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Thread: solving a diffusion pde

  1. #1
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    solving a diffusion pde

    Hallo everyone

    Please may I ask for some tips to help me solving this question.

    I have this equation:

    $\displaystyle \frac{\partial^2t}{\partial{x^2}}+\frac{\partial^2 t}{\partial{y^2}}=-2$


    this equation needs to be solved using the conditions where $\displaystyle t(x,y)=0$ with diffusion between circles $\displaystyle x^2+y^2=1$ and $\displaystyle x^2+y^2=e^2$

    we assume we can solve with form $\displaystyle t(x,y)=f(z)$ with $\displaystyle z=x^2+y^2$

    how can first equation with conditions now be written as:
    $\displaystyle
    z\frac{d^2{f}}{d{z^2}}+\frac{df}{dz}=-0.5,$ $\displaystyle \left\{\begin{array}{cc}f(1)=0,\\f(e^2)=0, \end{array}\right.$

    if someone can please give me tip with how i work out $\displaystyle \frac{df}{dz}$ (because i am confuse thinking what this means) maybe I can give question a go!

    Thank you
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi jescavez
    Quote Originally Posted by jescavez View Post
    I have this equation:

    $\displaystyle \frac{\partial^2t}{\partial{x^2}}+\frac{\partial^2 t}{\partial{y^2}}=-2$

    this equation needs to be solved using the conditions where $\displaystyle t(x,y)=0$ with diffusion between circles $\displaystyle x^2+y^2=1$ and $\displaystyle x^2+y^2=e^2$

    we assume we can solve with form $\displaystyle t(x,y)=f(z)$ with $\displaystyle z=x^2+y^2$

    how can first equation with conditions now be written as:
    $\displaystyle
    z\frac{d^2{f}}{d{z^2}}+\frac{df}{dz}=-0.5,$ $\displaystyle \left\{\begin{array}{cc}f(1)=0,\\f(e^2)=0, \end{array}\right.$

    if someone can please give me tip with how i work out $\displaystyle \frac{df}{dz}$ (because i am confuse thinking what this means) maybe I can give question a go!

    We have the equation $\displaystyle \frac{\partial^2t}{\partial{x^2}}+\frac{\partial^2 t}{\partial{y^2}}=-2$ and we want to replace $\displaystyle x$ and $\displaystyle y$ by $\displaystyle z=x^2+y^2$ so we need to compute $\displaystyle \frac{\partial^2t}{\partial{y^2}}=\frac{\partial^2 f}{\partial{y^2}}$ and $\displaystyle \frac{\partial^2t}{\partial{x^2}}=\frac{\partial^2 f}{\partial{x^2}}$ in terms of $\displaystyle f''(z)$ and $\displaystyle f'(z)$. Let's start with $\displaystyle \frac{\partial t(x,y)}{\partial x}=\frac{\partial f(z)}{\partial x}$ :

    The chain rule states that $\displaystyle \big( v\circ u\big) ' = u'\cdot v'\circ u$ hence $\displaystyle \frac{\partial f(z)}{\partial x}=\frac{\partial z}{\partial x}\cdot f'(z)=2x f'(z)$

    Let's differentiate $\displaystyle f$ a second time with respect to $\displaystyle x$ :

    $\displaystyle \begin{aligned}
    \frac{\partial ^2 f(z)}{\partial x^2}&=\frac{\partial }{\partial x}\Big[2x f'(z)\Big]\\
    &=2f'(z)+2x\frac{\partial f'(z)}{\partial x}
    \end{aligned}$

    Now what is $\displaystyle \frac{\partial f'(z)}{\partial x}$ ? Again, let's use the chain rule :

    $\displaystyle \frac{\partial f'(z)}{\partial x}=\frac{\partial z}{\partial x}\cdot f''(z)=2xf''(z)$

    hence $\displaystyle \frac{\partial ^2 f(z)}{\partial x^2}=2f'(z)+4x^2f''(z)$

    Following this idea you can compute $\displaystyle \frac{\partial ^2 t}{\partial y^2}= \frac{\partial ^2 f}{\partial y^2}$. Substituting $\displaystyle \frac{\partial ^2 t}{\partial y^2}$ and $\displaystyle \frac{\partial ^2 t}{\partial y^2}$ by their expressions in terms of $\displaystyle f''(z)$ and $\displaystyle f'(z)$ in the differential equation will give you the equation $\displaystyle zf''(z)+f'(z)=-1/2$ you're looking for.


    Hope it helps !
    Last edited by flyingsquirrel; Jun 25th 2008 at 05:47 AM.
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  3. #3
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    Thank you flying squirrel that was a great explanation I have it now.

