Euler's Method Global Error: How to calculate $C_1$ if $error = C_1 h$












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My textbook claims that, for small step size $h$, Euler's method has a global error which is at most proportional to $h$ such that error $= C_1h$. It is then claimed that $C_1$ depends on the initial value problem, but no explanation is given as to how one finds $C_1$.



So if I know $h$, then how can I deduce $C_1$ from the IVP?










share|cite|improve this question











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    1












    $begingroup$


    My textbook claims that, for small step size $h$, Euler's method has a global error which is at most proportional to $h$ such that error $= C_1h$. It is then claimed that $C_1$ depends on the initial value problem, but no explanation is given as to how one finds $C_1$.



    So if I know $h$, then how can I deduce $C_1$ from the IVP?










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      My textbook claims that, for small step size $h$, Euler's method has a global error which is at most proportional to $h$ such that error $= C_1h$. It is then claimed that $C_1$ depends on the initial value problem, but no explanation is given as to how one finds $C_1$.



      So if I know $h$, then how can I deduce $C_1$ from the IVP?










      share|cite|improve this question











      $endgroup$




      My textbook claims that, for small step size $h$, Euler's method has a global error which is at most proportional to $h$ such that error $= C_1h$. It is then claimed that $C_1$ depends on the initial value problem, but no explanation is given as to how one finds $C_1$.



      So if I know $h$, then how can I deduce $C_1$ from the IVP?







      ordinary-differential-equations numerical-methods eulers-method






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      share|cite|improve this question













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      edited Jul 21 '17 at 9:05







      The Pointer

















      asked Jul 21 '17 at 8:59









      The PointerThe Pointer

      2,66121639




      2,66121639






















          2 Answers
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          1












          $begingroup$

          Given an IVP:
          $$frac{dy}{dt}=f(t,y), y(a)=y_0, tin [a,b].$$
          Here is a Theorem from Numerical Analysis by Sauer:




          Assume that $f(t,y)$ has a Lipschitz constant $L$ for the variable $y$ and that the solution $y_i$ of the initial value problem at $t_i$ is approximated by $w_i$, using Euler's method. Let $M$ be an upper bound for $|y''(t)|$ on $[a,b]$. Then $$|w_i-y_i|le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$




          The proof is based on the following lemma:




          Assume that $f(t,y)$ is Lipschitz in the variable $y$ on the set $S=[a,b]times [alpha,beta]$. If $Y(t)$ and $Z(t)$ are solutions in $S$ of the differential equation $y'=f(t,y)$ with initial conditions $Y(a)$ and $Z(a)$ respectively, then $$|Y(t)-Z(t)|le e^{L(t-a)}|Y(a)-Z(a)|.$$




          Sketch of proof of the first theorem:



          Let $g_i$ be the global error, $e_i$ be the local truncation error, $z_i$ satisfy the local IVP:
          $$z_i'=f(t,z_i),z_i(t_i)=w_i, tin [t_i,t_{i+1}].$$



          Then
          $$g_i=|w_i-y_i|=|w_i-z_i(t)+z_i(t)-y_i|le |w_i-z_i(t)|+|z_i(t)-y_i|\
          le e_i+e^{Lh}g_{i-1}\
          le e_i+e^{Lh}(e_{i-1}+e^{Lh}g_{i-2})le cdots\
          le e_i+e^{Lh}e_{i-1}+e^{2Lh}e_{i-2}+cdots +e^{(i-1)Lh}e_1.$$
          Since each $e_ile frac{h^2M}{2}$, we have
          $$g_ile frac{h^2M}{2}(1+e^{Lh}+cdots+e^{(i-1)Lh})=frac{h^2M(e^{iLh}-1)}{2(e^{Lh}-1)}le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$



          Hope this helps.






          share|cite|improve this answer









          $endgroup$





















            -1












            $begingroup$

            Consider the IVP $y'=f(t,y)$, $y(t_0)=y_0$. Let $t_k=t_0+kh$, $y_k$ computed by the Euler method. We know that the error order of the Euler method is one. Thus the iterates $y_k$ have an error relative to the exact solution $y(t)$ of the form $$y_k=y(t_k)+c_kh$$ with some coefficients $c_k$ that will be closer determined in the course of this answer.





