Are characteristics the only solution to the advection equation in 1+1D?











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I'm currently reading about fluid dynamics and the Riemann problem, and a very commonly used equation to introduce the topic is the 1+1D advection equation with constant coefficient $v$:



$$ frac{partial u}{partial t} + v frac{partial u}{partial x} = 0tag{1}$$



for which a solution is
$$ u(x,t) = u(x-vt, 0) = u_0(x-vt) $$
where $u_0 = u(t=0)$ is some initial condition.



This can be easily derived using the method of separation of variables: Let $u(x,t) = f(x)g(y)$.
Then
$$ frac{partial u}{partial t} = f(x) frac{partial g}{partial t}$$



$$ frac{partial u}{partial x} = g(t) frac{partial f}{partial x}
$$

Inserting into the advection equation and restructuring a little, we get



$$frac{1}{g } frac{partial g}{partial t} = frac{1}{f}frac{partial f}{partial x} = -lambda $$



where $lambda$ is some constant. Solving each equation separately gives us



$$ g = K_1 e^{-lambda v t} $$
$$ f = K_2 e^{lambda x} $$
$$ Rightarrow u(x,t) = fg = K e^{lambda (x - vt)} $$



with $K_1$, $K_2$ and $K=K_1 K_2$ are constants stemming from integration.
With
$$u_0 = u(x,t=0) = K e^{lambda x}$$
one can easily see that the solution can be expressed as
$$u(x,t) = u_0(x-vt)$$



So far, so good. Here's my question: Is that the only solution of the 1+1D advection equation with constant coefficients? Is there a proof that this is the only solution?










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  • I voted to migrate this to Mathematics.
    – AccidentalFourierTransform
    Nov 9 at 17:19












  • Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
    – Kyle Kanos
    Nov 10 at 18:04















up vote
4
down vote

favorite
1












I'm currently reading about fluid dynamics and the Riemann problem, and a very commonly used equation to introduce the topic is the 1+1D advection equation with constant coefficient $v$:



$$ frac{partial u}{partial t} + v frac{partial u}{partial x} = 0tag{1}$$



for which a solution is
$$ u(x,t) = u(x-vt, 0) = u_0(x-vt) $$
where $u_0 = u(t=0)$ is some initial condition.



This can be easily derived using the method of separation of variables: Let $u(x,t) = f(x)g(y)$.
Then
$$ frac{partial u}{partial t} = f(x) frac{partial g}{partial t}$$



$$ frac{partial u}{partial x} = g(t) frac{partial f}{partial x}
$$

Inserting into the advection equation and restructuring a little, we get



$$frac{1}{g } frac{partial g}{partial t} = frac{1}{f}frac{partial f}{partial x} = -lambda $$



where $lambda$ is some constant. Solving each equation separately gives us



$$ g = K_1 e^{-lambda v t} $$
$$ f = K_2 e^{lambda x} $$
$$ Rightarrow u(x,t) = fg = K e^{lambda (x - vt)} $$



with $K_1$, $K_2$ and $K=K_1 K_2$ are constants stemming from integration.
With
$$u_0 = u(x,t=0) = K e^{lambda x}$$
one can easily see that the solution can be expressed as
$$u(x,t) = u_0(x-vt)$$



So far, so good. Here's my question: Is that the only solution of the 1+1D advection equation with constant coefficients? Is there a proof that this is the only solution?










share|cite|improve this question
























  • I voted to migrate this to Mathematics.
    – AccidentalFourierTransform
    Nov 9 at 17:19












  • Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
    – Kyle Kanos
    Nov 10 at 18:04













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





I'm currently reading about fluid dynamics and the Riemann problem, and a very commonly used equation to introduce the topic is the 1+1D advection equation with constant coefficient $v$:



$$ frac{partial u}{partial t} + v frac{partial u}{partial x} = 0tag{1}$$



for which a solution is
$$ u(x,t) = u(x-vt, 0) = u_0(x-vt) $$
where $u_0 = u(t=0)$ is some initial condition.



This can be easily derived using the method of separation of variables: Let $u(x,t) = f(x)g(y)$.
Then
$$ frac{partial u}{partial t} = f(x) frac{partial g}{partial t}$$



$$ frac{partial u}{partial x} = g(t) frac{partial f}{partial x}
$$

Inserting into the advection equation and restructuring a little, we get



$$frac{1}{g } frac{partial g}{partial t} = frac{1}{f}frac{partial f}{partial x} = -lambda $$



where $lambda$ is some constant. Solving each equation separately gives us



$$ g = K_1 e^{-lambda v t} $$
$$ f = K_2 e^{lambda x} $$
$$ Rightarrow u(x,t) = fg = K e^{lambda (x - vt)} $$



with $K_1$, $K_2$ and $K=K_1 K_2$ are constants stemming from integration.
With
$$u_0 = u(x,t=0) = K e^{lambda x}$$
one can easily see that the solution can be expressed as
$$u(x,t) = u_0(x-vt)$$



So far, so good. Here's my question: Is that the only solution of the 1+1D advection equation with constant coefficients? Is there a proof that this is the only solution?










share|cite|improve this question















I'm currently reading about fluid dynamics and the Riemann problem, and a very commonly used equation to introduce the topic is the 1+1D advection equation with constant coefficient $v$:



$$ frac{partial u}{partial t} + v frac{partial u}{partial x} = 0tag{1}$$



for which a solution is
$$ u(x,t) = u(x-vt, 0) = u_0(x-vt) $$
where $u_0 = u(t=0)$ is some initial condition.



