How to solve following multidimensional equation? Let's have an equation
$$
u_{t} = \Delta u ,\quad \mathbf x є R^{n},\quad t > 0, \quad u(\mathbf r , 0) = e^{-(x_{1} + ... + x_{n})^{2}}.
$$
How to solve it? I tried to reduce the equation to the form
$$
u_{t} = \tilde {\Delta} u ,\quad \mathbf x є R^{n},\quad t > 0, \quad u(\mathbf r , 0) = \prod_{i = 1}^{n}f_{k}(x_{i}),
$$
which I solved, but I failed.
Maybe, there be a good way to find solution in form 
$$
u(\mathbf r, t) = g(t)e^{-(x_{1} + ... + x_{n})^{2}f(t)}, \quad f(0) = g(0) = 1?
$$
After substituting it to the equation I got
$$
\partial_{t}u = \left( \dot {g} - g\dot {f}(\sum_{i, j}x_{i}x_{j})\right) exp(...)
$$
for time derivation and
$$
(-2nfg + 4ngf^{2}(\sum_{i, j}x_{i}x_{j}))exp(...)
$$
for the Laplacian.
So, I have a system
$$
\dot {g}(t) = -2nf(t)g(t), \quad -g(t)\dot {f}(t) = 4ng(t)f^{2}(t), \quad f(0) = g(0) = 1, 
$$
and I'll get a solution fast.
Is this method correct? 
How can I solve this equation more "strictly"?
 A: You are solving the heat equation in a half-space. The initial condition has a peculiar form: it depends only on $x_1+\dots+x_n$. This suggests changing the system of coordinates to make $x_1+\dots+x_n$ a new variable. For example:
$$\begin{split}
y_1&=n^{-1/2}(x_1+\dots+x_n) \\
y_2&=2^{-1/2}(x_1-x_2) \\
y_3&=2^{-1/2}(x_1-x_3) \\
\dots&\dots \\
y_n&=2^{-1/2}(x_1-x_n) \\
\end{split}
\tag1$$
Since (1)  is an orthogonal transformation, it commutes with taking the Laplacian. In the new coordinate system you are solving
$$u_{t} = \Delta u ,\quad \mathbf y \in \mathbb R^{n},\quad t > 0, \quad u(\mathbf r , 0) = e^{-n^2y_1^2}. \tag2$$
The initial data in (2) is independent of $y_2,\dots,y_n$. Therefore, the solution will not depend on them either. (Indeed, translating the solution in direction of $y_j$, $j>1$, you get another solution of the same problem; by uniqueness theorem it must be the same function). 
So, you end up solving the one-dimensional problem
$$u_{t} = \Delta u ,\quad   y \in \mathbb R,\quad t > 0, \quad u(y , 0) = e^{-n^2y^2}. \tag3$$
which is easy because the Gaussian function becomes a more diffuse Gaussian function under the heat flow. Then return to the original coordinates. 
Or you can take a shortcut: obtain a solution using the ansatz $$u(\mathbf r, t) = g(t)e^{-(x_{1} + ... + x_{n})^{2}f(t)}, \quad f(0) = g(0) = 1$$ and appeal to the uniqueness theorem. 
(Uniqueness for the heat equation applies to solutions that do not grow too rapidly; there are other, "unphysical" solutions, but nobody ever wants to look for them.)
