^{\infty } \mathcal{U}(k)t^{k}=1-t+\frac{t^{3}}{3}- \frac{t^{4}}{6}+ \frac{t^{5}}{30}-\cdots . \end{aligned}$$

(5.17)

To improve Eq. (5.12), we implement the (LPDTM) for the third-order approximation solution

$$\begin{aligned} \tilde{u}_{3}(t)=\sum_{k=0}^{3} \mathcal{U}(k)t^{k}=1-t+ \frac{t^{3}}{3}. \end{aligned}$$

(5.18)

Applying the Laplace transformation with respect to t for $u(t)$ yields

$$ \mathcal{L} \bigl[u(t) \bigr]=\frac{1}{s}-\frac{1}{s^{2}}+ \frac{2}{s^{4}}. $$

(5.19)

Let $s=\frac{1}{t}$, we have

$$ \mathcal{L} \bigl[u(t) \bigr]=t-t^{2}+2t^{4}. $$

(5.20)

For $m\geq 1$, $n\geq 1$, and $m+n\leq 4$, all of the $[\frac{m}{n} ](t)$-Padé approximants of Eq. (5.20) yield

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{t+t^{2}}{1+2t+2t^{2}}. $$

(5.21)

In this step, by writing $\frac{1}{s}$ instead of t in Eq. (5.21) we obtain

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{s+1}{s^{2}+2s+2}= \frac{s+1}{(s+1)^{2}+1}. $$

(5.22)

Finally, using the inverse Laplace transform on the Padé approximants (5.22), we arrive at an improved solution that corresponds to the exact solution $u(t)=e^{-t} \cos (t)$.

Example 5.3

As the third example, we consider the following nonlinear pantograph equation:

$$\begin{aligned} u^{\prime \prime }(t)=u(t)-\frac{8}{t^{2}}u^{2} \biggl(\frac{t}{2} \biggr),\quad 0 \leq t\leq 1, \end{aligned}$$

(5.23)

subject to the initial conditions

$$\begin{aligned} u(0)=0, \qquad u^{\prime }(0)=1. \end{aligned}$$

(5.24)

First, we rewrite the equation as follows:

$$\begin{aligned} t^{2} u^{\prime \prime }(t)=t^{2} u(t)-8u^{2} \biggl(\frac{t}{2} \biggr). \end{aligned}$$

(5.25)

By using Table 1 and Theorem 4.2, Eq. (5.15) transforms to the following recurrence relations:

$$\begin{aligned} &\sum_{r=0}^{k} \delta (r-2) (k-r+1) (k-r+2)\mathcal{U}(k-r+2) \\ &\quad =\sum_{r=0}^{k} \delta (r-2) \mathcal{U}(k-r)-8\sum_{r=0}^{k} \biggl( \frac{1}{2} \biggr)^{k}\mathcal{U}(r)\mathcal{U}(k-r), \end{aligned}$$

(5.26)

and from the initial conditions (5.24), we write

$$ \mathcal{U}(0)=0,\qquad \mathcal{U}^{\prime }(0)=1. $$

(5.27)

Substituting Eq. (5.27) in Eq. (5.26) recursively we derive the following results:

$$\begin{aligned} \begin{aligned}&\mathcal{U}(2)=-1,\qquad \mathcal{U}(3)=\frac{1}{2!},\qquad \mathcal{U}(4)= \frac{-1}{3!}, \\ &\mathcal{U}(5)=\frac{1}{4!}, \qquad \mathcal{U}(6)= \frac{-1}{5!}, \qquad \ldots . \end{aligned} \end{aligned}$$

(5.28)

Therefore, from (3.2) we have

$$\begin{aligned} u(t)=\sum_{k=0}^{\infty } \mathcal{U}(k)t^{k}=t-t^{2}+\frac{t^{3}}{2!}- \frac{t^{4}}{3!}+\frac{t^{5}}{4!}-\frac{t^{6}}{5!}+\cdots . \end{aligned}$$

(5.29)

To improve Eq. (5.23), we implement the (LPDTM) for the third-order approximation solution

$$\begin{aligned} \tilde{u}_{3}(t)=\sum_{k=0}^{3} \mathcal{U}(k)t^{k}=t-t^{2}+ \frac{t^{3}}{2!}. \end{aligned}$$

(5.30)

Applying the Laplace transformation with respect to t for $u(t)$ yields

$$ \mathcal{L} \bigl[u(t) \bigr]=\frac{1}{s^{2}}-\frac{2}{s^{3}}+ \frac{3}{s^{4}}. $$

