Given the linear equation $$u_t -xt u_x = x$$ $x\in\mathbb{R}$, $t>0$, with IVP $u(x,0)=u_0(x)$, my solution comes to $u(x,t) = u_0(xe^{t^2/2})+xt$, but Maple gives a much more complicated solution to this IVP. I would appreciate if someone could please point out what I might not be doing right.

Assuming the parametrization $x=x(s), t=t(s), u=u(s)$, and applying the method of characteristics, we get: $$t_s=1, x_s=-xt,u_s=x,$$ so $$u_x = -\frac{1}{t}, u_t=x,$$ thus

$$u=xt+c_2, x=e^{t^2/2}c_3,$$ where $c_2, c_2$ are constants.

So, $u-xt=c_2, c_3 = xe^{t^2/2}$, $G(xe^{t^2/2})=u-xt$, and $G(x) = u_0(x)$, and we get $$u(x,t) = u_0(xt^{t^2/2})+xt$$

But Maple gives me this:


  • $\begingroup$ I've just spotted an error: $u_x$ should be $-\frac{1}{t}$. But then, $u=G(xe^{t^2/2})-\frac{x}{t}$. How does one then apply the initial condition? $\endgroup$ – sequence Jun 2 '16 at 6:23
  • $\begingroup$ OK, I've resolved the issue. Used a slightly different approach. $\endgroup$ – sequence Jun 3 '16 at 6:16

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=-xt=-xs$ , letting $x(0)=x_0$ , we have $x=x_0e^{-\frac{s^2}{2}}=x_0e^{-\frac{t^2}{2}}$

$\dfrac{du}{ds}=x=x_0e^{-\frac{s^2}{2}}$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)+\int_0^sx_0e^{-\frac{\tau^2}{2}}~d\tau=f(xe^\frac{t^2}{2})+\int_0^txe^\frac{t^2-\tau^2}{2}~d\tau$

$u(x,0)=u_0(x)$ :


$\therefore u(x,t)=u_0(xe^\frac{t^2}{2})+\int_0^txe^\frac{t^2-\tau^2}{2}~d\tau$


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