# $u_t = t\Delta u$, solve with streching

Solve heat equation with time coefficient $$u_t = t\Delta u, u(x,0) = \delta(x)$$

The question asks using scaling and self-similarity argument, I know how to do the usual heat equation by sending $$x\to\lambda x$$ and $$t\to\lambda^2t$$. But for this one, I don't know how to deal with this coefficient $$t$$ before the Laplacian. Any ideas?

1. Equate dimensions. Dividing by $$t$$ gives $$\frac{1}{t}\frac{\partial u}{\partial t}=\sum_i\frac{\partial^2u}{\partial x_i^2}$$, or in terms of dimensions, $$\frac{U}{T^2}\sim \frac{U}{X^2}$$. This means $$t$$ and $$x$$ must have the same dimensions, so they scale with the same power, i.e. $$(x,t)\to(\lambda x,\lambda t)$$.
2. Find the independent variable scaling the direct way: let $$u(x,t)=v(\lambda x,\lambda^k t)=:v(y,s)$$. Then $$u_t=\lambda^kv_s$$ and $$\Delta_xu=\lambda^2\Delta_yv$$, so the heat equation becomes $$\lambda^kv_s=\lambda^2t\Delta_yv$$. Since $$t=s/\lambda^k$$, it reduces to $$v_s=\lambda^{2-2k}s\Delta_yv$$. Choosing $$k=1$$ gives the desired scaling invariance.
3. Reparametrize $$t$$. Let $$u(x,t)=v(x,f(t))$$. Then $$u_t=v_s\cdot f'(t)$$, so if you choose $$f(t)=t^2/2$$, the initial value problem becomes $$v_{s}=\Delta v$$, subject to $$v(x,0)=\delta(x)$$.