\(\((一)\), 
    Derived\ the\ funddamental\ solution\ of\ \(\(d\^4\) U \((x, s)\)\)\/dx\^4 + \(q\^4\) 
          U \((x, s)\) = δ \((x - s)\)\ , \ \(-∞\) < 
      x < ∞\ , \ 
    by\ using\ Fourier\ transform, 
\ \ \ \ \ \ \ \ \ \ \ \ \ inverse\ Fourier\ transform, residue\ theorem, \ 
    and\ limiting\ process\ of\ q -> 0\)
           (1) Is U (x, s) 
      singular?
      (2) 
      Is U (x, s) symmetric?
      (3) 
      Is U (x, s) degenerate form?

Sol : 
      U (k) = (∫_(-∞)^∞) 
          U (x) (E^(-Ikx)) \[DifferentialD]x
   ->(∫_(-∞)^∞) 
              U' (x) (E^(-Ikx)) \[DifferentialD]x = Ik*U (k)
   ->(∫_(-∞)^∞) 
                (U^((4))) (x) (E^(-Ikx)) \[DifferentialD]x = (((Ik))^4)*U (k)

\(\(Fourier\ \(Transform : 
\(\((Ik)\)\^4\) \(U\&_\) \((k, 
                        s)\) + \(q\^4\) \(U\&_\) \((k, 
                        s)\)\) = E\^\(-Iks\)

            \ \ \ \ \ \ \ \ \(k\^4\) \(U\&_\) \((k, 
                    s)\) + \(q\^4\) \(U\&_\) \((k, 
                    s)\) = E\^\(-Iks\)

            \ \ \ \ \ \ \ \ \(U\&_\) \((k, 
                  s)\) = \(1\/\(k\^4 + 
                      q\^4\)\) E\^\(-Iks\)

            \ 
            Inverse\ Fourier\ \(Transform : 
U \((x, 
                      s)\)\) = \(1\/\(2  π\)\) \(∫\_\(-∞\)\%∞\( 
                        1\/\(k\^4 + 
                          q\^4\)\) \(E\^\(-Iks\)\) \(E\^Ikx\) \[DifferentialD]k\)

            \ \ \ \ \ \ \ U \((x, 
                  s)\) = \(1\/\(2  π\)\) \(∫\_\(-∞\)\%∞\( 
                        1\/\(k\^4 + 
                          q\^4\)\) \(E\^\(Ik \((x - 
                              s)\)\)\) \[DifferentialD]k\)

            \ \ \ \ \ \ \ k\^4 = \(\(-q\^4\) = \(E\^\(\((1 + 
                            2  n)\) π\)\) q\^4\)\ , \ n = 0, 1, 2, 
            3

            \ \ \ \ \ \ \ k = \(E\^\(π\/4\)\) 
                q\ , \(E\^\(\(3  π\)\/4\)\) 
              q\ , \(E\^\(\(5  π\)\/4\)\) 
              q\ , \(E\^\(\(7  π\)\/4\)\) 
              q\ \ 

            \ \ \ \ \ \ \ U \((x, s)\) = 
              1\/\(2  π\)*2  πI \(∑Res \((\(1\/\(k\^4 + 
                              q\^4\)\) E\^\(Ik \((x - s)\)\))\)\)

            \ \ \ \ \ \ \ U \((x, 
                  s)\) = I \(∑Res \((\(1\/\(k\^4 + 
                              q\^4\)\) E\^\(Ik \((x - s)\)\))\)\)

            \((a)\) x > \(s : 
\ 
                  U \((x, 
                      s)\)\) = I \(∑\+上半平面 Res \((\(1\/\(k\^4 + 
                              q\^4\)\) E\^\(Ik \((x - s)\)\))\)\)

            \ \ \ \ \ \ \ \ k = \(E\^\(π\/4\)\) 
                q\ , \(E\^\(\(3  π\)\/4\)\) 
              q

