- #1
mhsd91
- 23
- 4
PROBLEM FORMULATION:
Considering the region [itex] \Omega [/itex] bounded as a square box within [itex] x \in [0,1], y \in [0,1] [/itex]. We wish to solve the 2D, stationary, advection-diffusion equation,
[itex]
0 = D\nabla^2 \rho(x,y) + \vec{V} \cdot \nabla \rho(x,y)
[/itex]
where [itex] D [/itex] is a scalar constant, and [itex] \vec{V} = V_1\hat{e}_x + V_2 \hat{e}_y [/itex] is a constant advection field vector. The problem has the following boundary conditions,
[itex]
\begin{align}
\nabla \rho(1,y) &= \nabla \rho(x,1) =& 0 \\
\rho(0,y) &= \rho(x,0) =& C = \textrm{known constant}
\end{align}
[/itex]
ATTEMPT AT SOLUTION: Method of Separation of Variables
Assuming [itex] \rho(x,y) = X(x) \cdot Y(y) [/itex], then
[itex]
\begin{align}
\nabla \rho &= Y X_{x} \hat{e}_x + X Y_{y} \hat{e}_y \\
\nabla^2 \rho &= Y X_{xx} + X Y_{yy}
\end{align}
[/itex]
where the subscript denotes partial derivation with respect to that particular variable, with the exception of [itex]\hat{e}[/itex] which represents the unit vector in either direction. Decomposing the advection field vector and inserting these results into the original problem, we get..
[itex]
0 = D (Y X_{xx} + X Y_{yy} ) + Y X_{x} v_{x} + X Y_{y} v_{y}
[/itex]
Multiply this eq. with [itex] 1/(XYD) [/itex] and moving all [itex] X [/itex] -expressions to the left hand side result in the separation of [itex] X [/itex] and [itex] Y [/itex] such that they both have to be equal some unknown constant [itex] \lambda [/itex] (we cannot have a small change in [itex] X [/itex], without the corrosponding change in [itex] Y [/itex] and vice versa). Thus,
\begin{align}
\lambda &= -\frac{X_{xx}}{X} - \frac{X_{x}}{X} \frac{V_{1}}{D} \\
\lambda &= +\frac{Y_{yy}}{Y} + \frac{Y_{y}}{Y} \frac{V_{2}}{D}
\end{align}
Which are two independent, linear, second order ODEs, with general solutions
[itex]
\begin{align}
X(x) &= C_1 \exp[-\alpha^{(+)}_{x} x] \\
&+ C_2 \exp[+\alpha^{(-)}_{x} x] \\
& \\
Y(y) &= C_3 \exp[-\alpha^{(+)}_{y} y] \\
&+ C_4 \exp[+\alpha^{(-)}_{y} y]
\end{align}
[/itex]
with
[itex]
\alpha^{(\pm)}_{m} = \frac{1}{2} \left( \sqrt{\frac{v_{m}^2}{D^2} +4\lambda} \pm \frac{v_{m}}{D}\right), \quad m=1 \vee 2
[/itex]
and [itex] C_i, i=1,2,3,4; [/itex] are constant coefficients. So far, so good! However, here is where my issues begin to pile up as I'm unable to sort out the [itex] C [/itex] -values with the boundary conditions.
Along the north and east boundaries, I'm able to write [itex] C_1 = (\textrm{some const expression}) \cdot C_2 [/itex] and similar for [itex] C_3, C_4 [/itex]. However, for the west and south bounds I end up with
[itex]
(C_1 + C_2)Y = C \\
(C_3 + C_4)X = C
[/itex]
Which will only be valid for [itex]C=0, C_1 = -C_2, C_3=-C_4[/itex]. This, evidently, results in the trivial solution [itex]\rho(x,y)=0[/itex], which obviously is not what we want ..
Any help is appreciated! I've also tried to solve it by integral transforms, but due to the stationarity (the problem-equation is homogenious), I fail as [itex] \rho [/itex] vanishes ..
