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Suppose among all the continuously differentiable functions $y(x), x\in \mathbb{R}$, with $y(0)=0$ and $y(1)=\frac{1}{2},$ the function $y_0(x)$ minimizes the functional,

$$\displaystyle\int_{0}^{1}e^{-(y'-x)}+(1+y)y'dx$$ Find the value of $y(\frac{1}{2}).$

My attempt: Using Euler's equation $F_y-\frac{d}{dx}F_{y'}=0$ here $F(x,y,y')=e^{-(y'-x)}+(1+y)y'$, we get

$$\implies y'-\frac{d}{dx}(-e^{x-y'}+1+y)=0$$ $$\implies y''-y'e^{y'-x}-1=0$$ (after simplification)

But now I am stuck as I don't know how to proceed from here? Please help me out. Thanks!

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Harry Potter
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1 Answers1

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I think you made a mistake going from the first line of your equation to the second:

$$ y' - \frac{d}{dx}(-e^{x-y'} + 1 + y) = 0 \\ \Rightarrow (1 - y'')e^{x-y'} = 0 $$

Which is only true if $y'' \equiv 1$. Which gives $y = \frac 12 x^2$ with the boundary conditions, and so $y\left(\frac 12\right) = \frac 18$. I haven't checked whether or not this is actually a minimum, you'll have to compute the second variation to get that.

David Kleiman
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