Original synthetic study
Same first year different long term decline tails
Curves matched at the initial rate and the first-year endpoint can still have different long-term tails. Matching two observations does not identify a unique decline model.
By PetroCalcHub Team · Updated October 24, 2026 · Free to read and use
Match the first-year condition explicitly
The study fixes qi = 1,000 STB/day and q at day 365 to 700 STB/day. For each selected b, Di is solved so that the endpoint condition is exactly satisfied.
This differs from the basic model comparison that holds Di fixed. Here the fitted initial decline changes with b, because matching the same endpoint requires it.
Compare what was not used for matching
At later times the models separate despite agreeing at the two matched points. Those later predictions reveal tail sensitivity that the first-year condition cannot constrain.
The table also reports cumulative production at the selected horizon. Rate endpoint agreement does not imply equal area under the curves even during the matched first year.
Interpret the study as nonuniqueness
No field observations are used and no family is declared more accurate. A real forecast would require additional history, physical context and a terminal rule.
The downloadable data include each b and its derived Di so the matched condition can be reproduced rather than assumed.
Explore the example
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
Select one of 5 recorded scenarios. Intermediate states are not interpolated.
Elapsed time (day) = 365. Exponential rate (STB/day): 700; Hyperbolic b=0.5 rate (STB/day): 700; Harmonic rate (STB/day): 700.
- Exponential rate (STB/day)
- Hyperbolic b=0.5 rate (STB/day)
- Harmonic rate (STB/day)
Worked example
Matched first-year endpoint, different tails
qi = 1,000 STB/day; all curves match 700 STB/day at day 365. Derived daily Di values for b 0, 0.5, 1: 0.000977191627, 0.0010697458, 0.0011741683.
Relation and unit basis
q = qi exp(−Dit) for b = 0; q = qi(1 + bDit)^(−1/b) otherwise
Time in days, rate in STB/day. b = 1 is harmonic. These are synthetic model curves, not field fits.
1. Define the baseline
Elapsed time (day) = 365
qi = 1,000 STB/day; all curves match 700 STB/day at day 365. Derived daily Di values for b 0, 0.5, 1: 0.000977191627, 0.0010697458, 0.0011741683.
2. Select the relation
q = qi exp(−Dit) for b = 0; q = qi(1 + bDit)^(−1/b) otherwise
Time in days, rate in STB/day. b = 1 is harmonic. These are synthetic model curves, not field fits.
3. Substitute the values
At day 365 each derived Di makes q/qi = 0.7, so q = 700 STB/day.
Retain intermediate precision. The displayed values are rounded only after evaluation.
4. Read the result
Exponential rate (STB/day) = 700; Hyperbolic b=0.5 rate (STB/day) = 700; Harmonic rate (STB/day) = 700
Matching one endpoint does not fix the intervening integral or the later tail.
Result and interpretation
Exponential rate (STB/day) = 700; Hyperbolic b=0.5 rate (STB/day) = 700; Harmonic rate (STB/day) = 700
Matching one endpoint does not fix the intervening integral or the later tail.
Compare later rates after checking that every model gives exactly 700 STB/day at day 365.
Study data and reproduction
Original synthetic teaching calculation. Equations and fixed controls are stated here; these are not field measurements.
| Elapsed time (day) | Exponential rate (STB/day) | Hyperbolic b=0.5 rate (STB/day) | Harmonic rate (STB/day) |
|---|---|---|---|
| 0 | 1,000 | 1,000 | 1,000 |
| 365 | 700 | 700 | 700 |
| 730 | 490 | 517.2312 | 538.4615 |
| 1,825 | 168.07 | 256.0727 | 318.1818 |
| 3,650 | 28.24752 | 114.7316 | 189.1892 |
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. If two curves match the first-year endpoint, must their first-year cumulative production agree?
Show the worked answer
No. Different shapes between the endpoints can have different integrals.
2. Reproduce the baseline Exponential rate (STB/day) without rounding intermediate values.
Show the worked answer
At day 365 each derived Di makes q/qi = 0.7, so q = 700 STB/day.. Exponential rate (STB/day) = 700. Matching one endpoint does not fix the intervening integral or the later tail.
3. Compare the first and last recorded scenarios. Which input changes?
Show the worked answer
Elapsed time (day) changes from 0 to 3650. Other controls remain as stated in the premise. Exponential rate (STB/day) changes from 1000 to 28.2475249.
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. Time in days, rate in STB/day. b = 1 is harmonic. These are synthetic model curves, not field fits. Matching one endpoint does not fix the intervening integral or the later tail.
Before using the result
- Solve Di separately for each b.
- Check the matched endpoint.
- Do not rank synthetic tails as field accuracy.
Matching one endpoint does not fix the intervening integral or the later tail.
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.