In this section we will look at trying to find a function $y(x)$ within $[a,b]$ which minimizes a quantity

💃 Example: the shortest distance from $(0,0)$ to $(1,1)$

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Functional: An operator which takes as input a function and outputs a number

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• For this we use the Length functional $\mathcal L$

  $$ \mathcal L [y]=\int^1_0 \sqrt{1+(y')^2} \,\text dx $$

  💎 Conclusion: here we get $\mathcal L [y_\text{min}]=\sqrt{2}$ since the shortest distance is a straight line

🗒️ Note: more generally to solve we apply small variations to minimise the path

💃 Example: Consider $y=x+\epsilon x(1-x)$ for the bounds $(0,0)$ to $(1,1)$ then here $\epsilon$ is our small varying parameter and $\eta(x)=x(1-x)$ is our varying function. If we derive the expression w.r.t to $\epsilon$ we see that setting $\epsilon$ to 0 recovers the differential equation for $y_\text{min}(x)$ which is of course our straight line from $(0,0)$ to $(1,1)$: $y=x$

Putting this into proper formalism we can look at the following

💼 Case: consider a path $y(x)$ which satisfies endpoint point constraints at $(a,y(a))$ and $(b,y(b))$ and another path $\eta (x)$ for which $\eta(a)=\eta(b)=0$

• Now in that case it is quite clear that $y(x)+\epsilon \eta (x)$ is another possible path

• We define the integrant $F(y,y',x)$ as follows

  $$ I[y+\epsilon \eta (x) ]=\int ^b_a F(y+\epsilon \eta , y'+\epsilon \eta',x)\,\text dx $$

  we can then Taylor expand it in $\epsilon$ giving

  $$ \begin{aligned} I[y+\epsilon \eta (x) ]&=I[y]+\epsilon \int ^b_a \left ( \eta(x)\frac{\partial F}{\partial y}+\eta'\frac{\partial F}{\partial y'}\right ) \,\text dx \\ \Rightarrow \qquad \delta I & = \epsilon \int ^ b_ a \eta (x) \left ( \frac{\partial F}{\partial y}- \frac{\text d }{\text dx} \left ( \frac{\partial F}{\partial y'} \right ) \right )\,\text dx + \epsilon \left [ \eta \frac{\partial F}{\partial y'} \right ]^b_a

  \end{aligned} $$

  🗒️ Note: the boundary term $\left [ \eta \frac{\partial F}{\partial y'} \right ]^b_a=0$ since $\eta(a)=\eta(b)=0$

• The condition for $y$ to be $y_\text{min}$ tis that $\text dI /\text d \epsilon$ to vanish. Since $\eta$ is arbitrary the term it multiples must vanish which gives us the Euler equation

  $$ \frac{\text d}{\text dx } \left (\frac{\partial F}{\partial y'} \right )-\frac{\partial F}{\partial y} =0 $$

• If $F$ has no explicit $x$ dependence then we switch to the second form of Euler equation

  $$ F-y'\frac{\partial F}{\partial y'}=c $$

💼 Case: going back to considering our path length problem $F$ doesn't\ depend on $x$ or $y$ so

• If we apply the first form of Euler's equation (which is not necessary here) we get

  $$ \frac{\text d}{\text dx} \frac{1}{\sqrt{1+(y')^2}}=0 \quad \Rightarrow \quad \frac{y'}{\sqrt{1+(y')^2}}=c $$

  which we can get to be expressed as $y'=\text{constant}$ or $y=Ax+B$

  💎 Conclusion: applying the boundary conditions we get $y=x$ as expected

Now this is very useful and we can use it for many problems, here is what to consider

🗒️ Notes: