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Given the Bateman equations $$ N_1(t)=N_1(0)e^{-\lambda_1 t} \\ N_2(t)=N_1(0) \frac{\lambda_1}{\lambda_2-\lambda_1} (e^{-\lambda_1 t}-e^{-\lambda_2 t}) $$

And given that the activities $A_1$ and $A_2$ are related by $A_2=A_1$ at $t=t_A$ and $A_2=3A_1$ at $t=2t_A$, I'm asked to compute $\lambda_1$ and $\lambda_2$ in terms of $t_A$.

I can compute the activity equations by

$$ A_1(t)=-\frac{d N_1}{dt} = \lambda_1 N_1(0)e^{-\lambda_1 t} \\ A_2(t)=-\frac{d N_2}{dt}=N_1(0) \frac{\lambda_1}{\lambda_2-\lambda_1} (\lambda_1 e^{-\lambda_1 t}- \lambda_2 e^{-\lambda_2 t}) $$

And so using these conditions leads to the equations

$$ \lambda_1 N_1(0)e^{-\lambda_1 t_A} = N_1(0) \frac{\lambda_1}{\lambda_2-\lambda_1} (\lambda_1 e^{-\lambda_1 t_A}- \lambda_2 e^{-\lambda_2 t_A}) \\ 3 \lambda_1 N_1(0)e^{-\lambda_1 2 t_A} = N_1(0) \frac{\lambda_1}{\lambda_2-\lambda_1} (\lambda_1 e^{-\lambda_1 2 t_A}- \lambda_2 e^{-\lambda_2 2 t_A}) $$

However, these equations are transcendental, and I cannot solve for $\lambda_1$ & $\lambda_2$ explicitly.

I was wondering, how would I avoid this problem? Are there ways of removing the transcendental nature of the equations? Are there simplifying assumptions that can be made to simplify the equations? How would I solve for $\lambda_1$ & $\lambda_2$ ?

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  • $\begingroup$ Why not solve for lambdas numerically? $\endgroup$ Apr 26, 2020 at 12:36
  • $\begingroup$ It is possible to solve for the lambdas exactly, even if it is a bit heavy. I will detail the calculations in an answer. $\endgroup$ Apr 26, 2020 at 14:24

1 Answer 1

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First, I believe you've made a mistake in the expression of $A_2$. Because of the minus sign in front, it should read:

$$A_2(t) = N_1(0) \frac{\lambda_1}{\lambda_2-\lambda_1} \left(\lambda_1 e^{-\lambda_1 t} - \lambda_2 e^{-\lambda_2 t} \right)$$

Note that the expression you got for $N_2$ is only valid for $\lambda_2 \neq \lambda_1$.

So the first condition can be written instead as:

$$\lambda_1 N_1(0) e^{-\lambda_1 t_A} = N_1(0) \frac{\lambda_1}{\lambda_2-\lambda_1} \left(\lambda_1 e^{-\lambda_1 t_A} - \lambda_2 e^{-\lambda_2 t_A} \right)$$

Simplifying by the common factors, and with a bit of algebra, this can be aranged into the following equation:

\begin{align} &\frac{\lambda_2}{\lambda_2-\lambda_1} e^{-\lambda_2 t_A} = \left(\frac{\lambda_1}{\lambda_2 - \lambda_1} -1 \right) e^{-\lambda_1 t_A}\\ \Leftrightarrow \quad &\frac{\lambda_2}{\lambda_2-\lambda_1} e^{-\lambda_2 t_A} = \frac{2\lambda_1 - \lambda_2}{\lambda_2 - \lambda_1} e^{-\lambda_1 t_A} \\ \Leftrightarrow \quad &\frac{\lambda_2}{2\lambda_1-\lambda_2} = e^{(\lambda_2-\lambda_1) t_A} \\ \end{align}

