Runge-Kutta method for higher-order differential equations











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I am studying Numerical Analysis with the book of Richard L.Burden.
A question which I'm struggling with right now is following.



Transform the second-order initial-value problem



$y'' - 2y' + 2y = e^{2t}sint$ for $0 leq t leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$



into a system of ifrst order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.



Then,
$$u_1(t) = y(t), u_2(t) = y'(t)$$
$$u_1'(t) = u_2(t)$$
$$u_2'(t) = e^{2t}sint - 2u_1(t) + u_2(t)$$
$$u_1(0) = -0.4, u_2(0) = -0.6$$



This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$



I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$



$f_1 = u_1'= u_2(t)$,
So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)



However, I can't understand the following.
$$k_{2,1} = hf_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2}) = hleft[w_{2,0} + frac{1}{2}k_{1,2}right]$$



Why does $f_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2})$ equal to $w_{2,0} + frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.






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    up vote
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    I am studying Numerical Analysis with the book of Richard L.Burden.
    A question which I'm struggling with right now is following.



    Transform the second-order initial-value problem



    $y'' - 2y' + 2y = e^{2t}sint$ for $0 leq t leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$



    into a system of ifrst order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.



    Then,
    $$u_1(t) = y(t), u_2(t) = y'(t)$$
    $$u_1'(t) = u_2(t)$$
    $$u_2'(t) = e^{2t}sint - 2u_1(t) + u_2(t)$$
    $$u_1(0) = -0.4, u_2(0) = -0.6$$



    This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$



    I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$



    $f_1 = u_1'= u_2(t)$,
    So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)



    However, I can't understand the following.
    $$k_{2,1} = hf_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2}) = hleft[w_{2,0} + frac{1}{2}k_{1,2}right]$$



    Why does $f_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2})$ equal to $w_{2,0} + frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.






    runge-kutta-methods







    share|cite|improve this question







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    James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am studying Numerical Analysis with the book of Richard L.Burden.
      A question which I'm struggling with right now is following.



      Transform the second-order initial-value problem



      $y'' - 2y' + 2y = e^{2t}sint$ for $0 leq t leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$



      into a system of ifrst order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.



      Then,
      $$u_1(t) = y(t), u_2(t) = y'(t)$$
      $$u_1'(t) = u_2(t)$$
      $$u_2'(t) = e^{2t}sint - 2u_1(t) + u_2(t)$$
      $$u_1(0) = -0.4, u_2(0) = -0.6$$



      This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$



      I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$



      $f_1 = u_1'= u_2(t)$,
      So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)



      However, I can't understand the following.
      $$k_{2,1} = hf_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2}) = hleft[w_{2,0} + frac{1}{2}k_{1,2}right]$$



      Why does $f_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2})$ equal to $w_{2,0} + frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.






      runge-kutta-methods







      share|cite|improve this question







      New contributor




      James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I am studying Numerical Analysis with the book of Richard L.Burden.
      A question which I'm struggling with right now is following.



      Transform the second-order initial-value problem



      $y'' - 2y' + 2y = e^{2t}sint$ for $0 leq t leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$



      into a system of ifrst order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.



      Then,
      $$u_1(t) = y(t), u_2(t) = y'(t)$$
      $$u_1'(t) = u_2(t)$$
      $$u_2'(t) = e^{2t}sint - 2u_1(t) + u_2(t)$$
      $$u_1(0) = -0.4, u_2(0) = -0.6$$



      This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$



      I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$



      $f_1 = u_1'= u_2(t)$,
      So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)



      However, I can't understand the following.
      $$k_{2,1} = hf_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2}) = hleft[w_{2,0} + frac{1}{2}k_{1,2}right]$$



      Why does $f_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2})$ equal to $w_{2,0} + frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.






      runge-kutta-methods




      runge-kutta-methods






      share|cite|improve this question







      New contributor




      James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







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      James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









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