Non-Conservative Forces

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

Then, $I={\int }_{x_1}^{x_2}\sqrt{(dx{)}^2+(dy{)}^2}={\int }_{x_1}^{x_2}\sqrt{1+(y^{\prime }(x){)}^2}dx$. 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 $I={\int }_{x_1}^{x_2}\varphi (y,y^{\prime },x)dx$ has a stationary value. Let $\bar{y}(x)=y(x)+\epsilon \eta (x)$ be a neighboring path, $\epsilon >0$ be small, and $\eta (x_1)=\eta (x_2)=0$.

So,

$$\begin{array}{rl}I(\bar{y},{\bar{y}}^{\prime },x)& =I(\epsilon )\\ & ={\int }_{x_1}^{x_2}\varphi (\bar{y},{\bar{y}}^{\prime },x)dx\\ & ={\int }_{x_1}^{x_2}\varphi (y(x)+\epsilon \eta (x),y^{\prime }(x),\epsilon {\eta }^{\prime }(x),x)dx\\ & ={\int }_{x_1}^{x_2}\varphi (y(x),y^{\prime }(x),x)+\frac{\partial \varphi }{\partial y}\epsilon \eta (x)+\frac{\partial \varphi }{\partial y^{\prime }}\epsilon {\eta }^{\prime }(x)dx\\ & =I+\epsilon {\int }_{x_1}^{x_2}\frac{\partial \varphi }{\partial y}\eta (x)+\frac{\partial \varphi }{\partial y^{\prime }}{\eta }^{\prime }(x)dx,\\ 0={\left[\frac{dI(\epsilon )}{d\epsilon }\right]}_{\epsilon =0}& ={\int }_{x_1}^{x_2}\frac{\partial \varphi }{\partial y}\eta (x)+\frac{\partial \varphi }{\partial y^{\prime }}{\eta }^{\prime }(x)dx\\ & ={\frac{\partial \varphi }{\partial y^{\prime }}\eta (x)|}_{x_1}^{x_2}+{\int }_{x_1}^{x_2}\frac{\partial \varphi }{\partial y}\eta (x)-\frac{d}{dx}\left(\frac{\partial \varphi }{\partial y^{\prime }}\right)\eta (x)dx\\ & ={\int }_{x_1}^{x_2}\eta (x)(\frac{\partial \varphi }{\partial y}-\frac{d}{dx}\left(\frac{\partial \varphi }{\partial y^{\prime }}\right))dx\end{array}$$

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

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

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

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

$\delta I[a,b]=\delta {\int }_a^b\varphi dx={\int }_a^b\delta \varphi dx={\int }_a^b\frac{\partial \varphi }{\partial y}\delta y+\frac{\partial \varphi }{\partial y^{\prime }}\delta y^{\prime }+\frac{\partial \varphi }{\partial x}\delta xdx={\int }_a^b\frac{\partial \varphi }{\partial y}\delta y+\frac{\partial \varphi }{\partial y^{\prime }}\delta y^{\prime }dx={\int }_a^b\frac{\partial \varphi }{\partial y}\delta y+\frac{\partial \varphi }{\partial y^{\prime }}\frac{d}{dx}\delta ydx={\frac{\partial \varphi }{\partial y^{\prime }}\delta y|}_a^b+{\int }_a^b(\frac{\partial \varphi }{\partial y}-\frac{d}{dx}\frac{\partial \varphi }{\partial y^{\prime }})\delta ydx={\int }_a^b(\frac{\partial \varphi }{\partial y}-\frac{d}{dx}\frac{\partial \varphi }{\partial y^{\prime }})\delta ydx$ 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: $\delta I[a,b]={\int }_a^b\sum _j^{n-m}(\frac{\partial \varphi }{\partial y_j}-\frac{d}{dx}\frac{\partial \varphi }{\partial y_j^{\prime }}-\sum _{\ell }{\lambda }_{\ell }\frac{\partial f_{\ell }}{\partial y_i})\delta y_{j}dx=0$ Suppose we have ${\varphi }^{\prime }=\varphi +\sum _{\ell }{\lambda }_{\ell }{f}_{\ell }$. We get,

$$\begin{array}{rl}0=\delta I[a,b]& ={\int }_a^b\delta \varphi (y,y^{\prime },x)dx+\sum _{\ell }{\lambda }_{\ell }{\int }_a^b\delta f_{\ell }(y,x)dx\\ & =\sum _j{\int }_a^b(\frac{\partial \varphi }{\partial y_j}-\frac{d}{dx}\frac{\partial \varphi }{\partial y_j^{\prime }})\delta y_{j}dx+\sum _{\ell }{\lambda }_{\ell }{\int }_a^b\sum _j\frac{\partial f_{\ell }}{\partial y_j}\delta y_{j}dx\\ & =\left[{\int }_a^b\sum _j(\frac{\partial \varphi }{\partial y_j}-\frac{d}{dx}\frac{\partial \varphi }{\partial y_j^{\prime }}+\sum _{\ell }{\lambda }_{\ell }\frac{\partial f_{\ell }}{\partial y_j})\delta y_{j}dx\right]\end{array}$$

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