I've carefully studied many resources (such as 1, 2, 3, 4 and etc.) where they discuss different forms of diffusion equations. However, my specific equation was not studied in there. I'm hoping to get help on solving a diffusion equation with spatially variable coefficient and source, as follows (1D):
$$\frac{\partial c}{\partial t}=\frac{\partial}{\partial x}J=\frac{\partial}{\partial x}(D\frac{\partial c}{\partial x} +E)$$ $$\sigma(x,0)=\sigma_0 \\ J(-L/2,t)=J_{-} \\ J(+L/2,t)=J_{+} $$
where $-L/2\leq x \leq L/2$ and, $D(x)$ and $E(x)$ are only spatially variable.
Separation of variables and Laplace's transforms are both welcome. Approximate form of solution with series analysis are highly appreciated.
More details as requested:
The original equation to solve is as follows ($-L/2\leq x \leq L/2$):
$$\frac{\partial{\sigma}}{\partial t}=\frac{\partial}{\partial x}J=B\frac{\partial}{\partial x}(-\frac{D}{kT}eZ\frac{\partial{V}}{\partial x}+\frac{D}{kT}\frac{Q}{T}\frac{\partial{T}}{\partial x}-\frac{D}{kT}\Omega\frac{\partial{\sigma}}{\partial x})$$ $$\sigma(x,0)=\sigma_0 \\ J(-L/2,t)=J_{-} \\ J(+L/2,t)=J_{+} $$
where
- Diffusivity: $D(x)=D_0e^{-\frac{E_a}{kT}}$ ☹️
- Voltage: $V(x)=-\rho j x$
- Temperature: $T(x)=T_0+T_m \big( 1-\cosh(\frac{L}{2\Gamma})\cosh(\frac{x}{\Gamma}) \big) +T_n \big( \sinh(\frac{L}{2\Gamma})\sinh(\frac{x}{\Gamma}) \big)$
- Mean hydro-static stress: $\sigma(x,t)$:
and constants are
- $B=3e10[Pa]$: Bulk modulus
- $k=1.38e-23[J/K]$: Boltzmann constant
- $e=1.6e-19[C]$: Electron charge
- $Z=1$: Charge number
- $Q=1.6e-19[J]$: Specific heat transfer
- $\Omega=1.66e-29[m^3]$: Atomic volume
- $D_0=7.56e-5[m^2/s]$ : Diffusion prefactor
- $E_a=1.6e-19[J]$ : Activation energy
- $\rho=1.67e-8[ohm.m]$ : Resistivity
- $T_0=\frac{T_-+T_+}{2}$ : Approx external temperature
- $T_n=\frac{T_--T_+}{2}$ : Ends Temperature diff
- $T_m=\frac{j^2\rho \Gamma^2}{k_{th}}$ : Max Temperature rise
- $\Gamma =6e-6[m] $ : Heating characteristic length
- $k_{th}=400[W/m.K]$ : Thermal conductance
But really, you only need to know the following inputs (roughly center ranges added):
- $L\sim 100[um]$ : Wire length
- $j\sim 5e10[A/m^2]$ : Current density
- $T_-=T(x=-L/2)\sim373[K]$ : Fixed temperature at $-L/2$ end
- $T_+=T(x=+L/2)\sim373[K]$ : ixed temperature at $+L/2$ end
- $\sigma(x,0) = \sigma_0$
The original equation can be seen as:
$$\sigma_t=(C_1\frac{D}{T}+C_2\frac{D}{T^2}T_x+C_3\frac{D}{T}\sigma_x)_x$$
By further encapsulation, it can be seen as follows (similar to what mentioned in the question):
$$\sigma_t=(F\sigma_x)_x+E_x$$
I would suggest the introduction of suitable functions $u$ and $v$ such that $c = u + v$ and $u$ satisfies the non-homogeneous boundary conditions
\begin{align*} x = -L/2: & \quad u_x = (J_o - E_o)/D_o, \\ x = L/2: & \quad u_x = (J_L - E_L)/D_L, \end{align*} where the subscripts for $D$, $J$ and $E$ indicate evaluation at the corresponding endpoint. The easiest function $u$ is $u(x) = \frac{1}{2} A ( x - B)^2$, where the constants $A$ and $B$ may be determined from the two conditions above.
Now, notice that the problem for $v$ is homogeneous in the spatial directions and given by
$$ v_t = ( D v_x)_x + Q(x), \quad -L/2 < x < L/2, $$
with BCs $v_x(-L/2,t) = v_x(L/2,t) = 0$ and IC $v(x,0) = c(x,0) - u(x)$. The source term is $Q(x) = (D u_x)_x + E_x$, which is also known.
Thanks to superposition, your problem has now homogeneous boundary conditions but contains a nonzero, in general, source term. You may find now nontrivial solutions for the problem with $Q = 0$ (eigenfunctions), and expand the solution to the problem in terms of these functions (basis).
Can you take it from here?
Follow up:
Let's consider for a moment the problem $w_t = (D w_x)_x$ with homogenous boundary conditions $w_x(-L/2,t) = w_x(L/2,t) = 0$. I don't care too much about the initial conditions for now. Since $D = D(x)$, the problem is separable and admits a combination of solutions of the form $w = X(x)T(t)$, where $X$ and $T$ are nonzero. Introduction of the ansatz yields
$$ X T' = T \, (D X')' \implies \frac{T'}{T} = \frac{1}{X} (D X')' = \lambda, $$ where the primes denote differentiation with respect the corresponding independent variable and $\lambda$ is a real (negative, positive or zero) constant. Now we need some information about $D$, since the shape of the final solution will strongly depend on the definition of $D$. Is it positive everywhere in the domain? Is it linear/quadratic/cubic...?
Note that the problem in $x$ is
$$ \frac{\mathrm{d}}{\mathrm{d}x} \left( D \frac{\mathrm{d}X}{\mathrm{d }x} \right) - \lambda X = 0, \quad -L/2 < x < L/2, \quad X'(-L/2) = X'(L/2) = 0, $$ where $\lambda$ is a priori unknown (it has to be chosen appropriately as an eigenvalue). This problem may not have an easy to obtain closed-form solution. For $D = 1$ (or any other positive constant) the solution take form of cosines and sines but for slightly more complicated choices of $D$ you arrive at Bessel, Airy, Legendre functions, etc.
If you give me a hint on what kind of $D$s you're trying to solve the problem for, I can elaborate on finding a basis for your eigenfunctions on your domain and expand $v$ in terms of this basis (the $w$ functions).
Hope this helps.