4
$\begingroup$

I've come across the following PDE problem: $$\frac{\partial u}{\partial t} + 2tx^2\frac{\partial u}{\partial x} = 0, \\ u = u(x,t) \\ u(x,0) = x^3$$

This is a first order, linear, homogeneous PDE. However, using the method of characteristics, when I solve the equation of characteristics, I get $x = 1/(-t^2 + constant)$, and when I plot that graph, the curves never intercept the $x$ axis. Being so, I can't pick a characteristic curve that passes through $(x_0, 1)$, and so I don't believe we can conclude that $u$ is constant along the characteristics.

What can we conclude about the PDE solution given the problem's condition? Will it be a weak (non-classical) solution?

$\endgroup$

1 Answer 1

2
$\begingroup$

The line $x=0$ is a characteristic of this PDE. Indeed, the equation says that when $x=0$, the derivative $u_t$ is zero, which means $u$ is constant on this line.

The reason you haven't found this characteristic when solving the ODE $$ \frac{dt}{dx} = \frac{1}{2tx^2} $$ is that by putting $x$ in the denominator, we already lost the solution $x=0$. This happens often when dealing with ODE. Always check for possible equilibrium solutions corresponding to denominator being zero; they may or may not be found otherwise.

I can't pick a characteristic curve that passes through $(x_0,1)$

It's $x=0$ if $x_0=0$, and the curve $\frac{1}{x}+t^2 = \frac{1}{x_0}+1$ otherwise.


You also ask about solutions. Any function of the form $u(x,t)=f(t^2+\frac1x)$ is a smooth solution provided $f$ is smooth on $\mathbb{R}$ and $z\mapsto f(1/z)$ is smooth at $0$. For example, $f(z) = z/(z^2+1)$ would work.

If instead we pick something like $f(z)=z$, then yes, the solution will be discontinuous across the $t$-axis.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .