Obtain the leading order uniform approximation of the solution












1














Obtain the leading order uniform approximation of the solution to $ epsilon y′′-x^2y′-y=0$.



The boundary conditions are $y(0)=y(1)=1$.



Since $a(x)<0$ the boundary layer is at $x=1$.



The outer solution will be of the form $y(x; epsilon) = y_0(x) + epsilon y_1(x) + ...$



So the leading order problem is $-x^2y_0' - y_0 =0$ and $y_0(0) = 1$.



Hence $y_0 = ke^{1/x}$.



Here's where the issue arises: to work out $k$ I get $y_0(0)=ke^{1/0}=1$ using the boundary condition at $x=0$, but $e^{1/0}$ is obviously undefined. How can I proceed form here?










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  • I think there is also a boundary layer of width $sqrt{epsilon}$ at $x=0$, since for $x~simsqrt{epsilon}$ you have $epsilonsim x^2$, so maybe the first term is important near $x=0$. If you can't satisfy the boundary condition this is often a sign that something is not working in your expansion.
    – David
    Dec 3 at 20:16












  • Ah I see, that would make sense, thank you! I took the boundary layer to be at x=1 as we were taught that if the coefficient of the y’ term is negative, the boundary layer is at x=1. How can you show that there is also a BL at x=0?
    – maria1991
    Dec 3 at 20:56










  • The coefficient is zero at $x=0$ which should make you pause to see if you have a dominant balance near $x=0$.
    – David
    Dec 3 at 21:22










  • This question asks about the same equation: math.stackexchange.com/q/2034260/131807.
    – David
    Dec 4 at 1:43
















1














Obtain the leading order uniform approximation of the solution to $ epsilon y′′-x^2y′-y=0$.



The boundary conditions are $y(0)=y(1)=1$.



Since $a(x)<0$ the boundary layer is at $x=1$.



The outer solution will be of the form $y(x; epsilon) = y_0(x) + epsilon y_1(x) + ...$



So the leading order problem is $-x^2y_0' - y_0 =0$ and $y_0(0) = 1$.



Hence $y_0 = ke^{1/x}$.



Here's where the issue arises: to work out $k$ I get $y_0(0)=ke^{1/0}=1$ using the boundary condition at $x=0$, but $e^{1/0}$ is obviously undefined. How can I proceed form here?










share|cite|improve this question
























  • I think there is also a boundary layer of width $sqrt{epsilon}$ at $x=0$, since for $x~simsqrt{epsilon}$ you have $epsilonsim x^2$, so maybe the first term is important near $x=0$. If you can't satisfy the boundary condition this is often a sign that something is not working in your expansion.
    – David
    Dec 3 at 20:16












  • Ah I see, that would make sense, thank you! I took the boundary layer to be at x=1 as we were taught that if the coefficient of the y’ term is negative, the boundary layer is at x=1. How can you show that there is also a BL at x=0?
    – maria1991
    Dec 3 at 20:56










  • The coefficient is zero at $x=0$ which should make you pause to see if you have a dominant balance near $x=0$.
    – David
    Dec 3 at 21:22










  • This question asks about the same equation: math.stackexchange.com/q/2034260/131807.
    – David
    Dec 4 at 1:43














1












1








1







Obtain the leading order uniform approximation of the solution to $ epsilon y′′-x^2y′-y=0$.



The boundary conditions are $y(0)=y(1)=1$.



Since $a(x)<0$ the boundary layer is at $x=1$.



The outer solution will be of the form $y(x; epsilon) = y_0(x) + epsilon y_1(x) + ...$



So the leading order problem is $-x^2y_0' - y_0 =0$ and $y_0(0) = 1$.



Hence $y_0 = ke^{1/x}$.



Here's where the issue arises: to work out $k$ I get $y_0(0)=ke^{1/0}=1$ using the boundary condition at $x=0$, but $e^{1/0}$ is obviously undefined. How can I proceed form here?










share|cite|improve this question















Obtain the leading order uniform approximation of the solution to $ epsilon y′′-x^2y′-y=0$.



The boundary conditions are $y(0)=y(1)=1$.



