I was reading about the Weierstrass transform $(W[\cdot])$, and how it's related to the difussion equation in one dimension. It's relation is given by that $W[f]$ is the convolution with the Heat kernel at $t=1$, and the general solution is given by the generalized Weierstrass transform $W_t[\cdot]$ as the exponential of $t\partial^2_x$. If someone applied $W_t[]$, it should satisfy the difussion equation according to the first source:
$$W_t[f](x)=\exp\left(t{\partial^2\over\partial x^2}\right)[f]$$
$$\Rightarrow {\partial\over\partial t}f= {\partial^2\over\partial x^2}f$$
$$\Rightarrow {\partial\over\partial t}W_t[f](x)= {\partial^2\over\partial x^2}W_t[f](x)$$
Given that $W_t[\cdot]$ evolve the initial function $f_0(x)=f(x,t)$ in time, and because of its exponential form, it seems to me that is just the Lagrange Shift operator acting in the time domain. My way of justify the property of $W_t[\cdot]$ as a time evolution operator would be the following.
- Write the shift operator acting on time by a displacement $\tau$, in exponential form.
- Replace $\partial_t$ by the relation given in the difussion equation, meaning $\partial_t=\partial^2_x$.
- Check if it commutes with every operator in the equation.
If we write the shift operator $T_\tau[\cdot]$:
$$T_\tau[f]=\exp\left(\tau{\partial\over\partial t}\right)=\exp\left(\tau{\partial^2\over\partial x^2}\right)$$
Then if $[\cdot,\cdot]$ its the commutator bracket, then $[T_\tau,\partial_t]=[T_\tau,\partial_x]=0$, because of symmetry of second deerivatives, and because polynomial of the same operator commute with each other. So its clear that $T_\tau$ satisfy the diffusion eq.
Now, I have to questions with this. The first is, given that $\tau$ is dummy variable and $t$ is a real variable of the problem, Is it my approach correct?
Second, Can we construct a solutions of any PDE of the form $\partial_t f=L[f]$, where $L$ is a linear operator only dependent of $x$, by applying $T_\tau[\cdot]$ to $f$, and setting $t=0$ to do a functional equation?
To give an example of how to solve this way, I'd solve the one-way wave equation:
$${\partial\over\partial t}f = \pm {\partial\over\partial x}f$$
Writing the time shift $T_\tau$:
$$T_\tau[f]=\exp\left(\tau{\partial\over\partial t}\right)=\exp\left(\pm\tau{\partial\over\partial x}\right)$$
Then applying to the equation:
$${\partial\over\partial t}\exp\left(\tau{\partial\over\partial t}\right)[f] = \pm {\partial\over\partial x}\exp\left(\pm\tau{\partial\over\partial x}\right)[f]$$
$${\partial\over\partial t}f(x,t+\tau)=\pm{\partial\over\partial x}f(x\pm\tau,t) $$
Setting $t=0$, then we have:
$$f(x,\tau)=f(x\pm\tau,0)=f_0(x\pm\tau) $$
Then replacing $\tau$ for $t$ gives the correct solution. This also works for $\partial_t f=x\partial_x f$, to give anothe example.