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Assume we have an equation which represents the flux of some quantity as $q = -D \dfrac{\partial T}{\partial x}$ (Eqn. 1), where the diffusion coefficient $D$, variable $T$, $x$ and $q$ have some dimensions.

If we use a fractional derivative of non-integer order $\alpha$ such that $\dfrac{\partial T}{\partial x}$ is replaced with $\dfrac{\partial^{\alpha} T}{\partial x^{\alpha}}$, the RHS must be multiplied by some characteristic length-scale $L^{\alpha-1}_c$ to recover the physical dimensions of $q$.

Let's say we non-dimensionalize (Eqn. 1) with some characteristic scales for $x, T$ and $q$, such that the dimensionless equation is $\tilde{q} = -R\dfrac{\partial \tilde{T}}{\partial \tilde{x}}$ (Eqn. 2), where tilded variables are dimensionless counterparts of the variables in Eqn. 1, and $R$ is a non-dimensional number which measures the ratio of the scales used to non-dimensionalize the problem.

Now I think that if we want to use fractional derivative in (Eqn. 2), we shouldn't worry about the dimensional mismatch that originates from the use of fractional spatial derivative, since the equation is dimensionless. Would that be reasonable?

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  • $\begingroup$ I'm a bit confused as to what you're trying to achieve here: In Eqn 1. for example, if you're applying a derivative (either fractional or not) on the RHS, you must do the same on the LHS to maintain equality. So when you write "$\dfrac{\partial T}{\partial x}$ is replaced with $\dfrac{\partial^{\alpha} T}{\partial x^{\alpha}}$", do you mean to say, assuming $q$ on the LHS was given by such a derivative ahead of time? $\endgroup$
    – Amit
    Commented Sep 14 at 14:25

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