    May I to clarify one other point? It is very relate to this, I have equation:

    $\displaystyle t(x,y)=\frac{1}{2}(1-(x^2+y^2))+\frac{1}{4}(e^2-1)log(x^2+y^2)$

    I see that $\displaystyle (1-(x^2+y^2))=0$ and $\displaystyle (e^2-1)=0$ and $\displaystyle log(x^2+y^2)=2$ and I think it is trying to say $\displaystyle t(x,y)=0$ but I don't understand why the $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{1}{4}$ are need or must to be there as I think they are multiplying by 0 anyway?

    Thank you and sorry for extra question!
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jescavez View Post
    May I to clarify one other point? It is very relate to this, I have equation:

    $\displaystyle t(x,y)=\frac{1}{2}(1-(x^2+y^2))+\frac{1}{4}(e^2-1)log(x^2+y^2)$

    I see that $\displaystyle (1-(x^2+y^2))=0$ and $\displaystyle (e^2-1)=0$ and $\displaystyle log(x^2+y^2)=2$ and I think it is trying to say $\displaystyle t(x,y)=0$ but I don't understand why the $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{1}{4}$ are need or must to be there as I think they are multiplying by 0 anyway?
    I'm not sure to understand what you mean : if one replaces $\displaystyle \frac{1}{2}$ by say $\displaystyle 42$, the new function is not a solution of our problem anymore :

    We know that the solutions of $\displaystyle zf''(z)+f'(z)=-\frac{1}{2}$ are $\displaystyle A\ln z-\frac{1}{2}z+B$. The two constants $\displaystyle A$ and $\displaystyle B$ are obtained with the two following conditions

    $\displaystyle
    \begin{cases}
    f(1)=0\\
    f(e^2)=0
    \end{cases}
    \implies\begin{cases}
    1\ln 1-\frac{1}{2}\cdot 1+B=0\\
    A\ln e^2-\frac{1}{2}e^2+B=0
    \end{cases}
    \implies
    \begin{cases}
    B=\displaystyle\frac{1}{2}\\
    A=\displaystyle\frac{1-e^2}{4{\color{red}\ln e}}
    \end{cases}$

    Changing the value of $\displaystyle A$ or the value $\displaystyle B$ makes at least one of the these conditions become false : the function with the new value of $\displaystyle A$ or $\displaystyle B$ can't be a solution of this problem.

    EDIT : If $\displaystyle e=\exp 1$ don't take into account what I wrote in red.
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  5. #5
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    Ah, ok I did not realise the solution would take this form. please could you explain me where $\displaystyle
    A\ln z-\frac{1}{2}z+B
    $ come from?
    Thank you.
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jescavez View Post
    Ah, ok I did not realise the solution would take this form. please could you explain me where $\displaystyle
    A\ln z-\frac{1}{2}z+B
    $ come from?
    I thought you'd solved it.

    Let $\displaystyle y=f'(z)$.

    The differential equation $\displaystyle zf''(z)+f'(z)=-\frac{1}{2}$ can be written $\displaystyle zy'+y=-\frac{1}{2} \implies y'=-\frac{1}{z}\left(y+\frac{1}{2}\right) \Longleftrightarrow \left( y+\frac{1}{2}\right)'=-\frac{1}{z}\left( y+\frac{1}{2}\right)$

    This is a first-order ODE which solutions are $\displaystyle y+\frac{1}{2}=\exp\left(\int -\frac{\mathrm{d}u}{u}\right)=\exp\left(-\ln z + C \right)=\frac{A}{z}$. Thus $\displaystyle f'(z)=\frac{A}{z}-\frac{1}{2}$ which gives us $\displaystyle f(z)=A\ln z-\frac{1}{2}z+B$ when integrated.
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  7. #7
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    Thanks flying squirrel. What is that technique called?

    a couple of Qs. why in:

    $\displaystyle
    zy'+y=-\frac{1}{2} \implies y'=-\frac{1}{z}\left(y+\frac{1}{2}\right) \Longleftrightarrow \left( y+\frac{1}{2}\right)'=-\frac{1}{z}\left( y+\frac{1}{2}\right)
    $

    does the implied bit work?
    and I'm not quite sure how you get to:

    $\displaystyle
    y+\frac{1}{2}=\exp\left(\int -\frac{\mathrm{d}u}{u}\right)
    $

    (And last question, hopefully, sorry!)For this bit:

    $\displaystyle
    \begin{cases}
    f(1)=0\\
    f(e^2)=0
    \end{cases}
    \implies\begin{cases}
    1\ln 1-\frac{1}{2}\cdot 1+B=0\\
    A\ln e^2-\frac{1}{2}e^2+B=0
    \end{cases}
    \implies
    \begin{cases}
    B=\displaystyle\frac{1}{2}\\
    A=\displaystyle\frac{e^2-1}{4{\color{red}\ln e}}
    \end{cases}
    $
    we are adding the solutions right to get to final equation? then what happens to the $\displaystyle -\frac{1}{2}e^2+B$ part of $\displaystyle f(e^2)$? we do not seem to use it?

