Engineering lesson
Modified hyperbolic decline and the terminal decline switch
Modified hyperbolic decline switches to an exponential tail when instantaneous nominal decline reaches a chosen terminal value. Continuity requires carrying the rate at the switch into the exponential segment.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Find the switch time on one basis
For D(t) = Di/(1 + bDit), the switch time is (Di/Dterm − 1)/(bDi) when Di exceeds Dterm and b is positive. All decline units must match.
The synthetic example uses daily declines throughout. A terminal percentage quoted effectively per year must first be converted according to its stated definition.
Preserve rate continuity
Evaluate the hyperbolic rate at ts, then use q(ts) exp[−Dterm(t − ts)] afterward. Restarting the tail at qi would create an artificial rate jump.
Cumulative production after the switch is the hyperbolic integral up to ts plus the exponential integral from that starting rate.
A terminal rule is an assumption
The terminal decline is selected for the scenario; it is not discovered from an equation alone. Record why it is appropriate to the intended forecast.
The explorer samples both sides of the switch. It checks the mathematics of the piecewise curve without claiming the selected tail predicts a field well.
Explore the example
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
Select one of 7 recorded scenarios. Intermediate states are not interpolated.
Elapsed time (day) = 1875. Modified rate (STB/day): 176.776695; Unswitched hyperbolic rate (STB/day): 176.776695.
- Modified rate (STB/day)
- Unswitched hyperbolic rate (STB/day)
Worked example
Continuous hyperbolic-to-exponential switch
qi = 1,000 STB/day, Di = 0.002/day, b = 0.8, terminal nominal decline 0.0005/day. Switch time = 1875 days.
Relation and unit basis
q = qi(1 + bDi min(t,ts))^(−1/b) exp[−Dt max(0,t−ts)]
Both decline coefficients are nominal per day. The tail starts at the rate reached at ts.
1. Define the baseline
Elapsed time (day) = 1875
qi = 1,000 STB/day, Di = 0.002/day, b = 0.8, terminal nominal decline 0.0005/day. Switch time = 1875 days.
2. Select the relation
q = qi(1 + bDi min(t,ts))^(−1/b) exp[−Dt max(0,t−ts)]
Both decline coefficients are nominal per day. The tail starts at the rate reached at ts.
3. Substitute the values
ts = (0.002/0.0005 − 1)/(0.8 × 0.002) = 1875 days; q(ts) = 1,000(1 + 0.8 × 0.002 × ts)^−1.25
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Modified rate (STB/day) = 176.776695; Unswitched hyperbolic rate (STB/day) = 176.776695
Rate is continuous at the switch; the later exponential tail no longer follows the unswitched hyperbolic decline.
Result and interpretation
Modified rate (STB/day) = 176.776695; Unswitched hyperbolic rate (STB/day) = 176.776695
Rate is continuous at the switch; the later exponential tail no longer follows the unswitched hyperbolic decline.
Read the samples immediately before and after the computed switch; the rate should remain continuous.
Exact scenario data
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Elapsed time (day) | Modified rate (STB/day) | Unswitched hyperbolic rate (STB/day) |
|---|---|---|
| 0 | 1,000 | 1,000 |
| 1,000 | 302.8889 | 302.8889 |
| 1,874 | 176.8651 | 176.8651 |
| 1,875 | 176.7767 | 176.7767 |
| 1,876 | 176.6883 | 176.6883 |
| 3,650 | 72.77608 | 90.40238 |
| 7,300 | 11.73279 | 41.79275 |
Scroll the table to see all columns. The highlighted row is the worked baseline. Downloads retain full numerical precision.
Test your understanding
Work through the question before revealing its answer. Use the stated assumptions and include units in every numerical step.
1. What initial rate belongs in the exponential tail?
Show the worked answer
The hyperbolic rate at the switch, q(ts), not the original qi.
2. Reproduce the baseline Modified rate (STB/day) without rounding intermediate values.
Show the worked answer
ts = (0.002/0.0005 − 1)/(0.8 × 0.002) = 1875 days; q(ts) = 1,000(1 + 0.8 × 0.002 × ts)^−1.25. Modified rate (STB/day) = 176.776695. Rate is continuous at the switch; the later exponential tail no longer follows the unswitched hyperbolic decline.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Elapsed time (day) changes from 0 to 7300. Other controls remain as stated in the premise. Modified rate (STB/day) changes from 1000 to 11.7327881.
4. What would be needed to apply this example to a different physical system?
Show the worked answer
Replace the synthetic controls with compatible documented inputs and check the relation's assumptions. Both decline coefficients are nominal per day. The tail starts at the rate reached at ts. Rate is continuous at the switch; the later exponential tail no longer follows the unswitched hyperbolic decline.
Before using the result
- Use nominal terminal decline on a matched time basis.
- Carry q(ts) into the tail.
- Add cumulative segment integrals.
Rate is continuous at the switch; the later exponential tail no longer follows the unswitched hyperbolic decline.
Continue learning
Start with the broader workflow guide
Sources and method
The explanation and synthetic examples are original PetroCalcHub teaching material. The references below support the definitions and model context. They do not supply field validation for the synthetic scenarios.