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Introduction
1D telegraph equation is used for signal investigation at transmission and proliferation of electrical signs. The telegraph equation also shows the mix of dissemination and wave proliferation by the properties of constrained velocity to standard warmth or mass transport condition. 1D telegraph equation is in the following form [1] for x ∈ [a,b],t ≥ 0:
{u_{tt}} + 2\alpha {u_t} + {\beta ^2}u = {u_{xx}} + f(x,t),
with initial conditions
u(x,t) = {g_1}(x),{u_t}(x,y) = {g_2}(x),
and boundary conditions
u(a,t) = {\gamma _1},u(b,t) = {\gamma _2},x \in [a,b],
where α,beta,a and b are constant and g1,g2,γ1,γ2 and f are known functions.
Several numerical methods were developed for solving 1D telegraph equation, for example, reduced differential transform method, differential quadrature method, new unconditionally stable difference schemes, Chebyshev tau method, dual reciprocity boundary integral equation, cubic B-spline collocation method, modified B-spline differential quadrature method, semi-discretization method, unconditionally stable ADI method, Rothe-wavelet method, collocation method, He’s variational iteration method, dual reciprocity boundary integral equation method, etc. [1, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]. The preferred hybrid method is used for solving many types of differential equations. These are differential-algebraic equations [21], partial differential equations [4, 6, 7, 8, 22, 23], and fractional differential equations [20]. Also, the other differential equation types can be solved by this hybrid method [25, 26, 27, 28, 29, 30, 31, 32].
The hybrid method is given as a blend of the differential transform and the finite difference methods. The differential transform method was proposed by Zhou for the solution of linear and nonlinear differential equations for electrical circuit analysis [24]. This method was developed by Chen and Ho for partial differential equations [5]. Ayaz applied differential transform method for the system of differential equations [3]. According to these studies, the transformation of the k-th derivative of u(x,t) based on t-time variable is a follows:
U(i,k) = \frac{1}{{k!}}{\left[ {\frac{{{d^k}}}{{d{t^k}}}u(x,t)} \right]_{t = 0}},k = 0,1,2, \ldots ,\;\;i = 0,1,2, \ldots ,
the inverse transformation of U(i,k) is identified by
u(x,t) = \sum\limits_{k = 0}^\infty U(i,k){t^k} = U(i,0) + U(i,1)t + U(i,2){t^2} + \cdots ,\;\;i = 0,1,2,3, \ldots ,
where U(i,k) = U(xi,k), xi = ih, i = 0,1,2,3,... and h is the finite difference step interval.
The theorems that can be deduced from Eqs (4) and (5) are as follows.
Let G be the differential transform of g, respectively.
The central difference derivative is given by the finite difference method, see Amirali and Amirali [2],
\frac{{{\partial ^2}u}}{{\partial {x^2}}} = \frac{{U(i + 1,k) - 2U(i,k) + U(i - 1,k)}}{{{h^2}}}
The truncation error constituted by the finite difference method is bigger than that constituted by the differential transform method. Hence, the aim of this study gives a hybrid method which is a mixture of these methods. Using the hybrid method, the approximate solution of the telegraph equation as different from the other methods is obtained from a very powerful iterative scheme.
Numerical Experiments of the Telegraph Equation with Hybrid Method
Here, we give three experiments of both linear and nonlinear 1D telegraph equations to approve the power and reliability of the hybrid method.
Example 1
We will consider the following linear telegraph equation with initial and boundary conditions [18]:
{u_{tt}} + 2{u_t} + 2u = {u_{xx}} + x{e^{ - t}},\;\;x \in [0,1],u(x,0) = x,{u_t}(x,0) = - x,u(0,t) = 0,\;u(1,t) = {e^{ - t}},
where f (x,t) = xe−t.
The exact solution to our test problem (6)–(8) is u(x,t) = xe−t.
