How can one prove that the function

$F: \mathbb{R}^n \times \mathbb{R}^+ \text{\ {0}} \to \mathbb{R}$ with

$$F(x,t) := t^{-n/2} \exp(-\frac{\Vert x \Vert_2^2}{4t})$$

is a solution of the partial differential equation

$$\Delta F - \frac{\partial F}{\partial t} = 0$$

I know that this can be done using the Laplace operator, which is given by $\Delta := \sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}$, but I don't know what the partial derivative of $\Delta F - \frac{\partial F}{\partial t} = 0$ looks like.

  • $\begingroup$ Try with $n=2$, $F(x_1,x_2,t) = t^{-1} \exp(-\frac{x_1^2+x_2^2}{4t})$. What are $\partial_{x_i}F$ and $\partial_{x_i}^2F$ and $\partial_t F$ and $\Delta F$ ? $\endgroup$
    – reuns
    Nov 23 '18 at 1:33

It's very long, so let me help you with some expressions

$$\Vert x \Vert _2^{2} = x_1^2 + x_2^2 + \cdots + x_n^2, $$

$$ \dfrac{\partial}{\partial x_i}(x_1^2 + x_2^2 + \cdots + x_n^2) = 2x_i, $$

$$\dfrac{\partial}{\partial x_i}\, \exp\! \Big(-\dfrac{\Vert x \Vert_2^2}{4t} \Big) = \exp\! \Big(-\dfrac{\Vert x \Vert_2^2}{4t} \Big) \cdot \dfrac{-1}{4t} \cdot 2x_i, $$

$$\dfrac{\partial^2}{\partial x_i^2}\, \exp\! \Big(-\dfrac{\Vert x \Vert_2^2}{4t} \Big) = \dfrac{-1}{4t} \Bigg[ \exp\! \Big(-\dfrac{\Vert x \Vert_2^2}{4t} \Big) \cdot \dfrac{-1}{4t} \cdot 2x_i \cdot 2 x_i + \exp\! \Big(-\dfrac{\Vert x \Vert_2^2}{4t} \Big) \cdot 2\Bigg] $$

$$\dfrac{\partial}{\partial t} \Big( t^{-n/2} \exp(-\frac{\Vert x \Vert_2^2}{4t}) \Big) = \dfrac{-n\ t^{n/2-1}}{2} \ \exp \!\Big(-\dfrac{\Vert x \Vert_2^2}{4t} \Big) + t^{-n/2} \exp\! \Big(-\frac{\Vert x \Vert_2^2}{4t} \Big) \cdot \dfrac{\Vert x \Vert_2^2}{4t^2}.$$

Can you continue with the calculations?


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