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

    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}}


    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.
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  3. #3
    Junior Member
    Mar 2011

    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^*}
    \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}
    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.
    Last edited by math2011; Aug 5th 2011 at 08:27 AM.
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