Why Use a Piecewise Exponential Model?
A single exponential model assumes one constant rate throughout the time period being studied. In practice, conditions can change: a population may grow quickly during one period and more slowly later, or a substance may be exposed to different decay conditions at different times.
A piecewise exponential model represents these phases with separate exponential formulas. Each phase has a constant rate of its own. The essential bookkeeping rule is that the amount at the end of one phase becomes the starting amount for the next. A rate can change abruptly while the amount itself remains continuous.
Here, \(A(t)\) denotes an amount at time \(t\), and each rate constant \(k_i\) is measured in inverse units of time. Positive \(k_i\) describes growth, negative \(k_i\) describes decay, and \(k_i=0\) describes a constant amount.
One Formula for Each Time Interval
Suppose the rate changes at times \(t_1,t_2,\ldots,t_{n-1}\), with \(t_0<t_1<\cdots<t_n\). On the interval from \(t_{i-1}\) to \(t_i\), assume the rate constant is \(k_i\). If the amount at the start of that interval is \(A(t_{i-1})\), the exponential model for that phase is
The time difference \(t-t_{i-1}\) is the elapsed time within this interval. The exponent is the rate for that interval multiplied by its elapsed duration. Because \(k_i\) has units of inverse time, the exponent is dimensionless, as required for an exponential.
For example, with two intervals and initial amount \(A_0=A(t_0)\), the model is
Writing the second formula this way matters: its starting amount is \(A(t_1)\), not the original amount \(A_0\). At \(t=t_1\), the first formula gives \(A(t_1)\), and the second gives \(A(t_1)e^0=A(t_1)\). Thus the formulas agree at the change time.
Calculating Across Several Phases
For a time \(t\) in the \(j\)th interval, the amount can also be written directly from the initial value. The amount is multiplied by the exponential growth or decay factor from every completed interval, followed by the factor for the elapsed part of the current interval:
Since a product of exponentials is the exponential of the sum of their exponents, the same calculation can be written as
Each term \(k_i(t_i-t_{i-1})\) uses the rate and duration from one phase. This form is useful for calculations, but the interval-by-interval form makes it especially clear how continuity is enforced.
Why Matching at the Change Time Works
The reason for carrying the ending amount into the next interval follows directly from the exponential formula. Suppose a phase begins at \(t_{i-1}\) with amount \(A(t_{i-1})\). Its amount at the end is
The next phase starts at \(t_i\) with this ending amount. If its rate is \(k_{i+1}\), then its formula is \(A(t_i)e^{k_{i+1}(t-t_i)}\). At the change time itself, this gives
The preceding phase also gives \(A(t_i)\) there. Therefore the two pieces have the same value at the boundary. Applying this argument at each change time shows that the constructed model is continuous across all phases.
Continuity does not require the rates on either side to be equal. The amount can pass smoothly through a change time while its growth or decay rate changes. In a model with distinct rates on the two sides, the slope of the amount-versus-time graph generally changes there as well.
Worked Examples
Worked Example: Growth Followed by Decay
An amount starts at \(100\) units at time \(t=0\). It grows with rate constant \(k_1=\ln 2\) per day for \(2\) days, then decays with rate constant \(k_2=-\ln 2\) per day for the next \(1.5\) days. Find the amount at \(t=3.5\), and check continuity at the change time.
First calculate the amount at the change time, \(t=2\):
The second interval starts from \(400\) units. Its duration is \(3.5-2=1.5\) days, so
Thus the amount is \(100\sqrt{2}\), approximately \(141.42\) units. At the boundary, the first piece gives \(400\), while the second gives \(400e^{(-\ln 2)(2-2)}=400\). The pieces agree, so the model is continuous at \(t=2\).
As a further check on the total calculation, the growth factor is \(e^{2\ln 2}=4\), and the decay factor is \(e^{-1.5\ln 2}=1/(2\sqrt{2})\). Their product is \(4/(2\sqrt{2})=\sqrt{2}\), giving \(100\sqrt{2}\) from the initial \(100\) units.
Worked Example: Three Intervals with Different Rates
An amount begins at \(80\) units. Its rate is \(\ln(3/2)\) per week from \(t=0\) to \(t=2\), \(\ln(2/3)\) per week from \(t=2\) to \(t=3\), and \(\ln 2\) per week from \(t=3\) to \(t=4.5\). Calculate the amount at each change time and at \(t=4.5\).
During the first two weeks,
The second interval lasts one week and starts at \(180\):
The third interval lasts \(4.5-3=1.5\) weeks and starts at \(120\):
The amount at \(t=4.5\) is \(240\sqrt{2}\), approximately \(339.41\) units. At \(t=2\), the first formula gives \(180\), and the second gives \(180e^0=180\). At \(t=3\), the second gives \(120\), and the third gives \(120e^0=120\). Both change-time checks confirm continuity.
Worked Example: Finding an Unknown Rate
An amount starts at \(200\) units and grows with rate constant \(k_1=\ln(3/2)/2\) per year for the first \(2\) years. It then follows a constant-rate exponential model for \(3\) more years. At \(t=5\), the amount is \(240\) units. Find the rate \(k_2\) for the second interval and check the resulting amount.
First determine the amount at \(t=2\):
Continuity means the second interval starts at \(300\). Its duration is \(5-2=3\), so the observation at \(t=5\) gives
Divide by \(300\), then take the natural logarithm of both positive sides:
Because \(4/5<1\), its logarithm is negative, so \(k_2\) describes decay. To check, substitute this rate back into the second-interval formula:
The model reproduces the observed amount. At the change time, the first interval ends at \(300\), and the second interval starts at \(300e^0=300\), as required.
A Common Mistake: Restarting from the Original Amount
A rate change does not normally reset the amount. For example, suppose an amount starts at \(100\), grows at rate \(\ln 2\) per day for \(2\) days, and then decays at rate \(-\ln 2\) per day for \(1\) day.
A wrong calculation applies the second rate to the original \(100\):
This skips the growth phase. The correct amount at the change time is
Now apply the second rate for one day starting from \(400\):
The result is also checked by multiplying the two phase factors: \(100\cdot4\cdot(1/2)=200\). The key is to use each rate for its own duration and carry the resulting amount forward.
Interpreting and Checking the Model
A piecewise exponential model is determined by its initial amount, its change times, and the rate on each interval. Before calculating, check that each rate is paired with the correct interval and that time units are consistent. Then calculate in order: find the amount at the first change, use it as the next starting amount, and continue.
The amount remains continuous under the matching rule, but the model does not claim that the rate is constant across a change time. A change from \(k_i\) to \(k_{i+1}\) changes the exponential factor used after that time. It does not create an instantaneous increase or decrease in the amount.
Check Your Understanding
For each question, identify the interval durations and carry the amount forward at every change time.
- An amount starts at \(50\) units, grows at rate \(\ln 2\) per hour for \(1\) hour, then decays at rate \(-\ln 2\) per hour for \(2\) hours. What is its final amount?
- Why must the second interval’s formula use the amount at the first change time rather than the original amount?
- An amount starts at \(120\) units and has rate \(k_1=\ln(3/2)\) per day for \(2\) days. What amount should be used to start the next interval?
- If the rate changes at \(t=4\), what does continuity require of the two formulas at \(t=4\)?
- An interval begins with \(300\) units and ends three years later with \(240\) units under a constant rate \(k\). Write an equation that can be used to find \(k\).
