Φ2(**x**, ¯*t*) = S *ξ δ* S − *∂***x** − 2 *∂***x** = S −1 " − *σ* 5 4 *ω*2*δ*+2 *ξ* 2*δ*+2 sec *h* 2 ` 2 + *σ***x** 2 + 3*σ* 5 4 *ω*2*δ*+<sup>2</sup> *ξ* 2*δ*+2 sin *h* 2 ` 2 + *σ***x** 2 sec *h* 4 ` 2 + *σ***x** 2 # + *σ* 7 4 S −1 " Γ(2*δ* + 1) Γ <sup>2</sup>(*δ* + 1) *ω*3*δ*+<sup>2</sup> *ξ* 3*δ*+2 sin *h* ` 2 + *σ***x** 2 sec *h* 5 ` 2 + *σ***x** 2 # = − *σ* 5 4 sec *h* 2 ` 2 + *σ***x** 2 + 3*σ* 5 4 sin *h* 2 ` 2 + *σ***x** 2 sec *h* 4 ` 2 + *σ***x** 2 ¯*t* 2*δ* Γ(2*δ* + 1) + *σ* 7 4 sin *h* ` 2 + *σ***x** 2 sec *h* 5 ` 2 + *σ***x** 2 Γ(2*δ* + 1)¯*t* 3*δ* Γ 2(*δ* + 1)Γ(3*δ* + 1) , Ψ2(**x**, ¯*t*) = S −1 " *ωδ ξ δ* − *∂*Φ1 *∂***x** − *∂* 3Φ1 *∂***x** 3 − *∂*Φ1Ψ1 *∂***x** # = *σ* 6 4 h 2 cosh2 ` 2 + *σ***x** 2 − 3 i sec *h* 4 ` 2 + *σ***x** 2 ¯*t* 2*δ* Γ(2*δ* + 1) , . . . Φ*m*(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − *∂*Ψ*m*−1 *∂***x** − 1 2 *∂*Φ2 *m*−1 *∂***x** # , Ψ*m*(**x**, ¯*t*) = S −1 " *ωδ ξ δ* − *∂*Φ*m*−1 *∂***x** − *∂* 3Φ*m*−1 *∂***x** 3 − *∂*Φ*m*−1Ψ*m*−1 *∂***x** # .

The series of solution for Problem 2 is presented as:

$$\begin{aligned} \Phi(\mathbf{x},\mathbf{\bar{t}}) &= \Phi\_0(\mathbf{x},\mathbf{\bar{t}}) + \Phi\_1(\mathbf{x},\mathbf{\bar{t}}) + \Phi\_2(\mathbf{x},\mathbf{\bar{t}}) + \dots \Phi\_m(\mathbf{x},\mathbf{\bar{t}}),\\ \Psi(\mathbf{x},\mathbf{\bar{t}}) &= \Psi\_0(\mathbf{x},\mathbf{\bar{t}}) + \Psi\_1(\mathbf{x},\mathbf{\bar{t}}) + \Psi\_2(\mathbf{x},\mathbf{\bar{t}}) + \dots \Psi\_m(\mathbf{x},\mathbf{\bar{t}}).\end{aligned}$$

Consequently, we have

Φ(**x**, ¯*t*) = *σ* tan *h* ` 2 + *σ***x** 2 + 1 − *σ* 2 2 sec *h* 2 ` 2 + *σ***x** 2 ¯*t δ* Γ(*δ* + 1) + − *σ* 5 4 sec *h* 2 ` 2 + *σ***x** 2 + 3*σ* 5 4 sin *h* 2 ` 2 + *σ***x** 2 sec *h* 4 ` 2 + *σ***x** 2 ¯*t* 2*δ* Γ(2*δ* + 1) + *σ* 7 4 sin *h* ` 2 + *σ***x** 2 sec *h* 5 ` 2 + *σ***x** 2 Γ(2*δ* + 1)¯*t* 3*δ* Γ <sup>2</sup>(*δ* + 1)Γ(3*δ* + 1) + ... , Ψ(**x**, ¯*t*) = <sup>−</sup><sup>1</sup> <sup>+</sup> *σ* 2 2 sec *h* 2 ` 2 + *σ***x** 2 + *σ* 3 2 sin *h* ` 2 + *σ***x** 2 sec *h* 3 ` 2 + *σ***x** 2 ¯*t δ* Γ(*δ* + 1) + *σ* 6 4 h 2 cosh2 ` 2 + *σ***x** 2 − 3 i sec *h* 4 ` 2 + *σ***x** 2 ¯*t* 2*δ* Γ(2*δ* + 1) + ... .

By setting *δ* = 1, we then obtain the exact solution of coupled KdV Equation (34)

$$\begin{aligned} \Phi(\mathbf{x},\mathbf{f}) &= \sigma \left( \tanh\left(\frac{\ell}{2} + \frac{\sigma \mathbf{x}}{2} - \frac{\sigma^2 \mathbf{f}}{2} \right) + 1 \right), \\ \Psi(\mathbf{x},\mathbf{f}) &= \frac{\sigma^2}{2} \sec h^2 \left(\frac{\ell}{2} + \frac{\sigma \mathbf{x}}{2} - \frac{\sigma^2 \mathbf{f}}{2} \right) - 1. \end{aligned}$$

In Figures 5 and 6, the exact and approximate results of Φ(**x**, ¯*t*) and Ψ(**x**, ¯*t*) are demonstrated at ` = 1, *σ* = 0.5 and *β* = 2. In Figures 7 and 8, the surface and 2D graph for Φ(**x**, ¯*t*) and Ψ(**x**, ¯*t*) for various fractional order are presented which shows that the SDM/SITM approximated results derived are in a strong agreement with the exact and the numerical ones. This comparison represents a strong correlation between the SDM and exact findings. Therefore, the SDM/SITM are reliable novel approaches which require less computation time and are quite straightforward and more flexible than the homotopy perturbation method and the homotopy analysis method.

**Figure 5.** The exact and approximate (SDM/SITM) solution graph at Φ(**x**, ¯*t*) of Problem 2 for ` = 1, *σ* = 0.5 and *β* = 2.

**Figure 6.** The exact and approximate (SDM/SITM) solution graph at Ψ(**x**, ¯*t*) of Problem 2 for ` = 1, *σ* = 0.5 and *β* = 2.

**Figure 7.** Numerical evaluation of graph of Ψ(**x**, ¯*t*) for Problem 2 for various fractional order *δ* = 0.4, 0.6, 0.8, 1, ` = 1, *σ* = 0.5 and *β* = 2.

**Figure 8.** Numerical evaluation of graph of Ψ(**x**, ¯*t*) for Problem 2 for various fractional order *δ* = 0.4, 0.6, 0.8, 1, ` = 1, *σ* = 0.5 and *β* = 2.

