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Given the following PDE: $u_t(x,t)=u_{xx}(x,t)$, where the subindices are partial differentiation. Using the Fourier transform $(\mathcal{F})$ for the spatial frequency domain $(\omega)$, in the eq. gives:

$$\mathcal{F}\{u_t(x,t)\} = \mathcal{F}\{u_{xx}(x,t)\}$$ $$\Rightarrow \hat{u_t}(\omega,t)=-\omega^2\hat u(\omega,t) $$ $$\Rightarrow{d\over dt}\hat u=-\omega^2\hat u $$

and then solving the last ODE...

$$\hat u(\omega,t)=e^{-\omega^2t}\ \hat u(\omega,0) $$

Using now the inverse Fourier transform $(\mathcal F^{-1})$...

$$\mathcal F^{-1}\{\hat u(\omega,t)\}=\mathcal F^{-1}\{e^{-\omega^2t}\ \hat u(\omega,0)\} $$ $$\Rightarrow u(x,t)=\mathcal{F}^{-1}\{e^{-\omega^2t}\}*u(x,0) $$ $$\Rightarrow u(x,t)={1\over 2\sqrt{\pi t}}e^{-x^2\over 4t}*u(x,0) $$

where the simbol $*$ indicates the convolution operation in the space domain. Now my question is how to solve for boundary conditions? If we solve for Neumann conditions with boundaries for a function $u:[a,b]\times \Bbb R^+\to\Bbb R$:

$$u_x(a,t)=u_x(b,t)=0$$

partial differentiate the convolution...

$${\partial u\over \partial x}={\partial\over \partial x}\left( {1\over 2\sqrt{\pi t}}e^{-x^2\over 4t}*u(x,0)\right) $$

$$\Rightarrow {1\over 2\sqrt{\pi t}}e^{-x^2\over 4t}*u_x(x,0)= {-x\over \sqrt{\pi t}}e^{-x^2\over 4t}*u(x,0) $$

but then I don't know how to proceed. I now that using Fourier series, for the Dirichlet boundary conditions, the solutios is writen as a series of sines because for the domain $[0,L]$, $\sin(0)=\sin(n\pi x/L)=0$; and for the Neumann boundary conditions, the solution is writen as a series of cosines because $(\cos(n\pi x/L))'=\sin(n\pi x/L)$ and then the same as above is true. But How can I relationate the solutions using Fourier series with the solution using the convolution with the gaussian?

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  • $\begingroup$ @Gonçalo very interesting. But this [Wikipedia's article][1], seems to only work for symmetrical initial functions. I was thinking of a general initial function [1]: en.wikipedia.org/wiki/Method_of_images $\endgroup$ Jun 24, 2023 at 22:28

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The usual Fourier transform is only defined for functions whose domain is the whole Euclidian space $\mathbb{R}^n$, so that method doesn't work when the domain of $u$ is $[a,b]$.

However, while the proof doesn't use Fourier transforms as far as I know, you can still write the solution as \begin{equation*} u(x,t) = {\int}_{\Omega} K(t,x,y)\ u(y,0)\ dy, \end{equation*} for a general domain $\Omega$. $K$ is called the heat kernel, and when $\Omega=\mathbb{R}$, it is indeed as you've written it : $$ K(t,x,y) = \frac{1}{2\sqrt{\pi t}} e^{-\frac{(x-y)^2}{4t}}. $$ As per Wikipedia,

On a more general domain Ω in $\mathbb{R}^d$, such an explicit formula is not generally possible. The next simplest cases of a disc or square involve, respectively, Bessel functions and Jacobi theta functions. Nevertheless, the heat kernel (for, say, the Dirichlet problem) still exists and is smooth for $t > 0$ on arbitrary domains and indeed on any Riemannian manifold with boundary, provided the boundary is sufficiently regular.

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Just as for the Laplace transform, boundary conditions can be added as terms in the Fourier transformed equation.

We consider $u$ to only live on the half-strip $[a,b] \times (0,\infty).$ Now let us extend $u$ to be defined for all $x\in(-\infty,\infty).$ Then, as a distribution, $$ u_{xx} - u_t = u_x(a) \delta_a - u_x(b) \delta_b + u(a) \delta_a' - u(b) \delta_b' . $$

Taking the Fourier transform ($x \to \xi$) gives $$ -\xi^2 \hat u - \hat u_t = u_x(a) e^{-i\xi a} - u_x(b) e^{-i\xi b} + u(a) (i\xi) e^{-i\xi a} - u(b) (i\xi) e^{-i\xi b} . $$ So you get an inhomogeneous first order ordinary differential equation to solve.

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  • $\begingroup$ From where comes the delta function? $\endgroup$ Jun 24, 2023 at 22:17
  • $\begingroup$ @DanielMuñoz. From the derivatives at the boundary of the strip. $\endgroup$
    – md2perpe
    Jun 25, 2023 at 5:24

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