Results 1 to 3 of 3

Thread: Similarity Solutions

  1. #1
    Junior Member
    Mar 2011

    Similarity Solutions

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

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

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

    Please tell me the procedure for finding similarity solution. Thank you in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Dec 2008
    Conway AR

    Re: Similarity Solutions

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

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


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

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

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

    You must now create IC's for this ODE from your given BCs/ICs.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Mar 2011

    Re: Similarity Solutions

    Thank you for your help. The question does require finding the ODE of $\displaystyle f(\eta)$ 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 $\displaystyle \displaystyle u^* = \frac{u}{u_0}$, $\displaystyle \displaystyle x^* = \frac{x}{x_0}$ and $\displaystyle \displaystyle t^* = \frac{t}{t_0}$ where $\displaystyle u_0,x_0,t_0$ are arbitrary and are to be determined.

    2. Then $\displaystyle u = u_0 u^*$,
    $\displaystyle \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^*} $
    $\displaystyle \frac{\partial u}{\partial x} = u_0 \frac{\partial u^*}{\partial x^*} \frac{\partial x^*}{\partial x} = u_0 \frac{\partial u^*}{\partial x^*} \frac{1}{x_0} = \frac{u_0}{x_0} \frac{\partial u^*}{\partial x^*}$
    $\displaystyle \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
    $\displaystyle \frac{u_0}{t_0} \frac{\partial u^*}{\partial t^*} =& k \frac{u_0}{x_0^2} \frac{\partial^2 u^*}{\partial {x^*}^2}$
    $\displaystyle \frac{\partial u^*}{\partial t^*} =& \frac{kt_0}{x_0^2} \frac{\partial^2 u^*}{\partial {x^*}^2} \text{.} $
    Because $\displaystyle x_0,t_0$ are arbitrary and undertermined, choose $\displaystyle \displaystyle \frac{kt_0}{x_0^2} = 1$.

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

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

    6. Substitute above values of $\displaystyle x_0$ and $\displaystyle u_0$ into $\displaystyle u = u_0u^*$ to get
    $\displaystyle u(x,t) = \sqrt{k t_0} u^*\left(\frac{x}{\sqrt{k t_0}}, \frac{t}{t_0}\right) $
    where $\displaystyle t_0$ is still arbitrary. Let $\displaystyle t_0 = t$, then
    $\displaystyle u(x,t) = \sqrt{k t} u^*\left(\frac{x}{\sqrt{k t}}, 1\right) \text{.} $
    Therefore define similarity variable $\displaystyle \displaystyle \eta = \frac{x}{\sqrt{kt}}$ and function $\displaystyle f(\eta) = u^*(\eta, 1)$ so that
    $\displaystyle 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.
    Last edited by math2011; Aug 5th 2011 at 07:27 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Similarity
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Nov 28th 2011, 12:38 AM
  2. Similarity
    Posted in the Geometry Forum
    Replies: 0
    Last Post: Feb 18th 2010, 09:03 AM
  3. Similarity.....
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Mar 3rd 2009, 06:51 AM
  4. Similarity
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Oct 17th 2008, 10:25 AM
  5. Similarity solutions:
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 3rd 2008, 08:58 AM

Search Tags

/mathhelpforum @mathhelpforum