**Problem 3** ([16])**.** *Assume the time-fractional coupled nonlinear MCKdV equations is presented as* (2) *subject to the condition*

$$\Phi(\mathbf{x},0) = \frac{2 + \tanh \mathbf{x}}{2}, \quad \Psi(\mathbf{x},0) = \frac{2 - \tanh \mathbf{x}}{4}, \quad \Psi(\mathbf{x},0) = 2 - \tanh \mathbf{x}. \tag{39}$$

**Case I.** First, we surmise the Shehu decomposition method for Problem 3. Employing the Shehu transformation to (2), we find

*ξ δ ωδ* U(*ξ*, *ω*) − *m*−1 ∑ *κ*=0 *ξ ω δ*−*κ*−1 Φ (*κ*) (0) = S 1 2 *∂* 3Φ *∂*¯*t* 3 − 3Φ 2 *∂*Φ *∂***x** + 3 2 Υ *∂* 2Ψ *∂***x** 2 + 3 *∂*Ψ *∂***x** *∂*Υ *∂***x** + 3 2 Ψ *∂* 2Υ *∂***x** 2 +3ΨΥ*∂*Φ *∂***x** + 3ΦΥ*∂*Ψ *∂***x** + 3ΦΨ*∂*Υ *∂***x** , *ξ δ ωδ* V(*ξ*, *ω*) − *m*−1 ∑ *κ*=0 *ξ ω δ*−*κ*−1 Ψ (*κ*) (0) = S − *∂* 3Ψ *∂***x** 3 − 3 *∂*Φ *∂***x** *∂*Ψ *∂***x** − 3Ψ *∂* 2Φ *∂***x** 2 − 3Ψ 2 *∂*Υ *∂***x** + 6ΦΨ*∂*Φ *∂***x** + 3Φ 2 *∂*Ψ *∂***x** , *ξ δ ωδ* W(*ξ*, *ω*) − *m*−1 ∑ *κ*=0 *ξ ω δ*−*κ*−1 Υ (*κ*) (0) = S − *∂* 3Υ *∂***x** 3 − 3 *∂*Φ *∂***x** *∂*Υ *∂***x** − 3Υ *∂* 2Φ *∂***x** 2 − 3Υ 2 *∂*Ψ *∂***x** + 6ΦΥ*∂*Φ *∂***x** + 3Φ 2 *∂*Υ *∂***x** . (40)

#### In view of (39) and simple computations yield

U(*ξ*, *ω*) = *ω ξ* Φ (0) (**x**, 0) + *ωδ ξ δ* S 1 2 *∂* 3Φ *∂*¯*t* 3 − 3Φ 2 *∂*Φ *∂***x** + 3 2 Υ *∂* 2Ψ *∂***x** 2 + 3 *∂*Ψ *∂***x** *∂*Υ *∂***x** + 3 2 Ψ *∂* 2Υ *∂***x** 2 +3ΨΥ*∂*Φ *∂***x** + 3ΦΥ*∂*Ψ *∂***x** + 3ΦΨ *∂*Υ *∂***x** , V(*ξ*, *ω*) = *ω ξ* Ψ (0) (**x**, 0) + *ωδ ξ δ* S − *∂* 3Ψ *∂***x** 3 − 3 *∂*Φ *∂***x** *∂*Ψ *∂***x** − 3Ψ *∂* 2Φ *∂***x** 2 − 3Ψ 2 *∂*Υ *∂***x** + 6ΦΨ *∂*Φ *∂***x** + 3Φ 2 *∂*Ψ *∂***x** , W(*ξ*, *ω*) = *ω ξ* Υ (0) (**x**, 0) + *ωδ ξ δ* S − *∂* 3Υ *∂***x** 3 − 3 *∂*Φ *∂***x** *∂*Υ *∂***x** − 3Υ *∂* 2Φ *∂***x** 2 − 3Υ 2 *∂*Ψ *∂***x** + 6ΦΥ*∂*Φ *∂***x** + 3Φ 2 *∂*Υ *∂***x** . (41)

Applying the inverse Shehu transform, we have

Φ(**x**, ¯*t*) = S −1 h *ω ξ* Φ(**x**, 0) i + S −1 " *ωδ ξ δ* S 1 2 *∂* 3Φ *∂*¯*t* 3 − 3Φ 2 *∂*Φ *∂***x** + 3 2 Υ *∂* 2Ψ *∂***x** 2 + 3 *∂*Ψ *∂***x** *∂*Υ *∂***x** + 3 2 Ψ *∂* 2Υ *∂***x** 2 +3ΨΥ*∂*Φ *∂***x** + 3ΦΥ*∂*Ψ *∂***x** + 3ΦΨ *∂*Υ *∂***x** # , Ψ(**x**, ¯*t*) = S −1 h *ω ξ* Ψ(**x**, 0) i + S −1 " *ωδ ξ δ* S − *∂* 3Ψ *∂***x** 3 − 3 *∂*Φ *∂***x** *∂*Ψ *∂***x** − 3Ψ *∂* 2Φ *∂***x** 2 − 3Ψ 2 *∂*Υ *∂***x** + 6ΦΨ *∂*Φ *∂***x** + 3Φ 2 *∂*Ψ *∂***x** # , Υ(**x**, ¯*t*) = S −1 h *ω ξ* Υ(**x**, 0) i + S −1 " *ωδ ξ δ* S − *∂* 3Υ *∂***x** 3 − 3 *∂*Φ *∂***x** *∂*Υ *∂***x** − 3Υ *∂* 2Φ *∂***x** 2 − 3Υ 2 *∂*Ψ *∂***x** + 6ΦΥ*∂*Φ *∂***x** + 3Φ 2 *∂*Υ *∂***x** # . (42)

#### By virtue of the Shehu decomposition method, we have

$$\begin{aligned} \Phi\_0(\mathbf{x}, \mathbb{F}) &= \mathbb{S}^{-1} \left[ \frac{\omega}{\tilde{\xi}} \Phi(\mathbf{x}, 0) \right] = \frac{1}{2} \mathbb{S}^{-1} \left[ \frac{\omega}{\tilde{\xi}} \left( 2 + \tanh \mathbf{x} \right) \right] \\ &= \frac{1}{2} \left( 2 + \tanh \mathbf{x} \right), \\ \Psi\_0(\mathbf{x}, \mathbb{F}) &= \mathbb{S}^{-1} \left[ \frac{\omega}{\tilde{\xi}} \Psi(\mathbf{x}, 0) \right] = \frac{1}{4} \left( 2 - \tanh \mathbf{x} \right), \\ \Psi\_0(\mathbf{x}, \mathbb{F}) &= \mathbb{S}^{-1} \left[ \frac{\omega}{\tilde{\xi}} \Psi(\mathbf{x}, 0) \right] = \left( 2 - \tanh \mathbf{x} \right). \end{aligned}$$