            Now insert this representation of $y_k$ into the Euler step and apply Taylor expansion where appropriate
            begin{align}
            [y(t_{k+1})+c_{k+1}h]&=[y(t_k)+c_kh]+hf(t_k,[y(t_k)+c_kh])\
            &=y(t_k)+c_kh+hBigl(f(t_k,y(t_k))+h,∂_yf(t_k,y(t_k)),c_k+O(h^2)Bigr)\
            y(t_k)+hy'(t_k)+tfrac12h^2y''(t_k)+O(h^3)&=y(t_k)+hy'(t_k)+hBigl[c_k+h,∂_yf(t_k,y(t_k)),c_kBigr]
            end{align}

            where $∂_y=frac{partial}{partial y}$ and later $∂_t=frac{partial}{partial t}$.





            In the Taylor series for $y(t_{k+1})=y(t_k+h)$ on the left side the first two terms cancel against the same terms on the right side. The second derivative can be written as
            $$
            y''(t)=frac{d}{dt}f(t,y(t))
            =∂_tf(t,y(t))+∂_yf(t,y(t)),f(t,y(t))
            overset{rm Def}=Df(t,y(t)).
            $$

            Divide the remaining equation by $h$ and re-arrange to get a difference equation for $c_k$
            $$
            c_{k+1}=c_k+hBigl[∂_yf(t_k,y(t_k))c_k-tfrac12Df(t_k,y(t_k))Bigr]+O(h^2).
            $$





            This looks like the Euler method for the linear ODE for a continuous differentiable function $c$,
            $$
            c'(t)=∂_yf(t,y(t))c(t)-tfrac12Df(t,y(t)),~~text{ with }~~c(t_0)=0.
            $$

            Again by the first order of the Euler method, $c_k$ and $c(t_k)$ will have a difference $O(h)$, so that the error we aim to estimate is
            $$y_k-y(t_k)=c(t_k)h+O(h^2).$$





            Now if $L$ is a bound for $∂_yf$, the $y$-Lipschitz constant, and $M$ is a bound for $Df=∂_tf+∂_yf,f$, or the second derivative, then by Grönwall's lemma
            $$
            |c'|le L|c|+frac12Mimplies |c(t)|le frac{M(e^{L|t-t_0|}-1)}{2L}
            $$

            which reproduces the usual specific estimate of the coefficient of the error term.






            share|cite|improve this answer











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              2 Answers
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              2 Answers
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              active

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              1












              $begingroup$

              Given an IVP:
              $$frac{dy}{dt}=f(t,y), y(a)=y_0, tin [a,b].$$
              Here is a Theorem from Numerical Analysis by Sauer:




              Assume that $f(t,y)$ has a Lipschitz constant $L$ for the variable $y$ and that the solution $y_i$ of the initial value problem at $t_i$ is approximated by $w_i$, using Euler's method. Let $M$ be an upper bound for $|y''(t)|$ on $[a,b]$. Then $$|w_i-y_i|le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$




              The proof is based on the following lemma:




              Assume that $f(t,y)$ is Lipschitz in the variable $y$ on the set $S=[a,b]times [alpha,beta]$. If $Y(t)$ and $Z(t)$ are solutions in $S$ of the differential equation $y'=f(t,y)$ with initial conditions $Y(a)$ and $Z(a)$ respectively, then $$|Y(t)-Z(t)|le e^{L(t-a)}|Y(a)-Z(a)|.$$




              Sketch of proof of the first theorem:



              Let $g_i$ be the global error, $e_i$ be the local truncation error, $z_i$ satisfy the local IVP:
              $$z_i'=f(t,z_i),z_i(t_i)=w_i, tin [t_i,t_{i+1}].$$



              Then
              $$g_i=|w_i-y_i|=|w_i-z_i(t)+z_i(t)-y_i|le |w_i-z_i(t)|+|z_i(t)-y_i|\
              le e_i+e^{Lh}g_{i-1}\
              le e_i+e^{Lh}(e_{i-1}+e^{Lh}g_{i-2})le cdots\
              le e_i+e^{Lh}e_{i-1}+e^{2Lh}e_{i-2}+cdots +e^{(i-1)Lh}e_1.$$
              Since each $e_ile frac{h^2M}{2}$, we have
              $$g_ile frac{h^2M}{2}(1+e^{Lh}+cdots+e^{(i-1)Lh})=frac{h^2M(e^{iLh}-1)}{2(e^{Lh}-1)}le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$



              Hope this helps.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Given an IVP:
                $$frac{dy}{dt}=f(t,y), y(a)=y_0, tin [a,b].$$
                Here is a Theorem from Numerical Analysis by Sauer:




