In 1977, Marchetti (1977) introduced a type of analysis of fuel source market penetration for the global energy market, characterized by the following equation (with \(F\) being the fraction of the market penetrated): $$ \frac{1}{F} \cdot \frac{d F}{d t}=\alpha \cdot(1-F) $$ a. What is \(F\) as a function of \(t\) ? b. Comment on the predictions made by this researcher concerning the current global energy source mix.

Short Answer

Expert verified
Answer: As time approaches infinity, the fraction of the market penetrated by the fuel source approaches 1, indicating that the fuel source will eventually saturate the market. However, this model is a simplification and does not account for the complexities of the actual global energy source mix.

Step by step solution

01

Identify the type of differential equation

We are given the following differential equation: $$ \frac{1}{F} \cdot \frac{dF}{dt} = \alpha \cdot (1 - F) $$ This is a first-order separable differential equation since we can separate the variables F and t on different sides of the equation.
02

Separate the variables

To separate the variables, we need to move all terms with F to the left side of the equation and all terms with t to the right side: $$ \frac{1}{F(1-F)} \cdot \frac{dF}{dt} = \alpha $$
03

Integrate both sides

Now, we will integrate both sides of the equation with respect to their respective variables: $$ \int \frac{1}{F(1-F)} dF = \int \alpha dt $$
04

Solve the left side integral

To solve the left side integral, we use partial fraction decomposition: $$ \frac{1}{F(1-F)} = \frac{A}{F} + \frac{B}{1-F} $$ By equating coefficients, we find that A = 1 and B = -1. Thus, integrating: $$ \int \left( \frac{1}{F} - \frac{1}{1-F} \right) dF = \ln\left(\frac{F}{1-F}\right) + C_1 $$
05

Solve the right side integral

The right side integral is straightforward: $$ \int \alpha dt = \alpha t + C_2 $$
06

Combine the results and solve for F(t)

Now, we can equate the results of the left and right side integrals: $$ \ln\left(\frac{F}{1-F}\right) + C_1 = \alpha t + C_2 $$ To solve for F(t), we first eliminate the constants C_1 and C_2 by defining a new constant C = C_1 - C_2: $$ \ln\left(\frac{F}{1-F}\right) = \alpha t + C $$ Next, we take the exponential of both sides: $$ \frac{F}{1-F} = e^{\alpha t + C} = e^C \cdot e^{\alpha t} $$ Let K = e^C, so: $$ \frac{F}{1-F} = K e^{\alpha t} $$ Finally, we solve for F(t): $$ F(t) = \frac{K e^{\alpha t}}{1 + K e^{\alpha t}} $$ Thus, the function F(t) is given by: $$ F(t) = \frac{K e^{\alpha t}}{1 + K e^{\alpha t}} $$ #b. Comment on the predictions made by this researcher concerning the current global energy source mix# When analyzing the model, we can notice that as time increases (t -> infinity), the term \(e^{\alpha t}\) approaches infinity, making the function F(t) approach 1. This would mean that, according to the model, the fuel source will eventually saturate the market. However, this model makes a simplification that there is only one competing fuel source and does not take into account the complex dynamics of multiple sources, the development of new technologies, and the influence of other external factors such as environmental policies and economic considerations. Therefore, while the model provides insight into the growth of a fuel source in the market, the current global energy source mix is more complex than the model suggests and is influenced by a wide range of factors.

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