It follows that

∞ ∑ *m*=0 Φ*m*+1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S 1 2 ∞ ∑ *m*=0 (Φ**xxx**)*m* − 3 ∞ ∑ *m*=0 E*m* + 3 2 ∞ ∑ *m*=0 F*m* + 3 ∞ ∑ *m*=0 G*m* + 3 2 ∞ ∑ *m*=0 H*m* +3 ∞ ∑ *m*=0 *Im* + 3 ∞ ∑ *m*=0 J*m* + 3 ∞ ∑ *m*=0 K*m* # , ∞ ∑ *m*=0 Ψ*m*+1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − ∞ ∑ *m*=0 (Ψ**xxx**)*m* − 3 ∞ ∑ *m*=0 M*m* − 3 ∞ ∑ *m*=0 N*m* − 3 ∞ ∑ *m*=0 O*m* + 6 ∞ ∑ *m*=0 P*m* + 3 ∞ ∑ *m*=0 Q*m* # , ∞ ∑ *m*=0 Υ*m*+1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − ∞ ∑ *m*=0 (Υ**xxx**)*m* − 3 ∞ ∑ *m*=0 R*m* − 3 ∞ ∑ *m*=0 *Sm* − 3 ∞ ∑ *m*=0 T*m* + 6 ∞ ∑ *m*=0 X*m* + 3 ∞ ∑ *m*=0 Y*m* # , *m* = 0, 1, 2, ... .

The first few Adomian polynomials are presented as follows:

E(Φ 2Φ**x**) = Φ2 0Φ0**x**, *f or* = 0 (2Φ0Φ1)Φ0**x** + Φ2 0Φ1**x**, *f or* = 1 (2Φ0Φ2 + Φ2 1 )Φ0**x** + (2Φ0Φ1)Φ1**x** + Φ2 0Φ2**x**, *f or* = 2 F(ΥΨ**xx**) = Υ0Ψ0**xx**, *f or* = 0 Υ1Ψ0**xx** + Υ0Ψ1**xx**, *f or* = 1 Υ2Ψ0**xx** + Υ1Ψ1**xx** + Υ0Ψ2**xx**, *f or* = 2 G(Ψ**x**Υ**x**) = Ψ0**x**Υ0**x**, *f or* = 0 Ψ0**x**Υ1**x** + Ψ1**x**Υ0**x** *f or* = 1 Ψ2**x**Υ0**x** + Ψ1**x**Υ1**x** + Ψ0**x**Υ2**x** *f or* = 2 H(Ψ**x**Υ**xx**) = Ψ0**x**Υ0**xx**, *f or* = 0 Ψ0**x**Υ1**xx** + Ψ1**x**Υ0**xx**, *f or* = 1 Ψ2**x**Υ0**xx** + Ψ1**x**Υ1**xx** + Ψ0**x**Υ2**xx**, *f or* = 2 *I*(Ψ*z*Φ**x**) = (ΨΥ)0Φ0**x**, *f or* = 0 (ΨΥ)0Φ1**x** + (ΨΥ)1Φ0**x**, *f or* = 1 (ΨΥ)0Φ2**x** + (ΨΥ)1Φ1**x** + (ΨΥ)2Φ0**x**, *f or* = 2 J(Φ*z*Ψ**x**) = (ΦΥ)0Ψ0**x**, *f or* = 0 (ΦΥ)0Ψ1**x** + (ΦΥ)1Ψ0**x**, *f or* = 1 (ΦΥ)0Ψ2**x** + (ΦΥ)1Ψ1**x** + (ΦΥ)2Ψ0**x**, *f or* = 2 K(ΦΨΥ**x**) = (ΦΨ)0Υ0**x**, *f or* = 0 (ΦΨ)0Υ1**x** + (ΦΨ)1Υ0**x**, *f or* = 1 (ΦΨ)0Υ2**x** + (ΦΨ)1Υ1**x** + (ΦΨ)2Υ0**x**, *f or* = 2 M(Φ**x**Ψ**x**) = Φ0**x**Ψ0**x**, *f or* = 0 Φ0**x**Ψ1**x** + Φ**x**Ψ0**x**, *f or* = 1 Φ2**x**Ψ0**x** + Φ1**x**Ψ1**x** + Φ0**x**Ψ1**x**, *f or* = 2 N(ΨΦ**xx**) = Ψ0Φ0**xx**, *f or* = 0 Ψ0Φ1**xx** + Ψ1Φ0**xx**, *f or* = 1 Ψ2Φ0**xx** + Ψ1Φ1**xx** + Ψ0Φ2**xx**, *f or* = 2 O(Ψ 2Υ**x**) = Ψ2 0Υ0**x**, *f or* = 0 (2Ψ0Ψ1)Υ0**x** + Ψ2 0Υ1**x**, *f or* = 1

$$\begin{aligned} \mathbf{\hat{x}}^{\prime} \qquad \left[ (2\mathbf{\hat{y}}\_{0}\mathbf{\hat{y}}\_{2} + \mathbf{\hat{y}}\_{1}^{2})\mathbf{\hat{y}}\_{0\mathbf{x}} + (2\mathbf{\hat{y}}\_{0}\mathbf{\hat{y}}\_{1})\mathbf{\hat{y}}\_{1\mathbf{x}} + \mathbf{\hat{y}}\_{0}^{2}\mathbf{\hat{y}}\_{2\mathbf{x}} \right] \text{ for } \mathbf{y} = \mathbf{2} \end{aligned} $$

