Solution for $u_{t} = alpha^{2}u_{xx}$ (problem with Fourier Series of inicial condition)












1















Solve the partial differential equation $u_{t} = alpha^{2}u_{xx}$ with conditions:
begin{cases}
u(x,0) = f(x)\
u(0,t) = u_{x}(L,t) = 0
end{cases}
where
$$f(x) = begin{cases} frac{2}{L}x,& x in left[0,frac{L}{2}right) \ 1,& x in left[frac{L}{2},Lright)end{cases}.$$
You must justify the convergence of the series, and justify aspects of odd and even extensions if necessary.




My attempt.



Step 1. Separation of variables:



Write $u(x,t) = F(x)G(t)$. Thus, $F(x)G'(t) = alpha^{2}F''(x)G(t)$, i.e,
$$frac{F''(x)}{F(x)} = y = frac{1}{alpha^{2}}frac{G'(t)}{G(t)}$$
where the parameter $y$ no depends of $x$ and $t$. We have two ODE's:
$$F''(x) - yF(x) = 0tag{1}$$
and
$$G'(t) - yalpha^{2}G(t) = 0tag{2}$$



Step 2: Bound conditions.



we have




  • If $y = 0$, $F equiv 0$,

  • If $y > 0$, $F equiv 0$,

  • If $y < 0$, take $y = -lambda^{2}$ then $F_{n}(x) = c_{2}sinleft(frac{npi x}{2L}right)$


where the solution for $y > 0$ is $F(x) = c_{1}cos(lambda x) + c_{2}sin(lambda x)$. The solution for $G'(t) - yalpha^{2}G(t) = 0$ is $G(t) = ce^{-yalpha^{2}t}$. So, for $n = pm 1, pm 2,dots$, we have
$$u_{n}(x,t) = Ke^{-frac{npi alpha}{2L}t}sinleft(frac{npi x}{2L}right).$$



Step 3. Find Fourier Series of $f$.



I tried to get an odd or even extension of $f$ and so, to find the Fourier Series of the extension (maybe is not possible to get a $sin$ or $cos$ Fourier Series). But I could not find a periodical extension of $f$. Precisely, I want to get a Fourier Series of the form
$$sum_{1}^{infty}c_{n}sinleft(frac{npi x}{2L}right).$$
With this,
$$u(x,t) = Ksum_{1}^{infty}c_{n}e^{-frac{npi alpha}{2L}t}sinleft(frac{npi x}{2L}right)$$
is a solution.



Can someone help me?






pde fourier-analysis even-and-odd-functions







share|cite|improve this question





























    1















    Solve the partial differential equation $u_{t} = alpha^{2}u_{xx}$ with conditions:
    begin{cases}
    u(x,0) = f(x)\
    u(0,t) = u_{x}(L,t) = 0
    end{cases}
    where
    $$f(x) = begin{cases} frac{2}{L}x,& x in left[0,frac{L}{2}right) \ 1,& x in left[frac{L}{2},Lright)end{cases}.$$
    You must justify the convergence of the series, and justify aspects of odd and even extensions if necessary.




    My attempt.



    Step 1. Separation of variables:



    Write $u(x,t) = F(x)G(t)$. Thus, $F(x)G'(t) = alpha^{2}F''(x)G(t)$, i.e,
    $$frac{F''(x)}{F(x)} = y = frac{1}{alpha^{2}}frac{G'(t)}{G(t)}$$
    where the parameter $y$ no depends of $x$ and $t$. We have two ODE's:
    $$F''(x) - yF(x) = 0tag{1}$$
    and
    $$G'(t) - yalpha^{2}G(t) = 0tag{2}$$



    Step 2: Bound conditions.



    we have




    • If $y = 0$, $F equiv 0$,

    • If $y > 0$, $F equiv 0$,

    • If $y < 0$, take $y = -lambda^{2}$ then $F_{n}(x) = c_{2}sinleft(frac{npi x}{2L}right)$


    where the solution for $y > 0$ is $F(x) = c_{1}cos(lambda x) + c_{2}sin(lambda x)$. The solution for $G'(t) - yalpha^{2}G(t) = 0$ is $G(t) = ce^{-yalpha^{2}t}$. So, for $n = pm 1, pm 2,dots$, we have
    $$u_{n}(x,t) = Ke^{-frac{npi alpha}{2L}t}sinleft(frac{npi x}{2L}right).$$



    Step 3. Find Fourier Series of $f$.



