Show that the boundary layers diffuse out from the plate with speed $\sqrt{\frac{\nu}{t}}$ I was wondering if somebody would be able to help me with this problem. I know how to solve it using dimension arguments but I'm unsure what is meant by transformation techniques. Any help would be greatly appreciated. 
An infinite horizontal plate moves with speed U in its own plane relative to surrounding fluid. The plate is initially at rest relative to the fluid. The equations for $u(y,t)$ govern by:
$$\partial_t u = \nu \partial_y^2 u$$
with $u(0,t)=U$ and $u(y,0)=0$. Show using transformation techniques that the boundary layers diffuse out from the plate with speed $\sqrt{\frac{\nu}{t}}$. 
 A: Solution using Laplace transforms
Using the definition of the Laplace transform:
$$\tilde{u}\left(y,s\right)=\int_{0}^{\infty}u\left(y,t\right)\exp\left(-st\right)dt$$
we can transform the PDE to an ODE:
$$s\tilde{u} - u\left(y,0\right)=\nu \frac{d^2\tilde{u}}{dy^2} \rightarrow \frac{d^2\tilde{u}}{dy^2}-\frac{s}{\nu}\tilde{u}=0$$
with transformed boundary conditions:
$$\tilde{u}\left(0,s\right)=\frac{U}{s}\quad \tilde{u}\left(\infty,s\right)=0$$
Taking a trial solution $\tilde{u}\left(y,s\right)=\exp\left(ky\right)$ and substituting into the ODE we find an equation for $k$:
$$k^2-\frac{s}{\nu}=0\rightarrow k=\pm\sqrt{\frac{s}{\nu}}$$
such that the general solution to the ODE is given by:
$$\tilde{u}\left(y,s\right)=A\exp\left(\sqrt{\frac{s}{\nu}}y\right)+B\exp\left(-\sqrt{\frac{s}{\nu}}y\right)$$
Applying the transformed boundary conditions yields $A=0$ and $B=\frac{U}{s}$ such that:
$$\tilde{u}\left(y,s\right)=\frac{U}{s}\exp\left(-\sqrt{\frac{s}{\nu}}y\right) \rightarrow u\left(y,t\right)=U\mathrm{erfc}\left(\frac{1}{2}\frac{y}{\sqrt{\nu t}}\right)$$
where the inverse Laplace transform was taken from here.
The implication of this solution is that the boundary layer grows as $\delta\left(t\right)=2\sqrt{\nu t}$ and the velocity with which it moves is:
$$v(t)=\frac{d\delta}{dt}=\sqrt{\frac{\nu}{t}}$$
Solution using similarity arguments
For diffusion problems where a scalar field is initially uniform and the scalar quantity starts diffusing from one boundary to another boundary very far away (e.g. $u(\infty,t)=0$), the profiles of the scalar are similar at each time step differing only by a 'stretching factor'. If the profiles are scaled by the 'stretching factor', all profiles collapse onto the same curve known as a similarity solution. This is qualitatively shown in the following figure:

Let's define a so-called similarity variable:
$$\eta=\frac{y}{\delta(t)}$$
where $y$ is scaled by a characteristic length scale $\delta(t)$ which is a function of time. This length scale is also known as the 'penetration length' and describes how far the momentum has diffused into the domain; we do not yet know what this length is. As the 'penetration length' can be assumed to increase with time, we can see this as the 'stretching factor' previously talked about.
Using:
$$\left.\frac{d\eta}{dy}\right|_t = \delta^{-1} \quad \left.\frac{d\eta}{dt}\right|_y = -y \delta^{-2} \frac{d\delta}{dt}=-\eta \delta^{-1} \frac{d\delta}{dt}$$
We use the chain rule on the diffusion equation to give:
$$\frac{\partial u}{\partial\eta}\left(\left.\frac{d\eta}{dt}\right|_y\right)=\nu \frac{\partial^2 u}{\partial\eta} \left(\left.\frac{d\eta}{dy}\right|_t\right)^2$$
which transforms the PDE to an ODE:
$$\frac{d^2u}{d\eta^2} + \left(\frac{\delta}{\nu}\frac{d\delta}{dt}\right) \eta \frac{du}{d\eta} = 0$$
If truly a similarity solution, then $u$ is a function of only $\eta$; this is only the case if $\frac{\delta}{\nu} \frac{d\delta}{dt}=n$ where $n$ is a constant to be determined.