$$\begin{array}{lcl} \mathcal{P}\_{\uparrow}(\boldsymbol{\Phi}\boldsymbol{\Psi}\boldsymbol{\Phi}\_{\mathbf{x}}) &=& \begin{cases} (\boldsymbol{\Phi}\boldsymbol{\Psi})\_{0}\boldsymbol{\Phi}\_{\mathbf{0}\mathbf{x}} & \text{for } \boldsymbol{\gamma} = \boldsymbol{0} \\ (\boldsymbol{\Phi}\boldsymbol{\Psi})\_{0}\boldsymbol{\Phi}\_{\mathbf{1}\mathbf{x}} + (\boldsymbol{\Phi}\boldsymbol{\Psi})\_{0}\boldsymbol{\Phi}\_{\mathbf{1}\mathbf{x}} & \text{for } \boldsymbol{\gamma} = \boldsymbol{1} \\ (\boldsymbol{\Phi}\boldsymbol{\Psi})\_{0}\boldsymbol{\Phi}\_{\mathbf{2}\mathbf{x}} + (\boldsymbol{\Phi}\boldsymbol{\Psi})\_{1}\boldsymbol{\Phi}\_{\mathbf{1}\mathbf{x}} + (\boldsymbol{\Phi}\boldsymbol{\Psi})\_{2}\boldsymbol{\Phi}\_{\mathbf{0}\mathbf{x}} & \text{for } \boldsymbol{\gamma} = \boldsymbol{2} \end{cases} \\ \mathcal{Q}\_{\uparrow}(\boldsymbol{\Phi}^{2}\boldsymbol{\Psi}\_{\mathbf{x}}) &=& \begin{cases} \boldsymbol{\Phi}\_{0}^{2}\boldsymbol{\Psi}\_{0\mathbf{x}} & \text{for } \boldsymbol{\gamma} = \boldsymbol{0} \\ (2\boldsymbol{\Phi}\_{0}\boldsymbol{\Phi}\_{1})\boldsymbol{\Psi}\_{0\mathbf{x}} + \boldsymbol{\Phi}\_{0}^{2}\boldsymbol{\Psi}\_{1\mathbf{x}} & \text{for } \boldsymbol{\gamma} = \boldsymbol{1} \\ (2\boldsymbol{\Phi}\_{0}\boldsymbol{\Phi}\_{2} + \boldsymbol{\Phi}\_{1}^{2})\boldsymbol{\Psi}\_{0\mathbf{x}} + (2\boldsymbol{\Phi}\_{0}\boldsymbol{\Phi}\_{1})\boldsymbol{\Psi}\_{1\mathbf{x}} + \boldsymbol{\Phi}\_{0}^{2}\boldsymbol{\Psi}\_{2\mathbf{x}} & \text{for } \boldsymbol{\gamma} = \boldsymbol{2} \end{cases} \end{array}$$

R0(Φ**x**Ψ**x**) = Φ0**x**Ψ0**x**, *f or* = 0 Φ0**x**Ψ1**x** + Φ1**x**Ψ0**x**, *f or* = 1 Φ2**x**Ψ0**x** + Φ1**x**Ψ1**x** + Φ0**x**Ψ2**x**, *f or* = 2 *S*(ΥΦ**xx**) = Υ0Φ0**xx**, *f or* = 0 Υ0Φ1**xx** + Υ1Φ0**xx**, *f or* = 1 Υ2Φ0**xx** + Υ1Φ1**xx** + Υ0Φ2**xx**, *f or* = 2 T(Υ 2Ψ**x**) = Υ 2 0Ψ0**x**, *f or* = 0 (2Υ0Υ1)Ψ0**x** + Υ 2 0Ψ1**x**, *f or* = 1 (2Υ0Υ2 + Υ 2 1 )Ψ0**x** + (2Υ0Υ1)Ψ1**x** + Υ 2 0Ψ2**x**, *f or* = 2 X(ΦΥ1Φ**x**) = (ΦΥ)0Φ0**x**, *f or* = 0 (ΦΥ)0Φ1**x** + (ΦΥ)1Φ0**x**, *f or* = 1 (ΦΥ)2Φ0**x** + (ΦΥ)1Φ1**x** + (ΦΥ)2Φ0**x**, *f or* = 2 Y(Φ 2Υ**x**) = Φ2 0Υ0**x**, *f or* = 0 (2Φ0Φ1)Υ0**x** + Φ2 0Υ1**x**, *f or* = 1 (2Φ0Φ2 + Φ2 1 )Υ0**x** + (2Φ0Φ1)Υ1**x** + Φ2 0Υ2**x**, *f or* = 2.

For *m* = 0, 1, 2, 3, ...

Φ1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S 1 2 (Φ**xxx**)0 − 3E0 + 3 2 F0 + 3G0 + 3 2 H0 + 3*I*0 + 3J0 + 3K0 # = 11 2 sec *h* 2 (**x**)S −1 " *ωδ*+2 *ξ δ*+2 # = 11 2 sec *h* 2 (**x**) ¯*t δ* Γ(*δ* + 1) , Ψ1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S (Ψ**xxx**)0 − 3M0 − 3N0 − 3O0 + 6P0 + 3Q0 # = − 11 8 sec *h* 2 (**x**) ¯*t δ* Γ(*δ* + 1) , Υ1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S (Υ**xxx**)0 − 3R0 − 3*S*b0 − 3T0 + 6X0 + 3Y0 # = − 11 2 sec *h* 2 (**x**) ¯*t δ* Γ(*δ* + 1) .

Φ2(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S 1 2 (Φ**xxx**)1 − 3E1 + 3 2 F1 + 3G1 + 3 2 H1 + 3*I*1 + 3J1 + 3K1 # = −121 8 tan *h*(**x**) sec *h* 2 (**x**) ¯*t* 2*δ* Γ(2*δ* + 1) , Ψ2(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S (Ψ**xxx**)1 − 3M1 − 3N1 − 3O1 + 6P1 + 3Q1 # = 121 8 tan *h*(**x**) sec *h* 2 (**x**) ¯*t* 2*δ* Γ(2*δ* + 1) , Υ2(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S (Υ**xxx**)1 − 3R1 − 3*S*b 1 − 3T1 + 6X1 + 3Y1 # = 242 8 tan *h*(**x**) sec *h* 2 (**x**) ¯*t* 2*δ* Γ(2*δ* + 1) ,

Φ3(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S 1 2 (Φ**xxx**)2 − 3E2 + 3 2 F2 + 3G2 + 3 2 H2 + 3*I*2 + 3J2 + 3K2 # = 1331 48 sec *h* 4 (**x**) - cosh(2**x**) − 2 ¯*t* 3*δ* Γ(3*δ* + 1) , Ψ3(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S (Ψ**xxx**)2 − 3M2 − 3N2 − 3O2 + 6P2 + 3Q2 # = 2662 96 sec *h* 4 (**x**) - cosh(2**x**) − 2 ¯*t* 3*δ* Γ(3*δ* + 1) , Υ3(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S (Υ**xxx**)2 − 3R2 − 3*S*b2 − 3T2 + 6X2 + 3Y2 # = −2662 48 sec *h* 4 (**x**) - cosh(2**x**) − 2 ¯*t* 3*δ* Γ(3*δ* + 1) , . . ..

