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.



An ODE for tensor - possibility of the equation together with the initial condition at $t=0$ deciding the solution for all $t>0$
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


$$A\_{111}=1-t-t^2/2,\;\;A\_{112}=t,$$
$$A\_{221}=t,\;\;A\_{222}=-1+2e^{-t}.$$




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More generally, you can take any $F\_{ij}(t)=\sum\_l A\_{ijl}(t)$, and define
$$A\_{ijl}=-\left(\sum\_{k=2}^n\lambda\_k^{-2}\right)^{-1}F'\_{ij},\;\;l\geq 2,$$
$$A\_{ij1}=F\_{ij}+(n-1)\left(\sum\_{k=2}^n\lambda\_k^{-2}\right)^{-1}F'\_{ij}.$$




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