1. ## Similarity Solutions

The original question is:
Try and apply the Similarity solution method to the following boundary value problems for $u(x,t)$.

$u_t = k u_{xx}$ for all $x > 0$ with boundary conditions
$u_x(0,t) = 1$
$u(x,t) \to 0$ as $x \to \infty$
$u(x,0) = 0$ for $x > 0$.

I know from my tutorial that I should first find $u = \sqrt{kt} f(\eta)$ where $f$ is an unknown function of similarity variable $\displaystyle \eta = \frac{x}{\sqrt{kt}}$. What I don't know is how to find the similarity variable $\eta$ and the formula $u = \sqrt{kt} f(\eta)$.

Please tell me the procedure for finding similarity solution. Thank you in advance.

2. ## Re: Similarity Solutions

You are givien the "similarity" variable. It's $\eta = \frac{x}{\sqrt{kt}}.$ What you need to do now is substitute the form of your solution into the PDE.

If $u = \sqrt{kt} f(\eta), \eta = \frac{x}{\sqrt{kt}}$

then

$u_t = \frac{k f(\eta)}{2\sqrt{kt}} - \frac{x f'(\eta)}{2t}$

$u_{xx} = \frac{f''(\eta)}{\sqrt{kt}}$
Your PDE becomes (after dividing by $\sqrt{k/t}$)

$\frac{1}{2} f(\eta) - \frac{1}{2} \eta f'(\eta) - f''(\eta) = 0$.

You must now create IC's for this ODE from your given BCs/ICs.

3. ## Re: Similarity Solutions

Thank you for your help. The question does require finding the ODE of $f(\eta)$ but also requires finding the similarity variable to start with. After getting the solution from my lecturer I typed up the solution below with slight variations according to my understanding.

1. Let $\displaystyle u^* = \frac{u}{u_0}$, $\displaystyle x^* = \frac{x}{x_0}$ and $\displaystyle t^* = \frac{t}{t_0}$ where $u_0,x_0,t_0$ are arbitrary and are to be determined.

2. Then $u = u_0 u^*$,
$\frac{\partial u}{\partial t} = u_0 \frac{\partial u^*}{\partial t^*} \frac{\partial t^*}{\partial t} = u_0 \frac{\partial u^*}{\partial t^*} \frac{1}{t_0} = \frac{u_0}{t_0} \frac{\partial u^*}{\partial t^*}$
$\frac{\partial u}{\partial x} = u_0 \frac{\partial u^*}{\partial x^*} \frac{\partial x^*}{\partial x} = u_0 \frac{\partial u^*}{\partial x^*} \frac{1}{x_0} = \frac{u_0}{x_0} \frac{\partial u^*}{\partial x^*}$
and
$\frac{\partial^2 u}{\partial x^2} =& \frac{\partial}{\partial x} \left( \frac{u_0}{x_0} \frac{\partial u^*}{\partial x^*} \right) = \frac{u_0}{x_0} \frac{\partial^2 u^*}{\partial {x^*}^2} \frac{\partial x^*}{\partial x} = \frac{u_0}{x_0^2} \frac{\partial^2 u^*}{\partial {x^*}^2} \text{.}$

3. Substituting above partial derivatives into original PDE results in
$\frac{u_0}{t_0} \frac{\partial u^*}{\partial t^*} =& k \frac{u_0}{x_0^2} \frac{\partial^2 u^*}{\partial {x^*}^2}$
$\frac{\partial u^*}{\partial t^*} =& \frac{kt_0}{x_0^2} \frac{\partial^2 u^*}{\partial {x^*}^2} \text{.}$
Because $x_0,t_0$ are arbitrary and undertermined, choose $\displaystyle \frac{kt_0}{x_0^2} = 1$.

4. The boundary conditions for $u$ are written into boundary condition for $u^*$ as
$\frac{\partial u}{\partial x}(0,t) =& 1$
$\frac{\partial u^*}{\partial x^*}(0, t^*) =& \frac{x_0}{u_0}$
and
$u(x,0) =& 0$
$u_0 u^*(x^*, 0) =& 0$
$u^*(x^*, 0) =& 0 \text{.}$
As $u_0,x_0$ are arbitrary and undetermined, choose $\displaystyle \frac{x_0}{u_0} = 1$.

5. Solving $\displaystyle \frac{k t_0}{x_0^2} = 1$ and $\displaystyle \frac{x_0}{u_0} = 1$ results in
$x_0 =& \sqrt{k t_0}$
$u_0 =& x_0 = \sqrt{k t_0} \text{.}$

6. Substitute above values of $x_0$ and $u_0$ into $u = u_0u^*$ to get
$u(x,t) = \sqrt{k t_0} u^*\left(\frac{x}{\sqrt{k t_0}}, \frac{t}{t_0}\right)$
where $t_0$ is still arbitrary. Let $t_0 = t$, then
$u(x,t) = \sqrt{k t} u^*\left(\frac{x}{\sqrt{k t}}, 1\right) \text{.}$
Therefore define similarity variable $\displaystyle \eta = \frac{x}{\sqrt{kt}}$ and function $f(\eta) = u^*(\eta, 1)$ so that
$u(x,t) = \sqrt{k t} f(\eta) \text{.}$

I am still having trouble with another similarity question. I will post it later in this thread.