Boundary derivative with one-sided differences

Derivation

Expand $f_1$ and $f_2$ about $x_0$:

$$f_1=f_0+hf'_0+\frac{h^2}{2}f''_0+\frac{h^3}{6}f'''_0$$ $$f_2=f_0+2hf'_0+2h^2f''_0+\frac{4h^3}{3}f'''_0$$

Eliminate $f''_0$: compute $4f_1-f_2$:

$$4f_1-f_2=3f_0+2hf'_0-\frac{2h^3}{3}f'''_0$$ $$\boxed{f'_0\approx\frac{-3f_0+4f_1-f_2}{2h}},\qquad f'_0=\text{approx}+\frac{h^2}{3}f'''_0+\cdots\ \Rightarrow\ O(h^2)$$

The first-order formula is $f'_0\approx(f_1-f_0)/h$. The second-order formula is important for evaluating wall shear stress and wall heat flux without needing a node outside the domain.

Numerical

Data: $f_0=50,\ f_1=46,\ f_2=40$ °C, $h=0.5\times10^{-3}$ m.

First-order:

$$\frac{dT}{dx}\approx\frac{46-50}{0.5\times10^{-3}}=-8000\ \text{K/m}$$

Second-order:

$$\frac{dT}{dx}\approx\frac{-3(50)+4(46)-40}{2\times0.5\times10^{-3}}=\frac{-150+184-40}{10^{-3}}=-6000\ \text{K/m}$$

The successive differences are -4 and -6, so the temperature curve is getting steeper away from the wall (curvature is present). The first-order formula effectively gives the slope at mid-interval (-8000 K/m), so it overestimates the magnitude of the wall slope; the second-order formula extrapolates to the wall itself.

Wall heat flux, $q''=-k\,dT/dx$:

$$q''=-0.6\times(-6000)=3600\ \text{W/m}^2$$

(first-order would give 4800 W/m², about 33% higher.)

Answer: gradient $-8000$ K/m (1st order), $-6000$ K/m (2nd order); $q''=3.6$ kW/m² (flux directed in the +x direction).