Consider two simple ODEs with identical right-hand sides, whose initial conditions are related by a simple formula.

$$x' = a \cdot x, \quad y' = a \cdot y, \quad x(0) = \alpha \cdot y (0)$$

The solutions to these are

$$y(t) = y(0)\exp(c \cdot t)$$

$$x(t) = x(0)\exp(c \cdot t) = \frac{\alpha \cdot y(0)}{y(0)} y(t).$$

In particular, this allows us to solve for $x(t)$ only by knowing $y(t)$, without having to integrate the right-hand side of the ODE describing $x$.

I would like to extend this idea to systems of ODEs. For example, consider the systems

$$ \begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, \quad \begin{bmatrix} y_1' \\ y_2' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} $$

with initial conditions related by

$$ \begin{bmatrix} x_1(0) \\ x_2(0) \end{bmatrix} = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} \begin{bmatrix} y_1 (0) \\ y_2(0) \end{bmatrix}. $$

Is it possible to express the solution to $\vec{x}(t)$ in terms of $\vec{y}(t)$, $\alpha$, $\beta$, $\gamma$, and $\delta$?

To preempt the obvious question, I need to solve many systems of this type with varying initial conditions, and would like to do the numerical integration only once.

  • $\begingroup$ For simple homogeneous D.E. you have the exact solution. Why need the numerical integration? $\endgroup$ – Abu Bakar Jul 20 '15 at 0:38
  • $\begingroup$ I have found it to be faster than calculating the matrix exponential when the system is large. Is there a better way? It's possible I'm doing something dumb. $\endgroup$ – rmccloskey Jul 20 '15 at 0:48
  • $\begingroup$ Shouldn't the equation-3 be: $x(t)=\alpha y(0)y(t)$ $\endgroup$ – Abu Bakar Jul 20 '15 at 1:04
  • $\begingroup$ I don't think so: I'm substituting $\exp(ct) = y(t) / y(0)$. $\endgroup$ – rmccloskey Jul 20 '15 at 1:09
  • $\begingroup$ Then $x(t) = \alpha y(t)$. According to the equation, wouldn't $x$ be independent of its initial conditions? $\endgroup$ – Abu Bakar Jul 20 '15 at 1:15

Something similar will work. You need to solve two ODEs $$ \begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} ,\quad \begin{bmatrix} y_1' \\ y_2' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} $$ with initial conditions $$ \begin{bmatrix} x_1(0) \\ x_2(0)\end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} y_1(0) \\ y_2(0)\end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} .$$ Then the solution to $$ \begin{bmatrix} z_1' \\ z_2' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} $$ is $$ \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = z_1(0) \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + z_2(0) \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} .$$

  • $\begingroup$ Also, this will work for non-homogeneous systems. All you need is linearity. $\endgroup$ – Stephen Montgomery-Smith Jul 20 '15 at 0:53
  • $\begingroup$ Should it be $z_1(0) [x_1, x_2] + z_2(0) [y_1, y_2]$? If not, where did the $z(1)$ come from? $\endgroup$ – rmccloskey Jul 20 '15 at 1:06
  • $\begingroup$ Yes, I fixed it. Thanks. $\endgroup$ – Stephen Montgomery-Smith Jul 20 '15 at 1:11
  • $\begingroup$ @StephenMontgomery-Smith Your answer is not being rendered correctly in my browser (chrome). There is no issue with other posts. $\endgroup$ – Abu Bakar Jul 21 '15 at 9:06
  • $\begingroup$ @Galaxy If you go to my profile, it will give you a link to my webpage, which has my email address. If you send me an email, I'll return you a PDF of my answer rendered correctly. (I also am using chrome.) $\endgroup$ – Stephen Montgomery-Smith Jul 21 '15 at 12:50

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