            \ \ \ \ \ \ \ U \((x, 
                  s)\) = \(I\ E\^\(I\ E\^\(\(π\ I\)\/4\)\ q\ \((x - s)\)\)\)\/\(q\^3\ \((E\^\(\(π\ I\)\/4\) - E\^\(1\/4\ \((3  π)\)\ I\))\)\ \((E\^\(\(π\ I\)\/4\) - E\^\(\(5\ π\)\/4\ I\))\)\ \((E\^\(\(π\ I\)\/4\) - E\^\(\(\(\ \)\(7\ π\)\)\/4\ I\))\)\) + 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(I\ E\^\(I\ E\^\(\(3  π\ I\)\/4\)\ q\ \((x - s)\)\)\)\/\(q\^3\ \((E\^\(\(3  π\ I\)\/4\) - E\^\(1\/4\ \((π)\)\ I\))\)\ \((E\^\(\(3  π\ I\)\/4\) - E\^\(\(\(5  π\)\/4\) I\))\)\ \((E\^\(\(3  π\ I\)\/4\) - E\^\(\(7  π\)\/4\ \ I\))\)\)\)
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(\(=\)\(\(\((\(-1\))\)\^\(1/4\)\ \((\ E\^\(\((\(-1\))\)\^\(1/4\)\ q\ \((s - x)\)\) - I\ E\^\(\((\(-1\))\)\^\(3/4\)\ q\ \((\(-s\) + x)\)\))\)\)\/\(4 
                     q\^3\)\)\)
\ \ \ \ \ \ \(特解 + 補解 = 特解 : \((創造\ \(0\/0\) 的型式)\)\)
\ \ \ \ \ \ \ \ \(U \((x, 
                  s)\) = \(\((\(-1\))\)\^\(1/4\)\ \((\ E\^\(\((\(-1\))\)\^\(1/4\)\ q\ \((s - x)\)\) - I\ E\^\(\((\(-1\))\)\^\(3/4\)\ q\ \((\(-s\) + x)\)\))\)\)\/\(4  q\^3\) - \(\((\(-1\))\)\^\(1/4\)\ \((1 - I\ )\)\)\/\(4\ q\^3\)\)
\ \ \ \ \ \ \ \ \(U \((x, s)\) = 
              Limit\+\(q -> 0\)[\(\((\(-1\))\)\^\(1/4\)\ \((\ E\^\(\((\(-1\))\)\^\(1/4\)\ q\ \((s - x)\)\) - I\ E\^\(\((\(-1\))\)\^\(3/4\)\ q\ \((\(-s\) + x)\)\))\)\)\/\(4  q\^3\) - \(\((\(-1\))\)\^\(1/4\)\ \((1 - I\ )\)\)\/\(4\ q\^3\)]\)
\ \ \ \ \ \ \(使用L' 
              Hospital\ rule\)
\ \ \ \ \ \ \ \ \(U \((x, 
                  s)\) = \(1\/12\) \((x - s)\)\^3, 
            x > s\)

