First a hydraulic jump is where fluid will transition from supercritical (having a Froude Number greater than one, to subcritical (having a Froude Number less than one). When the Froude number is near one, the jump is very weak and somewhat gradual such that you might not even be able to tell it is "jumping".
To figure out where the jump occurs you need to know when the Froude Number will transition from less than one to more than one. To help with this is an extremely useful equation for modeling open channel flow:
$$\frac{dy}{dx}=S\frac{1-\left(\frac{y_n}y\right)^3}{1-\left(\frac{y_c}y\right)^3}$$
Where $y$ represents the depth of flow in a riverbed, x is the distance along the flow, $S$ is the slope of the riverbed (positive being flowing downhill, negative, flowing uphill), $y_c$ is the depth of flow corresponding to $Fr=1$ and $y_n$ is the normal depth, the depth of flow corresponding to flow that is balancing the frictional losses with gravitation gains.
$$y_c=\sqrt[3]{\frac{q^2}g}$$
$$y_n=\sqrt[3]{\frac{q^2}{C^2S}}$$
Where $q$ is the flow rate per width of river, $C$ is Chezy's coefficient (approximated as constant), and $g$ is acceleration due to gravity.
A depth greater than $y_c$ corresponds to a slower velocity, and thus subcritical flow (having a Froude number less than 1). When this is the case, the bottom of the fraction is positive between 0 and 1.
Similarly, when the depth is less than $y_c$ the flow is supercritical and the bottom of the fraction is negative.
In this range therefore, if the top of the fraction is also negative, then the flow will be getting deeper, but if it is positive, the flow will be getting shallower. This corresponds to a depth of flow less than and greater than $y_n$ respectively. This means that while the flow is supercritical, the depth of flow will always be moving towards $y_n$
This means that if you have a steep slope such that $y_n$ is less than $y_c$ the depth of flow will move to and then stay at $y_n$. If you would like the flow to have a hydraulic jump simply decrease the slope till $y_n$ is greater than $y_c$ then the flow will deepen at progressively faster rates until you reach the hydraulic jump.
Note since $q$ is based on your width it's also possible to modify the width rather than slope to cause $y_n$ to be greater than $y_c$.
But wait! After the hydraulic jump the flow will be subcritical, which means the flow downstream can now influence the water level and push the hydraulic jump upstream. How far upstream? Depends on the flow downstream.
In subcritical flows it is easiest to work upstream. If you try to work downstream from an initial guess, you'll end up with non-physical flows unless you happen to guess perfectly. This is because in subcritical flows, as you move downstream, the depth moves away from the normal depth. If you're less than the normal depth, you can quickly end up with a negative depth, and if you're more than the normal depth you can end up with a depth the would overflow your walls quick quickly. However, if you work backwards, going upstream then the depth always moves towards the normal depth, which should always give a valid depth.
So then the question is where to get your initial condition? Downstream of subcritical flow, and upstream on subscrital flow. The opposite transition of hydraulic jump. This transition is always smooth and thus according to the equation always happens when $y=y_n=y_c$ with $y_n$ decreasing downstream. An extreme example would be the lip of a waterfall or edge of an overflowing dam. For given flow rate, the depth will always be easy to predict at these choke points. As an added bonus the slope at these choke points is largely flow rate independent:
$$y_c=y_n$$
$$\sqrt[3]{\frac{q^2}g}=\sqrt[3]{\frac{q^2}{C^2S}}$$
$$\frac1g=\frac1{C^2S}$$
$$S=\frac{g}{C^2}$$
So anytime the slope goes from shallower than $\frac{g}{C^2}$ to steeper there might be a transition to supercritical flow right there, and anytime the slope goes from steeper than $\frac{g}{C^2}$ to shallower there might be a hydraulic jump a bit upstream.
I say might because it is possible that a choke point further downstream is causing the depth to be too high to cause any transitions (you could be at the bottom of a lake for example.)
However, if the slope is constant and shallower than $\frac{g}{C^2}$ for a long time (many miles), then odds are, near the top of this section the depth will have converged to $y_n$. So by this shortcut you don't have to start all the way back from a choke point.
In either case, you can model the flow going upstream from the subcritical section and downstream from the supercritical section. Then having modeled the supercritical section you can plot vs position the depth of flow that would result from a hydraulic jump at that location. Any time that plot intersects the subcritical plot is a location where a hydraulic jump could be stable.
Equations are derived in full in MIT's open courseware chapter on open channels. My analysis of those equations was also guided by the analysis presented there.
