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I want to solve this system of advective-diffusive-reactive equations analytically:

$$\left(\alpha - k_0c_B\right)c_A+v\frac{dc_A}{dx}-D\frac{d^2c_A}{dx^2} = f_A $$ $$\left(\alpha - k_0c_A\right)c_B+v\frac{dc_B}{dx}-D\frac{d^2c_B}{dx^2} = f_B $$ $$k_0c_Ac_B+\alpha c_C+v\frac{dc_C}{dx}-D\frac{d^2c_C}{dx^2} = f_C $$ Where $\alpha$, $k_0$, $v$, $D$, and $f_{A/B/C}$ are nonzero constants. Here are the boundary conditions: $$c_A(x=0)=0,\;c_A(x=1)=0$$ $$c_B(x=0)=1,\;c_B(x=1)=0$$ $$c_C(x=0)=0,\;c_C(x=1)=0$$

if $k_0$ were zero, this becomes a system of independent and linear ODEs which I could solve using for example the Laplace Transform. I heard that for nonlinear ODEs, one cannot necessarily use Laplace Transforms without some modification to the original system (like linearization).

If there is a better way to solve the above system, I would like to know how, thanks!

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  • $\begingroup$ Is the third equation right, or did you mean to use $dc_C/dx$ and $d^2 c_C/dx^2$? $\endgroup$ – Robert Israel Oct 8 '15 at 19:39
  • $\begingroup$ Ah yes you're right, my bad. Fixed $\endgroup$ – Justin Oct 8 '15 at 19:40
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One thing you can try is expand each of $c_A$, $c_B$, $c_C$ in a series in powers of $k_0$. The equations for each power of $k_0$ will be linear in the coefficients of that power in $c_A$, $c_B$, $c_C$ given the lower-order coefficients.

EDIT: To reduce the number of subscripts, I'll write $k_0 = \epsilon$, $c_A = A$, $c_B = B$, $c_C = C$. Thus $$\eqalign{A &= A_0 + A_1 \epsilon + A_2 \epsilon^2 + \ldots\cr B &= B_0 + B_1 \epsilon + B_2 \epsilon^2 + \ldots\cr C &= C_0 + C_1 \epsilon + C_2 \epsilon^2 + \ldots\cr A_0(0) &= A_1(0) = A_2(0) = 0,\ A_0(1) = A_1(1) = A_2(1) = 0\cr B_0(0) &=1, B_1(0) = B_2(0) = 0,\ B_0(1) = B_1(1) = B_2(1) = 0\cr C_0(0) &= C_1(0) = C_2(0) = 0,\ C_0(1) = 1,\ C_1(1) = C_2(1) = 0\cr }$$ and the differential equations become $$ \eqalign{ \alpha\,A_{{0}} \left( x \right) &+v{\frac {\rm d}{{\rm d}x}}A_{{0}} \left( x \right) - D {\frac {{\rm d}^{2}}{{\rm d}{x}^{ 2}}}A_{{0}} \left( x \right) -f_{{A}} = 0\cr \alpha\,B_{{0}} \left( x \right) &+v{\frac {\rm d}{{\rm d}x}}B_{{0}} \left( x \right) - D {\frac {{\rm d}^{2}}{{\rm d}{x}^{ 2}}}B_{{0}} \left( x \right) -f_{{B}}=0\cr v{\frac {\rm d}{{\rm d}x}}C_{{0}} \left( x \right) &- D {\frac {{\rm d}^{2}}{{\rm d}{x}^{2}}}C_{{0}} \left( x \right) -f_{{C}} = 0\cr \alpha\,A_{{1}} \left( x \right) &-B_{{0}} \left( x \right) A_{{0}} \left( x \right) +v{\frac {\rm d}{{\rm d}x}}A_{{1}} \left( x \right) - D {\frac {{\rm d}^{2}}{{\rm d}{x}^{2}}}A_{{1}} \left( x \right)=0\cr \alpha\,B_{{1}} \left( x \right) &-B_{{0}} \left( x \right) A_{{0}} \left( x \right) +v{\frac {\rm d}{{\rm d}x}}B_{{1}} \left( x \right) - D {\frac {{\rm d}^{2}}{{\rm d}{x}^{2}}}B_{{1}} \left( x \right)=0\cr B_{{0}} \left( x \right) A_{{0}} \left( x \right) &+v{\frac {\rm d}{ {\rm d}x}}C_{{1}} \left( x \right) - D {\frac {{\rm d}^ {2}}{{\rm d}{x}^{2}}}C_{{1}} \left( x \right) =0\cr \alpha\,A_{{2}} \left( x \right) &-B_{{0}} \left( x \right) A_{{1}} \left( x \right) -B_{{1}} \left( x \right) A_{{0}} \left( x \right) + v{\frac {\rm d}{{\rm d}x}}A_{{2}} \left( x \right) - D {\frac {{\rm d}^{2}}{{\rm d}{x}^{2}}}A_{{2}} \left( x \right)=0 \cr \alpha\,B_{{2}} \left( x \right) &-B_{{1}} \left( x \right) A_{{0}} \left( x \right) -B_{{0}} \left( x \right) A_{{1}} \left( x \right) + v{\frac {\rm d}{{\rm d}x}}B_{{2}} \left( x \right) - D {\frac {{\rm d}^{2}}{{\rm d}{x}^{2}}}B_{{2}} \left( x \right) =0\cr B_{{1}} \left( x \right) A_{{0}} \left( x \right) &+B_{{0}} \left( x \right) A_{{1}} \left( x \right) +v{\frac {\rm d}{{\rm d}x}}C_{{2}} \left( x \right) - D {\frac {{\rm d}^{2}}{{\rm d}{x}^{ 2}}}C_{{2}} \left( x \right)=0 \cr}$$ etc. Each of these equations involves only one differentiated variable, occurring linearly and with constant coefficients, the other $A_i$, $B_i$ and $C_i$ occurring in the "inhomogeneous terms" being already known from the previous equations. Thus they may be solved one by one.

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  • $\begingroup$ What do you mean by this? Can you show me how this is done? $\endgroup$ – Justin Oct 8 '15 at 20:41
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There are a few things to notice about your system:

  • By shifting $c_{A/B/C}$, i.e. introducing $\hat{c}_A = c_A + \delta_A$ and similarly for $c_{B,C}$, you can get rid of the constant source terms $f_{A/B/C}$.
  • The first two equations are decoupled from the third one, so you can first focus on solving $c_{A}$ and $c_B$, and use that as input in the third equation.
  • You can remove all the nonlinear terms by a change of variables. Consider $u = c_A - c_B$; when you substract the second model equation from the first, you see that the resulting equation for $c_A - c_B = u$ is linear. The same holds for $v = c_A + c_C$ and for $w = c_B + c_C$. Using these new variables, you will end up with three decoupled, linear equations for $u,v,w$. Moreover, the equations for $u,v,w$ are identical up to the source term $f$. Since there is a linear relation between $u,v,w$ (in this case, $u = v - w$), this procedure does not solve the system completely. However, it will get you started.
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