    I tried to get an odd or even extension of $f$ and so, to find the Fourier Series of the extension (maybe is not possible to get a $sin$ or $cos$ Fourier Series). But I could not find a periodical extension of $f$. Precisely, I want to get a Fourier Series of the form
    $$sum_{1}^{infty}c_{n}sinleft(frac{npi x}{2L}right).$$
    With this,
    $$u(x,t) = Ksum_{1}^{infty}c_{n}e^{-frac{npi alpha}{2L}t}sinleft(frac{npi x}{2L}right)$$
    is a solution.



    Can someone help me?






    pde fourier-analysis even-and-odd-functions







    share|cite|improve this question



























      1












      1








      1


      1






      Solve the partial differential equation $u_{t} = alpha^{2}u_{xx}$ with conditions:
      begin{cases}
      u(x,0) = f(x)\
      u(0,t) = u_{x}(L,t) = 0
      end{cases}
      where
      $$f(x) = begin{cases} frac{2}{L}x,& x in left[0,frac{L}{2}right) \ 1,& x in left[frac{L}{2},Lright)end{cases}.$$
      You must justify the convergence of the series, and justify aspects of odd and even extensions if necessary.




      My attempt.



      Step 1. Separation of variables:



      Write $u(x,t) = F(x)G(t)$. Thus, $F(x)G'(t) = alpha^{2}F''(x)G(t)$, i.e,
      $$frac{F''(x)}{F(x)} = y = frac{1}{alpha^{2}}frac{G'(t)}{G(t)}$$
      where the parameter $y$ no depends of $x$ and $t$. We have two ODE's:
      $$F''(x) - yF(x) = 0tag{1}$$
      and
      $$G'(t) - yalpha^{2}G(t) = 0tag{2}$$



      Step 2: Bound conditions.



      we have




      • If $y = 0$, $F equiv 0$,

      • If $y > 0$, $F equiv 0$,

      • If $y < 0$, take $y = -lambda^{2}$ then $F_{n}(x) = c_{2}sinleft(frac{npi x}{2L}right)$


      where the solution for $y > 0$ is $F(x) = c_{1}cos(lambda x) + c_{2}sin(lambda x)$. The solution for $G'(t) - yalpha^{2}G(t) = 0$ is $G(t) = ce^{-yalpha^{2}t}$. So, for $n = pm 1, pm 2,dots$, we have
      $$u_{n}(x,t) = Ke^{-frac{npi alpha}{2L}t}sinleft(frac{npi x}{2L}right).$$



      Step 3. Find Fourier Series of $f$.



      I tried to get an odd or even extension of $f$ and so, to find the Fourier Series of the extension (maybe is not possible to get a $sin$ or $cos$ Fourier Series). But I could not find a periodical extension of $f$. Precisely, I want to get a Fourier Series of the form
      $$sum_{1}^{infty}c_{n}sinleft(frac{npi x}{2L}right).$$
      With this,
      $$u(x,t) = Ksum_{1}^{infty}c_{n}e^{-frac{npi alpha}{2L}t}sinleft(frac{npi x}{2L}right)$$
      is a solution.



      Can someone help me?






      pde fourier-analysis even-and-odd-functions







      share|cite|improve this question
















      Solve the partial differential equation $u_{t} = alpha^{2}u_{xx}$ with conditions:
      begin{cases}
      u(x,0) = f(x)\
      u(0,t) = u_{x}(L,t) = 0
      end{cases}
      where
      $$f(x) = begin{cases} frac{2}{L}x,& x in left[0,frac{L}{2}right) \ 1,& x in left[frac{L}{2},Lright)end{cases}.$$
      You must justify the convergence of the series, and justify aspects of odd and even extensions if necessary.




      My attempt.



