Solution of $u_t=mathcal{F}u$












5












$begingroup$


I tried to solve the equation $u_t=mathcal{F}u$, where $mathcal{F}$ denotes the Fourier transform, with initial data $u(x,0)=u_0(x)$. The solution should be given by
$$
u(x,t)=e^{mathcal{F}t}u_0=left(sum_{jgeq0}frac{mathcal{F}^jt^j}{j!}right)u_0(x)
$$
and I looked for an explicit expression of it in terms of $u_0$. Since $mathcal{F}^2=(cdot)^check{}$ (the operator that maps $v(x)mapstocheck{v}(x):=v(-x)$), we have that $mathcal{F}^3=mathcal{F}^{-1}$ and $mathcal{F}^4=id$. Suppose $u_0$ is a Schwartz function so all of this makes sense. Then we have
$$
u(x,t)=left(sum_{jin4mathbb{N}}frac{t^j}{j!}+sum_{jin1+4mathbb{N}}frac{t^j}{j!}mathcal{F}+sum_{jin2+4mathbb{N}}frac{t^j}{j!}(cdot)check{}+sum_{jin3+4mathbb{N}}frac{t^j}{j!}mathcal{F}^{-1}right)u_0(x)
$$

Next i wrote the function $u_0=phi+psi$, where $phi$ is even and $psi$ is odd, namely
$$
phi(x)=frac{1}{2}(u_0(x)+u_0(-x))quadtext{and}quadpsi(x)=frac{1}{2}(u_0(x)-u_0(-x))
$$
In this way (recalling that Fourier transform of an even function is odd and viceversa) one computes
$$
u(x,t)=sum_{jinmathbb{N}}frac{t^{2j}}{(2j)!}phi(x)+sum_{jinmathbb{N}}(-)^jfrac{t^{2j}}{(2j)!}psi(x)+sum_{jinmathbb{N}}(-)^jfrac{t^{2j+1}}{(2j+1)!}mathcal{F}phi(x)+sum_{jinmathbb{N}}frac{t^{2j+1}}{(2j+1)!}mathcal{F}psi(x)
$$
which equals
$$
u(x,t)=cosh(t)phi(x)+cos(t)psi(x)+sin(t)mathcal{F}phi(x)+sinh(t)mathcal{F}psi(x).
$$
The problem is that now, trying to check
$$
u_t(x,t)=sinh(t)phi(x)-sin(t)psi(x)+cos(t)mathcal{F}phi(x)+cosh(t)mathcal{F}psi(x)
$$
and
$$
mathcal{F}u(x,t)=cosh(t)mathcal{F}phi(x)+cos(t)mathcal{F}psi(x)+sin(t)phi(x)-sinh(t)psi(x)
$$
they are not equal. Is my calculation wrong? Every help is extremely appreciated.






pde fourier-analysis







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    I tried to solve the equation $u_t=mathcal{F}u$, where $mathcal{F}$ denotes the Fourier transform, with initial data $u(x,0)=u_0(x)$. The solution should be given by
    $$
    u(x,t)=e^{mathcal{F}t}u_0=left(sum_{jgeq0}frac{mathcal{F}^jt^j}{j!}right)u_0(x)
    $$
    and I looked for an explicit expression of it in terms of $u_0$. Since $mathcal{F}^2=(cdot)^check{}$ (the operator that maps $v(x)mapstocheck{v}(x):=v(-x)$), we have that $mathcal{F}^3=mathcal{F}^{-1}$ and $mathcal{F}^4=id$. Suppose $u_0$ is a Schwartz function so all of this makes sense. Then we have
    $$
    u(x,t)=left(sum_{jin4mathbb{N}}frac{t^j}{j!}+sum_{jin1+4mathbb{N}}frac{t^j}{j!}mathcal{F}+sum_{jin2+4mathbb{N}}frac{t^j}{j!}(cdot)check{}+sum_{jin3+4mathbb{N}}frac{t^j}{j!}mathcal{F}^{-1}right)u_0(x)
    $$

