# What does the exponential function mean in numerical ODE solving formulas?

I'm trying to read papers on numerical ODE algorithms and I always seem to stumble upon huge amounts of exponentials multiplied by each other. For example in New families of symplectic splitting methods for numerical integration in dynamical astronomy there are a few of them. Is it the ordinary exponential that is meant or is it just like a way of writing something else (just like with complex numbers)?

## 1 Answer

This is a way of writing the flow of a vector field. In other words, if you have an ordinary differential equation (ODE) given by $$\dot{x}(t) = v(x(t))$$ with initial value $$x(0) = x_0$$, you could formally write its solution as $$x(t) = \mathrm{e}^{t v} x_0$$. The formal solution coincides with the actual solution when $$v$$ is a linear mapping (think of the case in which $$v(x) = A x$$ for some matrix $$A$$).

Similarly, if we have another ODE $$\dot{y}(t) = w(y(t))$$ and an initial value $$y(0) = y_0$$, then we could write $$\mathrm{e}^{t v} \mathrm{e}^{s w} y_0$$ to represent the curve resulting from solving the second initial value problem up to time $$s$$, then using $$y(s; y_0)$$ as the initial value of the first ODE $$\dot{x} = v(x)$$ and solving that initial value problem up to time $$t$$. To reiterate, you can also think about the product of exponentials as meaning that you must first obtain the integral curve of the vector field $$w$$ that passes through $$y_0$$, follow said curve until it reaches $$y(s)$$, and then follow the integral curve of the vector field $$v$$ that starts at $$y(s)$$ and goes on for another $$t$$ units of time. I hope this helps clarify things a little bit more.

• What does multiplying these exponentials represent? – Hugo Gransträm Mar 23 '19 at 19:03
• I've edited my answer to explain it. – Juan M. Bello-Rivas Mar 23 '19 at 20:11
• Could that be thought of as composition? – Hugo Gransträm Mar 23 '19 at 20:14
• That's exactly what it is: composing the flow of one vector field with the flow of another. – Juan M. Bello-Rivas Mar 23 '19 at 20:18
• Think I'm starting to get the hang of this now :-) thank you very much Juan 😊 have a great day – Hugo Gransträm Mar 23 '19 at 20:22