Is this solution correct?
The heat equation is
$$\frac{\partial u(x,t)}{\partial t} = \frac{x_0^2}{t_0} \frac{\partial^2 u(x,t)}{\partial x^2}, \mbox{where } \frac{x_0^2}{t_0}=\mbox{"gamma"}$$
Suppose $u(x,t)$ is the temperature, measured in Kelvins. For dimensions' (physical units) sake, we insert 2 dimensional constants $x_0, t_0$ in the usual place where the $\gamma$ coefficient is. $\gamma$ can be thought of as the ratio, $\gamma=x_0^2/t_0$.
The PDE has scaling symmetry. We can do the scaling
$$x^\prime = k^A x$$ $$t^\prime = k^B t$$ $$u^\prime=k^C u$$
Apparently the values of $A,B,C$ do not matter as the scaling happens on both sides of the PDE and the scaling coefficients $k^A,k^B,k^C$ cancel out, leaving the PDE scale-invariant in all cases.
Since we have $u^\prime=k^C u$, this means that the solution $u(x,t)$ is a homogeneous function as follows:
$u$ is homogeneous in $x,x_0$ of degree $A$
$u$ is homogeneous in $t,t_0$ of degree $B$
$A+B=C$
We must always keep in mind that $x$ and $t$ have different physical dimensions (units). To maintain this important distinction, do the following:
Define isobaric weights $W$ as
$$W(x^A)=A$$ $$W(t^B)=B$$ $$W(u^C)=C$$ $$W(x^At^B)=A+B$$
This means $u(x,t)$ must be an integral rational function;
- $u(x,t)$ must be isobaric of weight C.
Then it must be true that
$$x\frac{\partial u(x,t)}{\partial x}+x_0\frac{\partial u(x,t)}{\partial x_0}=Au(x,t)$$ $$t\frac{\partial u(x,t)}{\partial t}+t_0\frac{\partial u(x,t)}{\partial t_0}=Bu(x,t)$$
Note that the homogeneity in $x$ and $t$ is written as two separate equations; because the isobaric weights of $x$ and $t$ are different; and they are measured in different physical units.
Use some sym calc to find the solution, it has to be
$$\boxed{u(x,t)=t_0^B x_0^A F\left(\frac{x}{x_0},\frac{t}{t_0}\right)}$$
where $F$ is an arbitrary function. This is the general solution of the heat equation.
The problem is in the factor $t_0^B x_0^A$: it has wrong units, while $F$ is dimensionless in the units of $x,x_0$ and $t,t_0$. We can allow $F$ to have dimensions of "$u$", by multiplying it with a constant $u_0$.
The only way to correct this problem is to set $A=B=0$. Then $u$ is homogeneous in $x,x_0$ of degree 0 and also homogeneous in $t,t_0$ of degree 0. This means $u$ is unaffected by scaling of $x$ or $t$.
Therefore, $u$ can look like $u_0 \phi\left(\frac{x}{x_0}\right) \psi\left(\frac{t}{t_0}\right)$, or $u_0 \phi\left(\frac{x}{x_0}\right) +u_0\psi\left(\frac{t}{t_0}\right)$ etc. For instance, solutions of the above heat equation derived with a symbolic package are
$$u(x,t)=\left[ C_1 e^{\frac{x}{x_0}} +C_2 e^{-\frac{x}{x_0}} \right] C_3 e^{\frac{t}{t_0}}$$
This solution is homogeneous as described above. I have looked up other solutions of this same heat equation on this forum and they all seem to comply with the rule for the solution being homogeneous (albeit for some of these examples it's hard to track the dimensional constants and the units within exponents or sines etc.).
And finally, because the heat equation also admits time translations and translations along $x$, then the general solution must have the general form
$$\boxed{u(x,t)=u_0F\left(\frac{x+x_0}{x_0},\frac{t+t_0}{t_0}\right)}$$
Is the general solution correct? Is it true that the solutions of the heat equation are homogeneous functions?
If this claim is false, please kindly point to the place in the argument where wrong assumptions are made; or point to mistakes in the argument.
Perhaps someone familiar with the heat equation and PDEs in general could verify these results. Does this function represent any and all solutions of the heat equation. Please comment on the scaling invariance of $u$.