The Seebeck effect can be stated mathematically as follows. The current density $\mathbf j$ in a conductor is related to the potential $V$ and temperature $T$ by
$$\mathbf j=-\sigma(\nabla V+S\nabla T),$$
Where $\sigma$ is the material's conductivity and $S$ is the Seebeck coefficient. Pretty much, this states that the temperature difference creates an electric field $$\mathbf E_{Seebeck}=-S\nabla T.$$
As such, in the stationary state we'd need the following relationship between the potential and the temperature
$$\nabla V=-S\nabla T,$$
Which allows us to write (for constant $S$)
$$V=-ST+V_0.$$
If we now suppose that everything happens in one dimension $x$ and suppose the effect happens between to infinite plates at $x=0$ and $x=L$ as to simplify our calculations and not have to consider field lines escaping from the wire, we can write the following relationship between the charge density $\rho$ and the potential (Poisson's equation)
$$\frac{d^2V}{d x^2}=-\frac{\rho}{\epsilon},$$
$$\frac{d^2T}{d x^2}=\frac{\rho}{S\epsilon}$$
With $\epsilon$ being the permitivitty of the material. If we now suppose the electron charge density is governed by a Boltzmann distribution (stationary thermal state) with respect to electric potential
$$\rho_e=-\rho_0\exp(\frac{eV}{k_BT}),$$
$$\rho_e=-\rho_0\exp(-\frac{eS}{k_B})\exp(\frac{eV_0}{k_BT}).$$
If we also add the (homogenous) proton density
$$\rho_p=-\frac{1}{L}\int\limits_0^L\rho_e\,dx,$$
Which is also a constant of the material, the charge density becomes
$$\rho=\rho_e+\rho_p$$
$$\rho=\rho_p\left(1-\frac{L\exp(\frac{eV_0}{k_BT})}{\int\limits_0^L\exp(\frac{eV_0}{k_BT})\,dx}\right)$$
The differential equation we need to solve is then
$$\frac{d^2T}{dx^2}= \frac{\rho_p}{S\epsilon}\left(1-\frac{L\exp(\frac{eV_0}{k_BT})}{\int\limits_0^L\exp(\frac{eV_0}{k_BT})\,dx}\right),$$
Which can be solved numerically with (I think) $V_0$ as a free parameter. You can try it with your favorite software now.
