Coupled partial differential equations












0












$begingroup$


I'm having trouble solving these coupled partial differential equations:




$$frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0,$$
$$frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0,$$

with $A,c$ real constants.




What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.






pde systems-of-equations partial-derivative







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  • $begingroup$
    I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
    $endgroup$
    – Mattos
    Mar 11 '16 at 16:03












  • $begingroup$
    @Mattos What do you mean by rearranging $partial_x p$?
    $endgroup$
    – new guy
    Mar 11 '16 at 16:05










  • $begingroup$
    From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
    $endgroup$
    – Mattos
    Mar 11 '16 at 16:06


















0












$begingroup$


I'm having trouble solving these coupled partial differential equations:




$$frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0,$$
$$frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0,$$

with $A,c$ real constants.




What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.






pde systems-of-equations partial-derivative







share|cite|improve this question











$endgroup$












  • $begingroup$
    I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
    $endgroup$
    – Mattos
    Mar 11 '16 at 16:03












  • $begingroup$
    @Mattos What do you mean by rearranging $partial_x p$?
    $endgroup$
    – new guy
    Mar 11 '16 at 16:05










  • $begingroup$
    From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
    $endgroup$
    – Mattos
    Mar 11 '16 at 16:06
















0












0








0





$begingroup$


I'm having trouble solving these coupled partial differential equations:




$$frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0,$$
$$frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0,$$

with $A,c$ real constants.




What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.






pde systems-of-equations partial-derivative







share|cite|improve this question











$endgroup$




I'm having trouble solving these coupled partial differential equations:




$$frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0,$$
$$frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0,$$

with $A,c$ real constants.




What is the "trick" to solve these? I haven't tried a lot since I wouldn't know where to start looking.






pde systems-of-equations partial-derivative




pde systems-of-equations partial-derivative






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edited Feb 25 at 11:36









Harry Peter

5,47911439




5,47911439










asked Mar 11 '16 at 15:58









new guynew guy

382




382












  • $begingroup$
    I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
    $endgroup$
    – Mattos
    Mar 11 '16 at 16:03












  • $begingroup$
    @Mattos What do you mean by rearranging $partial_x p$?
    $endgroup$
    – new guy
    Mar 11 '16 at 16:05










  • $begingroup$
    From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
    $endgroup$
    – Mattos
    Mar 11 '16 at 16:06




















  • $begingroup$
    I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
    $endgroup$
    – Mattos
    Mar 11 '16 at 16:03












  • $begingroup$
    @Mattos What do you mean by rearranging $partial_x p$?
    $endgroup$
    – new guy
    Mar 11 '16 at 16:05










  • $begingroup$
    From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
    $endgroup$
    – Mattos
    Mar 11 '16 at 16:06


















$begingroup$
I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
$endgroup$
– Mattos
Mar 11 '16 at 16:03






$begingroup$
I suppose you could take $partial_{x}$ (and $partial_{t}$) of the first equation, rearrange for $partial_{x} p$ and $partial_{t} p$, then substitute those results into your second equation which should give you everything in $f$.
$endgroup$
– Mattos
Mar 11 '16 at 16:03














$begingroup$
@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05




$begingroup$
@Mattos What do you mean by rearranging $partial_x p$?
$endgroup$
– new guy
Mar 11 '16 at 16:05












$begingroup$
From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06






$begingroup$
From the first equation $$partial_{x} p = frac{1}{A} partial_{x} bigg( partial_{t} f - c partial_{x} f bigg)$$ You can get something similar for $partial_{t} p$.
$endgroup$
– Mattos
Mar 11 '16 at 16:06












2 Answers
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$begingroup$

$$begin{cases}
frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
end{cases}$$
Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
begin{cases}
frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
end{cases}$



$$begin{cases}
f_T-f_X-p=0 \
p_T+p_X+f=0
end{cases}$$
$p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$



$$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$



Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$



For example of solving see : Finding the general solution of a second order PDE
This method leads to the integral form of solution :
$$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
$c(s)$ and $alpha(s)$ are arbitrary real or complex functions.






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    0












    $begingroup$

    Hint.