                Assume that $f(t,y)$ has a Lipschitz constant $L$ for the variable $y$ and that the solution $y_i$ of the initial value problem at $t_i$ is approximated by $w_i$, using Euler's method. Let $M$ be an upper bound for $|y''(t)|$ on $[a,b]$. Then $$|w_i-y_i|le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$




                The proof is based on the following lemma:




                Assume that $f(t,y)$ is Lipschitz in the variable $y$ on the set $S=[a,b]times [alpha,beta]$. If $Y(t)$ and $Z(t)$ are solutions in $S$ of the differential equation $y'=f(t,y)$ with initial conditions $Y(a)$ and $Z(a)$ respectively, then $$|Y(t)-Z(t)|le e^{L(t-a)}|Y(a)-Z(a)|.$$




                Sketch of proof of the first theorem:



                Let $g_i$ be the global error, $e_i$ be the local truncation error, $z_i$ satisfy the local IVP:
                $$z_i'=f(t,z_i),z_i(t_i)=w_i, tin [t_i,t_{i+1}].$$



                Then
                $$g_i=|w_i-y_i|=|w_i-z_i(t)+z_i(t)-y_i|le |w_i-z_i(t)|+|z_i(t)-y_i|\
                le e_i+e^{Lh}g_{i-1}\
                le e_i+e^{Lh}(e_{i-1}+e^{Lh}g_{i-2})le cdots\
                le e_i+e^{Lh}e_{i-1}+e^{2Lh}e_{i-2}+cdots +e^{(i-1)Lh}e_1.$$
                Since each $e_ile frac{h^2M}{2}$, we have
                $$g_ile frac{h^2M}{2}(1+e^{Lh}+cdots+e^{(i-1)Lh})=frac{h^2M(e^{iLh}-1)}{2(e^{Lh}-1)}le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$



                Hope this helps.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Given an IVP:
                  $$frac{dy}{dt}=f(t,y), y(a)=y_0, tin [a,b].$$
                  Here is a Theorem from Numerical Analysis by Sauer:




                  Assume that $f(t,y)$ has a Lipschitz constant $L$ for the variable $y$ and that the solution $y_i$ of the initial value problem at $t_i$ is approximated by $w_i$, using Euler's method. Let $M$ be an upper bound for $|y''(t)|$ on $[a,b]$. Then $$|w_i-y_i|le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$




                  The proof is based on the following lemma:




                  Assume that $f(t,y)$ is Lipschitz in the variable $y$ on the set $S=[a,b]times [alpha,beta]$. If $Y(t)$ and $Z(t)$ are solutions in $S$ of the differential equation $y'=f(t,y)$ with initial conditions $Y(a)$ and $Z(a)$ respectively, then $$|Y(t)-Z(t)|le e^{L(t-a)}|Y(a)-Z(a)|.$$




                  Sketch of proof of the first theorem:



                  Let $g_i$ be the global error, $e_i$ be the local truncation error, $z_i$ satisfy the local IVP:
                  $$z_i'=f(t,z_i),z_i(t_i)=w_i, tin [t_i,t_{i+1}].$$



                  Then
                  $$g_i=|w_i-y_i|=|w_i-z_i(t)+z_i(t)-y_i|le |w_i-z_i(t)|+|z_i(t)-y_i|\
                  le e_i+e^{Lh}g_{i-1}\
                  le e_i+e^{Lh}(e_{i-1}+e^{Lh}g_{i-2})le cdots\
                  le e_i+e^{Lh}e_{i-1}+e^{2Lh}e_{i-2}+cdots +e^{(i-1)Lh}e_1.$$
                  Since each $e_ile frac{h^2M}{2}$, we have
                  $$g_ile frac{h^2M}{2}(1+e^{Lh}+cdots+e^{(i-1)Lh})=frac{h^2M(e^{iLh}-1)}{2(e^{Lh}-1)}le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$



                  Hope this helps.






                  share|cite|improve this answer









                  $endgroup$



                  Given an IVP:
                  $$frac{dy}{dt}=f(t,y), y(a)=y_0, tin [a,b].$$
                  Here is a Theorem from Numerical Analysis by Sauer:




                  Assume that $f(t,y)$ has a Lipschitz constant $L$ for the variable $y$ and that the solution $y_i$ of the initial value problem at $t_i$ is approximated by $w_i$, using Euler's method. Let $M$ be an upper bound for $|y''(t)|$ on $[a,b]$. Then $$|w_i-y_i|le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$




                  The proof is based on the following lemma:




                  Assume that $f(t,y)$ is Lipschitz in the variable $y$ on the set $S=[a,b]times [alpha,beta]$. If $Y(t)$ and $Z(t)$ are solutions in $S$ of the differential equation $y'=f(t,y)$ with initial conditions $Y(a)$ and $Z(a)$ respectively, then $$|Y(t)-Z(t)|le e^{L(t-a)}|Y(a)-Z(a)|.$$




                  Sketch of proof of the first theorem:



                  Let $g_i$ be the global error, $e_i$ be the local truncation error, $z_i$ satisfy the local IVP:
                  $$z_i'=f(t,z_i),z_i(t_i)=w_i, tin [t_i,t_{i+1}].$$



                  Then
                  $$g_i=|w_i-y_i|=|w_i-z_i(t)+z_i(t)-y_i|le |w_i-z_i(t)|+|z_i(t)-y_i|\
                  le e_i+e^{Lh}g_{i-1}\
                  le e_i+e^{Lh}(e_{i-1}+e^{Lh}g_{i-2})le cdots\
                  le e_i+e^{Lh}e_{i-1}+e^{2Lh}e_{i-2}+cdots +e^{(i-1)Lh}e_1.$$
                  Since each $e_ile frac{h^2M}{2}$, we have
                  $$g_ile frac{h^2M}{2}(1+e^{Lh}+cdots+e^{(i-1)Lh})=frac{h^2M(e^{iLh}-1)}{2(e^{Lh}-1)}le frac{Mh}{2L}(e^{L(t_i-a)}-1).$$



                  Hope this helps.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 21 '17 at 10:17









                  KittyLKittyL

                  13.9k31534




                  13.9k31534























                      -1












                      $begingroup$

                      Consider the IVP $y'=f(t,y)$, $y(t_0)=y_0$. Let $t_k=t_0+kh$, $y_k$ computed by the Euler method. We know that the error order of the Euler method is one. Thus the iterates $y_k$ have an error relative to the exact solution $y(t)$ of the form $$y_k=y(t_k)+c_kh$$ with some coefficients $c_k$ that will be closer determined in the course of this answer.





                      Now insert this representation of $y_k$ into the Euler step and apply Taylor expansion where appropriate
                      begin{align}
                      [y(t_{k+1})+c_{k+1}h]&=[y(t_k)+c_kh]+hf(t_k,[y(t_k)+c_kh])\
                      &=y(t_k)+c_kh+hBigl(f(t_k,y(t_k))+h,∂_yf(t_k,y(t_k)),c_k+O(h^2)Bigr)\
                      y(t_k)+hy'(t_k)+tfrac12h^2y''(t_k)+O(h^3)&=y(t_k)+hy'(t_k)+hBigl[c_k+h,∂_yf(t_k,y(t_k)),c_kBigr]
                      end{align}

                      where $∂_y=frac{partial}{partial y}$ and later $∂_t=frac{partial}{partial t}$.





                      In the Taylor series for $y(t_{k+1})=y(t_k+h)$ on the left side the first two terms cancel against the same terms on the right side. The second derivative can be written as
                      $$
                      y''(t)=frac{d}{dt}f(t,y(t))
                      =∂_tf(t,y(t))+∂_yf(t,y(t)),f(t,y(t))
                      overset{rm Def}=Df(t,y(t)).
                      $$

                      Divide the remaining equation by $h$ and re-arrange to get a difference equation for $c_k$
                      $$
                      c_{k+1}=c_k+hBigl[∂_yf(t_k,y(t_k))c_k-tfrac12Df(t_k,y(t_k))Bigr]+O(h^2).
                      $$





                      This looks like the Euler method for the linear ODE for a continuous differentiable function $c$,
                      $$
                      c'(t)=∂_yf(t,y(t))c(t)-tfrac12Df(t,y(t)),~~text{ with }~~c(t_0)=0.
                      $$

                      Again by the first order of the Euler method, $c_k$ and $c(t_k)$ will have a difference $O(h)$, so that the error we aim to estimate is
                      $$y_k-y(t_k)=c(t_k)h+O(h^2).$$





                      Now if $L$ is a bound for $∂_yf$, the $y$-Lipschitz constant, and $M$ is a bound for $Df=∂_tf+∂_yf,f$, or the second derivative, then by Grönwall's lemma
                      $$
                      |c'|le L|c|+frac12Mimplies |c(t)|le frac{M(e^{L|t-t_0|}-1)}{2L}
                      $$

                      which reproduces the usual specific estimate of the coefficient of the error term.






                      share|cite|improve this answer











                      $endgroup$


















                        -1












                        $begingroup$

                        Consider the IVP $y'=f(t,y)$, $y(t_0)=y_0$. Let $t_k=t_0+kh$, $y_k$ computed by the Euler method. We know that the error order of the Euler method is one. Thus the iterates $y_k$ have an error relative to the exact solution $y(t)$ of the form $$y_k=y(t_k)+c_kh$$ with some coefficients $c_k$ that will be closer determined in the course of this answer.