    Next i wrote the function $u_0=phi+psi$, where $phi$ is even and $psi$ is odd, namely
    $$
    phi(x)=frac{1}{2}(u_0(x)+u_0(-x))quadtext{and}quadpsi(x)=frac{1}{2}(u_0(x)-u_0(-x))
    $$
    In this way (recalling that Fourier transform of an even function is odd and viceversa) one computes
    $$
    u(x,t)=sum_{jinmathbb{N}}frac{t^{2j}}{(2j)!}phi(x)+sum_{jinmathbb{N}}(-)^jfrac{t^{2j}}{(2j)!}psi(x)+sum_{jinmathbb{N}}(-)^jfrac{t^{2j+1}}{(2j+1)!}mathcal{F}phi(x)+sum_{jinmathbb{N}}frac{t^{2j+1}}{(2j+1)!}mathcal{F}psi(x)
    $$
    which equals
    $$
    u(x,t)=cosh(t)phi(x)+cos(t)psi(x)+sin(t)mathcal{F}phi(x)+sinh(t)mathcal{F}psi(x).
    $$
    The problem is that now, trying to check
    $$
    u_t(x,t)=sinh(t)phi(x)-sin(t)psi(x)+cos(t)mathcal{F}phi(x)+cosh(t)mathcal{F}psi(x)
    $$
    and
    $$
    mathcal{F}u(x,t)=cosh(t)mathcal{F}phi(x)+cos(t)mathcal{F}psi(x)+sin(t)phi(x)-sinh(t)psi(x)
    $$
    they are not equal. Is my calculation wrong? Every help is extremely appreciated.






    pde fourier-analysis







    share|cite|improve this question









    $endgroup$















      5












      5








      5


      2



      $begingroup$


      I tried to solve the equation $u_t=mathcal{F}u$, where $mathcal{F}$ denotes the Fourier transform, with initial data $u(x,0)=u_0(x)$. The solution should be given by
      $$
      u(x,t)=e^{mathcal{F}t}u_0=left(sum_{jgeq0}frac{mathcal{F}^jt^j}{j!}right)u_0(x)
      $$
      and I looked for an explicit expression of it in terms of $u_0$. Since $mathcal{F}^2=(cdot)^check{}$ (the operator that maps $v(x)mapstocheck{v}(x):=v(-x)$), we have that $mathcal{F}^3=mathcal{F}^{-1}$ and $mathcal{F}^4=id$. Suppose $u_0$ is a Schwartz function so all of this makes sense. Then we have
      $$
      u(x,t)=left(sum_{jin4mathbb{N}}frac{t^j}{j!}+sum_{jin1+4mathbb{N}}frac{t^j}{j!}mathcal{F}+sum_{jin2+4mathbb{N}}frac{t^j}{j!}(cdot)check{}+sum_{jin3+4mathbb{N}}frac{t^j}{j!}mathcal{F}^{-1}right)u_0(x)
      $$

      Next i wrote the function $u_0=phi+psi$, where $phi$ is even and $psi$ is odd, namely
      $$
      phi(x)=frac{1}{2}(u_0(x)+u_0(-x))quadtext{and}quadpsi(x)=frac{1}{2}(u_0(x)-u_0(-x))
      $$
      In this way (recalling that Fourier transform of an even function is odd and viceversa) one computes
      $$
      u(x,t)=sum_{jinmathbb{N}}frac{t^{2j}}{(2j)!}phi(x)+sum_{jinmathbb{N}}(-)^jfrac{t^{2j}}{(2j)!}psi(x)+sum_{jinmathbb{N}}(-)^jfrac{t^{2j+1}}{(2j+1)!}mathcal{F}phi(x)+sum_{jinmathbb{N}}frac{t^{2j+1}}{(2j+1)!}mathcal{F}psi(x)
      $$
      which equals
      $$
      u(x,t)=cosh(t)phi(x)+cos(t)psi(x)+sin(t)mathcal{F}phi(x)+sinh(t)mathcal{F}psi(x).
      $$
      The problem is that now, trying to check
      $$
      u_t(x,t)=sinh(t)phi(x)-sin(t)psi(x)+cos(t)mathcal{F}phi(x)+cosh(t)mathcal{F}psi(x)
      $$
      and
      $$
      mathcal{F}u(x,t)=cosh(t)mathcal{F}phi(x)+cos(t)mathcal{F}psi(x)+sin(t)phi(x)-sinh(t)psi(x)
      $$
      they are not equal. Is my calculation wrong? Every help is extremely appreciated.






