2
$\begingroup$

The non-linear convection equation

$$u_{t} +uu_{x}=0$$ admits implicit solutions of the form $$u=f(x-ut).$$ How does one interpret this solution intuitively? Is there an example of a solution of this equation that can be written in an explicit form? How does an initial function evolve over time?

$\endgroup$
0

1 Answer 1

3
$\begingroup$
  1. This quasi-linear 1st-order PDE $$\left(\frac{\partial u}{\partial t}\right)_{\!x} + u\left(\frac{\partial u}{\partial x}\right)_{\!t}~=~0\tag{1}$$ is the inviscid Burgers' equation.

  2. It can be solved via the method of characteristics. The ODE IVP $$ \frac{dx}{dt}~=~u, \qquad x(t\!=\!0)~=~\xi, \tag{2}$$ (where $u$ is treated as an external parameter and $\xi$ is an initial value) has integral curves $$ x(t) ~=~ u t + \xi. \tag{3}$$ The initial value $\xi$ and the external parameter $u$ label the integral curves (3) in an $(t,x)$-diagram.

  3. A solution $u(x,t)$ to the PDE is constant along above integral curves (3): $$ \frac{du(x(t),t)}{dt}~=~\left(\frac{\partial u}{\partial t}\right)_{\!x} + \left(\frac{\partial u}{\partial x}\right)_{\!t}\frac{dx}{dt}~\stackrel{(2)}{=}~ \left(\frac{\partial u}{\partial t}\right)_{\!x} + u\left(\frac{\partial u}{\partial x}\right)_{\!t}~\stackrel{(1)}{=}~0.\tag{4}$$ This leads to that the solution $u=f(\xi)$ should be a function of $\xi$.

  4. The function $$x\quad \mapsto\quad f(x)~=~u(x,t\!=\!0)\tag{5}$$ is the initial profile.

  5. The implicit solution $$u(x,t)~=~f(x-t u(x,t))~=~f(x-t f(x-t u(x,t)))~=~\ldots\tag{6}$$ is a fixed-point equation.

  6. Interestingly, the non-linear PDE can develop shock-wave singular solution $u$ from a smooth regular initial profile $f$.

  7. Example. The Wikipedia page lists the full solution $$u(x,t)~=~\frac{ax+b}{at+1}\tag{7}$$ for the initial profile $$f(x)~=~ax+b.\tag{8}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.