(5.31)

Let $s=\frac{1}{t}$, we have

$$ \mathcal{L} \bigl[u(t) \bigr]=t^{2}-2t^{3}+3t^{4}. $$

(5.32)

For $m\geq 1$, $n\geq 1$, and $m+n\leq 4$, all of the $[\frac{m}{n} ](t)$-Padé approximants of Eq. (5.32) yield

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{t^{2}}{1+2t+t^{2}}. $$

(5.33)

In this step, by writing $\frac{1}{s}$ instead of t in Eq. (5.33) we obtain

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{1}{s^{2}+2s+1}= \frac{1}{(s+1)^{2}}. $$

(5.34)

Finally, using the inverse Laplace transform on the Padé approximants (5.34), we arrive at an improved solution that corresponds to the exact solution $u(t)=te^{-t}$.

Example 5.4

As the fourth example, we consider the following nonlinear pantograph equation:

$$\begin{aligned} u^{\prime \prime \prime } \biggl(\frac{t}{2} \biggr)-2u^{\prime \prime } \biggl(\frac{t}{2} \biggr)u^{ \prime } \biggl(\frac{t}{2} \biggr)+\frac{1}{4}u(t)= \frac{1}{4}\cosh \biggl(\frac{t}{2} \biggr),\quad 0\leq t\leq 1, \end{aligned}$$

(5.35)

subject to the initial conditions

$$\begin{aligned} u(0)=1,\qquad u^{\prime }(0)=2,\qquad u^{\prime \prime }(0)=0. \end{aligned}$$

(5.36)

By using Table 1, Theorems 4.3 and 4.4, Eq. (5.35) transforms to the following recurrence relations:

$$\begin{aligned} & \biggl(\frac{1}{2} \biggr)^{k+3}(k+1) (k+2) (k+3)\mathcal{U}(k+3) \\ &\qquad {}-2\sum_{r=0}^{k} \biggl( \frac{1}{2} \biggr)^{k+3}(r+2) (r+1) (k-r+1)\mathcal{U}(r+2) \mathcal{U}(k-r+1)+\frac{1}{4}\mathcal{U}(k) \\ &\quad =\frac{1}{8} \biggl(\frac{(\frac{1}{2})^{k}+(\frac{-1}{2})^{k}}{k!} \biggr), \end{aligned}$$

(5.37)

and from the initial conditions (5.36), we write

$$ \mathcal{U}(0)=1,\qquad \mathcal{U}(1)=2,\qquad \mathcal{U}(2)=0. $$

(5.38)

Substituting Eq. (5.38) in Eq. (5.37) recursively we derive the following results:

$$\begin{aligned} \mathcal{U}(3)=\frac{1}{3}, \qquad \mathcal{U}(4)=0, \qquad \mathcal{U}(5)= \frac{1}{60},\qquad \mathcal{U}(6)=0,\qquad \ldots . \end{aligned}$$

(5.39)

Therefore, from (3.2) we have

$$\begin{aligned} u(t)=\sum_{k=0}^{\infty } \mathcal{U}(k)t^{k}=2t+\frac{t^{3}}{3}+ \frac{t^{5}}{60}+ \cdots . \end{aligned}$$

(5.40)

To improve Eq. (5.35), we implement the (LPDTM) for the third-order approximation solution

$$\begin{aligned} \tilde{u}_{3}(t)=\sum_{k=0}^{3} \mathcal{U}(k)t^{k}=2t+ \frac{t^{3}}{3}. \end{aligned}$$

(5.41)

Applying the Laplace transformation with respect to t for $u(t)$ yields

$$ \mathcal{L} \bigl[u(t) \bigr]=\frac{2}{s^{2}}+\frac{2}{s^{4}}. $$

(5.42)

Let $s=\frac{1}{t}$, we have

$$ \mathcal{L} \bigl[u(t) \bigr]=2t^{2}+2t^{4}. $$

(5.43)

For $m\geq 1$, $n\geq 1$, and $m+n\leq 6$, all of the $[\frac{m}{n} ](t)$-Padé approximants of Eq. (5.43) yield

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{2t^{2}}{1-t^{2}}. $$

(5.44)

In this step, by writing $\frac{1}{s}$ instead of t in Eq. (5.44) we obtain

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{2}{s^{2}-1}= \frac{1}{s-1}- \frac{1}{s+1}. $$

(5.45)

Finally, using the inverse Laplace transform on the Padé approximants (5.45), we arrive at an improved solution that corresponds to the exact solution $u(t)=e^{t}-e^{-t}$.