    We have



    $$
    mathcal{D}_1 f = A p\
    mathcal{D}_2 p = -A f
    $$



    then



    $$
    mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
    mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
    $$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      0












      $begingroup$

      $$begin{cases}
      frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
      frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
      end{cases}$$
      Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
      begin{cases}
      frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
      frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
      end{cases}$



      $$begin{cases}
      f_T-f_X-p=0 \
      p_T+p_X+f=0
      end{cases}$$
      $p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$



      $$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
      Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$



      Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$



      For example of solving see : Finding the general solution of a second order PDE
      This method leads to the integral form of solution :
      $$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
      $c(s)$ and $alpha(s)$ are arbitrary real or complex functions.






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        $$begin{cases}
        frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
        frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
        end{cases}$$
        Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
        begin{cases}
        frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
        frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
        end{cases}$



        $$begin{cases}
        f_T-f_X-p=0 \
        p_T+p_X+f=0
        end{cases}$$
        $p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$



        $$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
        Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$



        Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$



        For example of solving see : Finding the general solution of a second order PDE
        This method leads to the integral form of solution :
        $$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
        $c(s)$ and $alpha(s)$ are arbitrary real or complex functions.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          $$begin{cases}
          frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
          frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
          end{cases}$$
          Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
          begin{cases}
          frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
          frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
          end{cases}$



          $$begin{cases}
          f_T-f_X-p=0 \
          p_T+p_X+f=0
          end{cases}$$
          $p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$



          $$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
          Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$



          Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$



          For example of solving see : Finding the general solution of a second order PDE
          This method leads to the integral form of solution :
          $$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
          $c(s)$ and $alpha(s)$ are arbitrary real or complex functions.






          share|cite|improve this answer











          $endgroup$



          $$begin{cases}
          frac{partial}{partial t}f(x,t)-cfrac{partial}{partial x}f(x,t)-Ap(x,t)=0 \
          frac{partial}{partial t}p(x,t)+cfrac{partial}{partial x}p(x,t)+Af(x,t)=0
          end{cases}$$
          Regularised form with $begin{cases} T=At \ X=frac{A}{c}x end{cases} quadtoquad
          begin{cases}
          frac{partial}{partial T}f(X,T)-frac{partial}{partial X}f(X,T)-p(X,T)=0 \
          frac{partial}{partial T}p(X,T)+frac{partial}{partial X}p(X,T)+f(X,T)=0
          end{cases}$



          $$begin{cases}
          f_T-f_X-p=0 \
          p_T+p_X+f=0
          end{cases}$$
          $p=f_T-f_X quadtoquad (f_{TT}-f_{XT})+(f_{XT}-f_{XX})+f=0$



          $$frac{partial^2 f}{partial T^2}-frac{partial^2f}{partial X^2}+f(X,T)=0$$
          Solving this hyperbolic PDE leads to $f(X,T)=fleft(At:,:frac{A}{c}x right)$



          Then $quad p(X,T)=frac{partial f}{partial T}-frac{partial f}{partial X}=pleft(At:,:frac{A}{c}x right)$



          For example of solving see : Finding the general solution of a second order PDE
          This method leads to the integral form of solution :
          $$f(X,T)=int c(s)e^{sqrt{alpha(s)-frac{1}{2}}:X +sqrt{alpha(s)+frac{1}{2}} :T } ds$$
          $c(s)$ and $alpha(s)$ are arbitrary real or complex functions.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 13 '17 at 12:20









          Community♦

          1




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          answered Mar 16 '16 at 8:49









          JJacquelinJJacquelin

          44.5k21854




          44.5k21854























              0












              $begingroup$

              Hint.



              We have



              $$
              mathcal{D}_1 f = A p\
              mathcal{D}_2 p = -A f
              $$



              then



              $$
              mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
              mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
              $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint.



                We have



                $$
                mathcal{D}_1 f = A p\
                mathcal{D}_2 p = -A f
                $$



                then



                $$
                mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
                mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
                $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint.



                  We have



                  $$
                  mathcal{D}_1 f = A p\
                  mathcal{D}_2 p = -A f
                  $$



                  then



                  $$
                  mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
                  mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  Hint.



                  We have



                  $$
                  mathcal{D}_1 f = A p\
                  mathcal{D}_2 p = -A f
                  $$



                  then



                  $$
                  mathcal{D}_2mathcal{D}_1 f = Amathcal{D}_2 p = -A^2 f\
                  mathcal{D}_1mathcal{D}_2 p = -A mathcal{D}_1f = -A^2p
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 13:35









                  CesareoCesareo

                  9,2563517




                  9,2563517






























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