Since $a(x)<0$ the boundary layer is at $x=1$.



The outer solution will be of the form $y(x; epsilon) = y_0(x) + epsilon y_1(x) + ...$



So the leading order problem is $-x^2y_0' - y_0 =0$ and $y_0(0) = 1$.



Hence $y_0 = ke^{1/x}$.



Here's where the issue arises: to work out $k$ I get $y_0(0)=ke^{1/0}=1$ using the boundary condition at $x=0$, but $e^{1/0}$ is obviously undefined. How can I proceed form here?







asymptotics perturbation-theory boundary-layer






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edited Dec 3 at 22:57

























asked Dec 1 at 17:22









maria1991

104




104












  • I think there is also a boundary layer of width $sqrt{epsilon}$ at $x=0$, since for $x~simsqrt{epsilon}$ you have $epsilonsim x^2$, so maybe the first term is important near $x=0$. If you can't satisfy the boundary condition this is often a sign that something is not working in your expansion.
    – David
    Dec 3 at 20:16












  • Ah I see, that would make sense, thank you! I took the boundary layer to be at x=1 as we were taught that if the coefficient of the y’ term is negative, the boundary layer is at x=1. How can you show that there is also a BL at x=0?
    – maria1991
    Dec 3 at 20:56










  • The coefficient is zero at $x=0$ which should make you pause to see if you have a dominant balance near $x=0$.
    – David
    Dec 3 at 21:22










  • This question asks about the same equation: math.stackexchange.com/q/2034260/131807.
    – David
    Dec 4 at 1:43


















  • I think there is also a boundary layer of width $sqrt{epsilon}$ at $x=0$, since for $x~simsqrt{epsilon}$ you have $epsilonsim x^2$, so maybe the first term is important near $x=0$. If you can't satisfy the boundary condition this is often a sign that something is not working in your expansion.
    – David
    Dec 3 at 20:16












  • Ah I see, that would make sense, thank you! I took the boundary layer to be at x=1 as we were taught that if the coefficient of the y’ term is negative, the boundary layer is at x=1. How can you show that there is also a BL at x=0?
    – maria1991
    Dec 3 at 20:56










  • The coefficient is zero at $x=0$ which should make you pause to see if you have a dominant balance near $x=0$.
    – David
    Dec 3 at 21:22










  • This question asks about the same equation: math.stackexchange.com/q/2034260/131807.
    – David
    Dec 4 at 1:43
















I think there is also a boundary layer of width $sqrt{epsilon}$ at $x=0$, since for $x~simsqrt{epsilon}$ you have $epsilonsim x^2$, so maybe the first term is important near $x=0$. If you can't satisfy the boundary condition this is often a sign that something is not working in your expansion.
– David
Dec 3 at 20:16






I think there is also a boundary layer of width $sqrt{epsilon}$ at $x=0$, since for $x~simsqrt{epsilon}$ you have $epsilonsim x^2$, so maybe the first term is important near $x=0$. If you can't satisfy the boundary condition this is often a sign that something is not working in your expansion.
– David
Dec 3 at 20:16














Ah I see, that would make sense, thank you! I took the boundary layer to be at x=1 as we were taught that if the coefficient of the y’ term is negative, the boundary layer is at x=1. How can you show that there is also a BL at x=0?
– maria1991
Dec 3 at 20:56




Ah I see, that would make sense, thank you! I took the boundary layer to be at x=1 as we were taught that if the coefficient of the y’ term is negative, the boundary layer is at x=1. How can you show that there is also a BL at x=0?
– maria1991
Dec 3 at 20:56












The coefficient is zero at $x=0$ which should make you pause to see if you have a dominant balance near $x=0$.
– David
Dec 3 at 21:22




The coefficient is zero at $x=0$ which should make you pause to see if you have a dominant balance near $x=0$.
– David
Dec 3 at 21:22












This question asks about the same equation: math.stackexchange.com/q/2034260/131807.
– David
Dec 4 at 1:43




This question asks about the same equation: math.stackexchange.com/q/2034260/131807.
– David
Dec 4 at 1:43










