Let's consider a shallow-water free-surface jet exhausting in the X direction from a thin rectangular slot into a quiescent fluid. Within the jet the mean velocity field can be expressed as follows: $$u(x)= \frac{n}{kx+s}$$
where $$u(x = 0) = U_0$$
Jets exhausting into quiescent fluid spread linearly due to entrainment of ambient fluid so their velocity decay (conservation of momentum) but let's not deepen the subject about jets since that is out of the scope of the present question.
To consider instead the path of a fluid particle that moves within the jet under the approximation that the particle moves always with a velocity that equals the mean field, one must resolve the differential equation: $$\frac{dx}{dt}= \frac{n}{kx+s}$$ to yield (only positive solution): $$ x(t) = \sqrt{s^2+c_1k+2nkt}-s$$
which can be solved knowing that $$x(t = 0) = 0$$
to obtain $$c_1 = 0$$ and $$x(t) = \sqrt{s^2+2nkt}-s$$
Let's now add a force acting in the Y direction perpendicular to the jet direction of propagation. As a result, the jet curves in the direction of the force and its trajectory will depend on the force in the Y direction. we could consider the problem, from the particle point of view, as a bullet shot from a certain high parallel to the ground under the effect of gravity but with the difference of a velocity decay in the X direction as seen above.
Let's consider that the particle trajectory is a polynomial of a certain order (no higher than 2), so it can be expressed in the plane X-Y as:$$ y = Ax^f+Bx^{f-1}+C$$
and its derivative $$\frac{dy}{dx} = fAx^{f-1}+(f-1)B$$
differentiating with respect with t yields: $$\frac{dy}{dx}\frac{dt}{dt} = fAx^{f-1}+(f-1)B$$ which means $$\frac{dy}{dt} = [fAx^{f-1}+(f-1)B]\frac{dx}{dt}$$
now we consider separately the 2 cases: f = 1 and f = 2.
- f = 1 so the trajectory is a straight line
to compute the force one must derive $$\frac{dy}{dt} = A\frac{dx}{dt}$$ with respect to t: $$\frac{d^2y}{dt^2} = A\frac{d^2x}{dt^2}$$ considering the previous expression of x(t) one obtain: $$ \frac{d^2y}{dt^2} = -A\frac{kn^2}{\sqrt{(s^2+2nkt)^3}}$$ and substituting x to t (to go back to the jet instead of the particle):
$$ F_y \sim -A\frac{kn^2}{\sqrt{(x^2k^2+2xks+s^2)^3}}$$
which means that $$ F_y \sim x^{-\frac{3}{2}}$$
Let's now consider the case with f = 2 so that the trajectory is a parabola.
$$\frac{dy}{dt} = (2Ax+B)\frac{dx}{dt}$$
and deriving with respect to t:
$$\frac{d^2y}{dt^2} = \frac{d(2Ax+B)}{dt}\frac{dx}{dt}+ (2Ax+B)\frac{d^2x}{dt^2}$$
$$\frac{d^2y}{dt^2} = \frac{2An}{\sqrt{2nkt+s^2}}*\frac{n}{\sqrt{2nkt+s^2}}-(\frac{2A(\sqrt{s^2+2nkt}-s)}{k}+B)*\frac{kn^2}{\sqrt{(2nkt+s^2)^3}}$$
$$\frac{d^2y}{dt^2} = \frac{2An^2}{2nkt+s^2}-(\frac{2An^2}{2nkt+s^2}+\frac{-2Asn^2}{\sqrt{(2nkt+s^2)^3}}+\frac{Bkn^2}{\sqrt{(2nkt+s^2)^3}})$$
$$\frac{d^2y}{dt^2} = n^2(\frac{2As-Bk}{\sqrt{(2nkt+s^2)^3}})$$
and replacing x to t:
$$F_y \sim n^2(\frac{2As-Bk}{\sqrt{(x^2k^2+2kxs+s^2)^3}})$$
which means that $$ F_y \sim x^{-\frac{3}{2}}$$ which is the same result that I get for the line-trajectory.
Now I think it's impossible to get different trajectories when the force is basically the same and I expected a different dependance of $$F_y(x)$$ in the case of a parabola.
it seems that in the case of the parabola, this equation $$\frac{d^2y}{dt^2} = \frac{d(2Ax+B)}{dt}\frac{dx}{dt}+ (2Ax+B)\frac{d^2x}{dt^2}$$ is equivalent to (except for B) $$ 2Af^{'}*f^{'}+2Af*f^{''}$$ which $$ f^{'}*f^{'}+f*f^{''} = 0$$ for f = x(t). due to the presence of B, I get the same results as per the straight line.
Can anyone enlighten me, please?
** I apologize in advance for any calculation errors I could have committed while typing but the problem stands