Why is the inverse Laplace Transform of sinh ⁡ ( x s ) s ⋅ sinh ⁡ s equal to x + 2 π ∑ n = 1 ∞ ( − 1 ) n n e − ( n π ) 2 t sin ⁡ n π x ?

Name: Why is the inverse Laplace Transform of sinh ⁡ ( x s ) s ⋅ sinh ⁡ s equal to x + 2 π ∑ n = 1 ∞ ( − 1 ) n n e − ( n π ) 2 t sin ⁡ n π x ?
Uploaded: 2022-02-23T17:22:57+00:00
Description: Why is the inverse Laplace Transform of sinh ⁡ ( x s ) s ⋅ sinh ⁡ s equal to x + 2 π ∑ n = 1 ∞ ( − 1 ) n n e − ( n π ) 2 t sin ⁡ n π x ?

Question

Why is the inverse Laplace Transform of            sinh      ⁡      (      x              s            )              s      ⋅      sinh      ⁡                        s                    equal to  x  +      2    π        ∑          n      =      1              ∞                  (      −      1              )        n              n        e          −      (      n      π              )        2            t        sin  ⁡      n    π    x    ?

Waldruhylm · Accepted Answer

Consider the poles of  F  (  s  )  =            sinh      ⁡      (      x              s            )              s      ⋅      sinh      ⁡                        s                          e          s      t      First we need to see if s=0 is a branch point or not  F  (  s  )  =      1    s        (                  (        x                  s                )        +        (        x                  s                          )          3                          /                3        !        +        ⋯                    (                  s                )        +        (                  s                          )          3                          /                3        !        +        ⋯              )        (    1    +    (    s    t    )    +                  (        s        t                  )          2                            2        !              +    O    (          s      3        )    )    F  (  s  )  =      x    s        (                  1        +                  x          2                s                  /                3        !        +        ⋯                    1        +        s                  /                3        !        +        ⋯              )        (    1    +    (    s    t    )    +                  (        s        t                  )          2                            2        !              +    O    (          s      3        )    )  Hence we see that s=0 is a pole of order 1 with residue x. Now consider the other poles where the deliminator   sinh  ⁡  (      s    ) has zeros when       s    =  n  π  i hence   s  =  −  (  n  π      )    2   are the poles.      ∑                  s        0                  R    e    s    (  F  ,      s    0    )  =  x  +      ∑          n      =      1        ∞        R    e    s    (  F  ,  −  (  n  π      )    2    )      R    e    s    (  F  ,  −  (  n  π      )    2    )  =      lim          s      →      −      (      n      π              )        2              (  s  +  (  n  π      )    2    )            sinh      ⁡      (      x              s            )              s      ⋅      sinh      ⁡                        s                          e          s      t        =  2      lim          s      →      −      (      n      π              )        2              (  s  +  (  n  π      )    2    )            sinh      ⁡      (      x              s            )                      s            ⋅      cosh      ⁡                        s                          e          s      t            R    e    s    (  F  ,  −  (  n  π      )    2    )  =            i      sin      ⁡      (      n      π      x      )              i      (      n      π      )      ⋅      cos      ⁡              (        n        π        )                  e          −      (      n      π              )        2            t        =            (      −      1              )        n                    n      π        sin  ⁡  (  n  π  x  )      e          −      (      n      π              )        2            t      Collecting the results together we have            L              −      1        (  f  (  s  )  )  =  x  +      ∑          n      ≥      1                  (      −      1              )        n                    n      π        sin  ⁡  (  n  π  x  )      e          −      (      n      π              )        2            t

