What does d f(t,x) = 0 mean?

Question

A differential equation that can be written in the form $d\phi(t, x) = 0$ for some continuous and differentiable function $\phi(t, x)$ is called exact.

What does $d\phi(t, x) = 0$ mean?

Answer 1

A curve $$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a<t<b)$$ is a solution curve of the differential equation $$\omega:=f(x,y)\>dx+ g(x,y)\>dy=0\qquad\bigl((x,y)\in\Omega\bigr)\tag{1}$$ if $$f\bigl(x(t),y(t)\bigr)\>\dot x(t)+g\bigl(x(t),y(t)\bigr)\>\dot y(t)\equiv 0\qquad(a<t<b)\ .$$ This means that the solution curves of $(1)$ coincide with the solution curves of the differential equation $$y'=-{f(x,y)\over g(x,y)}\ .$$

Sometimes it is the case that the vector data ${\bf v}(x,y):=\bigl(f(x,y),g(x,y)\bigr)$ is the gradient field of a scalar function $\phi:\ \Omega\to{\mathbb R}$, in other words, that $$f(x,y)=\phi_x(x,y),\quad g(x,y)=\phi_y(x,y)\qquad\bigl((x,y)\in\Omega\bigr)\ .$$ When such a $\phi$ exists (at least locally) the differential equation $(1)$ is called exact, and one writes $\omega=d\phi$. The following computation shows that $\phi$ is constant along the solution curves: $$\eqalign{{d\over dt}\phi\bigl(x(t),y(t)\bigr)&=\phi_x\bigl(x(t),y(t)\bigr)\dot x(t)+\phi_y\bigl(x(t),y(t)\bigr)\dot y(t)\cr &=f\bigl(x(t),y(t)\bigr)\>\dot x(t)+g\bigl(x(t),y(t)\bigr)\>\dot y(t)\equiv0\ .\cr}$$ Therefore we can conclude with the following: The $1$-form $d\phi$, resp., the gradient $\nabla\phi$ is by no means identically zero in $\Omega$, but the pullback of $d\phi$ along the solution curves vanishes. Geometrically this says that the tangent vector to the solution curve through a point $(x_0,y_0)\in\Omega$ is in the kernel of the form $d\phi(x_0,y_0)$.

Answer 2

I've never seen this notation before, but based on what I know about exact equations, it looks like it means the sum of the two partials: $$d\phi(t,x) = \frac{\partial\phi}{\partial t} + \frac{\partial\phi}{\partial x}\frac{dx}{dt}.$$