O. Gross and R. Wagner, 1950. "A Continuous Colonel Blotto Game," Rand Research Memorandum RM-408 covers this case (pages 2-4) and much more.
If the players have equal resources, then it's always a tie regardless of strategy (since you didn't assign different values to the different battlefields).
If one player has more than twice the resources of the other, splitting them evenly guarantees a complete victory.
Suppose the players have $B$ and $E$ resources, and without loss of generality assume $2E \gt B \gt E$.
Let $m = \lfloor B/(B-E)\rfloor \ge 2$.
The value of the game with optimal play is $2/m$. This is the difference between wins and losses, and twice the probability of a win in both battlefields, $1/m$.
Let $d= B-E, B = md + r$. Let $p$ be any number in $(r, d)$. Let $q = E/(m-1)$.
An optimal strategy for Blotto, the player with $B$, is to have a $1/m$ chance to allocate each of $p, p + d, p + 2d, ... p+(m-1)d$ in the first battlefield.
An optimal strategy for the enemy with $E$ is to have a $1/m$ chance to allocate each of $0, q, 2q, ... (m-1)q$ in the first battlefield.
For example, if $E=1$ and $B=1.6$, then $m=2$ and the enemy can do no better than to commit everything to a random battlefield. Colonel Blotto tries to guess which one, and succeeds with probability $1/2$ (or else they split the battlefields). If $E=1$ and $B=1.4$, then $m=3$ and the enemy should allocate its forces as $(0,1)$, $(1/2,1/2)$, or $(1,0)$ with equal probabilities. Colonel Blotto can only win both battlefields in one of these situations, picks one, and guesses correctly with probability $1/3$.