Trying to prove, that $\frac{\partial f}{\partial t} (x,t) - \sum_{i=0}^{n} {\frac{\partial^2 f}{\partial^2 x_{i}}}(x,t) = 0$

Question

I currently have a little bit of a problem, that i figured out a while ago. I have the function:

$$f (x,t) =t^{-\frac{n}{2}} e^{\frac{\sum_{i=0}^{n}{x_{i}^2}}{4t}}$$

And i am trying to prove that this function solves the equation.

$$\frac{\partial f}{\partial t} (x,t) - \sum_{i=0}^{n} {\frac{\partial^2 f}{\partial^2 x_{i}}}(x,t) = 0$$

I have seen the exact same question over on this website, though after the derivatives were mentioned, i got no idea on how to show that $\displaystyle\frac{\partial f}{\partial t} (x,t) = \sum_{i=0}^{n} {\frac{\partial^2 f}{\partial^2 x_{i}}}(x,t)$, simply put, i do not know what to do with the Laplace Operator to which i can get to the solution of this problem.

Thanks for the help in advance.

Edit: I know this is pretty much the same question as was posed about 2 years ago, just me not knowing what the Laplace Operator does makes it kind of hard to continue my calculations.

Answer 1

I have edited this post after seeing in another answer that I had a big mistake.

Here there is an approach using the Fourier transform that may help you. Consider the heat equation $$u_{t}= \Delta u \hspace{0.9 cm} (x,t) \in \mathbb{R}^n \times \mathbb{R}^{+}$$ Taking the Fourier transform of the above expression you get $$\frac{\partial \hat{u}}{\partial t}(\zeta,t) + 4\pi^{2}|\zeta|^{2}\hat{u}(\zeta,t)=0$$ This is an ODE that you can solve to obtain \begin{eqnarray*} \hat{u}(\zeta,t)&=&Ce^{-4\pi^{2}|\zeta|^{2}t} \end{eqnarray*} with $C$ constant. Taking the inverse Fourier trasnform you finally get $$u(x,t) = C(4\pi t)^{-n/2} e^{\frac{-|x|^2}{4t}}$$ Now, if you take $f(x,t)= t^{-n/2} e^{-\frac{|x|^2}{4t}}$ (here I use $|x|= \sqrt{\sum_{i=0}^{n} x_{i}^{2}}$) using the chain and product rule you might realize that: $$ \frac{\partial f}{\partial t}(x,t) = e^{\frac{-|x|^2}{4t}}\left( -\left(\frac{n}{2} \right)t^{-\frac{n}{2}-1} + \frac{t^{-\frac{n}{2}-2}}{4} |x|^2 \right)$$ and $$ \frac{\partial f}{\partial x_i}(x,t) = -e^{\frac{-|x|^2}{4t}} \left( \frac{t^{-\frac{n}{2}-1}}{2} x_i\right) $$ $$ \frac{\partial^2 f}{\partial x_i^2}(x,t) = e^{\frac{-|x|^2}{4t}} \left( -\frac{ t^{-\frac{n}{2}-1}}{2}+ \frac{t^{-\frac{n}{2}-2}}{4}x_{i}^2\right)$$ and then, summing up you obtain: $$\Delta f(x,t) = \frac{\partial f}{\partial t}(x,t)$$

Answer 2

The claimed equality is wrong for your function $f$. It may be that you actually wanted the function $$f(x,t):=t^{-n/2}\exp\left(-{\sum_{i=1}^n x_i^2\over 4t}\right)\ .$$ The space dimension then is $n$, and $f$ decreases to $0$ when $|x|\to\infty$. This $f$ in fact depends only on $|x|$, so that we have $$f(x,t)=g\bigl(|x|,t\bigr),\qquad g(r,t)=t^{-n/2}\exp\left(-{r^2\over4t}\right)\ .\tag{1}$$ In this rotationally symmetric case the Laplace operator can be written in the form $$\Delta f(r,t)={\partial^2 g\over\partial r^2}+(n-1){\partial g\over\partial r}\ .$$ Doing this computation for the $g$ in $(1)$ one obtains $$\Delta f(r,t)={1\over4}\exp\left(-{r^2\over4t}\right)t^{-2-n/2}(r^2-2nt)\ .$$ From $(1)$ one also obtains $${\partial f\over\partial t}(x,t)={\partial g\over\partial t}(r,t)={1\over4}\exp\left(-{r^2\over4t}\right)t^{-2-n/2}(r^2-2nt)\ ,$$ so that we indeed have $${\partial f\over\partial t}(x,t)=\Delta f(x,t)\ .$$