# Smooth solutions of $u_t - x u u_x = 0$ deduced from characteristics

Consider the equation $$u_t - x u u_x = 0$$. with cauchy data $$u(x,0) = x$$. Solving this equation I see the characteristics are given by $$x= r e^{-rt}$$ for some $$r$$ and the solution is defined implicitly by

$$x = u e^{-ut}$$

Using lambert function, one can solve for $$u$$. When plotting the characteristics, I see that after $$t=4$$ there are no characteristics. How do we find analytically the values of $$t$$ for which we have a smooth solution?

• I don't see anything remarkable for $t>4$. What do you mean? – Rafa Budría Feb 18 at 14:37
• @Jimmy Sabater . You wrote : " I see that after $t=4$ there are no characteristics". Would you mind show us where exactly and how you saw that after $t=4$ there are no characteristics. – JJacquelin Feb 19 at 10:43

The characteristics are the curves $$x = x_0 \exp (-x_0 t)$$ for $$x_0\in\Bbb R$$, along which $$u=x_0$$ is constant (see e.g. this related post). Differentiating $$x$$ w.r.t. $$x_0$$, we find that $$\frac{\text d x}{\text d x_0} = (1 -t x_0) \exp (-x_0 t) ,$$ which vanishes at $$t=1/x_0$$. The smallest such positive value defines the breaking time $$t_b = \inf 1/x_0 = 0$$, which is only reached as $$x_0 \to \pm\infty$$. Hence, the solution keeps smooth. The solution can be expressed as $$x_0 = -W(-xt)/t = x\, e^{-W(-xt)} = u(x,t),$$ where $$W$$ is the Lambert W-function. This special function is real-valued if its argument $$-xt$$ is larger than $$-1/e$$. Hence, as represented in the $$x$$-$$t$$ plane below, the domain is bounded:

For positive abscissas $$x>0$$, we must have $$t<1/(ex)$$. Thus, if $$t=4$$, the solution is located at abscissas $$x<1/(4e) \approx 0.092$$, which may explain that you don't see any solution elsewhere. For negative abscissas $$x<0$$, there is no restriction at positive times $$t>0>1/(ex)$$.

I agree with your result : $$x=u\,e^{-u\,t}$$ This is the solution on the form of implicit equation.

Of course, if one want the explicit form the Lambert W function is required

$$u(x,t)=-\frac{1}{t} W\left(-t\,x \right)\quad$$ But we don't need it to plot $$u(x,t)$$.

Drawing the curves $$u(x)$$ for various $$t$$ is very easy, without special function. Plot the inverse function $$x(u)$$ , which is explicit. Then symmetry/rotation make the figure on the usual orientation.

The solutions remain smooth from $$t=0$$ up to large $$t$$.

Nothing special appears at $$t=4$$ and latter.

If you draw the curves with the Lambert W function instead of the above very simple method you avoid possible numerical artefact (Of course, with a sufficient accuracy of the numerical computation of the Lambert W function the results must be the same for the two methods).