Determine conditions on parameters (for consistency) on RK method $y_{n+1} = y_n + ha_1f(t_n,y_n) + ha_2f(t_n...












0












$begingroup$


I'm asked to find the conditions on the coefficients $a_1,a_2,b_1,b_2$ in the RK method
$$y_{n+1} = y_n + ha_1f(t_n,y_n) + ha_2f(t_n + b_1h, y_n + b_2hf(t_n,y_n))$$
such that is consistent of (a) first order and (b) second order.



The first assumption I make is that $b_1 = b_2 := b$ otherwise I have no idea how to solve it. I know it should be like this, but why is this reasonable? I know it follows if you look at the butcher tablaeu, but why is a method with $b_1 neq b_2$ "impossible"?



I taylor expand $y(t_{n+1})$,
$$y(t_{n+1}) = y(t_n) + hy'(t_n) + O(h^2).$$
The order of consistency is $p$ if $l = O(h^{p+1})$ where $l$ is the local residual.



The local residual takes the form
$$y(t_n) + hy'(t_n) + O(h^2) - y(t_n) - ha_1f(t_n,y_n) - ha_2f(t_{n+b}, y_{n+b}) $$
$$= hy'(t_n) - ha_1y'(t_n) - ha_2y'(t_{n+b}) + O(h^2).$$
So I need to find $a_1,a_2,b$ so that every term except the $O(h^2)$ remains. I get for (a)
$$a_1 + a_2 = 1, b=0.$$



For (b) I expand one term more
$$y(t_{n+1}) = y(t_n) + hy'(t_n) + frac{h^2y''(t_n)}{2} + O(h^3)$$
The local residual becomes
$$y(t_n) + hy'(t_n) + frac{h^2y''(t_n)}{2} + O(h^3) - y(t_n) - ha_1y(t_n) - ha_2y(t_{n+b}) $$



But here I am stuck. How am I supposed to choose the coefficients as to remove the $frac{h^2y''(t_n)}{2}$ term?



How do I solve (b)? And is my answer on (a) correct?






numerical-methods taylor-expansion runge-kutta-methods







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    – LutzL
    Dec 16 '18 at 10:26










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    – LutzL
    Dec 16 '18 at 10:30
















0












$begingroup$


I'm asked to find the conditions on the coefficients $a_1,a_2,b_1,b_2$ in the RK method
$$y_{n+1} = y_n + ha_1f(t_n,y_n) + ha_2f(t_n + b_1h, y_n + b_2hf(t_n,y_n))$$
such that is consistent of (a) first order and (b) second order.



The first assumption I make is that $b_1 = b_2 := b$ otherwise I have no idea how to solve it. I know it should be like this, but why is this reasonable? I know it follows if you look at the butcher tablaeu, but why is a method with $b_1 neq b_2$ "impossible"?



I taylor expand $y(t_{n+1})$,
$$y(t_{n+1}) = y(t_n) + hy'(t_n) + O(h^2).$$
The order of consistency is $p$ if $l = O(h^{p+1})$ where $l$ is the local residual.



The local residual takes the form
$$y(t_n) + hy'(t_n) + O(h^2) - y(t_n) - ha_1f(t_n,y_n) - ha_2f(t_{n+b}, y_{n+b}) $$
$$= hy'(t_n) - ha_1y'(t_n) - ha_2y'(t_{n+b}) + O(h^2).$$
So I need to find $a_1,a_2,b$ so that every term except the $O(h^2)$ remains. I get for (a)
$$a_1 + a_2 = 1, b=0.$$



For (b) I expand one term more
$$y(t_{n+1}) = y(t_n) + hy'(t_n) + frac{h^2y''(t_n)}{2} + O(h^3)$$
The local residual becomes
$$y(t_n) + hy'(t_n) + frac{h^2y''(t_n)}{2} + O(h^3) - y(t_n) - ha_1y(t_n) - ha_2y(t_{n+b}) $$



But here I am stuck. How am I supposed to choose the coefficients as to remove the $frac{h^2y''(t_n)}{2}$ term?



How do I solve (b)? And is my answer on (a) correct?






numerical-methods taylor-expansion runge-kutta-methods







share|cite|improve this question









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  • $begingroup$
    Please indicate cross posts of the same problem: scicomp.stackexchange.com/q/30721/6839
    $endgroup$
    – LutzL
    Dec 16 '18 at 10:26










  • $begingroup$
    Related with partial solution: math.stackexchange.com/questions/2715967/…
    $endgroup$
    – LutzL
    Dec 16 '18 at 10:30














0












0








0





$begingroup$


I'm asked to find the conditions on the coefficients $a_1,a_2,b_1,b_2$ in the RK method
$$y_{n+1} = y_n + ha_1f(t_n,y_n) + ha_2f(t_n + b_1h, y_n + b_2hf(t_n,y_n))$$
such that is consistent of (a) first order and (b) second order.



