In short terms, I must solve a certain first-order PDE. I applied the method of caracteristics to find the integral curves and found an answer that indeed satisfies the equation. However, Wolfram Mathematica gives a different solution. Comments on my solution and possible causes for the discrepancies would be the most appreciated. I remark that I follow Folland's Introduction to Partial Differential Equations in the nomenclature and notations.
Problem:
Given an open set $\Omega \subset \mathbb{R}^2$ and $\phi(x) \in C^1(\Omega)$, solve:
$\partial_tu + x\partial_xu = t^3 \\ u(x, 0) = \phi(x)$
My solution:
Since we have a linear first-order problem, the method of characteristics is the natural alternative. At first, we shall note that the hipersurface $S$ in which the initial conditions are given is $S = \{(x,y) \in \mathbb{R}; y = 0 \}$, and that the characteristic variety is ${\rm char} = \{ (\xi_x, \xi_t) \in \mathbb{R}^2 \setminus \{0\}; \xi_t = -x\xi_x \}$, so that the normal unit vector to $S$, namely $(0, 1)$, is not in $\rm char$, and hence we can proceed to integrate the vector field of the coefficients to find the (unique) solution to the problem (according to theorem 1.7 in Folland's book).
The associated system of ODEs is the following:
$\frac{dt}{d\alpha} = 1 \\ \frac{dx}{d\alpha} = x \\ \frac{du}{d\alpha} = t^3 \\ (x, t, u)|_{\alpha = 0} = (s,0,\phi(x)) $
being $(s, 0)$ the initial point in $S$ and $\alpha$ a parameter.
Solving the ODEs, we get:
$t(\alpha) = \alpha \\ x(\alpha) = se^\alpha \\ u(\alpha) = \frac{\alpha^4}{4} + \phi(s) $
At last, eliminating $\alpha$ and $s$, we get:
$u(x, t) = \frac{t^4}{4} + \phi(\frac{x}{e^t})$
We note that applying the operator $L = \partial_t + x\partial_x$ to the $u$ found above, we get $t^3$, as expected. However, Wolfram Mathematica gives the following: Mathematica inputs and solution.
Thank you for your help.
