$(1)$ A differential eq. (e.g. the heat eq. $u_t=\gamma u_{xx}$) admits $\frac{\partial}{\partial x}$, translations along $x$.
$(2)$ It then follows that the equation has solutions $\varphi(x,t)$ and $\psi(x,t)$ such, that $\varphi(x,t)=\psi(x-a,t)$, $a=$const
I tried reading Olver's 'Intro to PDEs' (see pages 305, 306) but still cannot derive $(2).$
How do we start at $(1)$ and arrive at $(2)$? Could someone derive $(2)$?
(at the intro level, please)
Here is a paragraph from the book, added as per your request:
The heat equation serves as an excellent testing ground for the general methodology, since it admits a rich variety of symmetry transformations that take solutions to solutions. The simplest are the translations. Moving the space and time coordinates by a fixed amount, $t \rightarrow t+a, x\rightarrow x+b $ where a, b are constants, changes the function u(t, x) into the translated function U(t, x) = u(t − a, x − b). A simple application of the chain rule proves that the partial derivatives of U with respect to t and x agree with the corresponding partial derivatives of u, so $U_t=u_t, U_x=u_x, U_{xx}=u_{xx}$ and so on. In particular, the function U(t, x) is a solution to the heat equation $U_t = \gamma U_{xx}$ whenever u(t, x) also solves $u_t = \gamma u_{xx}$. Physically, translation symmetry formalizes the property that the heat equation models a homogeneous medium, and hence the solution does not depend on the choice of reference point or origin of our coordinate system. As a consequence, each solution to the heat equation will produce an infinite family of translated solutions.
What the book is saying is different from what you're saying. The book is saying that if $\phi(x,t)$ is a solution, then $\phi(x-a,t)$ is also a solution. Your claim, if interpreted literally, is a special case of the book's claim, but I think you're probably meant to write something else and got confused because that claim would be weirdly specific.
The heat equation $u_{t}-\gamma u_{xx}=0$ admit $\frac{\partial}{\partial x}$ as a symmetry mean that, intuitively, the law of physics doesn't prioritize any position in space. Not surprisingly, the law of physics also doesn't prioritize your position in time and your choice of unit for heat, so this leads to time translation symmetry and scaling symmetry as well (scaling symmetry is part of what make this equation linear).
But let's talk about space translation symmetry. Mathematically, it mean that when you apply $\frac{\partial}{\partial x}$ to a small translation of the equation itself, the rate of change to the form of the equation is $0$. To avoid confusion, let's denote translation using variable $h$.
A translation of $u$ by $h$ is $u(x-h,t)$. Then $\frac{\partial}{\partial h}(u(x-h,t)_{t}-\gamma u(x-h,t)_{xx})=\frac{\partial}{\partial h}u(x-h,t)_{t}-\frac{\partial\gamma}{\partial h}u(x-h,t)_{xx}-\gamma\frac{\partial}{\partial h}u(x-h,t)_{xx}=\frac{\partial}{\partial x}u(x-h,t)_{t}\frac{\partial(x-h)}{\partial h}-\gamma\frac{\partial}{\partial x}u(x-h,t)_{xx}\frac{\partial(x-h)}{\partial h}=\frac{\partial}{\partial x}(u(x-h,t)_{t}-\gamma u(x-h,t)_{xx})\frac{\partial(x-h)}{\partial h}=\frac{\partial}{\partial x}(0)(-1)=0$.
Note that this is true for all $u$, whether it's a solution or not.
So for any translation amount $a$ and solution $\phi$ let $\psi(x,t)=\phi(x-a,t)$ then $\psi_{t}-\gamma\psi_{xx}=(\phi_{t}-\gamma\phi_{xx})+\int_{h=0}^{a}\frac{\partial}{\partial h}(\phi(x-h,t)_{t}-\gamma\phi(x-h,t)_{xx})dh=0+0=0$, so $\psi$ is also a solution.
Basically, what this is saying is that if you move the solution curve along the horizontal axis it's always a solution curve. Mathematically, if $\phi$ is a solution then $\phi(x-a,t)$ is also a solution.