Function extremal - calculus of variations

Question

Find a curve passing through (1,2) and (2,4) that is an extremal of the function:

$J(x,y')=\int_1^2 xy'(x)+(y'(x))^2dx$

I don't know what methods to use at all.

Answer 1

Hint: In general, a necessary condition for $$J = \int_A^B f(x, y, y') dx$$ to attain an extremum is that $y(x)$ must satisfy the Euler-Lagrange differential equation: $$\frac{\partial f}{\partial y} - \frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right) = 0.$$

In your case, $$\frac{\partial f}{\partial y} = 0 \quad \text{and} \quad \frac{\partial f}{\partial y'} = x + 2y',$$ so differentiating $x + 2y'$ with respect to $x$ and equating it to zero we obtain $$1 + 2y'' = 0,$$ which is a simple homogeneous second order differential equation. It can be shown (by integrating $1 + 2y'' = 0$ twice) that $$y(x) = -\frac{x^2}{4} + Ax + B.$$

Since $y(x)$ passes through $(1, 2)$ and $(2, 4)$, we must have $$y(1) = 2 = -\frac{1}{4} + A + B \quad \text{and} \quad y(2) = 4 = -1 + 2A + B.$$ The first equation yields $A = 9/4 - B$, which we substitute into the second equation to get $$4 = -1 + 2\left(\frac{9}{4} - B\right) + B.$$ Solving for $B$ we find $B = -1/2$, which means $A = 9/4 + 1/2 = 11/4$.

Answer 2

This is a very simple problem. You are trying to find the function $y$ which gives a minimum of the following integral:

$$ J = \int\left(xy' + y'^2\right)dx = \int\mathcal{L}(y, y', x)dx $$

Where $y$ is a function of $x$. The very first thing to notice is that $\mathcal{L}$ has no dependence on $y$! Therefore $\frac{\partial \mathcal{L}}{\partial y} = 0$. This leads to the following Euler-Lagrange equation:

$$ \frac{d}{dx}\left(\frac{\partial \mathcal{L}}{\partial y'}\right) = 0 $$

If the full derivative with respect to $x$ is $0$, then the this partial must be a constant. Let me write this out in two steps:

\begin{align} \frac{\partial \mathcal{L}}{\partial y'} =& \alpha \\ \frac{d}{dx}\left(\frac{\partial \mathcal{L}}{\partial y'}\right) =& \frac{d}{dx}\left(\alpha\right) = 0 \end{align}

Now simply set the partial with respect to $y'$ to a constant:

$$ \frac{\partial \mathcal{L}}{\partial y'} = x + 2y' = \alpha $$

This is a separable equation, just integrate both sides:

$$ 2y' = \alpha - x \\ 2y = \alpha x - \frac{x^2}{2} + \beta \\ y = \alpha x - \frac{x^2}{4} + \beta \\ y = -\frac{1}{4}\left(x^2 - 2\alpha x\right) + \beta \\ y = -\frac{1}{4}\left(x - \alpha\right)^2 + \beta - \alpha^2 \\ y = -\frac{1}{4}\left(x - \alpha\right)^2 + \beta $$

Notice that $\alpha$ and $\beta$ changed a lot in the last few steps. Rememeber these are constants, so I can make them whatever I like. Just as one example, in the last step I had $\beta - \alpha^2$ then changed to just $\beta$. This is a constant minus another constant which gives some other constant. That is, the new $\beta$ is whatever the old one was minus $\alpha^2$.

Now find $\alpha$ and $\beta$ for your problem:

$$ y(1) = 2 = -\frac{1}{4}(1 - \alpha)^2 + \beta \\ y(2) = 4 = -\frac{1}{4}(2 - \alpha)^2 + \beta \\ $$

Subtract the equations to find $\alpha$:

$$ 2 = -\frac{1}{4}\left((2 - \alpha)^2 - (1 - \alpha)^2\right) \\ -8 = 4 + \alpha^2 - 4\alpha - (1 + \alpha^2 - 2\alpha) \\ -8 = 3 - 2\alpha \rightarrow \alpha = \frac{11}{2} $$

Now use either equation to find $\beta$:

$$ 2 = -\frac{1}{4}\left(1 - \frac{11}{2}\right)^2 + \beta \\ \beta = 2 + \frac{1}{4}\cdot\frac{9^2}{2^2} = \frac{113}{16} $$

Giving:

$$ y(x) = -\frac{1}{4}\left(x - \frac{11}{2}\right)^2 + \frac{113}{16} = \frac{113 - (2x - 11)^2}{16}\\ y' = -\frac{2x - 11}{4} \rightarrow y'^2 = \frac{(2x - 11)^2}{16} $$

Finally, if you want to find the value of the integral:

\begin{align} J =& \int\limits_1^2\left(-x\frac{2x - 11}{4} + \frac{(2x - 11)^2}{16}\right)dx = \frac{1}{16}\int\limits_1^2\left(44x - 8x^2 + (2x - 11)^2\right)dx \\ J =& \frac{1}{16}\left.\left(22x^2 - \frac{8}{3}x^3 + \frac{1}{2\cdot 3}(2x - 11)^3\right)\right|_1^2 \\ J=& \frac{1}{16}\left(22\cdot3 - \frac{8}{3}\cdot7 + \frac{1}{6}\cdot386\right) = \frac{1}{16}\left(\frac{198 - 56 + 193}{3}\right) = \frac{335}{48} \end{align}