Does the exponential of a nonlinear operator solve the Cauchy problem for an ODE of say, this form \begin{align*} &\frac{dy}{dt}=f(t,y(t))\\ &y(0)=y_0 \end{align*} so is this true? \begin{align} y(t)=\sum_{k=0}^\infty \frac{(tf(t,.))^k}{k!}y_0 \end{align} or \begin{align} y(t)=\sum_{k=0}^\infty \frac{t^k f(t,.)^k}{k!}y_0 \end{align} I don't know if one is right. where $f(t,.)$ is acting on what's on the right of it (so acting $k$ times iteratively over $y_0$). Can't do the derivative myself. If I were to interpret https://en.wikipedia.org/wiki/C0-semigroup I'd say it only should work for linear operators, if not, why don't they say? EDIT: I know stuff like \begin{equation} y'=y^2,\quad y(0)=1 \end{equation} which has closed form $$ y(t)=\frac{1}{1-t} $$ should be equal in the second formulation $$ y(t)=\sum_{k=0}^\infty \frac{t^k}{k!} $$ which is not true, and in the first one to something extremely weird to which I don't have a closed solution.
1 Answer
I'm not sure how you ended up with this formula, but it doesn't work at all. Nonetheless, it is possible to recast the original nonlinear differential equation as a (pseudo-)linear one, in order to use the $C_0$-semigroup formalism, as you intended to do.
Indeed, considering your example, namely $\dot{y} = y^2$, it may be rewritten as $\dot{y} = Ay$, where the vector field $A = y^2\frac{\partial}{\partial y}$ can be exponentiated, so that the solution is given by $y(t) = e^{tA}y_0$.
More generally, this treatment can be extended straightforwardly to autonomous cases, i.e. $\dot{y} = f(y)$, hence $y(t) = e^{tf(y)\partial_y}y_0$. However, in non-autonomous cases, that is when $f$ depends on time $t$ explicitly, this resolution has to be adapted, due to the form of the derivative of the exponential map.