Example 5.5

As the fifth example, we consider the following linear VIDEs with proportional delay:

$$\begin{aligned} &u^{\prime \prime }(t)+u \biggl(\frac{t}{2} \biggr)- \frac{3}{4}u(t)- \int _{0}^{t} \xi u(\xi ) \,\mathrm{d}\xi \\ &\quad =- \frac{11}{4} \sin (t)+t\cos (t)+\sin { \biggl( \frac{t}{2} \biggr)},\quad 0\leq t \leq 1, \end{aligned}$$

(5.46)

subject to the initial conditions

$$\begin{aligned} u(0)=0,\qquad u^{\prime }(0)=1. \end{aligned}$$

(5.47)

By using Table 1, Theorems 4.1, 4.5 and 4.6, Eq. (5.46) transforms to the following recurrence relations:

$$\begin{aligned} &(k+1) (k+2)\mathcal{U}(k+2)+ \biggl(\frac{1}{2} \biggr)^{k}\mathcal{U}(k)- \frac{3}{4}\mathcal{U}(k)- \frac{1}{k}\sum_{r=0}^{k-1} \delta (r-1) \mathcal{U}(k-r) \\ &\quad =-\frac{11}{4} \frac{1}{k!}\sin \biggl({\frac{k\pi }{2}} \biggr)+\sum_{r=0}^{k} \delta (r-1) \frac{1}{(k-r)!}\cos { \biggl(\frac{(k-r)\pi }{2} \biggr)} \\ &\qquad {}+ \frac{(\frac{1}{2})^{k}}{k!}\sin \biggl({\frac{k\pi }{2}} \biggr), \end{aligned}$$

(5.48)

and from the initial conditions (5.47), we write

$$ \mathcal{U}(0)=0,\qquad \mathcal{U}(1)=1. $$

(5.49)

Substituting Eq. (5.49) in Eq. (5.48) recursively we derive the following results:

$$\begin{aligned} \mathcal{U}(2)=0,\qquad \mathcal{U}(3)=\frac{-1}{3!},\qquad \mathcal{U}(4)=0,\qquad \mathcal{U}(5)=\frac{1}{5!},\qquad \ldots . \end{aligned}$$

(5.50)

Therefore, from (3.2) we have

$$\begin{aligned} u(t)=\sum_{k=0}^{\infty } \mathcal{U}(k)t^{k}=t-\frac{t^{3}}{3!}+ \frac{t^{5}}{5!}+ \cdots . \end{aligned}$$

(5.51)

To improve Eq. (5.46), we implement the (LPDTM) for the fourth-order approximation solution

$$\begin{aligned} \tilde{u}_{4}(t)=\sum_{k=0}^{4} \mathcal{U}(k)t^{k}=t- \frac{t^{3}}{3!}. \end{aligned}$$

(5.52)

Applying the Laplace transformation with respect to t for $u(t)$ yields

$$ \mathcal{L} \bigl[u(t) \bigr]=\frac{1}{s^{2}}-\frac{1}{s^{4}}. $$

(5.53)

Let $s=\frac{1}{t}$, we have

$$ \mathcal{L} \bigl[u(t) \bigr]=t-t^{4}. $$

(5.54)

For $m\geq 1$, $n\geq 1$, and $m+n\leq 4$, all of the $[\frac{m}{n} ](t)$-Padé approximants of Eq. (5.54) yield

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{t^{2}}{1+t^{2}}. $$

(5.55)

In this step, by writing $\frac{1}{s}$ instead of t in Eq. (5.55) we obtain

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{1}{s^{2}+1}. $$

(5.56)

Finally, using the inverse Laplace transform on the Padé approximants (5.56), we arrive at an improved solution that corresponds to the exact solution $u(t)=\sin (t)$.

Example 5.6

As the sixth example, we consider the following nonlinear VIDEs with proportional delay:

$$\begin{aligned} u^{\prime }(t)u(t)-u \biggl(\frac{t}{2} \biggr)- \frac{3}{2}u \biggl(\frac{t}{2} \biggr) \int _{0}^{t} u(\xi )u \biggl( \frac{\xi }{2} \biggr) \,\mathrm{d}\xi =0,\qquad 0\leq t\leq 1, \end{aligned}$$

(5.57)

subject to the initial condition

$$\begin{aligned} u(0)=1. \end{aligned}$$

(5.58)

