Parallelogram identity in the wave equation












3














Using the parallelogram identity, I need to solve the following initial boundary value problem
for a vibrating semi-infinite string with a nonhomogeneous boundary condition:




$$ u_{tt} − u_{xx} = 0 , 0 < x < infty, t > 0 $$
$$u(0,t) = h(t)$$
$$u(x,0) = f(x), u_{t}(x,0) = g(x)$$
where $f, g, h ∈ C_2{[0, ∞)}$




I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.



Edit: The parallelogram identity is



$u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
$










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    3














    Using the parallelogram identity, I need to solve the following initial boundary value problem
    for a vibrating semi-infinite string with a nonhomogeneous boundary condition:




    $$ u_{tt} − u_{xx} = 0 , 0 < x < infty, t > 0 $$
    $$u(0,t) = h(t)$$
    $$u(x,0) = f(x), u_{t}(x,0) = g(x)$$
    where $f, g, h ∈ C_2{[0, ∞)}$




    I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.



    Edit: The parallelogram identity is



    $u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
    $










    share|cite|improve this question



























      3












      3








      3







      Using the parallelogram identity, I need to solve the following initial boundary value problem
      for a vibrating semi-infinite string with a nonhomogeneous boundary condition:




      $$ u_{tt} − u_{xx} = 0 , 0 < x < infty, t > 0 $$
      $$u(0,t) = h(t)$$
      $$u(x,0) = f(x), u_{t}(x,0) = g(x)$$
      where $f, g, h ∈ C_2{[0, ∞)}$




      I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.



      Edit: The parallelogram identity is



      $u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
      $










      share|cite|improve this question















      Using the parallelogram identity, I need to solve the following initial boundary value problem
      for a vibrating semi-infinite string with a nonhomogeneous boundary condition:




      $$ u_{tt} − u_{xx} = 0 , 0 < x < infty, t > 0 $$
      $$u(0,t) = h(t)$$
      $$u(x,0) = f(x), u_{t}(x,0) = g(x)$$
      where $f, g, h ∈ C_2{[0, ∞)}$




      I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.



      Edit: The parallelogram identity is



      $u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
      $







      pde






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      edited May 24 '18 at 10:32









      Dylan

      12.4k31026




      12.4k31026










      asked Oct 29 '12 at 22:39









      MariaMaria

      262




      262






















          1 Answer
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          I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.



          Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$



          $u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$



          Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$



          $u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$






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            0














            I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.



            Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$



            $u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$



            Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$



            $u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$






            share|cite|improve this answer




























              0














              I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.



              Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$



              $u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$



              Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$



              $u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$






              share|cite|improve this answer


























                0












                0








                0






                I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.



                Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$



                $u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$



                Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$



                $u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$






                share|cite|improve this answer














                I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.



                Case $1$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~ds=h(t)$



                $u(x,t)=dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds$



                Case $2$: $dfrac{f(t)+f(-t)}{2}+dfrac{1}{2}int_{-t}^t g(s)~dsneq h(t)$



                $u(x,t)=begin{cases}dfrac{f(x+t)+f(x-t)}{2}+dfrac{1}{2}int_{x-t}^{x+t}g(s)~ds&text{when}~x>t\dfrac{f(x+t)-f(t-x)}{2}+dfrac{1}{2}int_{t-x}^{x+t}g(s)~ds+h(t-x)&text{when}~x<tend{cases}$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 5 '13 at 22:02

























                answered Oct 29 '12 at 23:28









                doraemonpauldoraemonpaul

                12.5k31660




                12.5k31660






























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