The analytical solution for advection-diffusion equation with source term.

Question

We have: $$\frac{\partial w}{\partial t} + a(x) \frac{\partial w}{\partial x} - v \frac{\partial^2 w}{\partial x^2} = f(t)$$ within a domain $x \in [0,1]$

Simplest Sample is $a(x) = 1$ (constant) and $v = 0.01$ (constant) and $f(t) = 1$

with the initial condition

$w(x,0) = 0$ and Boundary condition

$w(0,t)=w(1,t) = 0$

I am stuck in solving this Advection-Diffusion equation with a constant source term. Hopf–Cole transformation could be used to solve the one without the source term, if it has the source term how can I solve that.

Could anyone show the paper or the method how to solve it? Thanks very much.

Answer 1

Consider the special case you mention, constant $a$ and $v$ and $f=1$.In view of the condition $w(0,t)=w(1,t)=0$ you can make an expansion in a Fourier series $$w(x,t)=∑_{n}d_{n}(t)exp[i2πnx]=∑_{n}d_{n}(t)exp[ik_{n}x]$$. Then you can obtain the values of the differential operators acting on the basis functions and the equations for the time-dependent Fourier coefficients.

Answer 2

Let's consider the PDE $$ w_t + aw_x - vw_{xx} = 1, \tag{1} $$ where $a$ and $v$ are constants. To remove the source term, we decompose $w$ as $w(x,t)=\varphi(x,t)+\psi(x)$, where $\psi(x)$ satisfies the ODE $a\psi'-v\psi''=1$ and the boundary conditions $\psi(0)=\psi(1)=0$. The general solution to the ODE is $\psi(x)=\frac{x}{a}+c_1+c_2e^{ax/v}$. To determine the coefficients $c_1$ and $c_2$, we apply the boundary conditions, and finally obtain $$ \psi(x)=\frac{1}{a}\left(x-\frac{e^{ax/v}-1}{e^{a/v}-1}\right). \tag{2} $$ With this choice of $\psi$, one can easily verify that $\varphi$ must satisfy the homogeneous PDE $$ \varphi_t + a\varphi_x - v\varphi_{xx} = 0, \tag{3} $$ subject to the boundary conditions $\varphi(0,t)=\varphi(1,t)=0$ and the initial condition $\varphi(x,0)=-\psi(x)$.

We can solve $(3)$ using the method of separation of variables. Substituting $\varphi(x,t)=F(x)G(t)$ in $(3)$, we obtain $$ F(x)G'(t)+aF'(x)G(t)-vF''(x)G(t)=0 $$ $$ \implies\frac{G'(t)}{G(t)}=-\lambda=\frac{vF''(x)-aF'(x)}{F(x)}. \tag{4} $$ The solution to the ODE satisfied by $F$ is given by $$ F(x)=Ae^{k_+x}+Be^{k_-x}\qquad\left(k_{\pm}=\frac{a\pm\sqrt{a^2-4\lambda v}}{2v}\right). \tag{5} $$ The boundary conditions $F(0)=F(1)=0$ yield the pair of equations $A+B=0$ and $Ae^{k_+}+Be^{k_-}=0$, which has nontrivial solution iff \begin{align} e^{k_+}=e^{k_-}&\implies e^{k_+-k_-}=1 \implies k_+-k_- = 2n\pi i \\ &\implies \frac{\sqrt{a^2-4\lambda v}}{v} = 2n\pi i \\ &\implies \lambda = \frac{a^2}{4v}+n^2\pi^2v \qquad(n\in\mathbb{N}^{*}). \tag{6} \end{align} The eigenfunction $F_n$ corresponding to the eigenvalue $(6)$ is given by $$ F_n(x)=A_n\exp\left(\frac{ax}{2v}\right)\sin(n\pi x). \tag{7} $$ In addition, solving the ODE for $G$ --- see Eq. $(4)$ ---, we obtain $$ G_n(t)=C_ne^{-\lambda_nt}=C_n\exp\left\{-\left(\frac{a^2}{4v}+n^2\pi^2v\right)t \right\}. \tag{8} $$ Combining $(7)$ and $(8)$ we can write the general solution to the PDE $(3)$ as $$ \varphi(x,t)=\exp\left(\frac{ax}{2v}-\frac{a^2t}{4v}\right) \sum_{n=1}^{\infty}a_n e^{-n^2\pi^2vt}\sin(n\pi x). \tag{9} $$ The coefficients $a_n$ are determined by the initial condition $\varphi(x,0)=-\psi(x)$: $$ e^{\frac{ax}{2v}}\sum_{n=1}^{\infty}a_n \sin(n\pi x)=-\psi(x) \implies a_n=-2\int_0^1\psi(x)e^{-\frac{ax}{2v}}\sin(n\pi x)\,dx. \tag{10} $$ Therefore, the solution to the PDE $(1)$ can be written as $w(x,t)=\varphi(x,t)+\psi(x)$, where $\psi(x)$ is given by $(2)$ and $\varphi(x,t)$ is given by $(9)$ and $(10)$.