                        Now insert this representation of $y_k$ into the Euler step and apply Taylor expansion where appropriate
                        begin{align}
                        [y(t_{k+1})+c_{k+1}h]&=[y(t_k)+c_kh]+hf(t_k,[y(t_k)+c_kh])\
                        &=y(t_k)+c_kh+hBigl(f(t_k,y(t_k))+h,∂_yf(t_k,y(t_k)),c_k+O(h^2)Bigr)\
                        y(t_k)+hy'(t_k)+tfrac12h^2y''(t_k)+O(h^3)&=y(t_k)+hy'(t_k)+hBigl[c_k+h,∂_yf(t_k,y(t_k)),c_kBigr]
                        end{align}

                        where $∂_y=frac{partial}{partial y}$ and later $∂_t=frac{partial}{partial t}$.





                        In the Taylor series for $y(t_{k+1})=y(t_k+h)$ on the left side the first two terms cancel against the same terms on the right side. The second derivative can be written as
                        $$
                        y''(t)=frac{d}{dt}f(t,y(t))
                        =∂_tf(t,y(t))+∂_yf(t,y(t)),f(t,y(t))
                        overset{rm Def}=Df(t,y(t)).
                        $$

                        Divide the remaining equation by $h$ and re-arrange to get a difference equation for $c_k$
                        $$
                        c_{k+1}=c_k+hBigl[∂_yf(t_k,y(t_k))c_k-tfrac12Df(t_k,y(t_k))Bigr]+O(h^2).
                        $$





                        This looks like the Euler method for the linear ODE for a continuous differentiable function $c$,
                        $$
                        c'(t)=∂_yf(t,y(t))c(t)-tfrac12Df(t,y(t)),~~text{ with }~~c(t_0)=0.
                        $$

                        Again by the first order of the Euler method, $c_k$ and $c(t_k)$ will have a difference $O(h)$, so that the error we aim to estimate is
                        $$y_k-y(t_k)=c(t_k)h+O(h^2).$$





                        Now if $L$ is a bound for $∂_yf$, the $y$-Lipschitz constant, and $M$ is a bound for $Df=∂_tf+∂_yf,f$, or the second derivative, then by Grönwall's lemma
                        $$
                        |c'|le L|c|+frac12Mimplies |c(t)|le frac{M(e^{L|t-t_0|}-1)}{2L}
                        $$

                        which reproduces the usual specific estimate of the coefficient of the error term.






                        share|cite|improve this answer











                        $endgroup$
















                          -1












                          -1








                          -1





                          $begingroup$

                          Consider the IVP $y'=f(t,y)$, $y(t_0)=y_0$. Let $t_k=t_0+kh$, $y_k$ computed by the Euler method. We know that the error order of the Euler method is one. Thus the iterates $y_k$ have an error relative to the exact solution $y(t)$ of the form $$y_k=y(t_k)+c_kh$$ with some coefficients $c_k$ that will be closer determined in the course of this answer.





                          Now insert this representation of $y_k$ into the Euler step and apply Taylor expansion where appropriate
                          begin{align}
                          [y(t_{k+1})+c_{k+1}h]&=[y(t_k)+c_kh]+hf(t_k,[y(t_k)+c_kh])\
                          &=y(t_k)+c_kh+hBigl(f(t_k,y(t_k))+h,∂_yf(t_k,y(t_k)),c_k+O(h^2)Bigr)\
                          y(t_k)+hy'(t_k)+tfrac12h^2y''(t_k)+O(h^3)&=y(t_k)+hy'(t_k)+hBigl[c_k+h,∂_yf(t_k,y(t_k)),c_kBigr]
                          end{align}

                          where $∂_y=frac{partial}{partial y}$ and later $∂_t=frac{partial}{partial t}$.





