Parametrizing solution surface in PDE $v_t-xv_x=1-v$

Question

The task is as follows:

You are given the Cauchy problem for $v(x, t)$: $$ v_t(x, t) − xv_x(x, t) = 1 − v(x, t),\qquad −∞ < x < ∞,\; t > 0, $$ $$ v(x, 0) = \sin x, \qquad −∞ < x < ∞. $$ (a) Determine the solution $v(x, t)$.

I parametrized with $t(s,0) = 0$, $x(s,0) = s$, and $v(s,0) = \sin(s)$. Basically I end up with

$v(s,τ) = [τ+\sin(s)]/(1+τ)$

$t(s,τ) = τ$, and $x(s,t) = s/(1+τ)$

I need $v$ in terms of the original $x$ and $t$, but can't figure out how to sub them back in. Could I have parametrized $v(s,0)$ differently perhaps?

Answer 1

Let us apply the method of characteristics to the Cauchy problem of the present PDE $v_t-xv_x=1-v$.

• $\text dt/\text ds = 1$, letting $t(0)=0$ we know $t=s$;
• $\text dx/\text ds = -x$, letting $x(0)=x_0$ we know $x = x_0e^{-s}$;
• $\text dv/\text d s = 1-v$, letting $v(0)=\sin x_0$ we know $v = 1 + (\sin x_0 -1)\, e^{-s}$

Combining everything, we find $$ v(x,t) = 1 + (\sin (xe^t) -1)\, e^{-t} . $$