Computational and mathematical physics Stationary action principle

Question

This is a computational and mathematical physics problem that has me frustrated. Been trying for a few hours to no avail. Any help is appreciated

The functional is of the form:

$ I\left [ f \right ]=\int_{0}^{1}F\left [ f,f',x \right ]dx=\int_{0}^{1}\left ( \frac{1}{2}f'\left ( x \right )^{2} +\frac{1}{2}xf\left ( x \right )^{2}+xf\left ( x \right )\right )$

I am asked to check that

$f_{\alpha}\left ( x \right )=\alpha x\left ( 1-x \right )$ satisfy the boundary condition $f_{\alpha}\left ( 0 \right )=f_{\alpha}\left ( 1 \right )=0$

Indeed, it does.

Now, I am asked to plot the exact solution along with $f_{\alpha}$ for various $\alpha$.

I don't understand what this exact solution is but it could be some non-unique trial solution that satisfy the boundary condition. Any clarification at this point might be helpful. Using Mathematica, the various value of $\alpha$ for the function $f_{\alpha}$ is

Then, I am now tasked to compute the stationary value of the functional

$I\left [ f_{\alpha}\left ( x \right ) \right ]$.

The stationary action of any functional I is really just a function f(x) such that the functional I equals to zero. This is equivalent to solving for the Euler-Lagrange equation which is

$\frac{\partial I}{\partial f}-\frac{d}{dt}\left ( \frac{\partial I}{\partial }\dot{f}\right )=0$

This is $\ddot{f}\left ( x \right )+xf\left ( x \right )+x=0$

Mathematica tells me there is no analytical solution to this ODE (weird).

This gives

$\ddot{f_{\alpha}\left ( x \right )}+xf_{\alpha}\left ( x \right )+x=0$

The above ODE has the same form as the one I made the claim that Mathematica has no analytical solution. Then what?

Edit:

solution plot (of airy function) and the blue plot represents a 'trial' solution $f_\alpha(x)$. Geometrically, the values for which the trial solution approximates the exact solution is $\alpha$=-0.5... We now have the trial solution $f_\alpha=-0.5(x)$

In computing the action functional of the trial solution and determining the values for $\alpha$, $\alpha=-10/21$ and $\alpha=0$. We ignore the trivial solution.

Plotting the trial solution $f_\alpha=-10/21(x)$ and $f_\alpha=-0.5(x)$:

Which seems like a pretty good approximation.

Edit:

For a two bifurcation parameter $\alpha$ and $\beta$ case:

For a more general function: $f_\alpha,\beta \left ( x \right )=x\left ( 1-x \right )\left ( \alpha+\beta x \right )$ has two bifurcation parameters.

As usual the boundary condition $f_\alpha,\beta \left ( 0 \right )$=$f_\alpha,\beta \left ( 1 \right )$=0

In computing the stationary values of the functional $I\left [ f_\alpha,\beta \left ( x \right ) \right ]$, it is easy to show that the the values are

$\frac{7\alpha^{2}}{40}+\frac{37\alpha \beta}{21}+\frac{\alpha}{12}+\frac{39 \beta^{2}}{560}+\frac{\beta}{20}$

as verified by Mathematica.

Here is where things starts getting mysterious:

In using the Solve function on Mathematica, there are two $\alpha$ and two $\beta$ values. How do I proceed from here.

Answer 1

I think I understand what the question is getting at.

Below, I plotted the exact solution to the Euler-Lagrange equation $f''(x) - xf(x)-x = 0$ with $f(0) = f(1) = 0$ in blue.

I also plotted the function $f_\alpha(x)$ with $\alpha = -\frac 5 {21}$ in pink.

Notice that $f_{\alpha = - \frac 5 {21}}(x)$ is a fairly good approximation to the exact solution $f(x)$!

So why is this the case? Well, $f(x)$ is the exact solution to the Euler-Lagrange equation, so it is the function that minimises $I[f]$ among the class of all functions obeying the boundary conditions $f(0) = f(1) = 0$.

Unfortunately, as you have noticed, it's very hard to work with the true solution $f(x)$, as we can't write down a simple analytical formula for it. Indeed, I had to calculate $f(x)$ numerically using NDSolve (as @Semiclassical suggested).

However, since we expect the true solution $f(x)$ to be some sort of convex "U-shaped" function, it's reasonable to guess that $f(x)$ can be roughly approximated by $f_\alpha(x) = \alpha x (1-x)$ for some value of $\alpha$. Since $f(x)$ minimises $I[f]$, we expect that the best approximation to $f(x)$ will be given by choosing the value of $\alpha$ such that $I[f_\alpha]$ is a small as possible.

You should now check (by explicitly plugging in the expression for $f_\alpha(x)$ into the expression for $I$) that $I[f_\alpha] = \frac 1 {12} \alpha + \frac {7}{40} \alpha^2$ for any given $\alpha$. You should then verify that the choice of $\alpha$ that makes $I[f_\alpha]$ as small as possible is $\alpha = - \frac 5 {21}$ (plotted in pink above).

Finally, the value of $I[f_{\alpha = \frac{5}{21}}]$ is $ -\frac {5}{504}$, which again you should check. This is an approximation (more specifically, an upper bound) for $I[f]$ evaluated on the true solution $f(x)$.