Let $A_{ijl}(t,x) : [0,\infty) \times \mathbb{R}^n \to [\mathbb{R}^n]^3$ be a smooth tensor field. That is, $i,j,l \in \{1,2,3, \cdots, n\}$
Further assume that $A_{ijl}(t,x)=A_{jil}(t,x)$ for all $(t,x) \in [0,\infty) \times \mathbb{R}^n$.
Also, let $0=\lambda_1< \lambda_2 < \cdots < \lambda_n$ be some fixed numbers and finally suppose that $A_{ijl}(t,x)$ satisfies the following ODE:
\begin{equation}
\partial_t \sum_{l=1}^n A_{ijl}(t,x)= -\sum_{l=1}^n \lambda^2_l A_{ijl}(t,x) \text{ with } \sum_{l=1}^n A_{ijl}(0,x)=\delta_{ij}
\end{equation}
Then, at least when $t=0$, we have
\begin{equation}
\partial_t \sum_{l=1}^n A_{ssl}(t,x)\mid_{t=0}=-\lambda^2_s
\end{equation}
for all $s=1,2, \cdots, n$ while
\begin{equation}
\partial_t \sum_{l=1}^n A_{ijl}(t,x)\mid_{t=0}=0
\end{equation}
for $i \neq j$.
From such information, is it possible to conclude, together with the smoothness assumption, that
\begin{equation}
\sum_{l=1}^n A_{ijl}(t,x)=e^{-\lambda_i^2 t} \text{ if } i=j \text{ and } 0 \text{ otherwise}
\end{equation}
for all $t \geq 0$?
It seems quite plausible for me, but I cannot really justify my guess for extrapolation to $t>0$. Could anyone please help me?
Edit : I add one more condition that $A_{ijl}(0,x)=\delta_{il}\delta_{jl}$ so that the above explanation makes sense.
Best Answer
These equations are not sufficient to determine $\sum_l A_{ijl}$ for $n>1$.
Here is a counterexample of a solution of the problem in the OP which contradicts the conjectured solution:
set $n=2$, $\lambda_1=0$, $\lambda_2=1$; all elements of $A_{ijl}$ are identically zero, except
$$A_{111}=1-t-t^2/2,\;\;A_{112}=t,$$ $$A_{221}=t,\;\;A_{222}=-1+2e^{-t}.$$
More generally, you can take any $F_{ij}(t)=\sum_l A_{ijl}(t)$, and define $$A_{ijl}=-\left(\sum_{k=2}^n\lambda_k^{-2}\right)^{-1}F'_{ij},\;\;l\geq 2,$$ $$A_{ij1}=F_{ij}+(n-1)\left(\sum_{k=2}^n\lambda_k^{-2}\right)^{-1}F'_{ij}.$$