Specific energy from alternate depths

Set-up

For a rectangular channel with discharge $q$ per unit width, two alternate depths $y_1$ and $y_2$ have the same specific energy $E$:

$$E=y_1+\frac{q^2}{2gy_1^2}=y_2+\frac{q^2}{2gy_2^2}\qquad(1)$$

Eliminate $q$

From the equality in (1):

$$y_2-y_1=\frac{q^2}{2g}\left(\frac{1}{y_1^2}-\frac{1}{y_2^2}\right)=\frac{q^2}{2g}\,\frac{(y_2-y_1)(y_2+y_1)}{y_1^2y_2^2}$$

Cancelling $(y_2-y_1)\neq0$:

$$\frac{q^2}{2g}=\frac{y_1^2y_2^2}{y_1+y_2}\qquad(2)$$

Specific energy

Substitute (2) into $E=y_1+\dfrac{q^2}{2gy_1^2}$:

$$E=y_1+\frac{y_1^2y_2^2}{(y_1+y_2)y_1^2}=y_1+\frac{y_2^2}{y_1+y_2}$$ $$E=\frac{y_1(y_1+y_2)+y_2^2}{y_1+y_2}$$ $$\boxed{E=\frac{y_1^2+y_1y_2+y_2^2}{y_1+y_2}}$$

Hence proved.