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Math Help - 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:

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


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

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

    how can first equation with conditions now be written as:
    <br />
z\frac{d^2{f}}{d{z^2}}+\frac{df}{dz}=-0.5, \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 \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:

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

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

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

    how can first equation with conditions now be written as:
    <br />
z\frac{d^2{f}}{d{z^2}}+\frac{df}{dz}=-0.5, \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 \frac{df}{dz} (because i am confuse thinking what this means) maybe I can give question a go!

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

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

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

    \begin{aligned}<br />
\frac{\partial ^2 f(z)}{\partial x^2}&=\frac{\partial }{\partial x}\Big[2x f'(z)\Big]\\<br />
&=2f'(z)+2x\frac{\partial f'(z)}{\partial x}<br />
\end{aligned}

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

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

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

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


    Hope it helps !
    Last edited by flyingsquirrel; June 25th 2008 at 06:47 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:

    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 (1-(x^2+y^2))=0 and (e^2-1)=0 and log(x^2+y^2)=2 and I think it is trying to say t(x,y)=0 but I don't understand why the \frac{1}{2} and \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:

    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 (1-(x^2+y^2))=0 and (e^2-1)=0 and log(x^2+y^2)=2 and I think it is trying to say t(x,y)=0 but I don't understand why the \frac{1}{2} and \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 \frac{1}{2} by say 42, the new function is not a solution of our problem anymore :

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

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

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

    EDIT : If 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 <br />
A\ln z-\frac{1}{2}z+B<br />
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 <br />
A\ln z-\frac{1}{2}z+B<br />
come from?
    I thought you'd solved it.

    Let y=f'(z).

    The differential equation zf''(z)+f'(z)=-\frac{1}{2} can be written 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 y+\frac{1}{2}=\exp\left(\int -\frac{\mathrm{d}u}{u}\right)=\exp\left(-\ln z + C \right)=\frac{A}{z}. Thus f'(z)=\frac{A}{z}-\frac{1}{2} which gives us 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:

    <br />
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)<br />

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

    <br />
y+\frac{1}{2}=\exp\left(\int -\frac{\mathrm{d}u}{u}\right)<br />

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

    <br />
\begin{cases}<br />
f(1)=0\\<br />
f(e^2)=0<br />
\end{cases}<br />
\implies\begin{cases}<br />
1\ln 1-\frac{1}{2}\cdot 1+B=0\\<br />
A\ln e^2-\frac{1}{2}e^2+B=0<br />
\end{cases}<br />
\implies<br />
\begin{cases}<br />
B=\displaystyle\frac{1}{2}\\<br />
A=\displaystyle\frac{e^2-1}{4{\color{red}\ln e}}<br />
\end{cases}<br />
    we are adding the solutions right to get to final equation? then what happens to the -\frac{1}{2}e^2+B part of 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:

    <br />
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)<br />

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

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

    <br />
y+\frac{1}{2}=\exp\left(\int -\frac{\mathrm{d}u}{u}\right)<br />
    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:

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

    <br />
\begin{aligned}<br />
A\ln e^2-\frac{1}{2}e^2+B=0 <br />
&\implies A\ln e^2-\frac{1}{2}e^2+\frac{1}{2}=0\\<br />
&\implies 2A\ln e = \frac{e^2-1}{2}\\<br />
&\implies A=\frac{e^2-1}{4\ln e}\\<br />
\end{aligned}

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

    <br />
\begin{aligned}<br />
\begin{cases}<br />
1\ln 1-\frac{1}{2}\cdot 1+B=0 & (1)\\<br />
A\ln e^2-\frac{1}{2}e^2+B=0 &(2)<br />
\end{cases} &\implies (2)-(1) : A\ln e^2-\frac{1}{2}e^2+\frac{1}{2}=0\\<br />
 &\implies A=\frac{\frac{1}{2}e^2-\frac{1}{2}}{2\ln e}\\<br />
&\implies A=\frac{e^2-1}{4\ln e}<br />
\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 y' going to 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 y' going to 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 \frac{1}{2} allows one to write the differential equation \left(y+\frac{1}{2}\right)'=-\frac{1}{z}\left(y+\frac{1}{2}\right) as  Y'=-\frac{Y}{z} with Y=y+\frac{1}{2}. I found this interesting since the equation in terms of Y is a little bit easier to solve than zy'+y=-\frac{1}{2} : with the notations used in the link given in my previous post, for the equation Y'=-\frac{Y}{z} we have q(z)=0 hence Y is simply \frac{C}{u(z)} whereas for the equation zy'+y=-\frac{1}{2} we have q(z)=-\frac{1}{2z} so one has to compute y=-\frac{\int \frac{1}{2z} u(z)\,\mathrm{d}z+C'}{u(z)} ...
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