We are seeking a general solution to the heat equation in the form $$u(t,x)=t^{-\alpha}v\left(\frac{\|x\|^2}{t}\right), x \in \mathbb{R}^n,\alpha>0,t>0$$ for an appropriate function $v:(0,+\infty) \rightarrow \mathbb{R}.$
- Plug into $u(t,x)$ into the heat equation and rearrange it so that it depends only on $\frac{\|x\|^2}{t}$.
- Substitute $\frac{\|x\|^2}{t}$ with new variable $s$ and multiply the new equation with the integration factor of $s^{\beta}$, choose $\alpha,\beta$ so that the new equation is an equation of derivatives.
I have calculated the derivatives using chain rule for $v(f(t,x)),\;f(t,x)=\frac{\|x\|^2}{t}$ $$u_t(t,x)=-\alpha t^{-1-\alpha}v\left(\frac{\|x\|^2}{t}\right)-t^{-\alpha - 2}v_t\left(\frac{\|x\|^2}{t}\right)$$ $$\frac{\partial u}{\partial x_i}=t^{-\alpha}\frac{\partial v}{\partial f}(\frac{\partial f}{\partial x})=t^{-\alpha}\frac{2x_i}{t}\frac{\partial v}{\partial f}$$ $$\frac{\partial^2 u}{\partial x_i^2}(t,x)=2t^{-\alpha-1}\frac{\partial v}{\partial f} +4t^{-\alpha-2}x_i^2\frac{\partial^2 v}{\partial x_i^2}\left(\frac{\|x\|^2}{t}\right),$$ which yields $$\triangle_x u=t^{-\alpha}\sum_{i=1}^n(\frac{\partial^2 v}{\partial^2 f}\frac{4x_i^2}{t^2}+\frac{\partial v}{\partial f}\frac{2}{t})=2nt^{-\alpha-1}\frac{\partial v}{\partial v}+t^{-\alpha-2}4\|x\|^2\frac{\partial^2 v}{\partial^2 f}.$$
Plugging into the heat equation yields
$$u_t-\Delta u= -\alpha t^{-\alpha-1}v-t^{-\alpha-2}\frac{\partial v}{\partial f}-2nt^{-\alpha-1}\frac{\partial v}{\partial f}-4t^{-\alpha-2}\|x\|^2\frac{\partial^2 v}{\partial^2 f}=0.$$
We multiply by $t^{-\alpha-1}$ $$-\alpha v-\frac{\|x\|^2}{t}\frac{\partial v}{\partial f}-2n\frac{\partial v}{\partial f}-4\frac{\|x\|^2}{t}\frac{\partial^2 v}{\partial^2 f}=0,$$ satisfying 1).Substituting $s=\frac{\|x\|^2}{t}$ and multiplying the equation by integral factor $s^{\beta}$ yields $$-\alpha s^{\beta}v-s^{\beta+1}\frac{\partial v}{\partial s}-s^{\beta}2n\frac{\partial v}{\partial s}-4s^{\beta+1}\frac{\partial^2 v}{\partial^2 s}=0$$ I'm not sure how to proceed from here.