1 Answer
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This is a complicated problem, but if you're careful it does work out. Firstly, let's write out the equation assuming there is a boundary layer or width $epsilon^alpha$ at $x_0$, so let $X=(x-x_0)/epsilon^alpha$ and $Y(X)=y(x)$. We get
$$ epsilon^{1-2alpha}Y_{XX}-epsilon^{-alpha}(epsilon^alpha X+x_0)^2Y_X-Y=0,$$
expanding
$$ epsilon^{1-2alpha}Y_{XX}-epsilon^{alpha}X^2Y_X-2Xx_0Y_X-epsilon^{-alpha}x_0^2Y_X-Y=0.$$



Now for dominant balance. If $x_0neq0$, then we would balance $epsilon^{1-2alpha}$ with $epsilon^{-alpha}$ (you can check that this is the only dominant balance) to give $alpha=1$. You can leave $x_0$ unknown and determine that it must be 1, but since you already know that we'll use the fact and write our leading-order inner equation at $x=1$ as
$$Y_{XX}-Y_X=0,$$
with boundary condition $Y(0)=1$ (since $x=1$ corresponds to $X=0$). The solution is $Y(X)=A+Be^X$ where $A+B=1$.



If $x_0=0$ then we have a different balance, with $alpha=1/2$. Let $mathsf Y(chi)=y(x)$ with $chi=x/sqrt{epsilon}$ and the leading-order inner equation at $x=0$ is
$$mathsf Y_{chichi}-mathsf Y=0,$$
with $mathsf Y(0)=1$. The solution is $mathsf Y=Ce^chi+De^{-chi}$, and since our solution must be bounded as we exit the boundary layer, we need $A=0$, and the boundary condition gives $D=1$.



The leading order outer solution is, as you found, $y=ke^{1/x}$. You can say here that $k=0$ since the solution must be bounded as you enter each boundary layer, or do it through asymptotic matching.



To fix the remaining constants $k$ and $A$ (or $B$), we need to match all the parts of the solution. To do this, we need the outer solution to be bounded, so we need $k=0$. Alternatively, consider matching the outer solution to the inner solution at $x=0$,
$$lim_{chirightarrowinfty}e^{-chi}=lim_{xrightarrow0}ke^{1/x}Rightarrow0=lim_{xrightarrow0}ke^{1/x}Rightarrow k=0.$$ Then the match between the outer layer and the inner layer at $x=1$ gives
$$lim{xrightarrow1}0=lim{Xrightarrow-infty}A+Be^XRightarrow A=0$$
and hence $B=1$.



So the inner solution at $x=0$ is $mathsf Y(chi)=e^{-chi}$, the outer solution is $y(x)=0$ and the inner solution at $x=1$ is $Y(X)=e^X$. We can find a uniformly valid approximation by adding the three equations (writing them all in terms of $x$) and subtracting off the matching constants (all zero) as,
$$y_{unif}(x)=e^{x/sqrt{epsilon}}+e^{(x-1)/epsilon}.$$



This will not be particularly accurate, it's an $O(sqrt{epsilon})$ approximation, but I did verify that numerically. I've made a plot of the solution for $epsilon=0.01$ first and $epsilon=0.02$ second.



enter image description hereenter image description here






share|cite|improve this answer





















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    1 Answer
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    This is a complicated problem, but if you're careful it does work out. Firstly, let's write out the equation assuming there is a boundary layer or width $epsilon^alpha$ at $x_0$, so let $X=(x-x_0)/epsilon^alpha$ and $Y(X)=y(x)$. We get
    $$ epsilon^{1-2alpha}Y_{XX}-epsilon^{-alpha}(epsilon^alpha X+x_0)^2Y_X-Y=0,$$
    expanding
    $$ epsilon^{1-2alpha}Y_{XX}-epsilon^{alpha}X^2Y_X-2Xx_0Y_X-epsilon^{-alpha}x_0^2Y_X-Y=0.$$



    Now for dominant balance. If $x_0neq0$, then we would balance $epsilon^{1-2alpha}$ with $epsilon^{-alpha}$ (you can check that this is the only dominant balance) to give $alpha=1$. You can leave $x_0$ unknown and determine that it must be 1, but since you already know that we'll use the fact and write our leading-order inner equation at $x=1$ as
    $$Y_{XX}-Y_X=0,$$
    with boundary condition $Y(0)=1$ (since $x=1$ corresponds to $X=0$). The solution is $Y(X)=A+Be^X$ where $A+B=1$.