                          In the Taylor series for $y(t_{k+1})=y(t_k+h)$ on the left side the first two terms cancel against the same terms on the right side. The second derivative can be written as
                          $$
                          y''(t)=frac{d}{dt}f(t,y(t))
                          =∂_tf(t,y(t))+∂_yf(t,y(t)),f(t,y(t))
                          overset{rm Def}=Df(t,y(t)).
                          $$

                          Divide the remaining equation by $h$ and re-arrange to get a difference equation for $c_k$
                          $$
                          c_{k+1}=c_k+hBigl[∂_yf(t_k,y(t_k))c_k-tfrac12Df(t_k,y(t_k))Bigr]+O(h^2).
                          $$





                          This looks like the Euler method for the linear ODE for a continuous differentiable function $c$,
                          $$
                          c'(t)=∂_yf(t,y(t))c(t)-tfrac12Df(t,y(t)),~~text{ with }~~c(t_0)=0.
                          $$

                          Again by the first order of the Euler method, $c_k$ and $c(t_k)$ will have a difference $O(h)$, so that the error we aim to estimate is
                          $$y_k-y(t_k)=c(t_k)h+O(h^2).$$





                          Now if $L$ is a bound for $∂_yf$, the $y$-Lipschitz constant, and $M$ is a bound for $Df=∂_tf+∂_yf,f$, or the second derivative, then by Grönwall's lemma
                          $$
                          |c'|le L|c|+frac12Mimplies |c(t)|le frac{M(e^{L|t-t_0|}-1)}{2L}
                          $$

                          which reproduces the usual specific estimate of the coefficient of the error term.






                          share|cite|improve this answer











                          $endgroup$



                          Consider the IVP $y'=f(t,y)$, $y(t_0)=y_0$. Let $t_k=t_0+kh$, $y_k$ computed by the Euler method. We know that the error order of the Euler method is one. Thus the iterates $y_k$ have an error relative to the exact solution $y(t)$ of the form $$y_k=y(t_k)+c_kh$$ with some coefficients $c_k$ that will be closer determined in the course of this answer.





                          Now insert this representation of $y_k$ into the Euler step and apply Taylor expansion where appropriate
                          begin{align}
                          [y(t_{k+1})+c_{k+1}h]&=[y(t_k)+c_kh]+hf(t_k,[y(t_k)+c_kh])\
                          &=y(t_k)+c_kh+hBigl(f(t_k,y(t_k))+h,∂_yf(t_k,y(t_k)),c_k+O(h^2)Bigr)\
                          y(t_k)+hy'(t_k)+tfrac12h^2y''(t_k)+O(h^3)&=y(t_k)+hy'(t_k)+hBigl[c_k+h,∂_yf(t_k,y(t_k)),c_kBigr]
                          end{align}

                          where $∂_y=frac{partial}{partial y}$ and later $∂_t=frac{partial}{partial t}$.





                          In the Taylor series for $y(t_{k+1})=y(t_k+h)$ on the left side the first two terms cancel against the same terms on the right side. The second derivative can be written as
                          $$
                          y''(t)=frac{d}{dt}f(t,y(t))
                          =∂_tf(t,y(t))+∂_yf(t,y(t)),f(t,y(t))
                          overset{rm Def}=Df(t,y(t)).
                          $$

                          Divide the remaining equation by $h$ and re-arrange to get a difference equation for $c_k$
                          $$
                          c_{k+1}=c_k+hBigl[∂_yf(t_k,y(t_k))c_k-tfrac12Df(t_k,y(t_k))Bigr]+O(h^2).
                          $$





                          This looks like the Euler method for the linear ODE for a continuous differentiable function $c$,
                          $$
                          c'(t)=∂_yf(t,y(t))c(t)-tfrac12Df(t,y(t)),~~text{ with }~~c(t_0)=0.
                          $$

                          Again by the first order of the Euler method, $c_k$ and $c(t_k)$ will have a difference $O(h)$, so that the error we aim to estimate is
                          $$y_k-y(t_k)=c(t_k)h+O(h^2).$$





                          Now if $L$ is a bound for $∂_yf$, the $y$-Lipschitz constant, and $M$ is a bound for $Df=∂_tf+∂_yf,f$, or the second derivative, then by Grönwall's lemma
                          $$
                          |c'|le L|c|+frac12Mimplies |c(t)|le frac{M(e^{L|t-t_0|}-1)}{2L}
                          $$

                          which reproduces the usual specific estimate of the coefficient of the error term.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 16 at 18:23

























                          answered Jan 15 at 18:49









                          LutzLLutzL

                          60.1k42057




                          60.1k42057






























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