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 $v$:

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

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?

  • $\begingroup$ I voted to migrate this to Mathematics. $\endgroup$ Commented Nov 9, 2018 at 17:19
  • $\begingroup$ Fluid Dynamics and solutions thereof may require maths, but certainly is a physics question... $\endgroup$
    – Kyle Kanos
    Commented Nov 10, 2018 at 18:04

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – mivkov
    Commented Nov 9, 2018 at 14:12
  • $\begingroup$ There are only 2 coordinates $(x^+,x^-)$ in 1+1D and $u$ cannot depend on $x^+$. So the above conclusion follows. $\endgroup$
    – Qmechanic
    Commented Nov 9, 2018 at 17:44

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


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