This can be easily derived using the method of separation of variables: Let $u(x,t) = f(x)g(y)$.
Then
$$ frac{partial u}{partial t} = f(x) frac{partial g}{partial t}$$



$$ frac{partial u}{partial x} = g(t) frac{partial f}{partial x}
$$

Inserting into the advection equation and restructuring a little, we get



$$frac{1}{g } frac{partial g}{partial t} = frac{1}{f}frac{partial f}{partial x} = -lambda $$



where $lambda$ is some constant. Solving each equation separately gives us



$$ g = K_1 e^{-lambda v t} $$
$$ f = K_2 e^{lambda x} $$
$$ Rightarrow u(x,t) = fg = K e^{lambda (x - vt)} $$



with $K_1$, $K_2$ and $K=K_1 K_2$ are constants stemming from integration.
With
$$u_0 = u(x,t=0) = K e^{lambda x}$$
one can easily see that the solution can be expressed as
$$u(x,t) = u_0(x-vt)$$



So far, so good. Here's my question: Is that the only solution of the 1+1D advection equation with constant coefficients? Is there a proof that this is the only solution?







fluid-dynamics waves mathematics differential-equations






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edited Nov 9 at 17:45









Qmechanic

99.7k121781119




99.7k121781119










asked Nov 9 at 12:20









lemdan

949




949












  • I voted to migrate this to Mathematics.
    – AccidentalFourierTransform
    Nov 9 at 17:19












  • Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
    – Kyle Kanos
    Nov 10 at 18:04


















  • I voted to migrate this to Mathematics.
    – AccidentalFourierTransform
    Nov 9 at 17:19












  • Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
    – Kyle Kanos
    Nov 10 at 18:04
















I voted to migrate this to Mathematics.
– AccidentalFourierTransform
Nov 9 at 17:19






I voted to migrate this to Mathematics.
– AccidentalFourierTransform
Nov 9 at 17:19














Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
– Kyle Kanos
Nov 10 at 18:04




Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question...
– Kyle Kanos
Nov 10 at 18:04










2 Answers
2






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up vote
4
down vote



accepted










Yes, it is the only solution. Hints for proof:




  1. Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.


  2. Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.


  3. Deduce that $u=u(x^-)$ is a function of $x^-$ only.







share|cite|improve this answer























  • I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
    – lemdan
    Nov 9 at 14:12










  • There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
    – Qmechanic
    Nov 9 at 17:44




















up vote
2
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The equation is linear, and the solution to a linear equation in one unknown is always unique.






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    2 Answers
    2






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    2 Answers
    2






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    active

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    up vote
    4
    down vote



    accepted










    Yes, it is the only solution. Hints for proof:




    1. Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.


    2. Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.


    3. Deduce that $u=u(x^-)$ is a function of $x^-$ only.







    share|cite|improve this answer























    • I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
      – lemdan
      Nov 9 at 14:12










    • There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
      – Qmechanic
      Nov 9 at 17:44

















    up vote
    4
    down vote



    accepted










    Yes, it is the only solution. Hints for proof:




    1. Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.


    2. Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.


    3. Deduce that $u=u(x^-)$ is a function of $x^-$ only.







    share|cite|improve this answer























    • I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
      – lemdan
      Nov 9 at 14:12










    • There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
      – Qmechanic
      Nov 9 at 17:44















    up vote
    4
    down vote



    accepted







    up vote
    4
    down vote



    accepted






    Yes, it is the only solution. Hints for proof:




    1. Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.


    2. Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.


    3. Deduce that $u=u(x^-)$ is a function of $x^-$ only.







    share|cite|improve this answer














    Yes, it is the only solution. Hints for proof:




    1. Go to lightcone coordinates: $x^{pm}~:=~x pm vt$.


    2. Show that OP's eq. (1) in 1+1D becomes $frac{partial u}{partial x^+}~=~0$.


    3. Deduce that $u=u(x^-)$ is a function of $x^-$ only.








    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 9 at 17:45

























    answered Nov 9 at 12:34









    Qmechanic

    99.7k121781119




    99.7k121781119












    • I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
      – lemdan
      Nov 9 at 14:12










    • There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
      – Qmechanic
      Nov 9 at 17:44




















    • I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
      – lemdan
      Nov 9 at 14:12










    • There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
      – Qmechanic
      Nov 9 at 17:44


















    I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
    – lemdan
    Nov 9 at 14:12




    I see that using $frac{partial u}{partial x^+} = 0$ implies that $u = u(x^ -)$, but I don't see how that excludes any other solution?
    – lemdan
    Nov 9 at 14:12












    There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
    – Qmechanic
    Nov 9 at 17:44






    There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows.
    – Qmechanic
    Nov 9 at 17:44












    up vote
    2
    down vote













    The equation is linear, and the solution to a linear equation in one unknown is always unique.






    share|cite|improve this answer

























      up vote
      2
      down vote













      The equation is linear, and the solution to a linear equation in one unknown is always unique.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        The equation is linear, and the solution to a linear equation in one unknown is always unique.






        share|cite|improve this answer












        The equation is linear, and the solution to a linear equation in one unknown is always unique.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 9 at 15:24









        Chester Miller

        13.9k2623




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