Debye–Hückel equation

Nondimensionalization

The differential equation is ready for solution (as stated above, the equation only holds for low concentrations): $${\displaystyle \frac{{\mathrm{\partial }}^2\phi (r)}{\mathrm{\partial }{r}^2}+\frac{2}{r}\frac{\mathrm{\partial }\phi (r)}{\mathrm{\partial }r}=\frac{Iq\phi (r)}{{\epsilon }_r{\epsilon }_0{k}_{\text{B}}T}={\kappa }^2\phi (r).}$$

Using the Buckingham π theorem on this problem results in the following dimensionless groups: $${\displaystyle \begin{array}{cc}{\pi }_1& =\frac{q\phi (r)}{k_{\text{B}}T}=\mathrm{\Phi }(R(r))\\ {\pi }_2& ={\epsilon }_r\\ {\pi }_3& =\frac{a{k}_{\text{B}}T{\epsilon }_0}{q^2}\\ {\pi }_4& =a^{3}I\\ {\pi }_5& =z_0\\ {\pi }_6& =\frac{r}{a}=R(r).\end{array}}$$ ${\displaystyle \mathrm{\Phi }}$ is called the reduced scalar electric potential field. ${\displaystyle R}$ is called the reduced radius. The existing groups may be recombined to form two other dimensionless groups for substitution into the differential equation. The first is what could be called the square of the reduced inverse screening length, ${\displaystyle (\kappa a{)}^2}$. The second could be called the reduced central ion charge, ${\displaystyle Z_0}$ (with a capital Z). Note that, though ${\displaystyle z_0}$ is already dimensionless, without the substitution given below, the differential equation would still be dimensional.

$${\displaystyle \frac{{\pi }_4}{{\pi }_2{\pi }_3}=\frac{a^2{q}^{2}I}{{\epsilon }_r{\epsilon }_0{k}_{\text{B}}T}=(\kappa a{)}^2}$$ $${\displaystyle \frac{{\pi }_5}{{\pi }_2{\pi }_3}=\frac{z_0{q}^2}{4\pi a{\epsilon }_r{\epsilon }_0{k}_{\text{B}}T}=Z_0}$$

To obtain the nondimensionalized differential equation and initial conditions, use the ${\displaystyle \pi }$ groups to eliminate ${\displaystyle \phi (r)}$ in favor of ${\displaystyle \mathrm{\Phi }(R(r))}$, then eliminate ${\displaystyle R(r)}$ in favor of ${\displaystyle r}$ while carrying out the chain rule and substituting ${\displaystyle R^{\mathrm{\prime }}(r)=a}$, then eliminate ${\displaystyle r}$ in favor of ${\displaystyle R}$ (no chain rule needed), then eliminate ${\displaystyle I}$ in favor of ${\displaystyle (\kappa a{)}^2}$, then eliminate ${\displaystyle z_0}$ in favor of ${\displaystyle Z_0}$. The resulting equations are as follows: $${\displaystyle \frac{\mathrm{\partial }\mathrm{\Phi }(R)}{\mathrm{\partial }R}{|}_{R=1}=-Z_0}$$ $${\displaystyle \mathrm{\Phi }(\mathrm{\infty })=0}$$ $${\displaystyle \frac{{\mathrm{\partial }}^2\mathrm{\Phi }(R)}{\mathrm{\partial }{R}^2}+\frac{2}{R}\frac{\mathrm{\partial }\mathrm{\Phi }(R)}{\mathrm{\partial }R}=(\kappa a{)}^2\mathrm{\Phi }(R).}$$

For table salt in 0.01 M solution at 25 °C, a typical value of ${\displaystyle (\kappa a{)}^2}$ is 0.0005636, while a typical value of ${\displaystyle Z_0}$ is 7.017, highlighting the fact that, in low concentrations, ${\displaystyle (\kappa a{)}^2}$ is a target for a zero order of magnitude approximation such as perturbation analysis. Unfortunately, because of the boundary condition at infinity, regular perturbation does not work. The same boundary condition prevents us from finding the exact solution to the equations. Singular perturbation may work, however.