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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"?

John Taylor
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1 Answers1

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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.)

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