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today I have a question in PDE. It concerns the heat equation:

  • Formulate the (homogeneous) heat equation for functions $f:(0,\infty)\times\mathbb{R^n} \longrightarrow\mathbb{C}$. Derive an equation for the Fourier transform with respect to the space variables $\hat{f}$ of $f$ that is equivalent to the heat equation (provided the function $x \mapsto f(t,x)$ belongs to the Schwartz space $S(\mathbb{R^n})$. Solve this equation. Find solutions (not only one) of the original heat equation by using the relation between the Fourier transform and convolution.

I transformed $\frac{\partial f}{\partial t}(t,x)=\Delta_xf(t,x)$ into $\frac{\partial \hat{f}}{\partial t}(t,\xi)=-|\xi|^2\hat{f}(t,\xi)$. Hence for fixed $\xi \in \mathbb{R}^n$, $\hat{f}(t,\xi)=C(\xi)e^{-t|\xi|^2}. $ Now I am stuck, I do not know how to continue. In fact this convolution formula states that if $\phi \in L^1(\mathbb{R^n})$ and $f\in L^p(\mathbb{R^n})$ with $p\in {1,2}$ then $\widehat{\phi*f}=(2\pi)^\frac{n}{2} \hat{\phi} \hat{f}$, where $\hat{f(y)}=\frac{1}{(2\pi)^{\frac{n}{2}}}\int_\mathbb{R^n} \hat{f(x)}e^{-i<x,y>}dy$. But I do not see, how to use this here.

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I will answer question 1) very quickly. I think you are overthinking the convolution formula; it really isn't difficult given what you have already done. Just note that

$$C(\xi) = \hat{f}(0,\xi)$$

the RHS being the FT of your initial condition:

$$\hat{f}(0,\xi) = \int_{\mathbb{R}^n} d^n x \: f(0,x) \, \exp{(i x \cdot \xi)} $$

All this means is that the solution you derived may be written as the inverse FT of $\hat{f}(0,\xi) \exp{(-|\xi|^2 t)}$...and you know the IFT of the gaussian term. Your solution will end up looking something like

$$\int_{\mathbb{R}^n} d^n x' \: f(0,x') \, \exp{\left(-\frac{|x-x'|^2}{4 t}\right )} $$

(I may not have the constants right in the exponential, but something like that.)

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