\(\((b)\) 
                  x < \(s : 
\ U \((x, s)\)\) = 
              1\/\(2  π\)*\(-2\) πI \(∑Res \((\(1\/\(k\^4 + 
                              q\^4\)\) E\^\(Ik \((x - s)\)\))\)\)\)
\ \ \ \ \ \ \ \ \(U \((x, 
                  s)\) = \(-I\) \(∑\+下半平面 
                    Res \((\(1\/\(k\^4 + 
                              q\^4\)\) E\^\(Ik \((x - s)\)\))\)\)\)
\ \ \ \ \ \ \ \ \(k = \(E\^\(\(5  π\)\/4\)\) q\ , \(E\^\(\(7  π\)\/4\)\) 
              q\)
\ \ \ \ \ \ \ \ \(U \((x, 
                s)\)\) = \(\(-I\)\ E\^\(I\ E\^\(\(5  π\ I\)\/4\)\ q\ \((x - s)\)\)\)\/\(q\^3\ \((E\^\(\(5  π\ I\)\/4\) - E\^\(π\/4\ \ I\))\)\ \((E\^\(\(5  π\ I\)\/4\) - E\^\(\(3\ π\)\/4\ \ I\))\)\ \((E\^\(\(5  π\ I\)\/4\) - E\^\(\(7\ π\)\/4\ \ I\))\)\) - 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \(I\ E\^\(I\ E\^\(\(7  π\ I\)\/4\)\ q\ \((x - s)\)\)\)\/\(\(q\^3\) \((E\^\(\(7  π\ I\)\/4\) - E\^\(π\/4\ \ I\))\)\ \((E\^\(\(7  π\ I\)\/4\) - E\^\(\(3\ π\)\/4\ \ I\))\) \((E\^\(\(7  π\ I\)\/4\) - E\^\(\(5\ π\)\/4\ \ I\))\)\)\)
                         =-(((((-1))^(3/4)) ((E^((((-1))^(3/4)) q (s - x))) + I (E^((((-1))^(1/4)) q (-s + x)))))/(4 (q^3)))
        \(特解 + 補解 = 特解 : \((創造\ \(0\/0\) 的型式)\)\)
        U (x, 
              s) = -(((((-1))^(3/4)) ((E^((((-1))^(3/4)) q (s - x))) + I (E^((((-1))^(1/4)) q (-s + x)))))/(4 (q^3))) + (((((-1))^(3/4)) (1 + I ))/(4  (q^3)))
        U (x, s) = 
          (Underscript[Limit, q -> 0])[-(((((-1))^(3/4)) ((E^((((-1))^(3/4)) q (s - x))) + I (E^((((-1))^(1/4)) q (-s + x)))))/(4 (q^3))) + (((((-1))^(3/4)) (1 + I ))/(4  (q^3)))]
      \(使用L' 
            Hospital\ \(rule : 
U \((x, 
                  s)\)\) = \(1\/12\) \((s - x)\)\^3, \ x < s\)
       U (x, s)={({{(1/12) (((x - s))^3), x > s}, {(1/12) (((s - x))^3), x < s}})={({{-((s^3)/12) + (((s^2) x)/4) - ((s (x^2))/4) + ((x^3)/12), x > s}, {((s^3)/12) - (((s^2) x)/4) + ((s (x^2))/4) - 
                                ((x^3)/12), x < s}})
 So U (x, s) 
          is regular . 
  U (x, s) 
          is symmetric . 
  U (x, 
            s) is degenerate form.

\(\((二)\), 
    Please\ extend\ to\ the\ case\ of\ cantilever\ rod . \((\(\(d\^2\) G \((x, s)\)\)\/dx\^2 = δ \((x - s)\))\)\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ G \((L, s)\) = 
      0, \(\(\(dG \((x, s)\)\)\/dx\)\( | \_\(x = 0\)\)\) = 0, \ 
    for\ odd\ number\ student . 
\ \ \ \ \ \ \ \((1)\) 
      Is\ G \((x, s)\) singular?
\ \ \ \ \ \ \ \((2)\) 
      Is\ G \((x, s)\)\ symmetric?
\ \ \ \ \ \ \ \((3)\) 
      Is\ G \((x, s)\)\ degenerate\ \(\(form\)\(?\)\)\)

\(Sol : 
\ \ \ \ \ \(令過左邊界 : a + bx\)\)
            \(過右邊界 : 
        c + dx\)
      \(代入左右邊界條件 : \(\(\(dG \((x, s)\)\)\/dx\)\( | \_\(x = 0\)\)\) = 0, 
      G \((L, s)\) = 0\)
      得{({{a              }, {c (x - l)}})
      由位移連續及兩邊斜率差1:
    得{({{a = c (s - l)}, {c=1              }})⟹{({{a = s - l}, {c=1      }})
      \(故\ G \((x, s)\)\)={({{s - l ,  0 < x < s}, {x - l ,  s < x < l}})
\(SO\ G \((x, 
        s)\)\ is\ regular . 
\ G \((x, 
        s)\)\ is\ symmetric . 
\ G \((x, 
        s)\)\ is\ degenerate\ form . 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 參考解答\ BY\ 周克勳\)