By using Table 1 and Theorems 4.1, 4.2 and 4.7, Eq. (5.57) transforms to the following recurrence relations:

$$\begin{aligned} &\sum_{r=0}^{k} (r+1) \mathcal{U}(r+1)\mathcal{U}(k-r)- \biggl(\frac{1}{2} \biggr)^{k} \mathcal{U}(k) \\ &\qquad {}-\frac{3}{2} \sum_{r=0}^{k-1} \sum_{s=0}^{k-r-1} \frac{1}{k-r} \biggl( \frac{1}{2} \biggr)^{k-s-1}\mathcal{U}(r)\mathcal{U}(s) \mathcal{U}(k-r-s-1)=0, \end{aligned}$$

(5.59)

and from the initial condition (5.58), we write

$$ \mathcal{U}(0)=1. $$

(5.60)

Substituting Eq. (5.60) in Eq. (5.59) recursively we derive the following results:

$$\begin{aligned} \begin{aligned}&\mathcal{U}(1)=1,\qquad \mathcal{U}(2)=\frac{1}{2!},\qquad \mathcal{U}(3)= \frac{1}{3!}, \\ & \mathcal{U}(4)=\frac{1}{4!},\qquad \mathcal{U}(5)= \frac{1}{5!},\qquad \ldots . \end{aligned} \end{aligned}$$

(5.61)

Therefore, from (3.2) we have

$$\begin{aligned} u(t)=\sum_{k=0}^{\infty } \mathcal{U}(k)t^{k}=1+t+\frac{t^{2}}{2!}+ \frac{t^{3}}{3!}+ \frac{t^{4}}{4!}+\frac{t^{5}}{5!}+\cdots . \end{aligned}$$

(5.62)

To improve Eq. (5.57), we implement the (LPDTM) for the third-order approximation solution

$$\begin{aligned} \tilde{u}_{3}(t)=\sum_{k=0}^{3} \mathcal{U}(k)t^{k}=1+t+ \frac{t^{2}}{2!}+\frac{t^{3}}{3!}. \end{aligned}$$

(5.63)

Applying the Laplace transformation with respect to t for $u(t)$ yields

$$ \mathcal{L} \bigl[u(t) \bigr]=\frac{1}{s}+\frac{1}{s^{2}}+ \frac{1}{s^{3}}+ \frac{1}{s^{4}}. $$

(5.64)

Let $s=\frac{1}{t}$, we have

$$ \mathcal{L} \bigl[u(t) \bigr]=t+t^{2}+t^{3}+t^{4}. $$

(5.65)

For $m\geq 1$, $n\geq 1$, and $m+n\leq 4$, all of the $[\frac{m}{n} ](t)$-Padé approximants of Eq. (5.65) yield

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{t}{1-t}. $$

(5.66)

In this step, by writing $\frac{1}{s}$ instead of t in Eq. (5.66) we obtain

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{1}{s-1}. $$

(5.67)

Finally, using the inverse Laplace transform on the Padé approximants (5.67), we arrive at an improved solution that corresponds to the exact solution $u(t)=e^{t}$.

Example 5.7

Lastly, we consider the following nonlinear VIDEs with proportional delay:

$$\begin{aligned} \frac{1}{4}u^{\prime } \biggl( \frac{t}{2} \biggr)+\frac{1}{8} u(t)u \biggl( \frac{t}{2} \biggr)-u^{ \prime \prime } \biggl(\frac{t}{2} \biggr) \int _{0}^{\frac{t}{2}} u(\xi )u^{\prime }( \xi ) \,\mathrm{d}\xi =0,\quad 0\leq t\leq 1, \end{aligned}$$

(5.68)

subject to the initial condition

$$\begin{aligned} u(0)=1,\qquad u^{\prime }(0)=-1. \end{aligned}$$

(5.69)

By using Table 1, Theorems 4.2, 4.3, 4.8 and 4.9 Eq. (5.68) transforms to the following recurrence relations:

$$\begin{aligned} & \biggl(\frac{1}{2} \biggr)^{k+3}(k+1) \mathcal{U}(k+1)+\sum_{r=0}^{k} \biggl( \frac{1}{2} \biggr)^{k-r+3} \mathcal{U}(r)\mathcal{U}(k-r) \\ &\quad {}-\sum_{r=0}^{k-1}\sum _{s=0}^{k-r-1} \frac{(r+2)! (k-r-s)!}{(k-r)! r!} \biggl( \frac{1}{2} \biggr)^{k+2}\mathcal{U}(r+2) \mathcal{U}(s) \mathcal{U}(k-r-s)=0, \end{aligned}$$

(5.70)

and from the initial conditions (5.69), we write

$$ \mathcal{U}(0)=1,\qquad \mathcal{U}(1)=-1. $$

(5.71)