      pde fourier-analysis







      share|cite|improve this question









      $endgroup$




      I tried to solve the equation $u_t=mathcal{F}u$, where $mathcal{F}$ denotes the Fourier transform, with initial data $u(x,0)=u_0(x)$. The solution should be given by
      $$
      u(x,t)=e^{mathcal{F}t}u_0=left(sum_{jgeq0}frac{mathcal{F}^jt^j}{j!}right)u_0(x)
      $$
      and I looked for an explicit expression of it in terms of $u_0$. Since $mathcal{F}^2=(cdot)^check{}$ (the operator that maps $v(x)mapstocheck{v}(x):=v(-x)$), we have that $mathcal{F}^3=mathcal{F}^{-1}$ and $mathcal{F}^4=id$. Suppose $u_0$ is a Schwartz function so all of this makes sense. Then we have
      $$
      u(x,t)=left(sum_{jin4mathbb{N}}frac{t^j}{j!}+sum_{jin1+4mathbb{N}}frac{t^j}{j!}mathcal{F}+sum_{jin2+4mathbb{N}}frac{t^j}{j!}(cdot)check{}+sum_{jin3+4mathbb{N}}frac{t^j}{j!}mathcal{F}^{-1}right)u_0(x)
      $$

      Next i wrote the function $u_0=phi+psi$, where $phi$ is even and $psi$ is odd, namely
      $$
      phi(x)=frac{1}{2}(u_0(x)+u_0(-x))quadtext{and}quadpsi(x)=frac{1}{2}(u_0(x)-u_0(-x))
      $$
      In this way (recalling that Fourier transform of an even function is odd and viceversa) one computes
      $$
      u(x,t)=sum_{jinmathbb{N}}frac{t^{2j}}{(2j)!}phi(x)+sum_{jinmathbb{N}}(-)^jfrac{t^{2j}}{(2j)!}psi(x)+sum_{jinmathbb{N}}(-)^jfrac{t^{2j+1}}{(2j+1)!}mathcal{F}phi(x)+sum_{jinmathbb{N}}frac{t^{2j+1}}{(2j+1)!}mathcal{F}psi(x)
      $$
      which equals
      $$
      u(x,t)=cosh(t)phi(x)+cos(t)psi(x)+sin(t)mathcal{F}phi(x)+sinh(t)mathcal{F}psi(x).
      $$
      The problem is that now, trying to check
      $$
      u_t(x,t)=sinh(t)phi(x)-sin(t)psi(x)+cos(t)mathcal{F}phi(x)+cosh(t)mathcal{F}psi(x)
      $$
      and
      $$
      mathcal{F}u(x,t)=cosh(t)mathcal{F}phi(x)+cos(t)mathcal{F}psi(x)+sin(t)phi(x)-sinh(t)psi(x)
      $$
      they are not equal. Is my calculation wrong? Every help is extremely appreciated.






      pde fourier-analysis




      pde fourier-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 14 '18 at 19:17









      MarcoMarco

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      336110






















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          $begingroup$

          As you noted, $mathcal{F}^4=I$. So
          $$
          (1-lambda^4)I=(mathcal{F}^4-lambda^4 I)=(mathcal{F}-lambda I)(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)
          $$
          which gives
          $$
          (lambda I-mathcal{F})^{-1}=frac{1}{lambda^4-1}(mathcal{F}^3+lambda mathcal{F}^2+lambda^2mathcal{F}+lambda^3 I)
          $$
          So, the following is evaluated by computing residues at $1,i,-1,-i$:
          $$
          e^{tmathcal{F}}=frac{1}{2pi i}oint_{C}e^{lambda t}(lambda I-mathcal{F})^{-1}dlambda \
          =frac{1}{2pi i}oint_{C}frac{e^{lambda t}(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)}{(lambda-1)(lambda-i)(lambda+1)(lambda+i)}dlambda
          $$