Similarly, using the second equation, we find:

$$\frac{\lambda_2}{4 \lambda_1 - 3\lambda_2} = e^{2(\lambda_2-\lambda_1) t_A}$$

But the RHS here is just the square of the first RHS, meaning that we must have:

\begin{align} &\left(\frac{\lambda_2}{2\lambda_1-\lambda_2} \right)^2 = \frac{\lambda_2}{4 \lambda_1 - 3\lambda_2} \quad \quad \quad (1)\\ \Leftrightarrow \quad &\lambda_2 (4\lambda_1 - 3\lambda_2) = (2 \lambda_1 - \lambda_2)^2 \\ \Leftrightarrow \quad &4 \lambda_2^2 - 8 \lambda_1 \lambda_2 + 4 \lambda_1^2 = 0 \\ \Leftrightarrow \quad &4 (\lambda_2 - \lambda_1)^2 = 0 \\ \Leftrightarrow \quad &\lambda_2 = \lambda_1 \end{align}

This is contradictory with the assumption that $\lambda_2 \neq \lambda_1$ to derive $N_2(t)$. In fact, if $\lambda_2 = \lambda_1 = \lambda$, it can be shown that the solutions to the Bateman equations are:

\begin{align} N_1(t) &= N_1(0) e^{-\lambda t} \\ N_2(t) &= N_1(0) \lambda t e^{-\lambda t} \end{align}

with the corresponding activities:

\begin{align} A_1(t) &= N_1(0) \lambda e^{-\lambda t} \\ A_2(t) &= N_1(0) \lambda (\lambda t - 1) e^{-\lambda t} \end{align}

Under these conditions, it can be shown easily that $A_1(t_A) = A_2(t_A)$ iff $\lambda = 2/t_A$. You can check that with this choice of $\lambda$, the second condition is automatically fulfilled and that $A_2(2 t_A) = 3 A_1(2 t_A)$, as expected.

Note that this particuliar solution with a linear dependency in front of the exponential for $N_2(t)$ only occurs because $\lambda_2 = \lambda_1$. If we would have chosen pretty much any other conditions, we would have found $\lambda_2 \neq \lambda_1$ by solving the equation equivalent to $(1)$. After finding the relative values of $\lambda_2$ and $\lambda_1$, you can find their absolute value (in terms of $1/t_A$) by injecting the equality between $\lambda_1$ and $\lambda_2$ into the first equation for instance.


Numerical check:

as it is always nice to check this kind of lengthy calculations, I have implemented numerically the Bateman equations in their differential form into Python. You can check that for $\lambda_1 = \lambda_2 = 2/t_A$ this yields the correct ratio of activities at $t = t_A$ and $t = 2t_A$.

general plot

zoom

Please find below the Python script used to generate these pictures:

import numpy as np



tA = 1.0 #arbitraty units
lambda_1 = 2/tA
lambda_2 = 1.*lambda_1
N0_1 = 1 #also arbitrary
N0_2 = 0 #assuming no atoms 2 at the beginning
t_max = 4*tA
N_steps = 1000
dt = t_max/N_steps

N_steps_tA = N_steps//4 #number of steps corresponding to t=tA
N_steps_2tA = 2*N_steps_tA #number of steps corresponding to t=2tA

N1 = np.zeros(N_steps+1)
N1[0] = N0_1
N2 = np.zeros(N_steps+1)
N2[0] = N0_2

for i in range(1, N_steps+1):
    N1[i] = (N1[i-1] - lambda_1*dt*N1[i-1]/2)/(1 + lambda_1*dt/2) #half step euler method
    N2[i] = (N2[i-1] - lambda_2*dt*N2[i-1]/2 + N1[i-1] - N1[i])/(1 + lambda_1*dt/2)

@np.vectorize    
def A1(N):
    return -(N1[N+1]-N1[N-1])/(2*dt)

@np.vectorize
def A2(N):
    return -(N2[N+1]-N2[N-1])/(2*dt)


print(A1(N=N_steps_tA), A2(N=N_steps_tA))
print(3*A1(N=N_steps_2tA), A2(N=N_steps_2tA))
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  • $\begingroup$ You're right about the incorrect A2 equation - I'll update it there. $\endgroup$
    – David
    Apr 26, 2020 at 20:10

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