Discretization of second order nonlinear ODE using finite difference approximation not correct












2














I have the differential equation
$$y'' + x(y^2)' - 2y^2 = g(x) Longleftrightarrow y'' + x2yy'-2y^2 = g(x).$$
Using finite-difference approximations
$$y''(x_m) approx frac{Y_{m-1} - 2Y_m + Y_{m+1}}{Delta x^2},$$
$$y'(x_m) approx frac{Y_{m+1} - Y_{m-1}}{2Delta x},$$
$$y(x_m) approx Y_m,$$
I get
$$frac{Y_{m-1} - 2Y_m + Y_{m+1}}{Delta x^2} + frac{x_m}{Delta x}Y_m(Y_{m+1}-Y_{m-1}) - 2Y_m^2 = g(x_m).$$
However, the answer is supposed to be
enter image description here



Why is the second term in my answer wrong?






differential-equations numerical-methods nonlinear-system finite-differences finite-difference-methods







share|cite|improve this question





























    2














    I have the differential equation
    $$y'' + x(y^2)' - 2y^2 = g(x) Longleftrightarrow y'' + x2yy'-2y^2 = g(x).$$
    Using finite-difference approximations
    $$y''(x_m) approx frac{Y_{m-1} - 2Y_m + Y_{m+1}}{Delta x^2},$$
    $$y'(x_m) approx frac{Y_{m+1} - Y_{m-1}}{2Delta x},$$
    $$y(x_m) approx Y_m,$$
    I get
    $$frac{Y_{m-1} - 2Y_m + Y_{m+1}}{Delta x^2} + frac{x_m}{Delta x}Y_m(Y_{m+1}-Y_{m-1}) - 2Y_m^2 = g(x_m).$$
    However, the answer is supposed to be
    enter image description here



    Why is the second term in my answer wrong?






    differential-equations numerical-methods nonlinear-system finite-differences finite-difference-methods







    share|cite|improve this question



























      2












      2








      2







      I have the differential equation
      $$y'' + x(y^2)' - 2y^2 = g(x) Longleftrightarrow y'' + x2yy'-2y^2 = g(x).$$
      Using finite-difference approximations
      $$y''(x_m) approx frac{Y_{m-1} - 2Y_m + Y_{m+1}}{Delta x^2},$$
      $$y'(x_m) approx frac{Y_{m+1} - Y_{m-1}}{2Delta x},$$
      $$y(x_m) approx Y_m,$$
      I get
      $$frac{Y_{m-1} - 2Y_m + Y_{m+1}}{Delta x^2} + frac{x_m}{Delta x}Y_m(Y_{m+1}-Y_{m-1}) - 2Y_m^2 = g(x_m).$$
      However, the answer is supposed to be
      enter image description here



      Why is the second term in my answer wrong?






      differential-equations numerical-methods nonlinear-system finite-differences finite-difference-methods







      share|cite|improve this question















      I have the differential equation
      $$y'' + x(y^2)' - 2y^2 = g(x) Longleftrightarrow y'' + x2yy'-2y^2 = g(x).$$
      Using finite-difference approximations
      $$y''(x_m) approx frac{Y_{m-1} - 2Y_m + Y_{m+1}}{Delta x^2},$$
      $$y'(x_m) approx frac{Y_{m+1} - Y_{m-1}}{2Delta x},$$
      $$y(x_m) approx Y_m,$$
      I get
      $$frac{Y_{m-1} - 2Y_m + Y_{m+1}}{Delta x^2} + frac{x_m}{Delta x}Y_m(Y_{m+1}-Y_{m-1}) - 2Y_m^2 = g(x_m).$$
      However, the answer is supposed to be
      enter image description here



      Why is the second term in my answer wrong?






      differential-equations numerical-methods nonlinear-system finite-differences finite-difference-methods




      differential-equations numerical-methods nonlinear-system finite-differences finite-difference-methods