The first assumption I make is that $b_1 = b_2 := b$ otherwise I have no idea how to solve it. I know it should be like this, but why is this reasonable? I know it follows if you look at the butcher tablaeu, but why is a method with $b_1 neq b_2$ "impossible"?



I taylor expand $y(t_{n+1})$,
$$y(t_{n+1}) = y(t_n) + hy'(t_n) + O(h^2).$$
The order of consistency is $p$ if $l = O(h^{p+1})$ where $l$ is the local residual.



The local residual takes the form
$$y(t_n) + hy'(t_n) + O(h^2) - y(t_n) - ha_1f(t_n,y_n) - ha_2f(t_{n+b}, y_{n+b}) $$
$$= hy'(t_n) - ha_1y'(t_n) - ha_2y'(t_{n+b}) + O(h^2).$$
So I need to find $a_1,a_2,b$ so that every term except the $O(h^2)$ remains. I get for (a)
$$a_1 + a_2 = 1, b=0.$$



For (b) I expand one term more
$$y(t_{n+1}) = y(t_n) + hy'(t_n) + frac{h^2y''(t_n)}{2} + O(h^3)$$
The local residual becomes
$$y(t_n) + hy'(t_n) + frac{h^2y''(t_n)}{2} + O(h^3) - y(t_n) - ha_1y(t_n) - ha_2y(t_{n+b}) $$



But here I am stuck. How am I supposed to choose the coefficients as to remove the $frac{h^2y''(t_n)}{2}$ term?



How do I solve (b)? And is my answer on (a) correct?






numerical-methods taylor-expansion runge-kutta-methods







share|cite|improve this question









$endgroup$




I'm asked to find the conditions on the coefficients $a_1,a_2,b_1,b_2$ in the RK method
$$y_{n+1} = y_n + ha_1f(t_n,y_n) + ha_2f(t_n + b_1h, y_n + b_2hf(t_n,y_n))$$
such that is consistent of (a) first order and (b) second order.



The first assumption I make is that $b_1 = b_2 := b$ otherwise I have no idea how to solve it. I know it should be like this, but why is this reasonable? I know it follows if you look at the butcher tablaeu, but why is a method with $b_1 neq b_2$ "impossible"?



I taylor expand $y(t_{n+1})$,
$$y(t_{n+1}) = y(t_n) + hy'(t_n) + O(h^2).$$
The order of consistency is $p$ if $l = O(h^{p+1})$ where $l$ is the local residual.



The local residual takes the form
$$y(t_n) + hy'(t_n) + O(h^2) - y(t_n) - ha_1f(t_n,y_n) - ha_2f(t_{n+b}, y_{n+b}) $$
$$= hy'(t_n) - ha_1y'(t_n) - ha_2y'(t_{n+b}) + O(h^2).$$
So I need to find $a_1,a_2,b$ so that every term except the $O(h^2)$ remains. I get for (a)
$$a_1 + a_2 = 1, b=0.$$



For (b) I expand one term more
$$y(t_{n+1}) = y(t_n) + hy'(t_n) + frac{h^2y''(t_n)}{2} + O(h^3)$$
The local residual becomes
$$y(t_n) + hy'(t_n) + frac{h^2y''(t_n)}{2} + O(h^3) - y(t_n) - ha_1y(t_n) - ha_2y(t_{n+b}) $$



But here I am stuck. How am I supposed to choose the coefficients as to remove the $frac{h^2y''(t_n)}{2}$ term?



How do I solve (b)? And is my answer on (a) correct?






numerical-methods taylor-expansion runge-kutta-methods




numerical-methods taylor-expansion runge-kutta-methods






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asked Dec 15 '18 at 20:16









HeuristicsHeuristics

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  • $begingroup$
    Please indicate cross posts of the same problem: scicomp.stackexchange.com/q/30721/6839
    $endgroup$
    – LutzL
    Dec 16 '18 at 10:26










  • $begingroup$
    Related with partial solution: math.stackexchange.com/questions/2715967/…
    $endgroup$
    – LutzL
    Dec 16 '18 at 10:30


















  • $begingroup$
    Please indicate cross posts of the same problem: scicomp.stackexchange.com/q/30721/6839
    $endgroup$
    – LutzL
    Dec 16 '18 at 10:26










  • $begingroup$
    Related with partial solution: math.stackexchange.com/questions/2715967/…
    $endgroup$
    – LutzL
    Dec 16 '18 at 10:30
















$begingroup$
Please indicate cross posts of the same problem: scicomp.stackexchange.com/q/30721/6839
$endgroup$
– LutzL
Dec 16 '18 at 10:26




