## 3 Optimal Design of Compliant Elements
### Inactive Constraint (\(\mu^{*}=0\))  
When the design for minimum energy consumption lies within the feasible design set, the constraint is inactive, i.e., \(g(\mathbf{x}_{j})<0\). In this case, the stationary point of the Lagrangian in Equation (28) takes the following form:  
\[\theta_{j}^{*(1)} = \frac{I_{\tau_{j}}I_{q_{j}^{2}}-I_{\tau q_{j}}I_{q_{j}}}{I_{\tau_{ j}}I_{q_{j}}-TI_{\tau q_{j}}} \tag{34}\] \[k_{j}^{*(1)} = \frac{I_{\tau_{j}}I_{q_{j}}-TI_{\tau q_{j}}}{\mathrm{K}_{r,j} \left(TI_{q_{j}^{2}}-I_{q_{j}}^{2}\right)}\] (35) \[\mu^{*(1)} = 0 \tag{36}\]  
The stationary point found is a minimum point if the condition 5 of Equation (33) is verified, i.e., if the Hessian matrix \(\mathbf{H}\) of the Lagrangian function with respect to the design variables is positive definite.  
\[\mathbf{H}=\
abla_{\mathbf{x}_{j}\mathbf{x}_{j}}^{2}\mathcal{L}(\mathbf{x}_{j }^{*(1)},\mu_{j}^{*(1)})=\begin{bmatrix}h_{11}&h_{12}\\ h_{21}&h_{22}\end{bmatrix} \tag{37}\]where  
\[h_{11} = \frac{2a_{j}T}{\text{K}_{r,j}}\left(\frac{I_{\tau_{j}}I_{q_{j}}-T1_{ \tau q_{j}}}{TI_{q_{j}^{2}}-I_{q_{j}}^{2}}\right)^{2} \tag{38}\] \[h_{12} = h_{21}=2a_{j}I_{\tau_{j}}\] (39) \[h_{22} = \frac{2a_{j}\text{K}_{r,j}\left(TI_{q_{j}^{2}}-I_{q_{j}}^{2} \right)\left(I_{\tau_{j}}^{2}I_{q_{j}^{2}}-2I_{\tau_{j}}I_{q_{j}}I_{\tau q_{j}}+ TI_{\tau q_{j}}^{2}\right)}{\left(I_{\tau_{j}}I_{q_{j}}-TI_{\tau q_{j}} \right)^{2}} \tag{40}\]  
The Hessian matrix \(\mathbf{H}\) is positive definite if the determinants of all the upper-left submatrices are positive, i.e., if the following conditions hold:  
\[\mathbf{1 st}\,\text{condition:}\quad\det(h_{11})=\frac{2a_{j}T}{ \text{K}_{r,j}}\left(\frac{I_{\tau_{j}}I_{q_{j}}-TI_{\tau q_{j}}}{TI_{q_{j}^{2 }}-I_{q_{j}}^{2}}\right)^{2}>0 \tag{41}\] \[\mathbf{2 nd}\,\text{condition:}\quad\quad\det(\mathbf{H})=\frac{4a _{j}^{2}\left(I_{\tau_{j}}I_{q_{j}}-TI_{\tau q_{j}}\right)^{2}}{TI_{q_{j}^{2} }-I_{q_{j}}^{2}}>0 \tag{42}\]  
The condition in Equation (41) is always satisfied, as \(a_{j}\), \(T\), and \(\text{K}_{r,j}\) are positive while the other term of the product is squared. The condition in Equation (42) is satisfied if the denominator is positive, i.e., if  
\[I_{q_{j}}^{2}<TI_{q_{j}^{2}} \tag{43}\]  
In order to demonstrate that this condition is satisfied, _[PERSON]'s inequality_ is applied.  
Theorem 1: _Let \(f\) be an integrable function defined on \([a,b]\) and let \(\phi\) be a convex function defined at least on the set \([m,M]\) where \(m\) is the integral of \(f\) and \(M\) is the upper bound of \(f\). Then,_  
\[\phi\left(\frac{1}{b-a}\int_{a}^{b}f(x)dx\right)\leq\frac{1}{b-a}\int_{a}^{b} \phi(f(x)dx) \tag{44}\]  
By applying the Jensen's inequality with \(\phi(y)=y^{2}\), \(f(x)=q(t)\), and \(t\in[t_{0},\,t_{f}]\) (\(t_{f}-t_{0}=T\)), the resulting equation is equivalent to Equation (43), less than it is not a strict inequality:  
\[\left(\frac{1}{T}\int_{t_{0}}^{t_{f}}q_{j}(t)dt\right)^{2}\leq\frac{1}{T}\int _{t_{0}}^{t_{f}}q_{j}^{2}(t)dt \tag{45}\]  
However, the equality of Equation (45) holds only if \(q_{j}(t)\) is a constant function, which contrasts with our assumption of analyzing the system in motion. Therefore, this is a strict inequality and both the conditions in Equations (41) and (42) are satisfied. The stationary point found in this case is a design of minimum energy for the system, corresponding to the global minimum of the unconstrained energy consumption function (Equation (13)). This design reduces the energy consumption of motor \(j\) of the quantity \(E_{s_{j}}\):  
\[E_{j}=E_{ws_{j}}-\underbrace{\frac{a_{j}}{\text{K}_{r,j}}\cdot\frac{I_{\tau_ {j}}^{2}I_{q_{j}}^{2}-2I_{\tau_{j}}I_{q_{j}}I_{\tau q_{j}}+TI_{\tau q_{j}}^{2}} {I_{q_{j}}^{2}-TI_{q_{j}^{2}}}}_{E_{s_{j}}} \tag{46}\]