Nola Aguilar · Accepted Answer

(1)                              Note that                                              1                          sinh              ⁡                              (                                                                                                                                  s                                                                                                                            )                                                    =                              1                                                                                          s                                                                                                  +        2                  ∑                      n            =            1                                ∞                                                (                                      −              1                                    )                                n                                                                                                                              s                                                                                                  s              +                              n                                  2                                                            π                                  2                                                                        and                                                                                                ∫                                                            0                                              +                                                              −                    ∞                                          i                                                                                                  0                                              +                                                              +                    ∞                                          i                                                                                                                                  sinh                      ⁡                                              (                                                                          x                                                                                                                                          s                                                                                                                                                                                    )                                                              s                                                                                                                                                                                                                              s                                                                                                                                                          s                      +                                              n                                                  2                                                                                            π                                                  2                                                                                                                                    exp                ⁡                                  (                                                        t                    s                                                      )                                                                                                                                    d                                            s                                                              2                      π                                              i                                                                                                                                                    =                             −                  ∫                      −            ∞                                0                                                              sinh              ⁡                              (                                                  x                                                                                                  −                      s                                                                                                                  i                                                                  )                                      s                                                                                                                                      −                  s                                                                                          i                                                    s              +                              n                                  2                                                            π                                  2                                            +                              i                                            0                                  +                                                                            exp        ⁡                  (                                t            s                              )                                                                            d                            s                                      2              π                              i                                                                                      −                  ∫                      0                                −            ∞                                                              sinh              ⁡                              (                                                  −                  x                                                                                                  −                      s                                                                                                                  i                                                                  )                                      s                                                                      −                                                                              −                  s                                                                                          i                                                    s              +                              n                                  2                                                            π                                  2                                            −                              i                                            0                                  +                                                                            exp        ⁡                  (                                t            s                              )                                                                            d                            s                                      2              π                              i                                                                        =                                       ∫                      0                                ∞                                                              sin              ⁡                              (                                                  x                                                                                                  s                                                                                                                            )                                      s                                                                                                                                      s                                                                                                  s              −                              n                                  2                                                            π                                  2                                            −                              i                                            0                                  +                                                                            exp        ⁡                  (                                −            t            s                              )                                                                            d                            s                                      2              π                              i                                                                                      −                  ∫                      0                                ∞                                                              sin              ⁡                              (                                                  x                                                                                                  s                                                                                                                            )                                      s                                                                                                                                      s                                                                                                  s              −                              n                                  2                                                            π                                  2                                            +                              i                                            0                                  +                                                                            exp        ⁡                  (                                −            t            s                              )                                                                            d                            s                                      2              π                              i                                                                                      =                  ∫                      0                                ∞                                                              sin              ⁡                              (                                                  x                                                                                                  s                                                                                                                            )                                                                                                                    s                                                                                                                                                      [                                                                                            1                                              s                        −                                                  n                                                      2                                                                                                    π                                                      2                                                                          −                                                  i                                                                          0                                                      +                                                                                                                                −                                                            1                                              s                        −                                                  n                                                      2                                                                                                    π                                                      2                                                                          +                                                  i                                                                          0                                                      +                                                                                                                                                              ]                            ⏞                                                                          2                π                                  i                                                δ                                  (                                                        s                    −                                          n                                              2                                                                                    π                                              2                                                                                                )                                                                            exp        ⁡                  (                                −            t            s                              )                                                                            d                            s                                      2              π                              i                                                                        (2)                    =                                                                                                                                             sin                      ⁡                                              (                                                                          n                          π                          x                                                                        )                                                                                    n                      π                                                                                                                            e                                                        −                                          n                                              2                                                                                    π                                              2                                                              t                                                                                                                                (                                                           note that                                                   lim                              n                →                0                                                    [                                                                                                        sin                      ⁡                                              (                                                                          n                          π                          x                                                                        )                                                                                    n                      π                                                                                                                            e                                                        −                                          n                                              2                                                                                    π                                              2                                                              t                                                                                            ]                        =                          x                                                       )                    With (1) and (2):      ∫                  0                  +                    −      ∞              i                            0                  +                    +      ∞              i                                sinh        ⁡                  (                                x                                                                    s                                                                                  )                            s        sinh        ⁡                  (                                                                                        s                                                                                  )                        exp  ⁡      (              t      s            )                                  d                s                    2        π                  i                      =                              x          +                                    2              π                                            ∑                          n              =              1                                      ∞                                                                          sin                ⁡                                  (                                                        n                    π                    x                                                      )                                            n                                          exp          ⁡                      (                                      −                                                [                                                        n                    π                                                      ]                                                  2                                            t                                    )

Why is the inverse Laplace Transform of (sin h(x sqrt(s)))/(s * sinh sqrt(s)) equal to x+2/(pi) sum_(n=1)^(oo)((-1)^n)/(n) e^(-(n pi)^2t)sin n pi x?

Answered question

Answer & Explanation

New Questions in Differential Equations