Let $a<b$ and let $f(x)$ be a continuously differentiable function on the interval $[a, b]$ with $f(x)>0$ for all $x\in [a, b]$. Let $A>0, B>0$ be constants. Show that the stationary path of the functional $$S[y]=\int_{a}^{b}f(x)\sqrt{1+y'^2}dx, y(a)=A, y(b)=B,$$ is given by $$y(x)=A+\beta\int_{a}^{x}\frac{1}{\sqrt{f(w)^2-\beta^2}}dw,$$ where $\beta$ is a constant satisfying $$B-A=\beta\int_{a}^{b}\frac{1}{\sqrt{f(w)^2-\beta^2}}dw.$$
Here's my work:
Let $A>0, B>0$ be constants.
Consider the functional $S[y]=\int_{a}^{b}f(x)\sqrt{1+y'^2}dx, y(a)=A, y(b)=B$.
By definition, the Euler-Lagrange equation for the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B$ is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B$.
Let $F(x, y, y')=f(x)\sqrt{1+y'^2}$.
Then $\frac{\partial F}{\partial y'}=\frac{f(x)y'}{\sqrt{1+y'^2}}$ and $\frac{\partial F}{\partial y}=0$.
This means $\frac{d}{dx}(\frac{\partial F}{\partial y'})=\frac{d}{dx}(\frac{f(x)y'}{\sqrt{1+y'^2}})=\frac{df(x)}{dx}\cdot \frac{y'}{\sqrt{1+y'^2}}+\frac{f(x)\cdot y''}{\sqrt{1+y'^2}}$.
Now we have $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0\implies \frac{df(x)}{dx}\cdot \frac{y'}{\sqrt{1+y'^2}}+\frac{f(x)\cdot y''}{\sqrt{1+y'^2}}-0=0\implies \frac{df(x)}{dx}\cdot \frac{y'}{\sqrt{1+y'^2}}+\frac{f(x)\cdot y''}{\sqrt{1+y'^2}}=0$.
Thus $\frac{1}{\sqrt{1+y'^2}}(\frac{df(x)}{dx}\cdot y'+f(x)\cdot y'')=0\implies \frac{df(x)}{dx}\cdot y'+f(x)\cdot y''=0$.
Now I'm stuck. How should I proceed from here by solving this second-order differential equation and find the stationary path?
Best Answer
You've made great progress in deriving the Euler-Lagrange equation for the functional $S[y] = \int_{a}^{b} f(x) \sqrt{1 + y'^2} \, dx$ with the given boundary conditions $y(a) = A$ and $y(b) = B$. Let's continue from where you left off.
From the Euler-Lagrange equation, you have: $$\frac{d}{dx} \left( \frac{f(x) y'}{\sqrt{1 + y'^2}} \right) = 0$$
This means that the quantity $\frac{f(x) y'}{\sqrt{1 + y'^2}}$ is constant along the stationary path. Let's denote this constant as $\beta$. So, we have: $$\frac{f(x) y'}{\sqrt{1 + y'^2}} = \beta$$
Now, let's solve for $y'$. Rearranging the equation, we get: $$y' = \beta \frac{\sqrt{1 + y'^2}}{f(x)}$$
Square both sides to eliminate the square root: $$y'^2 = \beta^2 \frac{1 + y'^2}{f(x)^2}$$
Rearrange and solve for $y'^2$: $$y'^2 = \frac{\beta^2}{f(x)^2 - \beta^2}$$
Taking the square root (and considering the positive root since $f(x) > 0$), we get: $$y' = \frac{\beta}{\sqrt{f(x)^2 - \beta^2}}$$
Now, integrate this expression to find $y(x)$. Remember, $y(a) = A$, so we will integrate from $a$ to $x$: $$y(x) = A + \int_{a}^{x} \frac{\beta}{\sqrt{f(w)^2 - \beta^2}} \, dw$$
The value of $\beta$ must be determined from the boundary condition $y(b) = B$. Substituting $x = b$ in the above equation, we get: $$B = A + \int_{a}^{b} \frac{\beta}{\sqrt{f(w)^2 - \beta^2}} \, dw$$
Rearranging, we find: $$B - A = \beta \int_{a}^{b} \frac{1}{\sqrt{f(w)^2 - \beta^2}} \, dw$$
This final equation gives us the condition that $\beta$ must satisfy. Thus, the stationary path $y(x)$ is indeed: $$y(x) = A + \beta \int_{a}^{x} \frac{1}{\sqrt{f(w)^2 - \beta^2}} \, dw$$ with $\beta$ satisfying $$B - A = \beta \int_{a}^{b} \frac{1}{\sqrt{f(w)^2 - \beta^2}} \, dw$$
This completes the derivation.