The Shehu decomposition method solution for Problem 3 is presented as:

$$\begin{split} \Phi(\mathbf{x},\mathbf{f}) &= \Phi\_0(\mathbf{x},\mathbf{f}) + \Phi\_1(\mathbf{x},\mathbf{f}) + \Phi\_2(\mathbf{x},\mathbf{f}) + \Phi\_3(\mathbf{x},\mathbf{f}) \dots \\ &= \frac{1}{2} \big{(}2 + \tanh\mathbf{x}\big{)} + \frac{11}{2} \operatorname{sech} h^2(\mathbf{x}) \frac{\mathbf{f}^{\delta}}{\Gamma(\delta+1)} - \frac{121}{8} \tan h(\mathbf{x}) \operatorname{sech} h^2(\mathbf{x}) \frac{\mathbf{f}^{2\delta}}{\Gamma(2\delta+1)} \\ &+ \frac{1331}{48} \operatorname{sech} h^4(\mathbf{x}) \Big{[}\cosh(2\mathbf{x}) - 2\Big{]} \frac{\mathbf{f}^{3\delta}}{\Gamma(3\delta+1)} + \dots \end{split}$$

Analogously, we have

$$\begin{split} \Psi(\mathbf{x},\bar{t}) &= \frac{1}{4} (2 - \tanh \mathbf{x}) - \frac{11}{8} \sec h^2(\mathbf{x}) \frac{\bar{t}^{\delta}}{\Gamma(\delta + 1)} + \frac{121}{8} \tan h(\mathbf{x}) \sec h^2(\mathbf{x}) \frac{\bar{t}^{2\delta}}{\Gamma(2\delta + 1)} \\ &- \frac{1331}{48} \sec h^4(\mathbf{x}) \left[ \cosh(2\mathbf{x}) - 2 \right] \frac{\bar{t}^{\delta}}{\Gamma(3\delta + 1)} + \dots \\ \Psi(\mathbf{x},\bar{t}) &= \left( 2 - \tanh \mathbf{x} \right) - \frac{11}{2} \sec h^2(\mathbf{x}) \frac{\bar{t}^{\delta}}{\Gamma(\delta + 1)} + \frac{121}{4} \tan h(\mathbf{x}) \sec h^2(\mathbf{x}) \frac{\bar{t}^{2\delta}}{\Gamma(2\delta + 1)} \\ &- \frac{2662}{48} \sec h^4(\mathbf{x}) \left[ \cosh(2\mathbf{x}) - 2 \right] \frac{\bar{t}^{3\delta}}{\Gamma(3\delta + 1)} + \dots \end{split}$$

By Setting *δ* = 1, we then obtain the exact solution of coupled KdV Equation (2)

$$\begin{aligned} \Phi(\mathbf{x},\bar{t}) &= \frac{1}{2} \Big( 2 + \tanh\left(\mathbf{x} - \frac{11\bar{t}}{2}\right) \Big), \; \Psi(\mathbf{x},\bar{t}) = \frac{1}{4} \Big( 2 - \tanh\left(\mathbf{x} - \frac{11\bar{t}}{2}\right) \Big), \\\ \Psi(\mathbf{x},\bar{t}) &= \Big( 2 - \tanh\left(\mathbf{x} - \frac{11\bar{t}}{2}\right) \Big). \end{aligned}$$

**Case II.** Now, we surmise the new iterative transform method for Problem 3. Applying the proposed analytical approach to (42) yields

Φ0(**x**, ¯*t*) = 1 2 2 + tanh **x** , Ψ0(**x**, ¯*t*) = 1 4 2 − tanh **x** , Υ0(**x**, ¯*t*) = 2 − tanh **x** , Φ1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S 1 2 *∂* 3Φ0 *∂*¯*t* 3 − 3Φ 2 0 *∂*Φ0 *∂***x** + 3 2 Υ0 *∂* 2Ψ0 *∂***x** 2 + 3 *∂*Ψ0 *∂***x** *∂*Υ0 *∂***x** + 3 2 Ψ0 *∂* 2Υ0 *∂***x** 2 +3Ψ0Υ0 *∂*Φ0 *∂***x** + 3Φ0Υ0 *∂*Ψ0 *∂***x** + 3Φ0Ψ0 *∂*Υ0 *∂***x** # = 11 2 sec *h* 2 (**x**)S −1 " *ωδ*+2 *ξ δ*+2 # = 11 2 sec *h* 2 (**x**) ¯*t δ* Γ(*δ* + 1) , Ψ1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − *∂* 3Ψ0 *∂***x** 3 − 3 *∂*Φ0 *∂***x** *∂*Ψ0 *∂***x** − 3Ψ0 *∂* 2Φ0 *∂***x** 2 − 3Ψ 2 0 *∂*Υ0 *∂***x** + 6Φ0Ψ0 *∂*Φ0 *∂***x** + 3Φ 2 0 *∂*Ψ0 *∂***x** # = − 11 8 sec *h* 2 (**x**) ¯*t δ* Γ(*δ* + 1) , Υ1(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − *∂* 3Υ0 *∂***x** 3 − 3 *∂*Φ0 *∂***x** *∂*Υ0 *∂***x** − 3Υ0 *∂* 2Φ0 *∂***x** 2 − 3Υ 2 0 *∂*Ψ0 *∂***x** + 6Φ0Υ0 *∂*Φ0 *∂***x** + 3Φ 2 0 *∂*Υ0 *∂***x** # = − 11 2 sec *h* 2 (**x**) ¯*t δ* Γ(*δ* + 1) , Φ2(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S 1 2 *∂* 3Φ1 *∂*¯*t* 3 − 3Φ 2 1 *∂*Φ1 *∂***x** + 3 2 Υ1 *∂* 2Ψ1 *∂***x** 2 + 3 *∂*Ψ1 *∂***x** *∂*Υ1 *∂***x** + 3 2 Ψ1 *∂* 2Υ1 *∂***x** 2 +3Ψ1Υ1 *∂*Φ1 *∂***x** + 3Φ1Υ1 *∂*Ψ1 *∂***x** + 3Φ1Ψ1 *∂*Υ1 *∂***x** # = −121 8 tan *h*(**x**) sec *h* 2 (**x**) ¯*t* 2*δ* Γ(2*δ* + 1) , Ψ2(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − *∂* 3Ψ1 *∂***x** 3 − 3 *∂*Φ1 *∂***x** *∂*Ψ1 *∂***x** − 3Ψ1 *∂* 2Φ1 *∂***x** 2 − 3Ψ 2 1 *∂*Υ1 *∂***x** + 6Φ1Ψ1 *∂*Φ1 *∂***x** + 3Φ 2 1 *∂*Ψ1 *∂***x** # = 121 8 tan *h*(**x**) sec *h* 2 (**x**) ¯*t* 2*δ* Γ(2*δ* + 1) ,

$$\begin{split} \mathcal{Y}\_{2}(\mathbf{x},\mathbf{f}) &= \mathbb{S}^{-1} \Big[ \frac{\omega^{\delta}}{\mathfrak{F}^{\delta}} \mathbb{S} \Big[ -\frac{\partial^{3} \mathcal{Y}\_{1}}{\partial \mathbf{x}^{3}} - 3 \frac{\partial \Phi\_{1}}{\partial \mathbf{x}} \frac{\partial \mathcal{Y}\_{1}}{\partial \mathbf{x}} - 3 \mathcal{Y}\_{1} \frac{\partial^{2} \Phi\_{1}}{\partial \mathbf{x}^{2}} - 3 \mathcal{Y}\_{1}^{2} \frac{\partial \Psi\_{1}}{\partial \mathbf{x}} + 6 \Phi\_{1} \mathcal{Y}\_{1} \frac{\partial \Phi\_{1}}{\partial \mathbf{x}} + 3 \Phi\_{1}^{2} \frac{\partial \mathcal{Y}\_{1}}{\partial \mathbf{x}} \Big] \Big] \\ &= \frac{242}{8} \tan h(\mathbf{x}) \sec h^{2}(\mathbf{x}) \frac{\mathfrak{f}^{2\delta}}{\Gamma(2\delta + 1)} \end{split}$$