    Thank you again.
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jescavez View Post
    What is that technique called?
    I don't think it has a name.

    why in:

    $\displaystyle
    zy'+y=-\frac{1}{2} \implies y'=-\frac{1}{z}\left(y+\frac{1}{2}\right) \Longleftrightarrow \left( y+\frac{1}{2}\right)'=-\frac{1}{z}\left( y+\frac{1}{2}\right)
    $

    does the implied bit work?
    I don't understand what you mean

    and I'm not quite sure how you get to:

    $\displaystyle
    y+\frac{1}{2}=\exp\left(\int -\frac{\mathrm{d}u}{u}\right)
    $
    You should take a look at this page : first-order ODE

    (And last question, hopefully, sorry!)
    There is no need to apologize each time you ask a question.
    For this bit:

    $\displaystyle
    \begin{cases}
    f(1)=0\\
    f(e^2)=0
    \end{cases}
    \implies\begin{cases}
    1\ln 1-\frac{1}{2}\cdot 1+B=0\\
    A\ln e^2-\frac{1}{2}e^2+B=0
    \end{cases}
    \implies
    \begin{cases}
    B=\displaystyle\frac{1}{2}\\
    A=\displaystyle\frac{e^2-1}{4{\color{red}\ln e}}
    \end{cases}
    $
    we are adding the solutions right to get to final equation? then what happens to the $\displaystyle -\frac{1}{2}e^2+B$ part of $\displaystyle f(e^2)$? we do not seem to use it?
    We do use this term :

    $\displaystyle
    \begin{aligned}
    A\ln e^2-\frac{1}{2}e^2+B=0
    &\implies A\ln e^2-\frac{1}{2}e^2+\frac{1}{2}=0\\
    &\implies 2A\ln e = \frac{e^2-1}{2}\\
    &\implies A=\frac{e^2-1}{4\ln e}\\
    \end{aligned}$

    If you want to do this by subtracting one equation to the other :

    $\displaystyle
    \begin{aligned}
    \begin{cases}
    1\ln 1-\frac{1}{2}\cdot 1+B=0 & (1)\\
    A\ln e^2-\frac{1}{2}e^2+B=0 &(2)
    \end{cases} &\implies (2)-(1) : A\ln e^2-\frac{1}{2}e^2+\frac{1}{2}=0\\
    &\implies A=\frac{\frac{1}{2}e^2-\frac{1}{2}}{2\ln e}\\
    &\implies A=\frac{e^2-1}{4\ln e}
    \end{aligned}$
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  9. #9
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    Quote Originally Posted by flyingsquirrel View Post


    I don't understand what you mean
    I mean the $\displaystyle y' $ going to $\displaystyle y' + \frac{1}{2}$ without anything else changing. I don't understand why that is ok to do (I see that once differentiated the 1/2 goes but I would have thought you can say same about any constant...)

    Thank you
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  10. #10
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jescavez View Post
    I mean the $\displaystyle y' $ going to $\displaystyle y' + \frac{1}{2}$ without anything else changing. I don't understand why that is ok to do (I see that once differentiated the 1/2 goes but I would have thought you can say same about any constant...)
    Yes, it would also have been right with any other constant. Choosing $\displaystyle \frac{1}{2}$ allows one to write the differential equation $\displaystyle \left(y+\frac{1}{2}\right)'=-\frac{1}{z}\left(y+\frac{1}{2}\right) $ as $\displaystyle Y'=-\frac{Y}{z}$ with $\displaystyle Y=y+\frac{1}{2}$. I found this interesting since the equation in terms of $\displaystyle Y$ is a little bit easier to solve than $\displaystyle zy'+y=-\frac{1}{2}$ : with the notations used in the link given in my previous post, for the equation $\displaystyle Y'=-\frac{Y}{z}$ we have $\displaystyle q(z)=0$ hence $\displaystyle Y$ is simply $\displaystyle \frac{C}{u(z)}$ whereas for the equation $\displaystyle zy'+y=-\frac{1}{2}$ we have $\displaystyle q(z)=-\frac{1}{2z}$ so one has to compute $\displaystyle y=-\frac{\int \frac{1}{2z} u(z)\,\mathrm{d}z+C'}{u(z)}$ ...
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