3x3 non-linear system of equations (Known and unkown isotope decay)

Question

TLDR version: is it possible to solve the following system for X, Y, Z?

$C_1=X+Y$

$C_2=C_3 X + Y Z^{C_4} $

$C_5=C_6 X + Y Z^{C_7} $

Background/Context: I have a mixture with a known isotope A (known decay constant $λ_Α$) and an unknown one B (unknown decay constant $λ_B$) in unknown proportions. When taking three activity measurements at different times $t_0$, $t_1$ and $t_2$ we get the following system of equations:

$Measurement(t_0)=A_0+B_0$

$Measurement(t_1)=A_0e^{-λ_At_1}+B_0e^{-λ_Bt_1} $

$Measurement(t_2)=A_0e^{-λ_At_2}+B_0e^{-λ_Bt_2} $

where $A_0$, $B_0$ and $λ_B$ are unknowns.

After bunching some known constants together, I reached the TLDR form of the system where X is $A_0$, Y is $B_0$ and Z is $e^{λ_B}$.

Thanks

Answer 1

I don't know if you can solve it, but the first steps seems pretty easy, Use the first equation to remove $X$ from the others:

\begin{align} C_1&=X+Y\\ C_2&=C_3 X + Y Z^{C_4} \\ C_5&=C_6 X + Y Z^{C_7} \end{align} becomes \begin{align} C_2&=C_3 (C_1 - Y) + Y Z^{C_4} \\ C_5&=C_6 (C_1 - Y) + Y Z^{C_7} \end{align} and now, to save ugliness, I'm going to change to mostly lowercase, and combine some constants into new ones.

\begin{align} u&=a - by + y Z^{C_4} \\ v&=c - dy + y Z^{C_7} \end{align} Letting $z = Z^{C_4}$, and $s = C_7/C_4$ \begin{align} u&=a - by + y z \\ v&=c - dy + y z^s \end{align} \begin{align} u-a&=y(z - b)\\ v-c&=y(z^s - d) \end{align} Now taking a quotient we get $$ \frac{u-a}{v-c} = \frac{z-b}{z^s - d} $$ which I'll write $$ K(z^s- 1) = z - b $$ so $$ Kz^s - z + b - K = 0 $$ Sadly, for various powers of $s$, such an equation can have many roots; any one of these will produce a value for $y$, and hence for $x$.

So...unless your $s$ value is nice, like $s = 1$, (or you have some bounded range for $z$, which might help), chances are good there's no unique solution.

Answer 2

Make sure to choose $t_2=2t_1$.

Then the system writes

$$\begin{cases}A+B&=M_0, \\Ar+Bs&=M_1, \\Ar^2+Bs^2&=M_2,\end{cases}$$

where $r$ and $M_k$ are known.

Solving the first two,

$$A=\frac{M_0s-M_1}{s-r},\\B=\frac{M_1-M_0r}{s-r}$$ and plugging in the third,

$$(M_0s-M_1)r^2+(M_1-M_0r)s^2=M_2(s-r)$$ is quadratic in $s$.

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Of course you can also work with other $\dfrac{t_2}{t_1}=\alpha$ ratios, but the equation gets a little harder, of the form

$$s^\alpha+ps+q=0.$$

Answer 3

Let's simplify a bit the notation and write $$ \left\{ \matrix{ C_{\,0} = A + B \hfill \cr C_{\,1} = Ae^{ - \alpha \,t} + Be^{ - \beta \,t} \hfill \cr C_{\,2} = Ae^{ - \alpha \,u} + Be^{ - \beta \,u} \hfill \cr} \right. $$

After that, let's continue and semplify further, by putting $$ \left\{ \matrix{ 1 = {A \over {C_{\,0} }} + {B \over {C_{\,0} }} = a + b \hfill \cr {{C_{\,1} } \over {C_{\,0} }} = c = ae^{ - \alpha \,t} + be^{ - \beta \,t} \hfill \cr {{C_{\,2} } \over {C_{\,0} }} = d = ae^{ - \alpha \,u} + be^{ - \beta \,u} \hfill \cr} \right. $$ and $$ \left\{ \matrix{ 1 = a + b \hfill \cr c = e^{ - \alpha \,t} + b\left( {e^{ - \beta \,t} - e^{ - \alpha \,t} } \right) = e^{ - \alpha \,t} \left( {1 + b\left( {e^{\left( {\alpha - \beta } \right)\,t} - 1} \right)} \right) \hfill \cr d = e^{ - \alpha \,u} + b\left( {e^{ - \beta \,u} - e^{ - \alpha \,u} } \right) = e^{ - \alpha \,u} \left( {1 + b\left( {e^{\left( {\alpha - \beta } \right)\,u} - 1} \right)} \right) \hfill \cr} \right. $$ and also $$ \left\{ \matrix{ 1 = a + b \hfill \cr \gamma = ce^{\alpha \,t} \hfill \cr \delta = de^{\alpha \,u} \hfill \cr \gamma - 1 = b\left( {e^{\left( {\alpha - \beta } \right)\,t} - 1} \right) \hfill \cr \delta - 1 = b\left( {e^{\left( {\alpha - \beta } \right)\,u} - 1} \right) \hfill \cr} \right. $$ and finally reach to $$ \left\{ \matrix{ 1 = a + b \hfill \cr \gamma = ce^{\alpha \,t} \hfill \cr \delta = de^{\alpha \,u} \hfill \cr \gamma - 1 = b\left( {e^{\left( {\alpha - \beta } \right)\,t} - 1} \right) \hfill \cr {{\delta - 1} \over {\gamma - 1}} = \left( {{{e^{\left( {\alpha - \beta } \right)\,u} - 1} \over {e^{\left( {\alpha - \beta } \right)\,t} - 1}}} \right) \hfill \cr} \right. $$ where $\gamma, \, \delta, u, t$ are known, so that:
- the 5th equation will give $\alpha -\beta$;
- the 4th will give $b$;
- the remaining will permit to go back to the original quantities $A,B$.

In the last equation , the function $$ {{e^{\,u\,x} - 1} \over {e^{\,t\,x} - 1}} $$ is steadily increasing and convex , for $t < u$, and therefore it allows to be easily solved numerically.