PX275 - H10 - FT and PDEs

the wave equation

for a multivariable function, $u (x, t)$ , FT can be done with respect to $x$ , $t$ or both
considering the wave equation:

\frac{\partial^{2} u}{\partial x^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} u}{\partial t^{2}}

\begin{aligned} \tilde{u} (k) & = \int_{- \infty}^{\infty} e^{i k x} u (x) d x \\ \tilde{u} (ω) & = \int_{- \infty}^{\infty} e^{i ω t} u (t) d t \end{aligned}

these can be combined to form a 2D FT:

\tilde{u} (k, ω) = \tilde{u} (k) = \iint_{- \infty}^{\infty} e^{i (k x + ω t)} u (x, t) d x d t

using the formula for the FT of the $n^{t h}$ derivative:

\begin{matrix} F (\frac{\partial^{n} f}{\partial x^{n}}) = (i k)^{n} \tilde{f} (k) \\ F (\frac{\partial^{n} f}{\partial t^{n}}) = (i k)^{n} \tilde{f} (ω) \end{matrix}

taking the FT of both sides of the wave equation:

(i k)^{2} \tilde{u} (k, ω) = \frac{1}{c^{2}} (i ω)^{2} \tilde{u} (k, ω) ⟹ (\frac{ω^{2}}{c^{2}} - k^{2}) \tilde{u} (k, ω) = 0

this yields the dispersion relation:

ω^{2} = k^{2} c^{2}

the diffusion relation

\frac{\partial u}{\partial t} = D \frac{\partial^{2} u}{\partial x^{2}}

taking the FT of both sides:

\begin{matrix} \int_{- \infty}^{\infty} e^{- i k x} \frac{\partial u}{\partial t} d x = D \int_{- \infty}^{\infty} e^{- i k x} \frac{\partial^{2} u}{\partial x^{2}} d x \\ \frac{\partial}{\partial t} \int_{- \infty}^{\infty} e^{- i k x} u d x = - D k^{2} \tilde{u} \\ \frac{\partial \tilde{u}}{\partial t} = - D k^{2} \tilde{u} \end{matrix}

this turned it into a first order PDE
the solutions for $\tilde{u}$ will have the form:

\tilde{u} (k, t) = \tilde{A} (k) e^{- D k^{2} t}

where, $\tilde{A} (k)$ is an undetermined function of only $k$

to find the solutions for $u$ , taking the inverse FT:

u = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{i k x} \tilde{A} (k) e^{- D k^{2} t}, d x

considering a special case of $\tilde{A} (k) = 1 :$

u = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{i k x} e^{- D k^{2} t} d x

this is the FT of a gaussian:

\begin{matrix} f (x) = \sqrt{\frac{α}{π}} e^{- α x^{2}} \\ \tilde{f} (k) = e^{- k^{2} / 4 α} \end{matrix}

∴ u (x, t) = \frac{1}{\sqrt{4 π D t}} \exp (\frac{- x^{2}}{4 D t})

for the general case, including $\tilde{A} (k)$ , this is a product of two functions, $\tilde{A} (k)$ and $\tilde{G} (k, t) = \exp (- D k^{2} t)$

u = F^{- 1} (\tilde{A} (k) \tilde{G} (k, t)) = A * G (x, t)

so, it is a convolution between $A (x)$ and $G (x, t)$
since $\tilde{G} = \exp (- D k^{2} t) :$

G = \frac{1}{\sqrt{4 π D t}} \exp (- \frac{x^{2}}{4 D t})

therefore, the solution:

u = A * G = \int_{- \infty}^{\infty} A (y) G (x - y, t) d y

considering an initial condition: $u (x, t = 0)$
recalling the definition of a $δ$ function: $$\lim_{\text{width}\to 0} G(x-y) = \delta(x-y)$$
it can be considered a 'local injection' at $x = y$
if $G$ is a $δ$ -function at $t = 0 :$

u (x, 0) = \int_{- \infty}^{\infty} A (y) δ (x - y) d y = A (x)

so, it can be seen how the initial condition settles $A (x)$ , and this $\tilde{A} (k)$
$\tilde{A} (k) = 1$ is the case where $u (x, 0) = δ (x)$
for $t > 0 :$

u (x, t) = \int_{- \infty}^{\infty} A (y) G (x - y, t) d y = \int_{- \infty}^{\infty} u (y, 0) G (x - y, t) d y

as seen before, convolutions with gaussians broaden and smear them
similarly, the initial condition is broadened over time as the particles diffuse