          share|cite|improve this answer









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            $begingroup$

            As you noted, $mathcal{F}^4=I$. So
            $$
            (1-lambda^4)I=(mathcal{F}^4-lambda^4 I)=(mathcal{F}-lambda I)(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)
            $$
            which gives
            $$
            (lambda I-mathcal{F})^{-1}=frac{1}{lambda^4-1}(mathcal{F}^3+lambda mathcal{F}^2+lambda^2mathcal{F}+lambda^3 I)
            $$
            So, the following is evaluated by computing residues at $1,i,-1,-i$:
            $$
            e^{tmathcal{F}}=frac{1}{2pi i}oint_{C}e^{lambda t}(lambda I-mathcal{F})^{-1}dlambda \
            =frac{1}{2pi i}oint_{C}frac{e^{lambda t}(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)}{(lambda-1)(lambda-i)(lambda+1)(lambda+i)}dlambda
            $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              As you noted, $mathcal{F}^4=I$. So
              $$
              (1-lambda^4)I=(mathcal{F}^4-lambda^4 I)=(mathcal{F}-lambda I)(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)
              $$
              which gives
              $$
              (lambda I-mathcal{F})^{-1}=frac{1}{lambda^4-1}(mathcal{F}^3+lambda mathcal{F}^2+lambda^2mathcal{F}+lambda^3 I)
              $$
              So, the following is evaluated by computing residues at $1,i,-1,-i$:
              $$
              e^{tmathcal{F}}=frac{1}{2pi i}oint_{C}e^{lambda t}(lambda I-mathcal{F})^{-1}dlambda \
              =frac{1}{2pi i}oint_{C}frac{e^{lambda t}(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)}{(lambda-1)(lambda-i)(lambda+1)(lambda+i)}dlambda
              $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                As you noted, $mathcal{F}^4=I$. So
                $$
                (1-lambda^4)I=(mathcal{F}^4-lambda^4 I)=(mathcal{F}-lambda I)(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)
                $$
                which gives
                $$
                (lambda I-mathcal{F})^{-1}=frac{1}{lambda^4-1}(mathcal{F}^3+lambda mathcal{F}^2+lambda^2mathcal{F}+lambda^3 I)
                $$
                So, the following is evaluated by computing residues at $1,i,-1,-i$:
                $$
                e^{tmathcal{F}}=frac{1}{2pi i}oint_{C}e^{lambda t}(lambda I-mathcal{F})^{-1}dlambda \
                =frac{1}{2pi i}oint_{C}frac{e^{lambda t}(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)}{(lambda-1)(lambda-i)(lambda+1)(lambda+i)}dlambda
                $$






                share|cite|improve this answer









                $endgroup$



                As you noted, $mathcal{F}^4=I$. So
                $$
                (1-lambda^4)I=(mathcal{F}^4-lambda^4 I)=(mathcal{F}-lambda I)(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)
                $$
                which gives
                $$
                (lambda I-mathcal{F})^{-1}=frac{1}{lambda^4-1}(mathcal{F}^3+lambda mathcal{F}^2+lambda^2mathcal{F}+lambda^3 I)
                $$
                So, the following is evaluated by computing residues at $1,i,-1,-i$:
                $$
                e^{tmathcal{F}}=frac{1}{2pi i}oint_{C}e^{lambda t}(lambda I-mathcal{F})^{-1}dlambda \
                =frac{1}{2pi i}oint_{C}frac{e^{lambda t}(mathcal{F}^3+lambdamathcal{F}^2+lambda^2mathcal{F}+lambda^3I)}{(lambda-1)(lambda-i)(lambda+1)(lambda+i)}dlambda
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 5:58









                DisintegratingByPartsDisintegratingByParts

                60.4k42681




                60.4k42681






























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