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      LutzL

      56.3k42054




      56.3k42054










      asked yesterday









      Heuristics

      470211




      470211






















          1 Answer
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          In the second version they are using a finite difference representation of the term $y^2$:



          $$
          (y^2)' approx frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} tag{1}
          $$



          While you are using a finite representation of $2y y'$:



          $$
          2y y' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} tag{2}
          $$



          But they are equivalent, if you take (if the solution $y$ is smooth enough you can do approximate the value at a node with the average of its neighbor nodes)



          $$
          y_m approx frac{y_{m + 1} + y_{m - 1}}{2} tag{3}
          $$



          Then Eq. (2) becomes



          $$
          2yy' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} approx (y_{m + 1} + y_{m - 1})frac{y_{m + 1} - y_{m - 1}}{2Delta x} = frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} approx (y^2)'
          $$






          share|cite|improve this answer























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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            In the second version they are using a finite difference representation of the term $y^2$:



            $$
            (y^2)' approx frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} tag{1}
            $$



            While you are using a finite representation of $2y y'$:



            $$
            2y y' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} tag{2}
            $$



            But they are equivalent, if you take (if the solution $y$ is smooth enough you can do approximate the value at a node with the average of its neighbor nodes)



            $$
            y_m approx frac{y_{m + 1} + y_{m - 1}}{2} tag{3}
            $$



            Then Eq. (2) becomes



            $$
            2yy' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} approx (y_{m + 1} + y_{m - 1})frac{y_{m + 1} - y_{m - 1}}{2Delta x} = frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} approx (y^2)'
            $$






            share|cite|improve this answer




























              2














              In the second version they are using a finite difference representation of the term $y^2$:



              $$
              (y^2)' approx frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} tag{1}
              $$



              While you are using a finite representation of $2y y'$:



              $$
              2y y' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} tag{2}
              $$



              But they are equivalent, if you take (if the solution $y$ is smooth enough you can do approximate the value at a node with the average of its neighbor nodes)



              $$
              y_m approx frac{y_{m + 1} + y_{m - 1}}{2} tag{3}
              $$



              Then Eq. (2) becomes



              $$
              2yy' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} approx (y_{m + 1} + y_{m - 1})frac{y_{m + 1} - y_{m - 1}}{2Delta x} = frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} approx (y^2)'
              $$






              share|cite|improve this answer


























                2












                2








                2






                In the second version they are using a finite difference representation of the term $y^2$:



                $$
                (y^2)' approx frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} tag{1}
                $$



                While you are using a finite representation of $2y y'$:



                $$
                2y y' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} tag{2}
                $$



                But they are equivalent, if you take (if the solution $y$ is smooth enough you can do approximate the value at a node with the average of its neighbor nodes)



                $$
                y_m approx frac{y_{m + 1} + y_{m - 1}}{2} tag{3}
                $$



                Then Eq. (2) becomes



                $$
                2yy' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} approx (y_{m + 1} + y_{m - 1})frac{y_{m + 1} - y_{m - 1}}{2Delta x} = frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} approx (y^2)'
                $$






                share|cite|improve this answer














                In the second version they are using a finite difference representation of the term $y^2$:



                $$
                (y^2)' approx frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} tag{1}
                $$



                While you are using a finite representation of $2y y'$:



                $$
                2y y' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} tag{2}
                $$



                But they are equivalent, if you take (if the solution $y$ is smooth enough you can do approximate the value at a node with the average of its neighbor nodes)



                $$
                y_m approx frac{y_{m + 1} + y_{m - 1}}{2} tag{3}
                $$



                Then Eq. (2) becomes



                $$
                2yy' approx 2y_mfrac{y_{m + 1} - y_{m - 1}}{2Delta x} approx (y_{m + 1} + y_{m - 1})frac{y_{m + 1} - y_{m - 1}}{2Delta x} = frac{y^2_{m + 1} - y^2_{m - 1}}{2Delta x} approx (y^2)'
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday

























                answered yesterday









                caverac

                13.8k21030




                13.8k21030






























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