Φ3(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S 1 2 *∂* 3Φ2 *∂*¯*t* 3 − 3Φ 2 2 *∂*Φ2 *∂***x** + 3 2 Υ2 *∂* 2Ψ2 *∂***x** 2 + 3 *∂*Ψ2 *∂***x** *∂*Υ2 *∂***x** + 3 2 Ψ2 *∂* 2Υ2 *∂***x** 2 +3Ψ2Υ2 *∂*Φ2 *∂***x** + 3Φ2Υ2 *∂*Ψ2 *∂***x** + 3Φ2Ψ2 *∂*Υ2 *∂***x** # = 1331 48 sec *h* 4 (**x**) - cosh(2**x**) − 2 ¯*t* 3*δ* Γ(3*δ* + 1) , Ψ3(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − *∂* 3Ψ2 *∂***x** 3 − 3 *∂*Φ2 *∂***x** *∂*Ψ2 *∂***x** − 3Ψ2 *∂* 2Φ2 *∂***x** 2 − 3Ψ 2 2 *∂*Υ2 *∂***x** + 6Φ2Ψ2 *∂*Φ2 *∂***x** + 3Φ 2 2 *∂*Ψ2 *∂***x** # = 2662 96 sec *h* 4 (**x**) - cosh(2**x**) − 2 ¯*t* 3*δ* Γ(3*δ* + 1) , Υ3(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − *∂* 3Υ2 *∂***x** 3 − 3 *∂*Φ2 *∂***x** *∂*Υ2 *∂***x** − 3Υ2 *∂* 2Φ2 *∂***x** 2 − 3Υ 2 2 *∂*Ψ2 *∂***x** + 6Φ2Υ2 *∂*Φ2 *∂***x** + 3Φ 2 2 *∂*Υ2 *∂***x** # = −2662 48 sec *h* 4 (**x**) - cosh(2**x**) − 2 ¯*t* 3*δ* Γ(3*δ* + 1) , . . . Φ*m*(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S 1 2 *∂* 3Φ*m*−1 *∂*¯*t* 3 − 3Φ 2 *m*−1 *∂*Φ*m*−1 *∂***x** + 3 2 Υ*m*−1 *∂* 2Ψ*m*−1 *∂***x** 2 + 3 *∂*Ψ*m*−1 *∂***x** *∂*Υ*m*−1 *∂***x** + 3 2 Ψ*m*−1 *∂* 2Υ*m*−1 *∂***x** 2 +3Ψ*m*−1Υ*m*−1 *∂*Φ*m*−1 *∂***x** + 3Φ*m*−1Υ*m*−1 *∂*Ψ*m*−1 *∂***x** + 3Φ*m*−1Ψ*m*−1 *∂*Υ*m*−1 *∂***x** # , Ψ*m*(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − *∂* 3Ψ*m*−1 *∂***x** 3 − 3 *∂*Φ*m*−1 *∂***x** *∂*Ψ*m*−1 *∂***x** − 3Ψ*m*−1 *∂* 2Φ*m*−1 *∂***x** 2 − 3Ψ 2 *m*−1 *∂*Υ*m*−1 *∂***x** +6Φ*m*−1Ψ*m*−1 *∂*Φ*m*−1 *∂***x** + 3Φ 2 *m*−1 *∂*Ψ*m*−1 *∂***x** # , Υ*m*(**x**, ¯*t*) = S −1 " *ωδ ξ δ* S − *∂* 3Υ*m*−1 *∂***x** 3 − 3 *∂*Φ*m*−1 *∂***x** *∂*Υ*m*−1 *∂***x** − 3Υ*m*−1 *∂* 2Φ*m*−1 *∂***x** 2 − 3Υ 2 *m*−1 *∂*Ψ*m*−1 *∂***x** +6Φ*m*−1Υ*m*−1 *∂*Φ*m*−1 *∂***x** + 3Φ 2 *m*−1 *∂*Υ*m*−1 *∂***x** # .

The series solution for Problem 3 is presented as:

$$\begin{split} \Phi(\mathbf{x},\mathbf{f}) &= \Phi\_0(\mathbf{x},\mathbf{f}) + \Phi\_1(\mathbf{x},\mathbf{f}) + \Phi\_2(\mathbf{x},\mathbf{f}) + \Phi\_3(\mathbf{x},\mathbf{f}) + \dots \Phi\_m(\mathbf{x},\mathbf{f}) \\ &= \frac{1}{2} \Big( 2 + \tanh \mathbf{x} \Big) + \frac{11}{2} \operatorname{sech}^2(\mathbf{x}) \frac{\overline{\mathbf{f}}^{\delta}}{\Gamma(\delta + 1)} - \frac{121}{8} \tan h(\mathbf{x}) \operatorname{sech}^2(\mathbf{x}) \frac{\overline{\mathbf{f}}^{2\delta}}{\Gamma(2\delta + 1)} \\ &+ \frac{1331}{48} \operatorname{sech}^4(\mathbf{x}) \Big[ \cosh(2\mathbf{x}) - 2 \Big] \frac{\overline{\mathbf{f}}^{3\delta}}{\Gamma(3\delta + 1)} + \dots \end{split}$$

Analogously, we have

$$\begin{split} \Psi(\mathbf{x},\mathbf{f}) &= \frac{1}{4} (2 - \tanh \mathbf{x}) - \frac{11}{8} \sec h^2(\mathbf{x}) \frac{\mathbf{f}^\delta}{\Gamma(\delta + 1)} + \frac{121}{8} \tan h(\mathbf{x}) \sec h^2(\mathbf{x}) \frac{\mathbf{f}^{2\delta}}{\Gamma(2\delta + 1)} \\ &- \frac{1331}{48} \sec h^4(\mathbf{x}) \left[ \cosh(2\mathbf{x}) - 2 \right] \frac{\mathbf{f}^{3\delta}}{\Gamma(3\delta + 1)} + \dots \\ \Psi(\mathbf{x},\mathbf{f}) &= \left( 2 - \tanh \mathbf{x} \right) - \frac{11}{2} \sec h^2(\mathbf{x}) \frac{\mathbf{f}^\delta}{\Gamma(\delta + 1)} + \frac{121}{4} \tan h(\mathbf{x}) \sec h^2(\mathbf{x}) \frac{\mathbf{f}^{2\delta}}{\Gamma(2\delta + 1)} \\ &- \frac{2662}{48} \sec h^4(\mathbf{x}) \left[ \cosh(2\mathbf{x}) - 2 \right] \frac{\mathbf{f}^{3\delta}}{\Gamma(3\delta + 1)} + \dots \end{split}$$

