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

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?

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

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