Considering the region [itex] \Omega [/itex] bounded as a square box within [itex] x \in [0,1], y \in [0,1] [/itex]. We wish to solve the 2D, stationary, advection-diffusion equation,
[itex]
0 = D\nabla^2 \rho(x,y) + \vec{V} \cdot \nabla \rho(x,y)
[/itex]
where [itex] D [/itex] is a scalar constant, and [itex] \vec{V} = V_1\hat{e}_x + V_2 \hat{e}_y [/itex] is a constant advection field vector. The problem has the following boundary conditions,
[itex]
\begin{align}
\nabla \rho(1,y) &= \nabla \rho(x,1) =& 0 \\
\rho(0,y) &= \rho(x,0) =& C = \textrm{known constant}
\end{align}
[/itex]
ATTEMPT AT SOLUTION: Method of Separation of Variables
Assuming [itex] \rho(x,y) = X(x) \cdot Y(y) [/itex], then
[itex]
\begin{align}
\nabla \rho &= Y X_{x} \hat{e}_x + X Y_{y} \hat{e}_y \\
\nabla^2 \rho &= Y X_{xx} + X Y_{yy}
\end{align}
[/itex]
where the subscript denotes partial derivation with respect to that particular variable, with the exception of [itex]\hat{e}[/itex] which represents the unit vector in either direction. Decomposing the advection field vector and inserting these results into the original problem, we get..
[itex]
0 = D (Y X_{xx} + X Y_{yy} ) + Y X_{x} v_{x} + X Y_{y} v_{y}
[/itex]
Multiply this eq. with [itex] 1/(XYD) [/itex] and moving all [itex] X [/itex] -expressions to the left hand side result in the separation of [itex] X [/itex] and [itex] Y [/itex] such that they both have to be equal some unknown constant [itex] \lambda [/itex] (we cannot have a small change in [itex] X [/itex], without the corrosponding change in [itex] Y [/itex] and vice versa). Thus,
\begin{align}
\lambda &= -\frac{X_{xx}}{X} - \frac{X_{x}}{X} \frac{V_{1}}{D} \\
\lambda &= +\frac{Y_{yy}}{Y} + \frac{Y_{y}}{Y} \frac{V_{2}}{D}
\end{align}
Which are two independent, linear, second order ODEs, with general solutions
[itex]
\begin{align}
X(x) &= C_1 \exp[-\alpha^{(+)}_{x} x] \\
&+ C_2 \exp[+\alpha^{(-)}_{x} x] \\
& \\
Y(y) &= C_3 \exp[-\alpha^{(+)}_{y} y] \\
&+ C_4 \exp[+\alpha^{(-)}_{y} y]
\end{align}
[/itex]
with
[itex]
\alpha^{(\pm)}_{m} = \frac{1}{2} \left( \sqrt{\frac{v_{m}^2}{D^2} +4\lambda} \pm \frac{v_{m}}{D}\right), \quad m=1 \vee 2
[/itex]
and [itex] C_i, i=1,2,3,4; [/itex] are constant coefficients. So far, so good! However, here is where my issues begin to pile up as I'm unable to sort out the [itex] C [/itex] -values with the boundary conditions.
Along the north and east boundaries, I'm able to write [itex] C_1 = (\textrm{some const expression}) \cdot C_2 [/itex] and similar for [itex] C_3, C_4 [/itex]. However, for the west and south bounds I end up with
[itex]
(C_1 + C_2)Y = C \\
(C_3 + C_4)X = C
[/itex]
Which will only be valid for [itex]C=0, C_1 = -C_2, C_3=-C_4[/itex]. This, evidently, results in the trivial solution [itex]\rho(x,y)=0[/itex], which obviously is not what we want ..
Any help is appreciated! I've also tried to solve it by integral transforms, but due to the stationarity (the problem-equation is homogenious), I fail as [itex] \rho [/itex] vanishes ..
Last edited: