When is diffusion steady-state?
There is no set answer to your problem, it depends on the geometry of the diffusional process. I give two examples.

(a) Suppose there is a volatile solute in a solvent placed in a beaker and the solute evaporates from the surface, then there is a concentration gradient of solute away from the surface as the solute is lost into the air. Similarly, if there is an electrode at which ions are being deposited as a metal, or a precipitate is being formed, then a concentration gradient is set up, but as material is removed no steady state can be set up as there is a finite amount of substance present.  

Solving Fick’s 2nd law produces an equation describing the concentration vs. time which is $c_{x,t}=c_0erf(x/(2\sqrt{Dt}) $ where erf is the error function, D the diffusion coefficient, t time and $C_0$ the initial concentration. (The initial conditions are $t=0, x\ge 0, c=c_0$;  $ t&gt;0,x=0, c=0$;$ x \rightarrow \infty, t=0, c=c_0$ ). 

The depletion distance, that is the distance from the sink that the concentration is just becoming less than than $c_0$ is given by $\delta = \sqrt { \pi Dt}$ at time t.

(b) In a closed system, for example, a cell with half of it containing a solution and the other half solvent, equilibrium will become established once diffusion into all the cell is allowed. Similarly an insulated bar half of which is initially hot will effectively come to an equilibrium at the average temperature. 

Consider a tube filled with solvent and then closed and some solute injected at some place, diffusion will ensure that equilibrium will eventually be reached. Because the initial distribution of concentration  must level out to a constant value, the solution to Fick’s equation must have both a time dependent and a constant part. 

At long times the temperature or concentration will become uniform and if $f(x)$ is the amount initially added at position x and $0 &lt; x &lt; L$, the long time value is the average value;  $\frac{1}{L}\int _0^Lf(x)dx$ . The initial condition is given by the shape of the concentration profile $f(x)$. 

The concentration of added solute is $c(x, t)$  at position x and time t. Because no material (or heat)  leaves the concentration gradients at the ends of the tube are zero at all times, or
$$\frac{dc(0,t)}{dx} = \frac{dC(L,t)}{dx} =0, ~~~t&gt;0$$
Starting with the diffusion equation
$$\frac{\partial c}{dt} = \frac{\partial ^2c}{dx^2} $$
and separating variables, gives
$$\frac{\partial c_t}{\partial t} = -Dk^2c_t$$
 and 
$$\frac{\partial ^2 c_x}{\partial x^2} =-c_xk^2 $$
where k is a separation parameter to be determined. 
The solution is $c=c_tc_x$.

The time dependent equation integrates to  $c_t=e^{-Dk^2t}$ but the spatial part needs special attention to incorporate the boundary conditions. 

The separation parameter k can have values of $0$, or it can be positive or positive and imaginary. Only when k is positive are the solutions non-trivial and are $c_x=a\sin (kx)+b\cos (kx)$.

The gradient boundary condition ensures that the solution is the cosine term because the derivative, $kb\sin (kx)$, is zero both at $x = 0$ and at L provided that $k=n\pi/L$, where n is an integer. 

Next we use k as if it were an integer and only substitute its actual value at the end. The solution is 
$$c_k=e^{-k^2Dt}cos(kx)  ~~~~~~~~  k=0,1,2…$$
and the final solution is a linear combination of the k terms including $k = 0$ and this will give a term that is independent of time. The initial condition ensures that 
$$ f(x)=\Sigma _{n=0}^{\infty}b_k\cos (kx)$$
giving 
$$ b_k=\frac{2}{L}\int _0^Lf(x)\cos (kx)dx$$
where $k=n\pi /L$. The complete solution is 
$$ c = \frac{b_0}{2}+\Sigma _{n=1}^\infty b_ke^{-(n\pi /L)^2Dt}\cos (n\pi x/L )  $$
The first term $b_0/2$ is divided by 2 to make it equal to the average concentration at long times, $\frac{1}{L}\int _0^Lf(x)dx$ because by its defining equation $b_0$ is otherwise 2 times too large. 
The results below show how the initial profile eventually reaches a constant, time-independent value, the total amount of material, or heat being conserved. Uniform concentrations are found when $ t &gt; \approx  5/ (n\pi /L)^2D $, n is found to be in the range $10$ to $20$. 



figure: One dimensional diffusion in a closed tube. The summation was taken to $n=10 $ terms (left) and $20$ (right).  (Left) The solute is injected in the centre and has a gaussian profile ($\exp (-5000(x/L-0.5)^2)$ at $t=0$  Right. Case where the solution is half the cell. The initial profile is a sigmoid curve $(1+\exp (-100(x/L-0.5))^{-1}$. The diffusion coefficient $D = 1$ and $L=1$. The profile at different times is shown.