I am solving the equation $X−A−B e^{−Xy}−C e^{−X z}=0$ where $X, A, B$ and $C$ are 2x2 matrices and $y$ and $z$ are scalars. What will be the closed form solution for $X$ without approximating the exponential term. If It were $X−A−B e^{−Xy}=0$ then I can write it as $X=\frac{1}{y} W(B y e^{−A y})+A$. Where $W$ is the Lambert W function. How to get the closed form solution for this any idea?
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$\begingroup$ Have you tried to get closed form solutions for the characteristic equations? Maybe if you set y=Pi/4? If I remember correctly Pontryagin did some work on that and th econclusion was that there is not much hope. $\endgroup$– Uwe StroinskiCommented Feb 17, 2015 at 8:22
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$\begingroup$ Actually your "If it were..." statement is wrong, if $A$ and $B$ don't commute. $\endgroup$– Robert IsraelCommented Feb 17, 2015 at 16:21
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1$\begingroup$ In fact, with $X = \dfrac{1}{y} W(Bye^{-Ay}) + A$, I get $X - A - B e^{-Xy} = (BAB - BA^2) y^2/2 + O(y^3)$ as $y \to 0$. $\endgroup$– Robert IsraelCommented Feb 17, 2015 at 18:28
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1$\begingroup$ Expand $\exp$ and $W$ in Maclaurin series in $y$, being careful about noncommuting variables. I actually used Maple's Physics package. $\endgroup$– Robert IsraelCommented Feb 17, 2015 at 20:05
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1$\begingroup$ Sorry, typo: that should be $(BAB - B^2 A) y^2/2$. $\endgroup$– Robert IsraelCommented Feb 17, 2015 at 20:34
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