    If $x_0=0$ then we have a different balance, with $alpha=1/2$. Let $mathsf Y(chi)=y(x)$ with $chi=x/sqrt{epsilon}$ and the leading-order inner equation at $x=0$ is
    $$mathsf Y_{chichi}-mathsf Y=0,$$
    with $mathsf Y(0)=1$. The solution is $mathsf Y=Ce^chi+De^{-chi}$, and since our solution must be bounded as we exit the boundary layer, we need $A=0$, and the boundary condition gives $D=1$.



    The leading order outer solution is, as you found, $y=ke^{1/x}$. You can say here that $k=0$ since the solution must be bounded as you enter each boundary layer, or do it through asymptotic matching.



    To fix the remaining constants $k$ and $A$ (or $B$), we need to match all the parts of the solution. To do this, we need the outer solution to be bounded, so we need $k=0$. Alternatively, consider matching the outer solution to the inner solution at $x=0$,
    $$lim_{chirightarrowinfty}e^{-chi}=lim_{xrightarrow0}ke^{1/x}Rightarrow0=lim_{xrightarrow0}ke^{1/x}Rightarrow k=0.$$ Then the match between the outer layer and the inner layer at $x=1$ gives
    $$lim{xrightarrow1}0=lim{Xrightarrow-infty}A+Be^XRightarrow A=0$$
    and hence $B=1$.



    So the inner solution at $x=0$ is $mathsf Y(chi)=e^{-chi}$, the outer solution is $y(x)=0$ and the inner solution at $x=1$ is $Y(X)=e^X$. We can find a uniformly valid approximation by adding the three equations (writing them all in terms of $x$) and subtracting off the matching constants (all zero) as,
    $$y_{unif}(x)=e^{x/sqrt{epsilon}}+e^{(x-1)/epsilon}.$$



    This will not be particularly accurate, it's an $O(sqrt{epsilon})$ approximation, but I did verify that numerically. I've made a plot of the solution for $epsilon=0.01$ first and $epsilon=0.02$ second.



    enter image description hereenter image description here






    share|cite|improve this answer


























      0














      This is a complicated problem, but if you're careful it does work out. Firstly, let's write out the equation assuming there is a boundary layer or width $epsilon^alpha$ at $x_0$, so let $X=(x-x_0)/epsilon^alpha$ and $Y(X)=y(x)$. We get
      $$ epsilon^{1-2alpha}Y_{XX}-epsilon^{-alpha}(epsilon^alpha X+x_0)^2Y_X-Y=0,$$
      expanding
      $$ epsilon^{1-2alpha}Y_{XX}-epsilon^{alpha}X^2Y_X-2Xx_0Y_X-epsilon^{-alpha}x_0^2Y_X-Y=0.$$



      Now for dominant balance. If $x_0neq0$, then we would balance $epsilon^{1-2alpha}$ with $epsilon^{-alpha}$ (you can check that this is the only dominant balance) to give $alpha=1$. You can leave $x_0$ unknown and determine that it must be 1, but since you already know that we'll use the fact and write our leading-order inner equation at $x=1$ as
      $$Y_{XX}-Y_X=0,$$
      with boundary condition $Y(0)=1$ (since $x=1$ corresponds to $X=0$). The solution is $Y(X)=A+Be^X$ where $A+B=1$.



      If $x_0=0$ then we have a different balance, with $alpha=1/2$. Let $mathsf Y(chi)=y(x)$ with $chi=x/sqrt{epsilon}$ and the leading-order inner equation at $x=0$ is
      $$mathsf Y_{chichi}-mathsf Y=0,$$
      with $mathsf Y(0)=1$. The solution is $mathsf Y=Ce^chi+De^{-chi}$, and since our solution must be bounded as we exit the boundary layer, we need $A=0$, and the boundary condition gives $D=1$.



