# Non-Conservative Forces

$$\sum_j\left(\frac{d}{dt}\frac{\partial T}{\partial \dot{q}_j} - \frac{\partial T}{\partial q_j} - Q_j\right)\delta q_j = 0$$.

Suppose we have a monogenic system: Is derivable from a poential function of the form $$Q_j^{(a)} = \frac{d}{dt}\frac{\partial V}{\partial q_j} - \frac{\partial V}{\partial q_j}$$ with $$V = V(q,\dot{q},t)$$.

Gives us our original lagrangian equations. Hence, all systems where the applied forces are conservative are monogenic, but not all monogenic systems have conserviative forces.

## Calculus of Variation

Suppose $$y = y(x)$$. If we want the stationary value, i.e. minimize the value for integration between two fixed values, $$x_1,x_2$$: $$I = \int_{x_1}^{x_2} f(y,y',x)dx$$

Then, $$I = \int_{x_1}^{x_2} \sqrt{(dx)^2 + (dy)^2} = \int_{x_1}^{x_2}\sqrt{1 + (y'(x))^2}dx$$. Now we can ask what we want $$y(x)$$ to be to minimize $$I$$.

Let $$x$$ be an independent variable on the interval $$[x_1,x_2]$$. Let $$y(x)$$ be some differentiable function defined on $$[x_1,x_2]$$ with fixed values at $$x_1$$ and $$x_2$$. Find $$y(x)$$ such that $$I = \int_{x_1}^{x_2}\phi(y,y',x)dx$$ has a stationary value. Let $$\overline{y}(x) = y(x) + \varepsilon\eta(x)$$ be a neighboring path, $$\varepsilon>0$$ be small, and $$\eta(x_1)=\eta(x_2)=0$$.

So,

\begin{align*} I(\overline{y}, \overline{y}', x) &= I(\varepsilon) \\ &= \int_{x_1}^{x_2}\phi(\overline{y}, \overline{y}', x)dx \\ &= \int_{x_1}^{x_2} \phi(y(x) + \varepsilon\eta(x), y'(x), \varepsilon\eta'(x), x)dx \\ &= \int_{x_1}^{x_2} \phi(y(x), y'(x), x) + \frac{\partial\phi}{\partial y}\varepsilon\eta(x) + \frac{\partial\phi}{\partial y'}\varepsilon\eta'(x)dx \\ &= I + \varepsilon\int_{x_1}^{x_2}\frac{\partial\phi}{\partial y}\eta(x) + \frac{\partial\phi}{\partial y'}\eta'(x)dx, \\ 0 = \left[\frac{dI(\varepsilon)}{d\varepsilon}\right]_{\varepsilon=0} &= \int_{x_1}^{x_2}\frac{\partial\phi}{\partial y}\eta(x) + \frac{\partial\phi}{\partial y'}\eta'(x)dx \\ &= \left.\frac{\partial\phi}{\partial y'}\eta(x)\right|_{x_1}^{x_2} + \int_{x_1}^{x_2}\frac{\partial\phi}{\partial y}\eta(x) - \frac{d}{dx}\left(\frac{\partial\phi}{\partial y'}\right)\eta(x)dx \\ &= \int_{x_1}^{x_2}\eta(x)\left(\frac{\partial\phi}{\partial y} - \frac{d}{dx}\left(\frac{\partial\phi}{\partial y'}\right)\right)dx \\ \end{align*}

Since it is zero for all $$\eta(x)$$ then, $$\frac{\partial\phi}{\partial y} - \frac{d}{dx}\left(\frac{\partial\phi}{\partial y'}\right) = 0$$, our condition on $$y$$.

$$\frac{d}{dx}\left(\frac{\partial}{\partial y'}\sqrt{1+y'(x)^2}\right) - \frac{\partial\phi}{\partial y} = \frac{d}{dx}(\frac{2y'(x)}{2\sqrt{1+y'(x)^2}}) = 0$$ Thus, $$y'(x)/\sqrt{1+y'^2}$$ is a constant so $$y''=0$$ hence a straight line.

Let $$I[a,b;\phi;y,y',x]$$ be the definite integral between two fixed points: $$\int_a^b\phi(y,y',x)dx$$. Written $$I[a,b]$$ when $$\phi,y,y',x$$ are understood. The stationary value of $$I[a,b]$$ occurs when $$\delta I[a,b] = 0$$.

• $$\delta y(x)|_{x=a} = 0$$
• $$\delta y(x)|_{x=b} = 0$$
• $$\frac{d}{dx}\delta y = \delta \frac{dy}{dx}$$
• $$\delta\int_a^b y(x)dx = \int_a^b\delta y(x)dx$$
• $$\delta x = 0$$

$$\delta I[a,b] = \delta \int_a^b\phi dx = \int_a^b \delta\phi dx = \int_a^b \frac{\partial\phi}{\partial y}\delta y + \frac{\partial\phi}{\partial y'}\delta y' + \frac{\partial \phi}{\partial x}\delta x dx = \int_a^b \frac{\partial\phi}{\partial y}\delta y + \frac{\partial\phi}{\partial y'}\delta y' dx = \int_a^b \frac{\partial\phi}{\partial y}\delta y + \frac{\partial\phi}{\partial y'}\frac{d}{dx}\delta y dx = \left.\frac{\partial \phi}{\partial y'}\delta y\right|_a^b + \int_a^b \left(\frac{\partial\phi}{\partial y} - \frac{d}{dx}\frac{\partial\phi}{\partial y'}\right)\delta y dx = \int_a^b \left(\frac{\partial\phi}{\partial y} - \frac{d}{dx}\frac{\partial\phi}{\partial y'}\right)\delta y dx$$ This arrives at the same Euler-Lagrange equation of the second kind.

What if we have holonomic constraints: $$f_\ell(y_1,\cdots,y_n,x) = 0$$? Ansatz: Then, we get: $$\delta I[a,b] = \int_a^b \sum_{j}^{n-m}\left(\frac{\partial\phi}{\partial y_j} - \frac{d}{dx}\frac{\partial\phi}{\partial y_j'} - \sum_\ell\lambda_\ell \frac{\partial f_\ell}{\partial y_i}\right)\delta y_j dx = 0$$ Suppose we have $$\phi' = \phi + \sum_\ell \lambda_\ell f_\ell$$. We get,

\begin{align*} 0=\delta I[a,b] &= \int_a^b \delta\phi(y,y',x)dx + \sum_\ell\lambda_\ell\int_a^b\delta f_\ell(y,x)dx \\ &= \sum_j\int_a^b \left(\frac{\partial\phi}{\partial y_j}-\frac{d}{dx}\frac{\partial\phi}{\partial y_j'}\right)\delta y_jdx + \sum_\ell\lambda_\ell\int_a^b \sum_j\frac{\partial f_\ell}{\partial y_j}\delta y_jdx \\ &= \left[\int_a^b \sum_j\left(\frac{\partial\phi}{\partial y_j}-\frac{d}{dx}\frac{\partial\phi}{\partial y_j'} + \sum_\ell\lambda_\ell\frac{\partial f_\ell}{\partial y_j}\right)\delta y_jdx\right] \end{align*}

Hence, we get an Euler-Lagrange Equation of the First Kind, for the covariate coordinates.

Created: 2024-05-30 Thu 21:20

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