Linear Form Involving Integration by Parts
The first example will be solving the following first-order linear ODE:
\begin{equation}
\frac{dy}{dx} + 2y = \sin x \tag{1}
\end{equation}
with the I.C. \( y(0) = 1/2 \). It has \( P(x) = 2 \) and hence the integrating factor is \( \exp(\int P(x) dx) = \exp(\int 2 dx) = e^{2x} \). Multiplying it on both sides leads to
\begin{align}
e^{2x}\frac{dy}{dx} + 2e^{2x}y &= e^{2x}\sin x \\
\frac{d}{dx}(e^{2x}y) &= e^{2x}\sin x \\
e^{2x}y &= \int e^{2x}\sin x dx \tag{2}
\end{align}
Set \( I = \int e^{2x}\sin x dx \) and this integral on the R.H.S. can be dealt with Integration by Parts (I.B.P.) in elementary Calculus. By I.B.P. twice, we can obtain
\begin{align}
I &= \int e^{2x}\sin x dx = \frac{1}{2} \int \sin x d(e^{2x}) \\
&= \frac{1}{2} [e^{2x}\sin x] -\frac{1}{2} \int e^{2x} d(\sin x) & & \text{(First I.B.P.)} \\
&= \frac{1}{2} [e^{2x}\sin x] -\frac{1}{2} \int e^{2x} \cos x dx \\
&= \frac{1}{2} [e^{2x}\sin x] -\frac{1}{4} \int \cos x d(e^{2x}) \\
&= \frac{1}{2} [e^{2x}\sin x] -\frac{1}{4} [e^{2x}\cos x] + \frac{1}{4} \int e^{2x} d(\cos x) & & \text{(Second I.B.P.)} \\
&= \frac{1}{2} [e^{2x}\sin x] -\frac{1}{4} [e^{2x}\cos x] -\frac{1}{4} \int e^{2x}\sin x dx \\
&= \frac{1}{2} [e^{2x}\sin x] -\frac{1}{4} [e^{2x}\cos x] -\frac{1}{4} I \tag{3}
\end{align}
Therefore
\begin{align}
I &= \frac{1}{2} [e^{2x}\sin x] -\frac{1}{4} [e^{2x}\cos x] -\frac{1}{4} I \\
\frac{5}{4} I &= \frac{1}{2} [e^{2x}\sin x] -\frac{1}{4} [e^{2x}\cos x] \\
I &= \frac{2}{5}e^{2x}\sin x -\frac{1}{5}e^{2x}\cos x + C \tag{4}
\end{align}
(Don’t forget to append the integration constant!) Substituting (4) back into (2) gives
\begin{align}
e^{2x}y &= \frac{2}{5}e^{2x}\sin x -\frac{1}{5}e^{2x}\cos x + C \\
y &= \frac{2}{5}\sin x -\frac{1}{5}\cos x + Ce^{-2x} \tag{5}
\end{align}
Using the I.C. yields
\begin{align}
(\frac{1}{2}) &= \frac{2}{5}(0) -\frac{1}{5}(1) + C(1) \\
\Rightarrow C &= \frac{7}{10} \tag{6}
\end{align}
So the full solution is
\begin{equation}
y = \frac{2}{5}\sin x -\frac{1}{5}\cos x + \frac{7}{10}e^{-2x} \tag{7}
\end{equation}
Logistic Population Model
The second example is the so-called Logistic Population Model in which the rate of change in population is simply given by the ODE
\begin{equation}
\frac{dN}{dt} = KN(M-N) \tag{8}
\end{equation}
where \( N = N(t) \) is the population at time \( t \), \( K \) is the natural growth factor, and \( M \) is the maximum sustainable population. \( KN \) represents the birth capacity, and \( M − N \) acts like a soft barrier to limit the population increase which corresponds to real-life restrictions like food consumption.