By setting *δ* = 1, we then obtain the exact solution of MCKdV Equation (2)

$$\begin{aligned} \Phi(\mathbf{x},\bar{t}) &= \frac{1}{2} \left( 2 + \tanh\left(\mathbf{x} - \frac{11\bar{t}}{2}\right) \right), \; \Psi(\mathbf{x},\bar{t}) = \frac{1}{4} \left( 2 - \tanh\left(\mathbf{x} - \frac{11\bar{t}}{2}\right) \right), \\\ \Psi(\mathbf{x},\bar{t}) &= \left( 2 - \tanh\left(\mathbf{x} - \frac{11\bar{t}}{2}\right) \right). \end{aligned}$$

In Figures 9–11 the exact and approximate results of Φ(**x**, ¯*t*), Ψ(**x**, ¯*t*) and Υ(**x**, ¯*t*) are demonstrated at ` = 1, *σ* = 0.5 and *β* = 2, respectively. In Figures 12–14, the surface and 2D graph for Φ(**x**, ¯*t*), Ψ(**x**, ¯*t*) and Υ(**x**, ¯*t*) for various fractional orders are presented which show that the SDM/SITM approximated results derived are in a strong agreement with the exact and the numerical ones. This comparison represents a strong correlation between the SDM and exact findings. Therefore, the SDM/SITM are reliable novel approaches which require less computation time and is quite straightforward and more flexible than the homotopy perturbation method or homotopy analysis method, because the ST permits one of several scenarios to reduce the deficiency mainly occurs because of unsatisfied initial conditions that appear in other semi-analytical methods such as the SDM/SITM.

**Figure 9.** The exact and analytical solution graph at Φ(**x**, ¯*t*) of Problem 3 for ` = 1, *σ* = 0.5 and *β* = 2.

**Figure 10.** The exact and analytical solution graph at Ψ(**x**, ¯*t*) of Problem 3 for ` = 1, *σ* = 0.5 and *β* = 2.

**Figure 11.** The exact and analytical solution graph at Υ(**x**, ¯*t*) of Problem 3 for ` = 1, *σ* = 0.5 and *β* = 2.

**Figure 12.** Numerical evaluation of graph at Ψ(**x**, ¯*t*) Problem 3 for various fractional order *δ* = 0.4, 0.6, 0.8, 1, ` = 1, *σ* = 0.5 and *β* = 2.

**Figure 13.** Numerical evaluation of graph at Φ(**x**, ¯*t*) Problem 3 for various fractional order *δ* = 0.4, 0.6, 0.8, 1, ` = 1, *σ* = 0.5 and *β* = 2.

**Figure 14.** Numerical evaluation of graph at Υ(**x**, ¯*t*) Problem 3 for various fractional order *δ* = 0.4, 0.6, 0.8, 1, ` = 1, *σ* = 0.5 and *β* = 2.

#### **7. Conclusions**

Understanding complex nonlinear PDEs remains a difficult challenge when their generative model is unknown. This challenge becomes more complex when it comes to evaluating time fractional nonlinear PDEs, surmising the model that governs their evolution. To cope with this difficulty, numerous numerical methods have been employed for dealing with nonlinear physical phenomena. Toward addressing this goal, in this paper, we have considered a time-fractional KdV equation and have developed effective, rigorous and robust algorithmic strategies (Shehu decompsition method and Shehu iterative transform method) to estimate approximate-analytical solutions and so identify the main numerical solutions appearing in the literature. In this approach, we do not need the Lagrange multiplier, correction functional, stationary conditions, or to calculate heavy integrals because the results established are noise free, which overcomes the shortcomings of existing methods. It is remarkable that the projected approaches are well-organized analytical methods for finding approximate-analytical solutions to complex nonlinear

PDEs. Finally, we conclude that this scheme will be taken into account in order to cope with other complex non-linear fractional order systems of equations.

**Author Contributions:** All authors contributed equally. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Acknowledgments:** This research was supported by Taif University Researchers Supporting Project Number (TURSP-2020/96), Taif University, Taif, Saudi Arabia.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**

MDPI St. Alban-Anlage 66 4052 Basel Switzerland Tel. +41 61 683 77 34 Fax +41 61 302 89 18 www.mdpi.com

*Symmetry* Editorial Office E-mail<EMAIL_ADDRESS>www.mdpi.com/journal/symmetry

MDPI St. Alban-Anlage 66 4052 Basel Switzerland

Tel: +41 61 683 77 34

www.mdpi.com

ISBN 978-3-0365-5147-0


*3.1. Design Based on Analytical Solution*

Birdcage RF coils are designed in various sizes with respect to their length and diameter in according with the specified applications. The main tasks in the designing of a birdcage RF coil is the choice of coil configurations, number of legs, dimensions of legs and end-ring segments, lumped capacitance, and port impedance matching at desired resonance frequency. Legs and end-ring segments conductors are the sources of inductance in the birdcage coil. Their count and dimensions are the controlling factors of the magnetic field homogeneity as well as SAR inside the coil. While on the other hand, the lumped capacitors are responsible for producing the resonance at required radio frequency without effecting the magnetic field distribution. A balance between the conductor size and lumped capacitance value is quite necessary for an appropriate design of the birdcage RF coil. In order to validate the proposed analytical technique, we designed the low pass, high pass and band pass configurations of the linearly polarized birdcage RF coil for 1H NMR imaging in small volume applications at 1.5 T and 3 T MRI systems as shown in Figure 7.

**Figure 7.** Configurations of the birdcage RF coil for the reception of 1H NMR signal at 1.5 T and 3 T MRI systems: (**a**) Low pass; (**b**) high pass; (**c**) band pass. (All dimensions are in mm).

The designed birdcage RF coils are linearly polarized having 8 legs and 2 end-rings which are made of 35 μm thick and 5 mm wide rectangular copper strips etched on a flexible FR4 substrate (εr = 4.4) of 0.2 mm thickness. A 2 mm gap is created in the respective conductor segments of the legs and/or end-rings for SMT lumped capacitors. Total length of a single leg of the birdcage RF coil including capacitor gap is 162 mm. In order to keep the permanent 3D symmetry, the birdcage conductor pattern etched flexible FR4 was further attached on a solid acryl (εr = 3.2) cylinder of 5 mm thickness and 40 mm internal diameter. The length and diameter of the coil along with the dimension of the legs and end-rings segments are assumed to be constant for all three configurations of the 1.5 T and 3 T birdcage RF coils.

