I'm trying to follow a video lecture on solving the heat equation.
$I) \space u_t = ku_{xx}, x \in \mathbb{R}, t > 0$
$II) \space u(x,0)=\phi (x), $
$k$ is const, $\phi (x) $ is a known function.
An earlier video in lecture series proved these properties:
If $u(x,t)$ is a solution to $I)$ then for each fixed $y$, the translate $u(x-y,t)$ is also a solution.
If $u$ is a solution then every derivative of $u$ is also a solution.
Every linear combination of solutions is also a solution.
If $S(x,t)$ is a solution then so is its integral. Thus $S(x-y,t)$ is a solution and so is $\int_{-\infty}^{\infty} S(x-y,t)g(y)dy$ for each function $g$ that makes the integral converge.
If $u(x,t)$ is a solution the for each $a>0$ the dilates function $u(\sqrt{a}x,at)$ is also a solution.
Using the the translate and integral properties, you can find this solution to $I)$ as $u(x,t)=\displaystyle \int_{-\infty}^{\infty} S(x-y,t)\phi(y)dy $
The first part of the video shows that $I)$ holds.
To show that the initial condition $II)$ holds, you begin by doing the following:
$S := \frac{\partial }{\partial x}Q$
$u(x,t)=\displaystyle \int_{-\infty}^{\infty} \frac{\partial }{\partial x} Q(x-y,t)\phi(y)dy $
$ \color{green}{\mathbb{\bf*}} \\ = \displaystyle \int_{-\infty}^{\infty} \frac{\partial }{\partial y}[-Q (x-y,t)]\phi(y)dy $
Using integration by parts:
$- \displaystyle \int_{-\infty}^{\infty} Q_y (x-y,t)\phi(y)dy $
$\begin{array}{ll} w &=\phi(y) & v_y = Q_y(x-y,t) \\ w_y &=\phi'(y) & v = -Q(x-y,t) \end{array}$
$=- \left[ -Q (x-y,t)\phi(y)\bigg |_{y \rightarrow -\infty}^{y \rightarrow \infty}- \displaystyle \int_{-\infty}^{\infty} -Q (x-y,t) \phi'(y)dy \right]$
$=Q (x-y,t)\phi(y)\bigg |_{y \rightarrow -\infty}^{y \rightarrow \infty}- \displaystyle \int_{-\infty}^{\infty} Q (x-y,t) \phi'(y)dy $
I don't understand what happens on the line labeled $\color{green}{\mathbb{\bf*}}$. Why can you switch from $Q_x$ to $-Q_y$?
The reasoning seems to be that, letting $z = x-y$, we can use the chain rule to say that $$\frac{\partial Q}{\partial y} = \frac{\partial Q}{\partial z} \frac{\partial z}{\partial y} = \frac{\partial Q}{\partial z} \left(\frac{\partial}{\partial y}\left(x - y \right)\right) = - \frac{\partial Q}{\partial z} \Longrightarrow \frac{\partial Q}{\partial y} = - \frac{\partial Q}{\partial z}$$ Similarly we can see that $$\frac{\partial Q}{\partial x} = \frac{\partial Q}{\partial z} \frac{\partial z}{\partial x} = \frac{\partial Q}{\partial z} \left(\frac{\partial}{\partial x}\left(x - y \right)\right) = \frac{\partial Q}{\partial z} \Longrightarrow \frac{\partial Q}{\partial x} = \frac{\partial Q}{\partial z} $$
We can combine the results of these two equations to deduce that $$ \frac{\partial Q}{\partial x} = - \frac{\partial Q}{\partial y}$$