Constraint functions

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Carlos Adir edited this page Oct 28, 2021 · 1 revision

For our case, we have the lagrange equation of frist kind given by

$\dfrac{d}{dt}\left(\dfrac{\partial E}{\partial \dot{x}_{i}}\right) - \dfrac{\partial E}{\partial x_{i}} + \sum_{j=1}^{n} \lambda_{j} \cdot \dfrac{\partial g_j}{\partial x_{i}} = 0$

As we call the lagrande operator L like

$L = \dfrac{d}{dt} \dfrac{\partial E}{\partial \dot{x}_i} - \dfrac{\partial E}{\partial x_{i}}$

We have the equation

$L_{x_{i}} + \sum_{j=1}^{n} \lambda_{j} \cdot \dfrac{\partial g_j}{\partial x_{i}} = 0$

As we have n variables and m contraint functions, we have that

$\bold{L} = \begin{bmatrix}L_{x_{1}} \\ \vdots \\ L_{x_{n}} \end{bmatrix}_{(n)} = \dfrac{d}{dt}\left(\dfrac{\partial E}{\partial \bold{\dot{x}}}\right) - \dfrac{\partial E}{\partial \bold{x}} \Rightarrow \left[\bold{L}\right]_{i} = L_{x_{i}}$

And the gradient:

$\nabla g_{j} = \begin{bmatrix} \dfrac{\partial g_{1}}{\partial x_{1}} \\ \vdots \\ \dfrac{\partial g_{1}}{\partial x_{n}} \end{bmatrix}_{(n)} \Rightarrow \nabla \bold{g} = \begin{bmatrix} \dfrac{\partial g_{1}}{\partial x_{1}} & \cdots & \dfrac{\partial g_{m}}{\partial x_{1}} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial g_{1}}{\partial x_{n}} & \cdots & \dfrac{\partial g_{m}}{\partial x_{n}} \end{bmatrix}_{(n \times m)} \Leftrightarrow \left[\nabla \bold{g}\right] _{ij} = \dfrac{\partial g_{j}}{\partial x_{i}}$

So, our equation become

$\left[\bold{L}\right] + \left[\nabla \bold{g}\right]\cdot \left[\lambda\right] = 0$

Now, we get a case where we can write the energy like

$E = \dfrac{1}{2} \cdot \bold{\dot{x}}^{T} \cdot \left[M\right] \cdot \bold{\dot{x}} + \dfrac{1}{2}\cdot \bold{x}^{T} \cdot \left[K\right] \cdot \bold{x} + \bold{\dot{x}} \cdot \left[V\right] \cdot \bold{x} + \left[A\right] \cdot \bold{\dot{x}} + \left[B\right] \cdot \bold{x} + \left[C\right]$

The lagrangian with these matrix are so

$\begin{align*} L_{x_{i}} & = \dfrac{1}{2}\left(M_{ij}+M_{ji}\right) \cdot \ddot{x}_{j} + \dfrac{1}{2}\left(\dfrac{\partial M_{ij}}{\partial x_{k}} + \dfrac{\partial M_{ji}}{\partial x_{k}} - \dfrac{\partial M_{kj}}{\partial x_{i}}\right)\dot{x}_{j}\dot{x}_{k} + \\ & +\left(-\dfrac{\partial V_{kj}}{\partial x_{i}}\right) \cdot \dot{x}_{k}\dot{x}_{j} + \left(-\dfrac{1}{2} \cdot \dfrac{\partial K_{kj}}{\partial x_{i}}\right)\cdot x_{k}x_{j} + \left(V_{ik}-V_{ki} - \dfrac{\partial A_{k}}{\partial x_{i}}\right) \dot{x}_{k} + \\ & - \dfrac{1}{2}\left(K_{ik}+K_{ki} + 2\dfrac{\partial B_{k}}{\partial x_{i}}\right)x_{k} + \left(B_{k} + \dfrac{\partial C}{\partial x_{i}}\right) \end{align*}$

With the matrix constants we get only

$\bold{L} = \dfrac{1}{2}\left(M+M^{T}\right) \cdot \bold{\ddot{x}} + \left(V-V^{T}\right) \bold{\dot{x}} - \dfrac{1}{2}\left(K+K^{T} \right) \cdot \bold{x} + B$