      The leading order outer solution is, as you found, $y=ke^{1/x}$. You can say here that $k=0$ since the solution must be bounded as you enter each boundary layer, or do it through asymptotic matching.



      To fix the remaining constants $k$ and $A$ (or $B$), we need to match all the parts of the solution. To do this, we need the outer solution to be bounded, so we need $k=0$. Alternatively, consider matching the outer solution to the inner solution at $x=0$,
      $$lim_{chirightarrowinfty}e^{-chi}=lim_{xrightarrow0}ke^{1/x}Rightarrow0=lim_{xrightarrow0}ke^{1/x}Rightarrow k=0.$$ Then the match between the outer layer and the inner layer at $x=1$ gives
      $$lim{xrightarrow1}0=lim{Xrightarrow-infty}A+Be^XRightarrow A=0$$
      and hence $B=1$.



      So the inner solution at $x=0$ is $mathsf Y(chi)=e^{-chi}$, the outer solution is $y(x)=0$ and the inner solution at $x=1$ is $Y(X)=e^X$. We can find a uniformly valid approximation by adding the three equations (writing them all in terms of $x$) and subtracting off the matching constants (all zero) as,
      $$y_{unif}(x)=e^{x/sqrt{epsilon}}+e^{(x-1)/epsilon}.$$



      This will not be particularly accurate, it's an $O(sqrt{epsilon})$ approximation, but I did verify that numerically. I've made a plot of the solution for $epsilon=0.01$ first and $epsilon=0.02$ second.



      enter image description hereenter image description here






      share|cite|improve this answer
























        0












        0








        0






        This is a complicated problem, but if you're careful it does work out. Firstly, let's write out the equation assuming there is a boundary layer or width $epsilon^alpha$ at $x_0$, so let $X=(x-x_0)/epsilon^alpha$ and $Y(X)=y(x)$. We get
        $$ epsilon^{1-2alpha}Y_{XX}-epsilon^{-alpha}(epsilon^alpha X+x_0)^2Y_X-Y=0,$$
        expanding
        $$ epsilon^{1-2alpha}Y_{XX}-epsilon^{alpha}X^2Y_X-2Xx_0Y_X-epsilon^{-alpha}x_0^2Y_X-Y=0.$$



        Now for dominant balance. If $x_0neq0$, then we would balance $epsilon^{1-2alpha}$ with $epsilon^{-alpha}$ (you can check that this is the only dominant balance) to give $alpha=1$. You can leave $x_0$ unknown and determine that it must be 1, but since you already know that we'll use the fact and write our leading-order inner equation at $x=1$ as
        $$Y_{XX}-Y_X=0,$$
        with boundary condition $Y(0)=1$ (since $x=1$ corresponds to $X=0$). The solution is $Y(X)=A+Be^X$ where $A+B=1$.



        If $x_0=0$ then we have a different balance, with $alpha=1/2$. Let $mathsf Y(chi)=y(x)$ with $chi=x/sqrt{epsilon}$ and the leading-order inner equation at $x=0$ is
        $$mathsf Y_{chichi}-mathsf Y=0,$$
        with $mathsf Y(0)=1$. The solution is $mathsf Y=Ce^chi+De^{-chi}$, and since our solution must be bounded as we exit the boundary layer, we need $A=0$, and the boundary condition gives $D=1$.



        The leading order outer solution is, as you found, $y=ke^{1/x}$. You can say here that $k=0$ since the solution must be bounded as you enter each boundary layer, or do it through asymptotic matching.



        To fix the remaining constants $k$ and $A$ (or $B$), we need to match all the parts of the solution. To do this, we need the outer solution to be bounded, so we need $k=0$. Alternatively, consider matching the outer solution to the inner solution at $x=0$,
        $$lim_{chirightarrowinfty}e^{-chi}=lim_{xrightarrow0}ke^{1/x}Rightarrow0=lim_{xrightarrow0}ke^{1/x}Rightarrow k=0.$$ Then the match between the outer layer and the inner layer at $x=1$ gives
        $$lim{xrightarrow1}0=lim{Xrightarrow-infty}A+Be^XRightarrow A=0$$
        and hence $B=1$.



