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Geometric optimal techniques to control the muscular force response to functional electrical stimulation using a non-isometric force-fatigue model

  • * Corresponding author: Jérémy Rouot

    * Corresponding author: Jérémy Rouot

Dedicated to Professor Tony Bloch on the occasion of his 65th birthday

This research paper benefited from the support of the FMJH Program PGMO and from the support of EDF, Thales, Orange and the authors are partially supported by the Labex AMIES

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  • A recent force-fatigue parameterized mathematical model, based on the seminal contributions of V. Hill to describe muscular activity, allows to predict the muscular force response to external electrical stimulation (FES) and it opens the road to optimize the FES-input to maximize the force response to a pulse train, to track a reference force while minimizing the fatigue for a sequence of pulse trains or to follow a reference joint angle trajectory to produce motion in the non-isometric case. In this article, we introduce the geometric frame to analyze the dynamics and we present Pontryagin types necessary optimality conditions adapted to digital controls, used in the experiments, vs permanent control and which fits in the optimal sampled-data control frame. This leads to Hamiltonian differential variational inequalities, which can be numerically implemented vs direct optimization schemes.

    Mathematics Subject Classification: 49K15, 93B07, 92B05.

    Citation:

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  • Figure 1.  Extremal of the Ding et al. model defined on $ [0,T] $, $ T = 100ms $, with four sampling times and maximizing the final force

    Figure 2.  Sensitivity analysis. Time evolution of the Jacobi fields component $ \delta F(\cdot) $ according to Definition 2.7 and associated to the trajectory of the force-fatigue model

    Figure 3.  Sensitivity analysis. Time evolution of the Jacobi fields component $ \delta F(\cdot) $ where their initializations are $ 10\% $ of the physical values $ A_{\text{rest}},K_{m,\text{rest}},\tau_{1,\text{rest}} $ associated to the trajectory of the force-fatigue model

    Figure 4.  Time evolution on $ [0,T],\, T = 0.4s $ of the state of the Ding et al. model computed with the indirect approach maximizing $ \varphi(x(T)) = F(T) $ with $ n = 6 $, $ I_m = 20 $ms. The optimal value is the same as the optimal value computed with the direct method (see Fig.5). The model parameters are given in Table 1

    Figure 5.  Time evolution on $ [0,T],\, T = 0.4s $ of the normalized state of the Ding et al. model computed with the direct approach maximizing $ \varphi(x(T)) = F(T) $ with $ n = 6 $, $ I_m = 20 $ms. The model parameters are given in Table 1

    Figure 6.  Time evolution on $ [0,T],\, T = 0.9s $ of the normalized state of the Marion et al. model computed with the direct approach maximizing $ \varphi(x(T)) = F(T)/F_{ext}+A_{90}(T)/A_{90,0} $ with $ n = 10 $, $ I_m = 20 $ms. The model parameters are given in Table 2

    Figure 7.  Time evolution on $ [0,T],\, T = 0.9s $ of the normalized state of the Marion et al. model with the direct method when considering $ t_i = T - (n-i+1)I_m,\ i = 1,\dots,n $, $ n = 10 $, $ I_m = 20ms $. The model parameters are given in Table 2

    Table 2.  Numerical values for parameters of the Marion et al. model

    Variable Value Unit Variable Value Unit
    $ F_{load} $ 45.4 N $ F_M $ 247.5 N
    $ \tau_c $ 20×-3 s $ \tau_{fat} $ 99.4 s
    $ A_{90,0} $ 2100 N s−1 $ K_{m,0} $ 3.52×-1
    $ \tau_{1,0} $ 3.61×-2 s $ \tau_2 $ 5.21×-2 s
    $ a $ 4.49×-4 deg−2 $ b $ 3.44×-2 deg−1
    $ v_1 $ 3.71×-1 N deg−2 $ v_2 $ 2.29×-2 deg−1
    $ \ell = L/I $ 9.85 kg m−1 $ \alpha_A $ -4.02×-1 s−2
    $ \alpha_{K_m} $ 1.36×-5 s−1 N−1 $ \alpha_{\tau_1} $ 2.93×-5 N−1
    $ \beta_{\tau_1} $ 8.54×-7 s deg−1 N−1 $ R_0 $ 1.143
    $ \beta_{90} $ 0 deg−1 s−1 $ \beta_{K_m} $ 0 deg−1 N−1
     | Show Table
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    Table 1.  Numerical values for parameters of the Ding et al. model

    Variable Value Unit Variable Value Unit
    $ \tau_c $ 20×-3 s $ \tau_{fat} $ 127 s
    $ A_{rest} $ 3.009 kN s−1 $ K_{m,rest} $ 1.03×-1
    $ \tau_{1,rest} $ 5.1×-2 s $ \tau_2 $ 1.24×-1 s
    $ \alpha_A $ -4×-1 s−2 $ \alpha_{K_m} $ 1.9×-2 s−1 kN−1
    $ \alpha_{\tau_1} $ 2.1×-2 kN−1 $ R_0 $ 2
     | Show Table
    DownLoad: CSV
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