Solving a PDE in $1$D

Question

I want to solve the below equation

$\partial_t f(x,t)=D \partial_x^2 f(x,t)$ with initial and boundary conditions:

$f(x,0)=\delta(x-x_0)$

$D\partial_x f(x,t)=0$ at $x=L$ and $x=0$;

Can anyone help me with solving it?

Answer 1

Let $f(x,t) = X(x)T(t)$ then we have (assuming $D\ne 0$)

$$ \frac{T'(t)}{DT(t)} = \frac{X''(x) + \frac{1}{D}X'(x)}{X(x)} = -\lambda $$

where $X(x)$ satisfies the BVP

\begin{cases} X'' + \frac{1}{D}X' + \lambda X = 0 \\ X'(0) + \frac{1}{D}X(0) = 0 \\ X'(L) + \frac{1}{D}X(L) = 0 \end{cases}

The characteristic equation is

$$ r^2 + \frac{1}{D}r + \lambda = \left(r+\frac{1}{2D}\right)^2 + \left(\lambda - \frac{1}{4D^2}\right) = 0 $$

Let $\lambda - \frac{1}{4D^2} = \mu^2$, then the solution is

$$ X(x) = e^{-x/2D}\left[A\cos(\mu x) + B\sin(\mu x) \right] $$

Matching up the boundary conditions gives $A=-2D\mu B$ and $\sin(\mu L) = 0$, so we have

$$ X_n(x) = e^{-x/2D}\left[\frac{n\pi}{L} \cos\left(\frac{n\pi}{L}x\right) - \frac{1}{2D}\sin\left(\frac{n\pi}{L}x\right) \right] $$

Along with

$$ T_n(t) = \exp\left(-\left(\frac{n^2\pi^2}{L^2}+\frac{1}{4D^2}\right)Dt\right) $$

Leading to the series solution $u(x,t) = \sum\limits_{n=1}^\infty c_n X_n(x)T_n(t)$. The initial condition requires

$$ u(x,0) = \sum_{n=1}^\infty c_nX_n(x) = \delta(x-x_0) $$

where the constants $c_n$ are given by orthogonality

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Edit 3/1: Proof for orthogonality

Here I will show that the eigenfunctions in $X(x)$ are orthogonal with respect to a weighting function. Note that the boundary conditions are equivalent to

$$ X' + \frac{1}{D}X = 0 \implies (e^{x/D}X)' = 0, \quad x = \{0,L\} $$

Let $Y(x) = e^{x/D}X(x)$, then the BVP is reduced to

$$ Y_n'' - \frac{1}{D}Y_n' = -\lambda Y_n, \quad Y_n'(0) = Y_n'(L) = 0$$

or

$$ (e^{-x/D}Y_n')' = -\lambda e^{-x/D}Y_n $$

Following the method similar to this answer we can show that

$$ \int_0^L (e^{-x/D}Y_n')'Y_m dx - \int_0^L (e^{-x/D}Y_m')'Y_n dx = 0 $$

$$ \implies \int_0^L e^{-x/D}Y_nY_m dx = 0 $$

$$ \implies \int_0^L e^{x/D}X_nX_m dx = 0 $$

Now we solve for the constants by

\begin{align} \int_0^L e^{x/D}\delta(x-x_0) X_m(x)\ dx &= \sum_{n=1}^\infty c_n \int_0^L e^{x/D} X_n(x)X_m(x)dx \\ \implies e^{x_0/D}X_m(x_0) &= c_m \int_0^L e^{x/D}X_m^2(x) dx \end{align}

Plugging in the expression for $X(x)$ gives

$$ c_m = \frac{\displaystyle e^{-x_0/D}\left[\frac{m\pi}{L} \cos\left(\frac{m\pi}{L}x_0\right) - \frac{1}{2D}\sin\left(\frac{m\pi}{L}x_0\right) \right]}{\displaystyle \int_0^L \left[\frac{m\pi}{L} \cos\left(\frac{m\pi}{L}x\right) - \frac{1}{2D}\sin\left(\frac{m\pi}{L}x\right) \right]^2dx} $$