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I'm using Fourier transformation to solve the pde $u_t-\Delta u +au=0$ on $\mathbb R\times (0,\infty)$ with boundary condition $u(x,0)=g(x)$ on $\mathbb R^n \times \{0\}$.

Fourier transformation gives an ode $\hat{u_t}+k^2 \hat u +a\hat u=0$ with initial condition $\hat u(k,0)= \hat g(k)$. The solution to the ode is $\hat u=e^{-(k^2+a)t}g(k)$. Now apply Inverse Fourier transformation I have $$u(x)=1/(2\pi)^{n/2}\int_{\mathbb R^n}e^{-(k^2+a)t}g(k)e^{ikx}dk$$

My question is how do I simplify this integral. In particular, how do I get rid of the $i$ in the exponential? Can anyone kindly help me with the calculation?

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1 Answer 1

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The key realization here is that $\mathcal F(u * v) = \hat u\cdot \hat v$ where $\mathcal F$ is the Fourier transform and $u *v$ is the (spatial) convolution of $u$ and $v$: $$(u*v)(x) = \int_{\mathbb R^n} u(x-y)v(y) dy.$$ Applying the inverse operator shows that $u* v = \mathcal F^{-1}(\hat u \cdot \hat v).$

You have $$\hat u(k,t) = e^{-(\lvert k \rvert^2 + a)t}\hat g(k).$$ Thus $$u(x,t) = \frac{e^{-at}}{(2\pi)^{n/2}} \int_{\mathbb R^n} e^{-\lvert k \rvert^2 t} \hat g(k) e^{ikx} dk = \frac{e^{-at}}{(2\pi)^{n/2}} \mathcal F^{-1}(e^{-\lvert k \rvert^2 t} \hat g(k)) = \frac{e^{-at}}{(2\pi)^{n/2}}(h(\cdot, t) * g)(x) $$ where $h(x,t)$ is the function such that $\hat h(k,t) = e^{-\lvert k \rvert^2t}$. However, it is well known that the Guassian is it's own Fourier/inverse Fourier transform (I can't be bothered to repeat the calculation here but this is the step you are missing and you can find the details easily online). Specifically (with the constants placed where they are), we have $$\mathcal F^{-1}(e^{-\lvert k \rvert^2 t}) = \frac{1}{(2t )^{n/2}} e^{-\lvert x \rvert^2/4t}$$ and so your solution is $$u(x,t) = \frac{e^{-at}}{(4\pi t)^{n/2}}\int_{\mathbb R^n} e^{-\lvert x -y \rvert^2/4t}. g(y) dy.$$ You may notice this is simply $e^{-at}$ times the fundamental solution to the heat equation. That's because $v(x,t) = e^{at} u(x,t)$ solves the heat equation with the same initial data.

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