Rayleigh problem A simple example is a semi-infinite domain bounded by a rigid wall and filled with viscous fluid. At time t=0 the wall is made to move with constant speed U in a fixed direction (for definiteness, say the x direction and consider only the x-y plane), one can see that there is no distinguished length scale given in the problem. This is known as the
Rayleigh problem. The boundary conditions of no-slip is u{(y\!=\!0)} = U Also, the condition that the plate has no effect on the fluid at infinity is enforced as u{(y\!\to\!\infty)} = 0. Now, from the Navier–Stokes equations \rho \left( \dfrac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \nabla \vec{u} \right) =- \nabla p + \mu \nabla^{2} \vec{u} one can observe that this flow will be
rectilinear, with gradients in the y direction and flow in the x direction, and that the pressure term will have no tangential component so that \dfrac{\partial p}{\partial y} = 0. The x component of the Navier–Stokes equations then becomes \dfrac{\partial \vec{u}}{\partial t} = \nu \frac{\partial^2 \vec{u}}{\partial y^2} and the scaling arguments can be applied to show that \frac{U}{t} \sim \nu \frac{U}{y^{2}} which gives the scaling of the y co-ordinate as y \sim (\nu t)^{1/2}. This allows one to pose a self-similar ansatz such that, with f and \eta dimensionless, u = U f{\left(\eta \equiv \dfrac{y}{(\nu t)^{1/2}}\right)} The above contains all the relevant physics and the next step is to solve the equations, which for many cases will include numerical methods. This equation is - \eta f'/2 = f'' with solution satisfying the boundary conditions that f = 1 - \operatorname{erf} (\eta / 2) \quad \text{ or } \quad u = U \left(1 - \operatorname{erf} \left(y / (4 \nu t)^{1/2} \right)\right) which is a self-similar solution of the first kind.
Semi-infinite solid approximation In transient
heat transfer applications, such as impingement heating on a ship deck during missile launches and the sizing of
thermal protection systems, self-similar solutions can be found for semi-infinite solids. The governing equation when
heat conduction is the primary heat transfer mechanism is the one-dimensional energy equation:\rho c_{p} \frac{\partial T}{\partial t} = \frac{\partial}{\partial x}\left( k \frac{\partial T}{\partial x} \right)where \rho is the material's
density, c_{p} is the material's
specific heat capacity, k is the material's
thermal conductivity. In the case when the material is assumed to be homogeneous and its properties constant, the energy equation is reduced to the
heat equation:\frac{\partial T}{\partial t} = \alpha \frac{\partial^{2}T}{\partial x^{2}}, \quad \alpha = \frac{k}{\rho c_{p}}with \alpha being the
thermal diffusivity. By introducing the similarity variable \eta = x/\sqrt{t} and assuming that T(t,x) = f(\eta), the PDE can be transformed into the ODE:f''(\eta) + \frac{1}{2\alpha}\eta f'(\eta) = 0If a simple model of thermal protection system sizing is assumed, where decomposition,
pyrolysis gas flow, and surface recession are ignored, with the initial temperature T(0,x) = f(\infty) = T_{i} and a constant surface temperature T(t,0) = f(0) = T_{s}, then the ODE can be solved for the temperature at a depth x and time t:T(t,x) = \text{erf}\left( \frac{x}{2\sqrt{\alpha t}} \right) \left( T_{i} - T_{s} \right) + T_{s}where \text{erf}(\cdot) is the
error function. ==References==