# 3D linear transport equation solution

If we consider the following transport equation with $$t>0$$ and $$x\in \mathbb{R^3}$$: $$\begin{cases} \partial_t f(t,x) + v(t,x). \nabla f(t,x)=0\\ f(0,x)=g(x) \end{cases}$$ And if we define the follwing function $$X$$ defined on $$(s,t,x)\in\mathbb{R^+}\times\mathbb{R^+}\times\mathbb{R^3}$$: $$\begin{cases} \frac{d}{ds}X(s,t,x)=v(t,X(s,t,x))\\ X(t,t,x)=x \end{cases}$$

I want to prove that the function: $$f(t,x)=g(X(0,t,x))$$ is a solution the firt PDE (transport equation).

My idea:

I tried to calculate $$\partial_t f(t,x) + v(t,x). \nabla f(t,x)=\partial_t g(X(0,t,x)) + v(t,x). \nabla g(X(0,t,x))$$ $$\partial_t f(t,x) + v(t,x). \nabla f(t,x)=\partial_t X(0,t,x) \nabla g(X(0,t,x)) + v(t,x). \nabla g(X(0,t,x))$$ $$\partial_t f(t,x) + v(t,x). \nabla f(t,x)=[\partial_t X(0,t,x) + v(t,x)] \nabla g(X(0,t,x))$$ But i dont what to do next to show that it equals to zero.

For the transport equation $$\partial_t f + v \cdot \nabla f = 0$$ with initial data $$f(0,x) = g(x)$$, the method of characteristics gives $$\frac{\text d}{\text d t} X = v$$ and $$\frac{\text d}{\text d t} f = 0$$. Along the characteristic curves $$t\mapsto X(t)$$, we have therefore $$f(t,X(t)) = g(x_0)$$, where the curve trajectory satisfies the differential equation $$\frac{\text d}{\text d t} X(t) = v(t,X(t)) \qquad\text{with}\qquad \quad X(t) = x,\quad X(0) = x_0 .$$ Thus, we may write $$x = x_0 + \int_0^t v(\tau ,X(\tau ))\, \text d \tau$$ by using the FTC. Injecting this in the expression of $$f$$ along characteristics, we have finally $$f(t,x) = g\left(x - \int_0^t v(\tau , X(\tau ))\, \text d \tau\right) ,$$ where the function $$\tau\mapsto X(\tau)$$ solves the above initial value problem. In the particular case where the velocity vector $$v$$ is constant in time and space, we recover the classical solution $$f(t,x) = g(x-tv)$$.
To show that the above expression works, let us compute partial derivatives in space and time. We have \begin{aligned} \partial_t f &= \partial_t x_0\cdot \nabla g(x_0) = -v\cdot \nabla g(x_0) \\ \nabla f &= \nabla x_0\cdot \nabla g(x_0) = \nabla g(x_0) \end{aligned} which ends the proof.