The self-inductance (referred as calculated inductance *Lcal* in this paper) of the leg and end-rings segments which is constant for all designed coils is determined using Equation (6). However, in the actual scenario there also exist very strong mutual inductance between the parallel conductor segments of the birdcage RF coil which affects the self-inductances of the coil. The self-inductance of the leg and end-ring segment which includes the effect of mutual-inductance was computed using the methods described by the birdcage builder [25]. This self-inductance is referred as effective inductance *Le*ff in this paper. For the given dimensions of the birdcage RF coil the inductance *Lcal* and *Le*ff for the leg and end-ring segment respectively are given in the Table 1 below.

**Table 1.** Birdcage coil segments dimensions and inductances.

These calculated and effective inductance values were further used to compute the lumped capacitances required in the legs and/or end-rings for all three configurations of the birdcage RF coil at the desired frequencies of 63.8 MHz and 127.7 MHz (for 1.5 T and 3 T MRI systems respectively) by using the Equations (7)–(9). In order to prove the validity of the proposed method, we also computed the lumped capacitances for all three configurations by using the conventional method described by Equation (1), 2D circuit simulation and 3D electromagnetic simulation. The actual values of the lumped capacitance used in the implementation of the prototypes of low pass, high pass and band pass configurations of 1.5 T and 3 T birdcage RF coils were also measured. A comparison among lumped capacitor values determined through different techniques for 1.5 T and 3 T birdcage coils are given below in the Tables 2 and 3 respectively.

**Table 2.** Lumped capacitor values in pF for 1.5 T birdcage coils.

**Table 3.** Lumped capacitor values in pF for 3 T birdcage coils.

The comparison provided in above Tables 2 and 3 concludes that the numerical technique does only provide an approximate value of the lumped capacitor which is quite close to the actual value that is deployed in the prototype implementation. The measured capacitance value for the desired dominant resonance frequency is obtained by slightly tuning the numerically obtained capacitance. It is evident from the information provided in Tables 2 and 3 that in comparison to the conventional methods, the effective lumped capacitance obtained through the proposed method exhibits least numerical error and requires lesser tuning to meet the desired resonance condition. Furthermore, the proposed analytic method described in this paper is quite simple and it does not require the hit and try method for capacitance computation which is commonly adopted in conventioanl 3D EM simulations. The results obtained through such analytical technique can directly be used in the coil implementation engineering which also includes the impedance matching operation at the desired resonance frequency.

#### *3.2. Resonance Frequency*

The proposed analytical solution given in Section 1 does also provide the information about the resonance frequency modes for all three configurations (low pass, high pass and band pass) of the birdcage RF coil by plotting its input impedance *ZER\_p* against frequency. We assumed the case where the receiver port is connected across the half end-ring segment and then determined the resonance frequencies spectrum of all three coil configurations using Equation (21). The effective inductance values of the leg and end-ring segments along with their corresponding effective capacitance values (with respect to the coil configuration) were used to compute the impedances *Zl* and *Zer*. A comparison between the input impedance obtained though the analytical solution proposed in this paper and the input impedance plot obtained through 2D circuit simulation with effective inductance and their corresponding capacitance values in ANSYS [26] for each configuration of 1.5 T and 3 T birdcage RF coils are given in the Figure 8 below.

(**a**) (**b**)

**Figure 8.** Input impedance obtained using proposed analytical solution and conventional 2D simulation for the different configurations of the birdcage RF coil: (**a**) 1.5T low pass, (**b**) 1.5 T high pass, (**c**) 1.5 T band pass, (**d**) 3 T low pass, (**e**) 3 T high pass, (**f**) 3 T band pass.

#### *3.3. Impedance Matching*

It can be observed in all cases of Figure 8 that the pattern of variations in the input impedance (at the dominant resonance mode as well as the higher order resonance modes) provided by the proposed analytical solution does perfectly coincide with the one which was obtained using conventional 2D circuit simulation technique. The numerical values of the input impedance *Zin* at the desired resonance frequency is further used to design the impedance matching circuit which serves as an interface between the RF coil and receiver circuit. A conventional "L-matching" network with lumped (series) inductance and (parallel) capacitance can be used to accomplish this task. The input impedance *Zin* serves as the load impedance which is used to obtain the required values of the lumped components of the impedance matching network connected between the desired segment (end-ring in this case) of the birdcage RF coil and the 50 ohm port of the receiver. The numerical values of the input impedance *ZER\_p* as obtained via proposed analytical solution along with the corresponding values of the lumped components of the matching circuit for all configurations of the designed 1.5 T and 3 T birdcage RF coils are provided in the Table 4.

**Table 4.** Input impedance *ZER\_p* of the designed birdcage RF coils at dominant resonance frequency and the corresponding lumped components of the matching circuit for 1.5 T and 3T MRI systems.

The reflection coefficient S11 which is a core parameter to explains the impedance matching of an RF circuit can easily be determined with the help of input impedance by using the following Equation (23).

$$\text{S11} = \frac{\left(Z\_{ER\_p} - 50\right)}{\left(Z\_{ER\_p} + 50\right)}\tag{23}$$

A comparison between the reflection coefficients obtained by using the proposed analytical solution in the absence as well as presence of the impedance matching circuit for all configurations of the implemented 1.5 T and 3 T birdcage RF coils is provided in Figure 9.

**Figure 9.** *Cont.*

**Figure 9.** Reflection coefficient S11 for the different configurations of the birdcage RF coil without and with impedance matching circuit: (**a**) 1.5 T low pass, (**b**) 1.5 T high pass, (**c**) 1.5 T band pass, (**d**) 3 T low pass, (**e**) 3 T high pass, (**f**) 3 T band pass.

#### *3.4. Magnetic Field Homogeneity*

Homogeneity of the magnetic field at the desired dominant resonance frequency (63.8 MHz for 1.5 T and 127.7 MHz for 3 T) everywhere inside the designed birdcage coils using the proposed analytical solution can be observed via 3D electromagnetic simulation. The simulation models of the birdcage RF coils were created in according with the dimensional information provided in Section 3.1 using 3D electromagnetic simulation software ANSYS HFSS [26]. The magnetic field distribution inside the coil under no load conditions (assuming air inside the coil) as determined in the transverse plans at different positions for each coil configuration is shown in Figure 10 below.

**Figure 10.** Magnetic field distribution at dominant resonance frequency in transvers planes at various positions for all configurations of 1.5 T and 3 T birdcage RF coils.

It can be observed in Figure 10 above that each coil configuration causes quite homogeneous magnetic field distribution at the dominant resonance frequency while the bandpass configuration produces rather strong magnetic field than others for both 1.5 T and 3 T cases.