I want to understand the explicit and the implicit Euler method better. Assume I have the initial value problem y ″ + y = 0, y ( 0 ) = 0, y ′ ( 0 ) = 1, which is of course solved by y = sin ⁡ ( x ), and I convert this into a system of first-degree linear equations ( y ′ z ′ ) = ( 0 1 − 1 0 ) ( y z ) .When I apply the explicit and the implicit Euler methods to the differential equation y ″ + y = 0, with stepwidth h &gt; 0, I get approximation points that satisfy: y k = i h 2 ( ( 1 − i h ) k − ( 1 + i h ) k ) for the explicit Euler method y ~ k = i h 2 ( 1 + h 2 ) k ( ( 1 − i h ) k − ( 1 + i h ) k ) for the implicit Euler method I am interested in seeing how fast these sequences grow with k. My guess is that the first sequence is unbounded, whereas the second converges to zero. At least that is what I get for , but I have no idea how to prove this. Any suggestions?

Question

I want to understand the explicit and the implicit Euler method better. Assume I have the initial value problem       y    ″    +  y  =  0,   y  (  0  )  =  0,       y    ′    (  0  )  =  1, which is of course solved by   y  =  sin  ⁡  (  x  ), and I convert this into a system of first-degree linear equations      (                                        y            ′                                                            z            ′                                )    =      (                            0                          1                                      −          1                          0                      )        (                            y                                      z                      )    .When I apply the explicit and the implicit Euler methods to the differential equation       y    ″    +  y  =  0, with stepwidth   h  &amp;gt;  0, I get approximation points that satisfy:                              y                      k                                              =                                          i            h                    2                (        (        1        −        i        h                  )                      k                          −        (        1        +        i        h                  )                      k                          )                    for the explicit Euler method                                                                y              ~                                            k                                              =                                          i            h                                2            (            1            +                          h                              2                                                    )                              k                                                    (        (        1        −        i        h                  )                      k                          −        (        1        +        i        h                  )                      k                          )                    for the implicit Euler method            I am interested in seeing how fast these sequences grow with k. My guess is that the first sequence is unbounded, whereas the second converges to zero. At least that is what I get for , but I have no idea how to prove this. Any suggestions?

prelatiuvq · Accepted Answer

Let   θ  =  arctan  ⁡  h  ,  r  =      1    +          h      2       and then  1  +  i  h  =  r  (  cos  ⁡  θ  +  i  sin  ⁡  θ  )  ,  1  −  i  h  =      1    +          h      2        (  cos  ⁡  θ  −  i  sin  ⁡  θ  )  .So                              y                      k                                              =                                          i            h                    2                (        (        1        −        i        h                  )                      k                          −        (        1        +        i        h                  )                      k                          )                                          =                    h                  r          k                sin        ⁡        (        k        θ        )        .            Since   h  &amp;gt;  0, you have   r  &amp;gt;  1 and   θ  ∈  (  0  ,      π    2    ) and hence   {      y    k    } diverges. Similarly                                                        y              ~                                            k                                              =                                          i            h                                2            (            1            +                          h                              2                                                    )                              k                                                    (        (        1        −        i        h                  )                      k                          −        (        1        +        i        h                  )                      k                          )                                          =                    h                  r                      −            k                          sin        ⁡        (        k        θ        )        .            So   {                    y        ~              k    } converges to 0 since   r  &amp;gt;  1.

patylomy7 · Accepted Answer

|    sin  ⁡  (  k  θ  )      |    ≤  1 and hence       |        r          −      k        sin  ⁡  (  k  θ  )      |    ≤      r          −      k        →  0 as   k  →  ∞.So now we know that   sin  ⁡  (  k  θ  ) does not converge to zero faster than       r          k       goes to infinity

I want to understand the explicit and the implicit Euler method better. Assume I have the initial value problem y′′+y=0, y(0)=0, y′(0)=1, which is of course solved by y=sin(x), and I convert this into a system of first-degree linear equations (y′ z′)=(0 −1 1 0)(y z).

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