Applying the hybrid method for Eqs (6)–(8), first, we have differential transforms of terms dependent on t-time variable in the telegraph Eq. (6), and then the central difference of derivative terms dependent on the x-position variable is obtained. The x-position variable is replaced with mesh positions in other x-terms in the telegraph equation.
\begin{array}{*{20}{c}}{\frac{{{\partial ^2}u}}{{\partial {t^2}}} \to (k + 1)(k + 2)U(i,k + 2),}\\{\frac{{\partial u}}{{\partial t}} \to (k + 1)U(i,k + 1),}\\{u(x,t) \to U(i,k),}\\{\frac{{{\partial ^2}u}}{{\partial {x^2}}} \to \frac{{U(i + 1,k) - 2U(i,k) + U(i - 1,k)}}{{{h^2}}},}\\{x{e^{ - t}} \to {x_i}\frac{{{{( - 1)}^k}}}{{k!}},}\\{u(x,0) \to U(i,0) = {x_i},}\\{{u_t}(x,0) \to U(i,0) = - {x_i},}\\{u(0,t) \to U(0,k) = 0,}\\{u(1,t) \to U(1,k) = \frac{{{{( - 1)}^k}}}{{k!}},}\\{{x_i} = ih,h = 0.1,i = 0,1,2, \ldots ,k = 0,1,2, \ldots .}\end{array}
If these differential transforms and the central difference equivalents are written in Eq. (6), we have recurrence correlation and the approximate solution is reached on mesh points xi as
\begin{array}{l}U(i,k + 2) = \frac{1}{{(k + 1)(k + 2)}}\left[ {\frac{{U(i + 1,k) - 2U(i,k) + U(i - 1,k)}}{{{h^2}}} - 2(k + 1)U(i,k + 1) - 2U(i,k) + {x_i}\frac{{{{( - 1)}^k}}}{{k!}}} \right],\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{{x_i}}&{ = 0,u(0,t) = \sum\limits_{k = 0}^\infty U(0,k){t^k} = U(0,0) + U(0,1)t + U(0,2){t^2} + \cdots }\\{}&{ = 0 + t + 0{t^2} + \cdots + 0{t^{10}} + \cdots }\\{{x_i}}&{ = 0.1,u(0.1,t) = \sum\limits_{k = 0}^\infty U(0.1,k){t^k} = U(0.1,0) + U(0.1,1)t + \cdots }\\{}&{ = 0.1 - 0.1t + 0.05{t^2} + \cdots + 0.2755731922 \cdot {{10}^{ - 7}}{t^{10}}}\\{{x_i}}&{ = 1,u(1,t) = \sum\limits_{k = 0}^\infty U(1,k){t^k} = U(1,0) + U(1,1)t + U(1,2){t^2} + \cdots }\\{}&{ = 1 - 1t + 0.5{t^2} + \cdots + 0.2755731922 \cdot {{10}^{ - 6}}{t^{10}} + \cdots .}\end{array}\end{array}
After we find these approximate solutions for x and t = 0.01, we give Table 1. Furthermore, in Figure 1 as 2D and 3D, we compare the results of exact and numerical solution of Example 1 for t = 0.01 and x = 0.1.
Fig. 1
Comparison of exact and approximate solutions for Example 1.
Exact solution, approximate solution, and error values for t = 0.01
x
Exact solution
Approximate solution
Error
0.0
0.0000000000
0.0000000000
0
0.1
0.0990049833
0.0990049833
0
0.2
0.1980099668
0.1980099667
0.1×10−9
0.3
0.2970149501
0.2970149501
0
0.4
0.3960199335
0.3960199335
0
0.5
0.4950249169
0.4950249168
0.1×10−9
0.6
0.5940299002
0.5940299002
0
0.7
0.6930348836
0.6930348836
0
0.8
0.7920398670
0.7920398670
0
0.9
0.8910448504
0.8910448503
0.1×10−9
1.0
0.9900498337
0.9900498337
0
Example 2
Let us consider the following linear telegraph equation [18]:
{u_{tt}} + {u_t} - u = {u_{xx}},\;\;x \in [0,1],u(x,0) = \sin x,{u_t}(x,0) = - \sin x,t \ge 0,u(0,t) = 0,u(1,t) = \sin (1){e^{ - t}},
where f (x,t) = 0.
