I am struggling with a proposition I found in a paper. It is said that if $f$ satisfies :
$$f'(t) = -f(\alpha t), ~0<\alpha<1$$
Then $f$ can be seen as a function of $t$ and $\alpha$ and also satisfies the heat equation :
$$\frac{df}{d\alpha} = \frac{t^2}{2\alpha}\frac{d^2f}{dt^2}.$$
I know that $f$ can be seen as function of both $\alpha$ and $t$ but I can't get the heat equation from there. Any help would be appreciated. Many thanks !
In the paper it is defined in equation (2)
$$y(t,q)=\sum_{n=0}^\infty \frac{(-1)^n q^{\frac{n(n-1)}{2}}}{n!}t^n$$
From which you can easily see that $$\frac{\partial y}{\partial q}(t,q)=\sum_{n=0}^\infty \frac{n(n-1)}{2} \frac{(-1)^n q^{\frac{n(n-1)}{2}-1}}{n!}t^n=\frac{1}{2}\sum_{n=2}^\infty n(n-1)~\frac{(-1)^n q^{\frac{n(n-1)}{2}-1}}{n!}t^n$$
$$\frac{\partial^2 y}{\partial t^2}(t,q)=\sum_{n=0}^\infty n(n-1)~\frac{(-1)^n q^{\frac{n(n-1)}{2}}}{n!}t^{n-2}=\sum_{n=2}^\infty n(n-1)~\frac{(-1)^n q^{\frac{n(n-1)}{2}}}{n!}t^{n-2}$$ From which we get the "heat equation" (It's not actually the heat equation but anyway): $$2\frac{\partial y}{\partial q}(t,q)=\frac{t^2}{q}\frac{\partial^2 y}{\partial t^2}(t,q)$$