Are there any results on proving the convexity of a function defined implicitly through an ode?
More specifically I'm interested in proving convexity of a function $f(t,u):\mathbb{R}_{+}^{2}\rightarrow\mathbb{R}_{+}$ in $u$ ($0 \leq u \leq 1$) for all $t$ given that it satisfies the following ode: $$ \frac{\partial f(t,u)}{\partial t}=a-\min(u,f(t,u)), $$ with initial condition $f(0)$.
Update 2: Here is a more basic question (which can help answering the original one): can we show that $f(t,u)$ is nonincreasing in $u$ for all $t$? (without solving the ode) This is very intuitive: one can think of $f$ as the content of a buffer/tank at time $t$ with input rate $a$ and output rate $\min(u,f(t,u))$. The output rate is at most $u$ but also cannot exceed the current content level. By increasing $u$ we are basically increasing the potential output rate and hence it must be that $f$ is nonincreasing in $u$.
Update 1: Eventually I'm interested in showing convexity when $a$ is also a function of $t$, i.e., $f$ solves $$ \frac{\partial f(t,u)}{\partial t}=a(t)-\min(u,f(t,u)). $$ This is why I'm looking for an implicit approach to prove convexity (rather than solving the ode). When $a(t)=a$ the ode can be solved (as @daw mentioned) by considering the two cases $u<f(t,u)$ and $u \geq f(t,u)$. This leads to two ode's with solutions given by $$ f_{1}(t,u)=f_{1}(s)+(t-s)(\lambda-u),\\ f_{2}(t,u)=(f_{2}(s)-a) \exp(-(t-s))+a, $$ which characterize the solution of the original ode: Given an initial condition $f(0)$ the solution is given by one of the equations above with the possibility of crossing over to the other equation at most once. Here is a plot of the trajectory of $f$ for a fixed $u$:
As you see given initial condition $f(0)=1$ the solution is given by the first (linear) equation $f_{1}$ but once it reaches $u=0.9$ it takes the exponential form of $f_{2}$.
The solution is also continuous in $u$ and I'm pretty sure convex in $u$ for all $t$ even when $a=a(t)$. Here is a plot of $f(t,u)$ as a function of $u$ for differnet values of $t$:
I believe something can be done by differentiating the ode with respect to $u$ and trying to show the monotonicity of the derivative, but I haven't had any luck yet.
We can construct solutions to your differential equation for which convexity in $u$ does not hold.
Take any continuous function $a(t)$ that satisfies $a(t) \geq 1$ for all $t \geq 0$. Take any (possibly nonconvex) function $g(u)$ that satisfies $g(u) \geq u$ for all $u \in [0,1]$. Consider the function: $$f(t,u) = g(u) -ut + \int_{0}^t a(\tau)d\tau$$
Claim: This function $f(t,u)$ satisfies the following for all $t \geq 0$ and all $u \in [0,1]$: $$ \frac{\partial f(t,u)}{\partial t} = a(t) - \min[u, f(t,u)] $$
Proof: We first show that $\min[u, f(t,u)]=u$. Fix $t \geq 0, u \in [0,1]$. Then:
\begin{align} f(t,u) &= g(u) - ut + \int_0^t a(\tau)d\tau \\ &\geq u - t + \int_0^t a(\tau)d\tau \quad \mbox{[since $g(u)\geq u$, and $u \in [0,1], t\geq 0$]} \\ &= u + \int_0^t [a(\tau)-1]d\tau \\ &\geq u \quad \mbox{[since $a(\tau)\geq 1$ for all $\tau \geq 0$]} \end{align}
Hence, $\min[u, f(t,u)]=u$ for all $t \geq 0$ and all $u \in [0,1]$. Hence, for $t \geq 0, u \in [0,1]$ we have \begin{align} \frac{\partial f (t,u)}{\partial t} &=\frac{\partial}{\partial t}\left[ g(u) -ut + \int_{0}^t a(\tau)d\tau \right] \\ &=-u + a(t) \\ &= a(t) - \min[u, f(t,u)] \end{align} $\Box$
Now observe that if our chosen function $g(u)$ is not convex over $u \in [0,1]$, then for any fixed $t\geq 0$ the function $f(t,u)$ defined above is not convex in $u$.