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.


1 Answer 1


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




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


(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 $y'(x)$.

  • $\begingroup$ Brilliant, very clear. Thank you! $\endgroup$
    – Maximaxmax
    Dec 16, 2022 at 17:32

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