Suppose I have the following nonlinear ODE:
$$\frac{dy}{dt}=F(y)$$ with $y(0) = z$. Suppose the solution of this nonlinear ODE is $y(t,z)$. Show the following:
$$\frac{\partial y(t,z)}{\partial t}\Bigr\rvert_{z \text{ fixed }} -F(z)\frac{\partial y(t,z)}{\partial z} \Bigr\rvert_{t \text{ fixed }} =0$$
We can consider the simplest case such as $F(y) = y$. And can solve $y(t) = ze^t$ and then plug into the partial differential equation. We have $F(z) = z$.
To formally prove this, I am not sure where to start with. I am also confused about the following,
$$\frac{\partial y(t,z)}{\partial t}\Bigr\rvert_{z \text{ fixed }} = F(z)??$$
If this is true, then $\frac{\partial y(t,z)}{\partial z} \Bigr\rvert_{t \text{ fixed }} = 1$, which does not make any sense obviously.
The following is my new work on this:
$$\int_z^y \frac{dy}{F(y)} = \int_{0}^t dt \Rightarrow G(y) - G(z) = t$$ where $$G(y) = \int \frac{dy}{F(y)}, \text{ which is a indefinite integral}$$
Now we have $$G(y) = t + G(z) \Rightarrow y(t,z) = G^{-1}(t + G(z))$$ where $$G^{-1}(y) = ?$$
I am a bit confused the following:
- Does the above derivation make sense?
- What is $G^{-1}$ here?
- I am also not quite sure the following? $$\frac{\partial G^{-1}(t+G(z))}{\partial t} = \frac{\partial G^{-1}(t+G(z))}{\partial (t+G(z))}\frac{\partial (t+G(z)))}{\partial t} = ? $$