      Step 1. Separation of variables:



      Write $u(x,t) = F(x)G(t)$. Thus, $F(x)G'(t) = alpha^{2}F''(x)G(t)$, i.e,
      $$frac{F''(x)}{F(x)} = y = frac{1}{alpha^{2}}frac{G'(t)}{G(t)}$$
      where the parameter $y$ no depends of $x$ and $t$. We have two ODE's:
      $$F''(x) - yF(x) = 0tag{1}$$
      and
      $$G'(t) - yalpha^{2}G(t) = 0tag{2}$$



      Step 2: Bound conditions.



      we have




      • If $y = 0$, $F equiv 0$,

      • If $y > 0$, $F equiv 0$,

      • If $y < 0$, take $y = -lambda^{2}$ then $F_{n}(x) = c_{2}sinleft(frac{npi x}{2L}right)$


      where the solution for $y > 0$ is $F(x) = c_{1}cos(lambda x) + c_{2}sin(lambda x)$. The solution for $G'(t) - yalpha^{2}G(t) = 0$ is $G(t) = ce^{-yalpha^{2}t}$. So, for $n = pm 1, pm 2,dots$, we have
      $$u_{n}(x,t) = Ke^{-frac{npi alpha}{2L}t}sinleft(frac{npi x}{2L}right).$$



      Step 3. Find Fourier Series of $f$.



      I tried to get an odd or even extension of $f$ and so, to find the Fourier Series of the extension (maybe is not possible to get a $sin$ or $cos$ Fourier Series). But I could not find a periodical extension of $f$. Precisely, I want to get a Fourier Series of the form
      $$sum_{1}^{infty}c_{n}sinleft(frac{npi x}{2L}right).$$
      With this,
      $$u(x,t) = Ksum_{1}^{infty}c_{n}e^{-frac{npi alpha}{2L}t}sinleft(frac{npi x}{2L}right)$$
      is a solution.



      Can someone help me?






      pde fourier-analysis even-and-odd-functions




      pde fourier-analysis even-and-odd-functions






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 25 at 2:31

























      asked Nov 25 at 2:21









      Lucas Corrêa

      1,4471321




      1,4471321






















          1 Answer
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          Your eigenfunction is almost, but not quite correct. You need



          $$ F_n'(L) = frac{npi}{2L}cosleft(frac{npi}{2}right) = 0 $$



          This is only valid if $n$ is odd, so $n=1,3,5,dots$



          Furthermore, since $sin$ is odd, you'll want to have an odd periodic extension of $f(x)$, i.e. $f(-x)_{xin(-L,0)} = -f(x)_{xin(0,L)}$ and so on. This gives a period of $2L$.



          Using the definition, we have



          $$ c_n = frac{int_{-L}^L f(x)sin left(frac{npi}{2L}xright) dx}{int_{-L}^Lsin^2 left(frac{npi}{2L}xright)dx} = frac{2}{L}int_0^L f(x) sin left(frac{npi}{2L}xright) dx $$



          You can integrate piecewise, i.e.



          $$ c_n = frac{2}{L} left(int_0^{L/2} frac{2}{L}x sin left(frac{npi}{2L}xright) dx + int_{L/2}^L sin left(frac{npi}{2L}xright) dx right) $$



          for odd $n$. I'll leave this to you.






          share|cite|improve this answer





















          • I got it! Thank you!
            – Lucas Corrêa
            Nov 30 at 20:56











          Your Answer





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          1 Answer
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          active

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          active

          oldest

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          active

          oldest

          votes









          1














          Your eigenfunction is almost, but not quite correct. You need



          $$ F_n'(L) = frac{npi}{2L}cosleft(frac{npi}{2}right) = 0 $$



          This is only valid if $n$ is odd, so $n=1,3,5,dots$



          Furthermore, since $sin$ is odd, you'll want to have an odd periodic extension of $f(x)$, i.e. $f(-x)_{xin(-L,0)} = -f(x)_{xin(0,L)}$ and so on. This gives a period of $2L$.