Now we want to solve the ODE, given that the I.C. at \( t = 0 \) is \( N(0) = M/4 \): the initial population is a quarter of the maximum population. We can approach this as a separable ODE:
\begin{align}
\frac{dN}{dt} &= KN(M-N) \\
\int \frac{dN}{N(M-N)} &= \int K dt \\
\int \frac{1}{M}(\frac{1}{N} + \frac{1}{M-N}) dN &= \int K dt \tag{9}
\end{align}
where we have utilized the partial fractions \( 1/(N(M-N)) = 1/M[1/N + 1/(M-N)] \). Continuing from (9) leads to
\begin{align}
\int \frac{1}{M}(\frac{1}{N} + \frac{1}{M-N}) dN &= \int K dt \\
\frac{1}{M} (\ln |N| -\ln |M-N|) &= Kt + C \\
\ln \left|\frac{N}{M-N}\right| &= MKt + MC \\
\left|\frac{N}{M-N}\right| &= e^{MKt + MC} \\
\frac{N}{M-N} &= \pm e^{MKt + MC} = Ae^{MKt} \tag{10}
\end{align}
where we let \( A = \pm e^{MC} \). Then
\begin{align}
N &= (M-N)A e^{MKt} \\
N (1+Ae^{MKt}) &= MAe^{MKt} \\
N &= \frac{MAe^{MKt}}{1+Ae^{MKt}} \tag{11}
\end{align}
Plugging in the I.C. \( N(0) = M/4 \) gives
\begin{align}
(\frac{M}{4}) &= \frac{MA(1)}{1+A(1)} = \frac{MA}{1+A} \\
\frac{1}{4} &= \frac{A}{1+A} \Rightarrow A = \frac{1}{3} \tag{12}
\end{align}
So the full solution is
\begin{align}
N &= \frac{M(1/3)e^{MKt}}{1+(1/3)e^{MKt}} \\
&= \frac{Me^{MKt}}{3+e^{MKt}} \tag{13}
\end{align}
Below is a schematic plot showing the case when \( K = 0.001, M = 100 \).
where the graph is a sigmoid function.
Exercise – Barometric Equation
Derive the Barometric Equation with an isothermal atmosphere (the temperature \(T = T_0\) being a constant), with the Ideal Gas Law
\begin{equation}
p = \rho R_{d}T \tag{14}
\end{equation}
and the Hydrostatic Equation
\begin{equation}
\frac{dp}{dz} = -\rho g \tag{15}
\end{equation}
where \( p \) is the pressure, \( \rho \) is the density, \( R_{d} = 287.0 \text{ J} \text{ kg}^{-1} \text{ K}^{-1} \) is the dry gas constant, and \( g = 9.81 \text{ m} \text{ s}^{-2} \) is the gravitational acceleration. You should be able to obtain
\begin{equation}
p = p_0e^{-(g/R_dT_0)z} = p_0e^{-z/H_0} \tag{16}
\end{equation}
where \( p_0 \) is the sea-level pressure. The quantity \( H_0 = R_dT_0/g \) is known as the scale height.
Then, re-do this with an atmospheric temperature profile that is linearly decreasing with height, \( T = T_0 -\Gamma z\). You should be able to get
\begin{equation}
p = p_0\left(\frac{T_0-\Gamma z}{T_0}\right)^{g/R_d\Gamma} \tag{17}
\end{equation}
If \( p_0 = 1000 \text{ hPa}\), \( T_0 = 20 \text{ }^\circ\text{C}\), and \( \Gamma = 6.5 \text{ K} \text{ km}^{-1} \), find the altitude at which the atmospheric pressure drops to \(30 \%\) of the surface value, i.e. \( 300 \text{ hPa} \), in both cases and explain the comparison based on atmospheric thermodynamics.
Answer
Combining (14) and (15):
\begin{align}
\frac{dp}{dz} &= -\rho g = -\frac{p}{R_dT_0} g \\
\int_{p_0}^p \frac{dp}{p} &= -\int_0^z \frac{g}{R_dT_0} dz \\
[\ln |p|]_{p_0}^p = \ln (\frac{p}{p_0}) &= -\frac{g}{R_dT_0}z \\
p &= p_0e^{-(g/R_dT_0)z}
\end{align}
where we have included the initial condition \( p(z=0) = p_0\) at the sea-level surface in the lower limit of integration.
For the second part, the modified hydrostatic equation is
\begin{align}
\frac{dp}{dz} &= -\frac{g}{R_d(T_0-\Gamma z)} p \\
\frac{dp}{p} &= -\frac{g}{R_d(T_0-\Gamma z)} dz
\end{align}
Integrating from the surface to some height again, we have
\begin{align}
\int_{p_0}^{p} \frac{dp}{p} &= -\int_0^z \frac{g}{R_d(T_0-\Gamma z)} dz \\
\ln |\frac{p}{p_0}| &= \frac{g}{R_d\Gamma} [\ln |T_0-\Gamma z|]_0^z \\
\ln (\frac{p}{p_0}) &= \frac{g}{R_d\Gamma} \ln(\frac{T_0-\Gamma z}{T_0}) \\
\ln (\frac{p}{p_0}) &= \ln\left\{\left(\frac{T_0-\Gamma z}{T_0}\right)^{g/R_d\Gamma}\right\} \\
p &= p_0\left(\frac{T_0-\Gamma z}{T_0}\right)^{g/R_d\Gamma}
\end{align}
Plugging in all the given values (convert the original units to SI units properly!), the pressure decreases to \( 30 \%\) of the surface value when \( z \approx 9.229 \text{ km} \) now, which is lower when compared to \( z \approx 10.326 \text{ km} \) for an isothermal atmosphere, because the decreasing temperature going up implies thinner layers.
A small extra challenge: Show that (17) is reduced to (16) in the limit of \( \Gamma \to 0 \).
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