Solving the following linear ODE by a numerical method












0












$begingroup$


ODE:
$$y'(x)+3y(x)=1$$
Initial condition: $y(0)=0$



We know that the exact solution is:
$y left( t right) =1/3-1/3,{{rm e}^{-3,t}}.$



My Objective:
I want to solve the ODE by Legendre wavelets method. The following solution is right? If else, what' is the wrong?



My try: (shortly)



Let $y'(t)=C^intercalpsi(t)$.
By integrating, we have
$y(t)=C^intercal P psi(t)+y(0).$



Substituting $y(t), y'(t)$ to the ode, we have
$$C^intercal(I+3P)psi(t)=F^intercalpsi(t)$$

where $F$ is legendre coefficient matrix of $1$ which is
$1=F^intercalpsi(t)$



So, equating the coefficients of $psi(t)$, we have



$$C^intercal=F^intercal(I+3P)^{-1}$$



Substituting $C^intercal$ to $y(t)=C^intercal P psi(t),$



the solution is
$$y(t)=F^intercal(I+3P)^{-1}P psi(t).$$
(In here, superscript $^intercal$ represents transpose of the matrices. $C,F,psi$ are $Ntimes1$ matrices.Identity matrix $I$ and $P$ are $Ntimes N$ matrices where $N=2^{k-1}M$. )



Many thanks.



Best regards.










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    ODE:
    $$y'(x)+3y(x)=1$$
    Initial condition: $y(0)=0$



    We know that the exact solution is:
    $y left( t right) =1/3-1/3,{{rm e}^{-3,t}}.$



    My Objective:
    I want to solve the ODE by Legendre wavelets method. The following solution is right? If else, what' is the wrong?



    My try: (shortly)



    Let $y'(t)=C^intercalpsi(t)$.
    By integrating, we have
    $y(t)=C^intercal P psi(t)+y(0).$



    Substituting $y(t), y'(t)$ to the ode, we have
    $$C^intercal(I+3P)psi(t)=F^intercalpsi(t)$$

    where $F$ is legendre coefficient matrix of $1$ which is
    $1=F^intercalpsi(t)$



    So, equating the coefficients of $psi(t)$, we have



    $$C^intercal=F^intercal(I+3P)^{-1}$$



    Substituting $C^intercal$ to $y(t)=C^intercal P psi(t),$



    the solution is
    $$y(t)=F^intercal(I+3P)^{-1}P psi(t).$$
    (In here, superscript $^intercal$ represents transpose of the matrices. $C,F,psi$ are $Ntimes1$ matrices.Identity matrix $I$ and $P$ are $Ntimes N$ matrices where $N=2^{k-1}M$. )



    Many thanks.



    Best regards.










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      ODE:
      $$y'(x)+3y(x)=1$$
      Initial condition: $y(0)=0$



      We know that the exact solution is:
      $y left( t right) =1/3-1/3,{{rm e}^{-3,t}}.$



      My Objective:
      I want to solve the ODE by Legendre wavelets method. The following solution is right? If else, what' is the wrong?



      My try: (shortly)



      Let $y'(t)=C^intercalpsi(t)$.
      By integrating, we have
      $y(t)=C^intercal P psi(t)+y(0).$



      Substituting $y(t), y'(t)$ to the ode, we have
      $$C^intercal(I+3P)psi(t)=F^intercalpsi(t)$$

      where $F$ is legendre coefficient matrix of $1$ which is
      $1=F^intercalpsi(t)$



      So, equating the coefficients of $psi(t)$, we have



      $$C^intercal=F^intercal(I+3P)^{-1}$$



      Substituting $C^intercal$ to $y(t)=C^intercal P psi(t),$



      the solution is
      $$y(t)=F^intercal(I+3P)^{-1}P psi(t).$$
      (In here, superscript $^intercal$ represents transpose of the matrices. $C,F,psi$ are $Ntimes1$ matrices.Identity matrix $I$ and $P$ are $Ntimes N$ matrices where $N=2^{k-1}M$. )



      Many thanks.



      Best regards.










      share|cite|improve this question











      $endgroup$




      ODE:
      $$y'(x)+3y(x)=1$$
      Initial condition: $y(0)=0$



      We know that the exact solution is:
      $y left( t right) =1/3-1/3,{{rm e}^{-3,t}}.$



      My Objective:
      I want to solve the ODE by Legendre wavelets method. The following solution is right? If else, what' is the wrong?



      My try: (shortly)



      Let $y'(t)=C^intercalpsi(t)$.
      By integrating, we have
      $y(t)=C^intercal P psi(t)+y(0).$



      Substituting $y(t), y'(t)$ to the ode, we have
      $$C^intercal(I+3P)psi(t)=F^intercalpsi(t)$$

      where $F$ is legendre coefficient matrix of $1$ which is
      $1=F^intercalpsi(t)$



      So, equating the coefficients of $psi(t)$, we have



      $$C^intercal=F^intercal(I+3P)^{-1}$$



      Substituting $C^intercal$ to $y(t)=C^intercal P psi(t),$



      the solution is
      $$y(t)=F^intercal(I+3P)^{-1}P psi(t).$$
      (In here, superscript $^intercal$ represents transpose of the matrices. $C,F,psi$ are $Ntimes1$ matrices.Identity matrix $I$ and $P$ are $Ntimes N$ matrices where $N=2^{k-1}M$. )



      Many thanks.



      Best regards.







      ordinary-differential-equations numerical-methods legendre-polynomials wavelets






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      edited Dec 21 '18 at 6:29







      HD239

















      asked Dec 20 '18 at 16:50









      HD239HD239

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