Eqs (9) and (10) have the following exact solution:
u(x,t) = {e^{ - t}}\sin x.
If Eq. (9) is separated by hybrid method, we obtain differential transforms of terms dependent on t-time variable. Then, the central difference of derivative terms dependent on the x-position variable is obtained. The other x-position variables in the problem are replaced with x:
\begin{array}{*{20}{c}}{\frac{{\partial u}}{{\partial t}} \to U(i,k) = (k + 1)U(i,k + 1),}\\{\frac{{{\partial ^2}u}}{{\partial {x^2}}} \to \frac{{U(i + 1,k) - 2U(i,k) + U(i - 1,k)}}{{{h^2}}},}\\{u(x,t) \to U(i,k),}\\{u(x,0) \to U(i,0) = \sin {x_i},}\\{{u_t}(x,0) \to U(i,0) = - \sin {x_i},}\\{u(0,t) \to U(0,k) = 0,}\\{u(1,t) \to U(1,k) = \sin (1)\frac{{{{( - 1)}^k}}}{{k!}},}\\{{x_i} = ih,h = 0.1,i = 0,1,2, \ldots ,k = 0,1,2, \ldots .}\end{array}
If these differential transforms and the central difference equivalents are written in Eq. (9), then the following recurrence relation is obtained:
U(i,k + 2) = \frac{1}{{(k + 1)(k + 2)}}\left[ {\frac{{U(i + 1,k) - 2U(i,k) + U(i - 1,k)}}{{{h^2}}} - (k + 1)U(i,k + 1) + U(i,k)} \right],
Now, we find approximate solutions on xi mesh points for t = 0.01, Numerical values are demonstrated in Table 2. To compare the results of exact and approximate solution of Example 2, Figure 2 is plotted.
\begin{array}{*{20}{l}}
{{x_i}}&{ = 0,u(0,t) = \sum\limits_{k = 0}^\infty U(0,k){t^k} = U(0,0) + U(0,1)t + U(0,2){t^2} + \cdots }\\
{}&{ = 0 + t + 0{t^2} + \cdots + 0{t^{10}} + \cdots }\\
{{x_i}}&{ = 0.1,u(0.1,t) = \sum\limits_{k = 0}^\infty U(0.1,k){t^k} = U(0.1,0) + U(0.1,1)t + \cdots }\\
{}&{ = 0.09983341665 - 0.09983341665t + 0.09933466549{t^2} + \cdots + 0.2751141332 \cdot {{10}^{ - 7}}{t^{10}} + \cdots .}\\
{{x_i}}&{ = 1,u(1,t) = \sum\limits_{k = 0}^\infty U(1,k){t^k} = U(1,0) + U(1,1)t + U(1,2){t^2} + \cdots }\\
{}&{ = 0.8414709848 - 0.8414709848t + 0.4207354924{t^2} + \cdots + 0.2318868455 \cdot {{10}^{ - 6}}{t^{10}} + \cdots .}
\end{array}
In the form of 2D and 3D, the results of exact and approximate solutions of Example 2 are compared for t = 0.01 and x = 0.1 as in Figure 2.
Fig. 2
Comparison of exact and approximate solutions for Example 2.
Exact solution, approximate solution, and error values for t = 0.01
x
Exact solution
Approximate solution
Error
0.0
0.0000000000
0.0000000000
0
0.1
0.9884005755
0.9884005755
0
0.2
0.1966925379
0.1980099667
0.1×10−9
0.3
0.2925797315
0.2925797314
0.1×10−9
0.4
0.3855435650
0.3855435651
0.1×10−9
0.5
0.4746551748
0.4746551748
0
0.6
0.5590241869
0.5590241869
0
0.7
0.6378076141
0.6378076141
0
0.8
0.7102182785
0.7102182785
0
0.9
0.7755326766
0.7755326765
0.1×10−9
1.0
0.8330982086
0.8330982087
0.1×10−9
Example 3
Considering the following nonlinear telegraph equation [19]:
{u_{tt}} + 2{u_t} = {u_{xx}} - {u^2} + {e^{2x + 4t}} - {e^{x - 2t}},x \in [0,1],
which has the following initial conditions, boundary conditions, and exact solution, respectively:
u(x,0) = {e^x},{u_t}(x,0) = - 2{e^x},t \ge 0,\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{u(0,t) = {e^{ - 2t}},u(1,t) = {e^{1 - 2t}},x \in [0,1],}\\
{u(x,t) = {e^{x - 2t}},}
\end{array}}
\end{array}
where f (x,t) = e2x+4t − ex−2t.
