So i've been given this job, i could use some help please.
This is what i have so far:
$$J(x,y,y') = \int_{a}^{b} \sqrt{1+\left|\frac{dy}{dx}\right|^{2}} dx \implies $$
$$J(t,x,y,\dot{x},\dot{y}) = \int_{0}^{1} \Phi (t,x,y,\dot{x},\dot{y})~dt=\int_{0}^{1} \sqrt{\dot{x}^2+\dot{y}^2}~dt$$
we have that $$\delta J(x,y;h_1,h_2) = \int_{0}^{1} h_1 \left[\frac{\partial \Phi}{\partial x} – \frac{d}{dt} \frac{d \Phi }{d\dot{x}}\right] + h_2 \left[\frac{\partial \Phi}{\partial y} – \frac{d}{dt} \frac{d \Phi }{d\dot{y}}\right]~dt$$
Based on the idea that $$\delta K(x,u,u';h) = \int_{a}^{b} \left[ \frac{\partial \Lambda}{\partial u}(x,u,u')v + \frac{\partial \Lambda}{\partial u'}(x,u,u')v'\right] = \int_{a}^{b} \left[ \frac{\partial \Lambda}{\partial u} – \frac{d}{dx} \frac{\partial \Lambda}{\partial u'}\right]~dx$$
$$\delta J(x,y;h_1,h_2) = \int_{0}^{1} h_1 \left[\frac{\partial \Phi}{\partial x} – \frac{d}{dt} \frac{d \Phi }{d\dot{x}}\right] + h_2 \left[\frac{\partial \Phi}{\partial y} – \frac{d}{dt} \frac{d \Phi }{d\dot{y}}\right]~dt $$
$$= \int_{a}^{b} h_1 \left[ – \frac{\dot{y}(\ddot{x}\dot{y}-\dot{x}\ddot{y})}{(\dot{x}^2 + \dot{y}^2)^{3/2}} \right] +h_2 \left[ – \frac{\dot{x}(\ddot{y}\dot{x}-\dot{y}\ddot{x})}{(\dot{x}^2 + \dot{y}^2)^{3/2}} \right] dt = \int_{a}^{b} \frac{\ddot{x}\dot{y}(h_2 \dot{x}-h_1 \dot{y}) + \ddot{y}\dot{x}(h_1\dot{y}-h_2 \dot{x})}{(\dot{x}^2 + \dot{y}^2)^{3/2}} dt$$
this entire thing seems needlessly complicated to me, where am i going wrong? is the above correct or should i just recalculate using
$$\delta j(\mathbf{u};\mathbf{v}) = \left.\frac{d}{dt} J(\mathbf{u}+t\mathbf{v})\right|_{t=0}$$?
thanks for the help.
Best Answer
Your calculation looks correct (assuming, of course, that your derivatives are fine, which I did not check). However, notice that there's no point in merging $h_1$ and $h_2$ inside a single integral. They are merely helper functions to aid your expansion.
What you need to calculate is $\frac{\delta J}{\delta x(t')}$ and $\frac{\delta J}{\delta y(t')}$, which you find by letting $h_1(t) \to \delta(t'-t)$ and $h_2(t) \to \delta(t'-t)$, so by removing these $h$'s your solution simplifies a bit.