Solution

Consider the boy's movements with the speed $\vec{v}$, when he is carried away by the river with the current $\vec{u}$

Let's write in projections on the horizontal and vertical axes, taking into account that $u = 2v$

$$v_x = 2v - v \cos\alpha$$

$$v_y = v \sin\alpha$$

Find the time it takes the boy to swim across the river

$$t = \frac{H}{v_y} = \frac{H}{v \sin\alpha}$$

During time $t$ it will be carried along the coast by an amount

$$L = v_xt = H\frac{2 - \cos\alpha}{\sin\alpha}\quad(1)$$

To find the minimum of $L$, it is necessary to find the extremum of the function $f(\alpha ) = \frac{2 - \cos\alpha}{\sin\alpha}$ on the interval $\alpha\in (0,\pi )$

Let's find $\alpha$, at which the derivative is equal to $0$

$$\frac{df}{d \alpha} = \frac{\sin^2 \alpha - \cos\alpha (2-\cos\alpha )}{\sin^2 \alpha}=0$$

$$1-2 \cos\alpha =0$$

$$\fbox{$\alpha = \pi /3$}$$

Substitute into $(1)$ and find the distance it will be carried away

$$\fbox{$L = H\sqrt{3}$}$$