        So the inner solution at $x=0$ is $mathsf Y(chi)=e^{-chi}$, the outer solution is $y(x)=0$ and the inner solution at $x=1$ is $Y(X)=e^X$. We can find a uniformly valid approximation by adding the three equations (writing them all in terms of $x$) and subtracting off the matching constants (all zero) as,
        $$y_{unif}(x)=e^{x/sqrt{epsilon}}+e^{(x-1)/epsilon}.$$



        This will not be particularly accurate, it's an $O(sqrt{epsilon})$ approximation, but I did verify that numerically. I've made a plot of the solution for $epsilon=0.01$ first and $epsilon=0.02$ second.



        enter image description hereenter image description here






        share|cite|improve this answer












        This is a complicated problem, but if you're careful it does work out. Firstly, let's write out the equation assuming there is a boundary layer or width $epsilon^alpha$ at $x_0$, so let $X=(x-x_0)/epsilon^alpha$ and $Y(X)=y(x)$. We get
        $$ epsilon^{1-2alpha}Y_{XX}-epsilon^{-alpha}(epsilon^alpha X+x_0)^2Y_X-Y=0,$$
        expanding
        $$ epsilon^{1-2alpha}Y_{XX}-epsilon^{alpha}X^2Y_X-2Xx_0Y_X-epsilon^{-alpha}x_0^2Y_X-Y=0.$$



        Now for dominant balance. If $x_0neq0$, then we would balance $epsilon^{1-2alpha}$ with $epsilon^{-alpha}$ (you can check that this is the only dominant balance) to give $alpha=1$. You can leave $x_0$ unknown and determine that it must be 1, but since you already know that we'll use the fact and write our leading-order inner equation at $x=1$ as
        $$Y_{XX}-Y_X=0,$$
        with boundary condition $Y(0)=1$ (since $x=1$ corresponds to $X=0$). The solution is $Y(X)=A+Be^X$ where $A+B=1$.



        If $x_0=0$ then we have a different balance, with $alpha=1/2$. Let $mathsf Y(chi)=y(x)$ with $chi=x/sqrt{epsilon}$ and the leading-order inner equation at $x=0$ is
        $$mathsf Y_{chichi}-mathsf Y=0,$$
        with $mathsf Y(0)=1$. The solution is $mathsf Y=Ce^chi+De^{-chi}$, and since our solution must be bounded as we exit the boundary layer, we need $A=0$, and the boundary condition gives $D=1$.



        The leading order outer solution is, as you found, $y=ke^{1/x}$. You can say here that $k=0$ since the solution must be bounded as you enter each boundary layer, or do it through asymptotic matching.



        To fix the remaining constants $k$ and $A$ (or $B$), we need to match all the parts of the solution. To do this, we need the outer solution to be bounded, so we need $k=0$. Alternatively, consider matching the outer solution to the inner solution at $x=0$,
        $$lim_{chirightarrowinfty}e^{-chi}=lim_{xrightarrow0}ke^{1/x}Rightarrow0=lim_{xrightarrow0}ke^{1/x}Rightarrow k=0.$$ Then the match between the outer layer and the inner layer at $x=1$ gives
        $$lim{xrightarrow1}0=lim{Xrightarrow-infty}A+Be^XRightarrow A=0$$
        and hence $B=1$.



        So the inner solution at $x=0$ is $mathsf Y(chi)=e^{-chi}$, the outer solution is $y(x)=0$ and the inner solution at $x=1$ is $Y(X)=e^X$. We can find a uniformly valid approximation by adding the three equations (writing them all in terms of $x$) and subtracting off the matching constants (all zero) as,
        $$y_{unif}(x)=e^{x/sqrt{epsilon}}+e^{(x-1)/epsilon}.$$



        This will not be particularly accurate, it's an $O(sqrt{epsilon})$ approximation, but I did verify that numerically. I've made a plot of the solution for $epsilon=0.01$ first and $epsilon=0.02$ second.



        enter image description hereenter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 at 2:57









        David

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