Since the PDE was transformed to a second-order ODE, the boundary and initial conditions are over-specified. However, when these are likewise transformed we see that in terms of $\eta$ we retrieve two unique boundary conditions:
$$u(0,t)=u(0)=U \quad u(\infty,t)=u(y,0)=u(\infty)=0$$
if we assume $\delta(0)=0$, i.e. at $t=0$ the momentum has not penetrated into the domain yet. This fully specifies the problem which indicates there is indeed a similarity solution possible.
Integrating the ODE we find:
$$u\left(\eta\right)=K_{2}+K_{1}\int_{0}^{\eta}\exp\left(-\frac{n}{2}\eta'^{2}\right)d\eta'$$
where $\eta'$ is a dummy integration variable. The unevaluated integral is related to the 'error function' and cannot be determined analytically; numerical approximations are available instead. However, it is known that:
$$\int_{0}^{\infty}\exp\left(-\eta'^{2}\right)d\eta'=\frac{\sqrt{\pi}}{2}$$
which, if we define $n=2$, is used to apply the boundary conditions to give the solution:
$$\frac{u\left(\eta\right)}{U}=1-\frac{2}{\sqrt{\pi}}\int_{0}^{\eta}\exp\left(-\eta'^{2}\right)d\eta'$$
What remains is to determine the 'penetration length' $\delta(t)$:
$$\delta \frac{d\delta}{dt}=\frac{1}{2}\frac{d\delta^2}{dt}=2\nu \rightarrow \delta(t)^2=4\nu t+K_3$$
Using the previously determined condition $\delta(0)=0$, we finally find that the 'penetration length' is:
$$\delta(t)=2\sqrt{\nu t}$$
The requested 'penetration velocity' is again found as:
$$v(t)=\frac{d\delta}{dt}=\sqrt{\frac{\nu}{t}}$$
Note: In transport phenomena, the 'penetration length' is usually defined as $\tilde{\delta}(t)=\sqrt{\pi\nu t}$. This can be found from the above analysis by taking the derivative at $y=0$:
$$\frac{du}{dy}(0,t) = -\frac{U}{\tilde{\delta}(t)}$$
which implies that a tangent at $y=0$ will cross the $y$-axis at $\tilde{\delta}(t)$. From the analysis we determine:
$$\frac{du}{dy}(0,t) = \frac{du}{d\eta}(0,t) \frac{d\eta}{dy} = -\frac{2}{\sqrt{\pi}}U\delta^{-1}=-\frac{U}{\sqrt{\pi\nu t}}$$
which shows that indeed $\tilde{\delta}(t)=\sqrt{\pi\nu t}$.
A: The equation $$\partial_tu=\nu\partial^2_yu$$ is a diffusion equation in one dimension. Its Green's function in infinite space is
$$G(y,t)=\frac{\mathrm e^{-y^2/4\nu t}}{\sqrt{4\pi\nu t}}.$$
For $t=0$, we have $G(y,0)=\delta(y)$. Note that any solution of your equation depends linearly on $G$. Notice also that the constant function is also a solution of the diffusion equation. So let us consider the problem with $v=U-u$.
The boundary conditions for $v$ are $v(0,t)=0$, $v(y,0)=U$. Using the Green's function, we find that 
$$v(y,t)=\int_0^\infty v(x,0)G(y-x,t)\mathrm dx=U\operatorname{erf}\left(\frac{y}{\sqrt{4\nu t}}\right),$$
where $\operatorname{erf}(x)=\frac1{\sqrt\pi}\int_{-\infty}^x\mathrm e^{-z^2}\mathrm dz$. 
The velocity field is therefore given by
$$u(y,t)=U\left[1-\operatorname{erf}\left(\frac{|y|}{\sqrt{4\nu t}}\right)\right].$$
It is a function of $y/\sqrt{\nu t}$ only. The layer with velocity $u_0$ is in one to one correspondence with $x_0=y_0/\sqrt{\nu t_0}=\text{constant}$. The layer with velocity $u_0$ moves at velocity $\mathrm dy/\mathrm dt$ that we obtain by derivating $x_0$ with respect to $t$.
$$\frac{\mathrm dx_0}{\mathrm dt}=0=\frac{1}{\sqrt{\nu t}}\frac{\mathrm dy}{\mathrm dt}-\frac{y}{2\sqrt{\nu t^3}}$$
from which we obtain
$$\frac{\mathrm dy}{\mathrm dt}=\frac y{2t}=\frac{x_0}{2}\sqrt{\frac{\nu}{t}}.$$
From this result, we deduce that the boundary moving with velocity $\sqrt{\frac\nu t}$ is the one with $x_0=2$, which permits to understand the question since no definition of the moving boundary was given. Any layer with a certain fixed velocity move at a speed given by the last equation. 
