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?


Three ways:

  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 $v(x,0)=\delta(x)$.


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