If these differential transforms and the central difference equivalents are written in the Eq. (12), then the following recurrence relation is obtained:
\begin{array}{l}
U(i,k + 2) = \frac{1}{{(k + 1)(k + 2)}}\\
\,\,\,\,\,\left[ {\frac{{U(i + 1,k) - 2U(i,k) + U(i - 1,k)}}{{{h^2}}} - 2(k + 1)U(i,k + 1) - {{(U(i,k))}^2} + {e^{2{x_i}}}\left( {\frac{{{4^k}}}{{k!}}} \right) - {e^{{x_i}}}\left( {\frac{{{{( - 2)}^k}}}{{k!}}} \right)\,} \right]
\end{array}
and then we have the approximate solutions on xi mesh points for t = 0.01 as:
\begin{array}{*{20}{l}}
{{x_i}}&{ = 0,u(0,t) = \sum\limits_{k = 0}^\infty U(0,k){t^k} = U(0,0) + U(0,1)t + U(0,2){t^2} + \cdots }\\
{}&{ = 1 - 2t + 2t + \cdots + 0.0002821869489{t^{10}} + \cdots }\\
{{x_i}}&{ = 0.1,u(0.1,t) = \sum\limits_{k = 0}^\infty U(0.1,k){t^k} = U(0.1,0) + U(0.1,1)t + \cdots }\\
{}&{ = 1.105170918 - 2.210341836t + 2.210341836{t^2} + \cdots + 0.0003118648093{t^{10}} + \cdots }\\
{{x_i}}&{ = 1,u(1,t) = \sum\limits_{k = 0}^\infty U(1,k){t^k} = U(1,0) + U(1,1)t + U(1,2){t^2} + \cdots }\\
{}&{ = 2.718281828 - 5.436563656t + 5.436563655{t^2} + \cdots + 0.0007670636553{t^{10}} + \cdots .}
\end{array}
The obtained data of Example 3 are given in Table 3, and the plot of corresponding exact and approximate solutions are demonstrated in Figure 3 for t = 0.01 and x = 0.1.
Briefly, throughout the study, each figure (Figures 1–3) consists of two images. The first picture is two dimensional and the second picture is three dimensional. They compare exact and approximate solutions for t = 0.01 and x = 0.1.
Fig. 3
Comparison of exact and approximate solutions for Example 3.
Exact solution, approximate solution, and error values for t = 0.01
x
Exact solution
Approximate solution
Error
0.0
0.980198673
0.980198673
0
0.1
1.083287067
1.083287068
0.1×10−8
0.2
1.197217363
1.197217363
0
0.3
1.323129812
1.323129813
0.1×10−8
0.4
1.462284589
1.462284590
0.1×10−8
0.5
1.616074402
1.616074403
0.1×10−8
0.6
1.786038431
1.786038431
0
0.7
1.973877732
1.973877732
0
0.8
2.181472265
2.181472265
0
0.9
2.410899706
2.410899707
0.1×10−8
1.0
2.664456242
2.664456241
0.1×10−8
Conclusion
A robust hybrid numerical method was presented for the numerical solution of the linear and nonlinear telegraph equations. The hybrid method was introduced. Three test problems were solved by using this method. The exact solution, approximate solution, and error values were computed for t = 0.01 and different x values. The results were demonstrated in Tables 1–3. 2D and 3D forms of graphics were plotted in Figures 1–3. It was seen that the curves of the exact and approximate solutions are almost identical. As a result, the numerical results proved that the proposed method was working very well. Therefore, we propose that the hybrid method can be used for the other partial differential equations as an alternative method.