          Using the definition, we have



          $$ c_n = frac{int_{-L}^L f(x)sin left(frac{npi}{2L}xright) dx}{int_{-L}^Lsin^2 left(frac{npi}{2L}xright)dx} = frac{2}{L}int_0^L f(x) sin left(frac{npi}{2L}xright) dx $$



          You can integrate piecewise, i.e.



          $$ c_n = frac{2}{L} left(int_0^{L/2} frac{2}{L}x sin left(frac{npi}{2L}xright) dx + int_{L/2}^L sin left(frac{npi}{2L}xright) dx right) $$



          for odd $n$. I'll leave this to you.






          share|cite|improve this answer





















          • I got it! Thank you!
            – Lucas Corrêa
            Nov 30 at 20:56
















          1














          Your eigenfunction is almost, but not quite correct. You need



          $$ F_n'(L) = frac{npi}{2L}cosleft(frac{npi}{2}right) = 0 $$



          This is only valid if $n$ is odd, so $n=1,3,5,dots$



          Furthermore, since $sin$ is odd, you'll want to have an odd periodic extension of $f(x)$, i.e. $f(-x)_{xin(-L,0)} = -f(x)_{xin(0,L)}$ and so on. This gives a period of $2L$.



          Using the definition, we have



          $$ c_n = frac{int_{-L}^L f(x)sin left(frac{npi}{2L}xright) dx}{int_{-L}^Lsin^2 left(frac{npi}{2L}xright)dx} = frac{2}{L}int_0^L f(x) sin left(frac{npi}{2L}xright) dx $$



          You can integrate piecewise, i.e.



          $$ c_n = frac{2}{L} left(int_0^{L/2} frac{2}{L}x sin left(frac{npi}{2L}xright) dx + int_{L/2}^L sin left(frac{npi}{2L}xright) dx right) $$



          for odd $n$. I'll leave this to you.






          share|cite|improve this answer





















          • I got it! Thank you!
            – Lucas Corrêa
            Nov 30 at 20:56














          1












          1








          1






          Your eigenfunction is almost, but not quite correct. You need



          $$ F_n'(L) = frac{npi}{2L}cosleft(frac{npi}{2}right) = 0 $$



          This is only valid if $n$ is odd, so $n=1,3,5,dots$



          Furthermore, since $sin$ is odd, you'll want to have an odd periodic extension of $f(x)$, i.e. $f(-x)_{xin(-L,0)} = -f(x)_{xin(0,L)}$ and so on. This gives a period of $2L$.



          Using the definition, we have



          $$ c_n = frac{int_{-L}^L f(x)sin left(frac{npi}{2L}xright) dx}{int_{-L}^Lsin^2 left(frac{npi}{2L}xright)dx} = frac{2}{L}int_0^L f(x) sin left(frac{npi}{2L}xright) dx $$



          You can integrate piecewise, i.e.



          $$ c_n = frac{2}{L} left(int_0^{L/2} frac{2}{L}x sin left(frac{npi}{2L}xright) dx + int_{L/2}^L sin left(frac{npi}{2L}xright) dx right) $$



          for odd $n$. I'll leave this to you.






          share|cite|improve this answer












          Your eigenfunction is almost, but not quite correct. You need



          $$ F_n'(L) = frac{npi}{2L}cosleft(frac{npi}{2}right) = 0 $$



          This is only valid if $n$ is odd, so $n=1,3,5,dots$



          Furthermore, since $sin$ is odd, you'll want to have an odd periodic extension of $f(x)$, i.e. $f(-x)_{xin(-L,0)} = -f(x)_{xin(0,L)}$ and so on. This gives a period of $2L$.



          Using the definition, we have



          $$ c_n = frac{int_{-L}^L f(x)sin left(frac{npi}{2L}xright) dx}{int_{-L}^Lsin^2 left(frac{npi}{2L}xright)dx} = frac{2}{L}int_0^L f(x) sin left(frac{npi}{2L}xright) dx $$



          You can integrate piecewise, i.e.



          $$ c_n = frac{2}{L} left(int_0^{L/2} frac{2}{L}x sin left(frac{npi}{2L}xright) dx + int_{L/2}^L sin left(frac{npi}{2L}xright) dx right) $$



          for odd $n$. I'll leave this to you.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 at 10:36









          Dylan

          12.2k31026




          12.2k31026












          • I got it! Thank you!
            – Lucas Corrêa
            Nov 30 at 20:56


















          • I got it! Thank you!
            – Lucas Corrêa
            Nov 30 at 20:56
















          I got it! Thank you!
          – Lucas Corrêa
          Nov 30 at 20:56




          I got it! Thank you!
          – Lucas Corrêa
          Nov 30 at 20:56


















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