Substituting Eq. (5.71) in Eq. (5.70) recursively we derive the following results:

$$\begin{aligned} \mathcal{U}(2)=\frac{1}{2!},\qquad \mathcal{U}(3)=\frac{-1}{3!},\qquad \mathcal{U}(4)=\frac{1}{4!},\qquad \mathcal{U}(5)=\frac{-1}{5!},\qquad \ldots . \end{aligned}$$

(5.72)

Therefore, from (3.2) we have

$$\begin{aligned} u(t)=\sum_{k=0}^{\infty } \mathcal{U}(k)t^{k}=1-t+\frac{t^{2}}{2!}- \frac{t^{3}}{3!}+ \frac{t^{4}}{4!}-\frac{t^{5}}{5!}+\cdots . \end{aligned}$$

(5.73)

To improve Eq. (5.68), we implement the (LPDTM) for the third-order approximation solution

$$\begin{aligned} \tilde{u}_{3}(t)=\sum_{k=0}^{3} \mathcal{U}(k)t^{k}=1-t+ \frac{t^{2}}{2!}-\frac{t^{3}}{3!}. \end{aligned}$$

(5.74)

Applying the Laplace transformation with respect to t for $u(t)$ yields

$$ \mathcal{L} \bigl[u(t) \bigr]=\frac{1}{s}-\frac{1}{s^{2}}+ \frac{1}{s^{3}}- \frac{1}{s^{4}}. $$

(5.75)

Let $s=\frac{1}{t}$, we have

$$ \mathcal{L} \bigl[u(t) \bigr]=t-t^{2}+t^{3}-t^{4}. $$

(5.76)

For $m\geq 1$, $n\geq 1$, and $m+n\leq 4$, all of the $[\frac{m}{n} ](t)$-Padé approximants of Eq. (5.76) yield

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{t}{1+t}. $$

(5.77)

In this step, by writing $\frac{1}{s}$ instead of t in Eq. (5.77) we obtain

$$ \biggl[\frac{m}{n} \biggr](t)=\frac{1}{s+1}. $$

(5.78)

Finally, using the inverse Laplace transform on the Padé approximants (5.78), we arrive at an improved solution that corresponds to the exact solution $u(t)=e^{-t}$.

Comparing our result with the solutions obtained in [13, 14, 17, 58, 59], we can see that the results are the same.

Results for Examples 5.1–5.7 are reported in Figs. 1–7 and Tables 2–8, respectively. In these tables, the terms $u_{E}$, $u_{n,R}$ and $e(u)$ stand for exact solution, nth order approximate solution of DTM and their absolute error, respectively.

Figure 1

Comparison of the exact solution (blue) and the approximate solutions (red) of Example 5.1

Figure 2

Comparison of the exact solution (blue) and the approximate solutions (red) of Example 5.2

Figure 3

Comparison of the exact solution (blue) and the approximate solutions (red) of Example 5.3

Figure 4

Comparison of the exact solution (blue) and the approximate solutions (red) of Example 5.4

Figure 5

Comparison of the exact solution (blue) and the approximate solutions (red) of Example 5.5

Figure 6

Comparison of the exact solution (blue) and the approximate solutions (red) of Example 5.6

Figure 7

Comparison of the exact solution (blue) and the approximate solutions (red) of Example 5.7

Table 2 Numerical result of Example 5.1

Table 3 Numerical result of Example 5.2

Table 4 Numerical result of Example 5.3

Table 5 Numerical result of Example 5.4

Table 6 Numerical result of Example 5.5

Table 7 Numerical result of Example 5.6

Table 8 Numerical result of Example 5.7

6 Conclusion

Most pantograph equations with proportional delays are usually difficult to solve analytically. In many cases, it is required to obtain approximate solutions. In this work, for this purpose, the modified differential transform method, a combined form of the differential transform method with Laplace transforms, and the Padé approximant (LPDTM) is effectively used to find the exact solution of linear and nonlinear pantograph type of differential and Volterra integro-differential equations (DEs and VIDEs) with proportional delays. In fact, the main advantage of this method is its capability of combining the two strongest methods for finding a fast convergent series solution of pantograph equations. Furthermore, as seen from the examples, the results indicate the reliability and efficiency of the method and show that it needs less effort to achieve the results, and is a promising and powerful method over other methods for solving many linear and nonlinear pantograph type of DEs and VIDEs with proportional delays arising in mathematical physics. The method can be extended easily with some modifications to the fractional delay differential equations. It is our aim for future work.