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Find $u(x)$ in the interval $x \in(0, l)$ such that $$ \begin{aligned} a \frac{\mathrm{d} u}{\mathrm{~d} x}-\kappa \frac{\mathrm{d}^2 u}{\mathrm{~d} x^2} & =f(x), \quad x \in(0, \ell) \\ u(0) & =0 \\ u(\ell) & =1 \end{aligned} $$ where $a$ denotes the advective velocity, $\kappa$ is the diffusion coefficient while $f(x)$ is the source.


The solution to this for $f \equiv 0$ can be analytically written as $$ u(x)=\frac{1-\exp (a x / \kappa)}{1-\exp (a \ell / \kappa)} $$ where $a \ell / \kappa$ is known as the Peclet number $(\mathrm{Pe})$ and measures the ratio of the strength of advection to the strength of diffusion.


Can one explain how analytical solution is coming as $\boxed{ u(x)=\frac{1-\exp (a x / \kappa)}{1-\exp (a \ell / \kappa)}} $

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If $f\equiv 0$, the ODE reduces to $$ au'-\kappa u''=0. \tag{1} $$ Integrating $(1)$, we obtain $$ au-\kappa u'=C_1. \tag{2} $$ Rearranging terms, we can rewrite $(2)$ as $$ \frac{\kappa u'}{au-C_1}=1. \tag{3} $$ Integrating both sides of $(3)$ with respect to $x$ we obtain $$ \frac{\kappa}{a}\ln|au-C_1| = x+C_2 \Rightarrow u(x)=C_1'+C_2'e^{ax/\kappa}, \tag{4} $$ where $C_1'$ and $C_2'$ are arbitrary constants. To determine them, we use the boundary conditions: \begin{align} u(0)=0&\Rightarrow C_1'+ C_2' = 0, \\ u(\ell)=1&\Rightarrow C_1'+ C_2'e^{a\ell/\kappa} = 1. \tag{5} \end{align} Solving $(5)$ we obtain $$ C_1'=-C_2'=-\frac{1}{e^{a\ell/\kappa}-1} \tag{6} $$ and, finally, $$ u(x)=\frac{e^{ax/\kappa}-1}{e^{a\ell/\kappa}-1}. \tag{7} $$

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Hint: the solution can be written as \begin{equation} u(x) = \alpha + \beta e^{\frac{a}{k}x}+ \int_0^\ell \varphi(x-t)f(t)dt \end{equation} where $\varphi(x)$ is the fundamental solution \begin{equation} \varphi(x) = \left\{ \begin{array} \\-\frac{1}{a}\quad\text{if}\quad x<0\\ -\frac{1}{a}e^{\frac{a}{k}x}\quad\text{if}\quad x>0 \end{array} \right. \end{equation} and $\alpha, \beta$ are suitable constants depending on $f$.

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  • $\begingroup$ can you give me a detailed solution? $\endgroup$ Commented Apr 24, 2023 at 6:00
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    $\begingroup$ This is detailed enough. You can determine $\alpha$ and $\beta$ by writing $u(0)$ and $u(\ell)$. $\endgroup$ Commented Apr 24, 2023 at 6:02

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