# Calculus of variations and integrating $\left(\frac{dy}{dx}\right)^2=\alpha^2(1+x)$

I'm trying to wrap my head around a step in the solution of an exercise that has me stumped. The exercise is the following :

Given the functional $$S[y]=\int_0^{1}dx\sqrt{1+x+y'^2}, \quad y(0)=x_0, \quad y(1)=x_1$$

it is asked to show that $$y(x)$$ defined by $$y'(x)=k\sqrt{1+x+y'(x)^2},$$ where $$k$$ is a constant, makes the functional stationary. Then, by expressing $$y'(x)$$ in terms of $$x$$, one should show that the solution is $$y(x)=x_0+\frac{x_1-x_0}{2^{3/2}-1}\left((1+x)^{3/2}-1\right).$$

It has been shown previously in the course that given the functional $$S[y]=\int_a^{b}dxF(x,y'),\quad y(a)=A, \quad y(b)=B,$$ we obtain the following differential equation for the stationary path $$y$$ of $$S$$: $$\frac{\partial}{\partial y'}F(x,y')=k, \quad y(a)=A, \quad y(b)=B,$$ where $$k$$ is a constant.

Now, the solution of the initial problem goes as follows:

We have $$F(x,v)=\sqrt{1+x+v^2}$$ and the general equation (above) becomes $$v=k\sqrt{1+x+v^2}, \quad \text{where}\quad v=y'(x).$$ Rearranging and squaring, we obtain $$\left(\frac{dy}{dx}\right)^2=\alpha^2(1+x), \quad \alpha^2=\frac{k^2}{1-k^2}.$$

Integrating gives the solution $$y(x)-x_0=\alpha\int_0^xdx\sqrt{1+x}=\frac{2\alpha}{3}\left((1+x)^{3/2}-1\right).$$

It then goes on to find the value of $$\alpha$$ with the boundary conditions.

My questioning mostly pertains to the integration part. I feel like it is very vague, and I'm not sure how to go about such an integration and come out with the desired result. What about the integration limits $$0$$ and $$x$$? Secondarily, I also wonder about the need to introduce the variable $$v$$. Any help would be tremendously appreciated.

If I understand you correctly, the part you're confused about is how we can go from

$$(y'(x))^2=\alpha^2(1+x)$$

to

$$y(x)-x_0=\alpha\int_0^x\sqrt{1+x}~\mathrm{d}x=\frac{2\alpha}{3}\left((1+x)^{3/2}-1\right).$$

So what we start by doing is to just rewrite the first equation as

$$y'(x)=\alpha\sqrt{1+x}$$

(note the absence of a $$\pm$$ in front of it; that is because we can just bake the sign into $$\alpha$$ and determine it later). Now recall that by the fundamental theorem of calculus,

$$\int_a^b y'(t)~\mathrm{d}t=y(b)-y(a),$$

and so in particular, since we have the value $$y(0)=x_0$$, we can choose $$a=0$$ (but we could just as well have chosen some other value if that was more convenient), and as we wish to find an expression for $$y(x)$$, we set $$b=x$$, so that

$$\int_0^x y'(t)~\mathrm{d}t=y(x)-x_0.$$

This means that, using the equality we had above,

$$y(x)-x_0=\int_0^x y'(t)~\mathrm{d}t=\int_0^x \alpha\sqrt{1+t}~\mathrm{d}t.$$

Now the right hand side we can calculate as

$$\int_0^x \alpha\sqrt{1+t}~\mathrm{d}t=\alpha\int_1^{x+1} u^{1/2}~\mathrm{d}u=\alpha\biggl[\frac{2u^{3/2}}{3}\biggr]_1^{x+1}=\frac{2\alpha}{3}\left((1+x)^{3/2}-1\right),$$

where we made the substitution $$u=1+t$$ as an intermediate step. This now gives the desired equality.

As for your second question, no, there is really no need to introduce $$v=y'(x)$$ other than, perhaps, to make it easier to solve $$y'(x)=\alpha\sqrt{1+x+y'(x)^2}$$ for $$y'(x)$$.

• Brilliant, very clear. Thank you! Dec 16, 2022 at 17:32