$begingroup$
Please indicate cross posts of the same problem: scicomp.stackexchange.com/q/30721/6839
$endgroup$
– LutzL
Dec 16 '18 at 10:26












$begingroup$
Related with partial solution: math.stackexchange.com/questions/2715967/…
$endgroup$
– LutzL
Dec 16 '18 at 10:30




$begingroup$
Related with partial solution: math.stackexchange.com/questions/2715967/…
$endgroup$
– LutzL
Dec 16 '18 at 10:30










1 Answer
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$begingroup$

Apply the method to $f(t,x)=1$, $f(t,x)=2t$ and $f(t,x)=x$ at $(t_0,x_0)=(0,1)$ to get
begin{align}
f(t,x)=1&:& 1+h &= 1+ha_1+ha_2 &implies& a_1+a_2=1\
f(t,x)=2t&:& 1+h^2 &= 1+ha_2(2(b_1h)) &implies& 2a_2b_1=1\
f(t,x)=x&:& e^h &= 1+ha_1+ha_2(1+b_2h)) + O(h^3)&implies& a_2b_2=frac12
end{align}
so one gets necessarily $a_1=1-a_2$ and $b_1=b_2=frac1{2a_2}$. Now insert these conditions into the Taylor expansion to find that they are also sufficient for all continuously differentiable $f$.






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    $begingroup$

    Apply the method to $f(t,x)=1$, $f(t,x)=2t$ and $f(t,x)=x$ at $(t_0,x_0)=(0,1)$ to get
    begin{align}
    f(t,x)=1&:& 1+h &= 1+ha_1+ha_2 &implies& a_1+a_2=1\
    f(t,x)=2t&:& 1+h^2 &= 1+ha_2(2(b_1h)) &implies& 2a_2b_1=1\
    f(t,x)=x&:& e^h &= 1+ha_1+ha_2(1+b_2h)) + O(h^3)&implies& a_2b_2=frac12
    end{align}
    so one gets necessarily $a_1=1-a_2$ and $b_1=b_2=frac1{2a_2}$. Now insert these conditions into the Taylor expansion to find that they are also sufficient for all continuously differentiable $f$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Apply the method to $f(t,x)=1$, $f(t,x)=2t$ and $f(t,x)=x$ at $(t_0,x_0)=(0,1)$ to get
      begin{align}
      f(t,x)=1&:& 1+h &= 1+ha_1+ha_2 &implies& a_1+a_2=1\
      f(t,x)=2t&:& 1+h^2 &= 1+ha_2(2(b_1h)) &implies& 2a_2b_1=1\
      f(t,x)=x&:& e^h &= 1+ha_1+ha_2(1+b_2h)) + O(h^3)&implies& a_2b_2=frac12
      end{align}
      so one gets necessarily $a_1=1-a_2$ and $b_1=b_2=frac1{2a_2}$. Now insert these conditions into the Taylor expansion to find that they are also sufficient for all continuously differentiable $f$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Apply the method to $f(t,x)=1$, $f(t,x)=2t$ and $f(t,x)=x$ at $(t_0,x_0)=(0,1)$ to get
        begin{align}
        f(t,x)=1&:& 1+h &= 1+ha_1+ha_2 &implies& a_1+a_2=1\
        f(t,x)=2t&:& 1+h^2 &= 1+ha_2(2(b_1h)) &implies& 2a_2b_1=1\
        f(t,x)=x&:& e^h &= 1+ha_1+ha_2(1+b_2h)) + O(h^3)&implies& a_2b_2=frac12
        end{align}
        so one gets necessarily $a_1=1-a_2$ and $b_1=b_2=frac1{2a_2}$. Now insert these conditions into the Taylor expansion to find that they are also sufficient for all continuously differentiable $f$.






        share|cite|improve this answer









        $endgroup$



        Apply the method to $f(t,x)=1$, $f(t,x)=2t$ and $f(t,x)=x$ at $(t_0,x_0)=(0,1)$ to get
        begin{align}
        f(t,x)=1&:& 1+h &= 1+ha_1+ha_2 &implies& a_1+a_2=1\
        f(t,x)=2t&:& 1+h^2 &= 1+ha_2(2(b_1h)) &implies& 2a_2b_1=1\
        f(t,x)=x&:& e^h &= 1+ha_1+ha_2(1+b_2h)) + O(h^3)&implies& a_2b_2=frac12
        end{align}
        so one gets necessarily $a_1=1-a_2$ and $b_1=b_2=frac1{2a_2}$. Now insert these conditions into the Taylor expansion to find that they are also sufficient for all continuously differentiable $f$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 10:42









        LutzLLutzL

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        57.1k42054






























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