# Solving $-\epsilon u''(x) + \beta u'(x) = 1$

Consider the equation: $$-\epsilon u''(x) + \beta u'(x) = 1, \;\; x \in (0,1)$$ $$u(0) = 0, \;\; u(1) = 0.$$ $\beta > 0, \;\; \epsilon > 0$.

Can someone please help me to solve this equation? (I don't know very well differential equations, so I need help).

Thank you!

First, you need to find the general solution of the associated homogeneous equation $$-\epsilon u''(x) + \beta u'(x) = 0\ ,$$ by setting $u'(x)=y(x)$. You have $$-\epsilon y^\prime(x)+\beta y(x)=0\ ,$$ which is separable $$\frac{\epsilon}{\beta}\frac{dy}{y}=dx\ ,$$ yielding $$y(x)=\kappa_1 e^{\beta x/\epsilon}\ ,$$ where $\kappa_1$ is an arbitrary constant. Therefore $$\frac{du}{dx}=\kappa_1 e^{\beta x/\epsilon}\Rightarrow u(x)=\kappa_1 \frac{\epsilon}{\beta}e^{\beta x/\epsilon}+\kappa_2\ .$$

Next, find a particular solution of the inhomogeneous equation $$-\epsilon u''(x) + \beta u'(x) = 1\ ,$$ for example take $u(x)=A x$, where you fix $A$ by requiring that $$\beta A=1\Rightarrow A=1/\beta\ .$$

Therefore the general solution of your original equation is $$u(x)=\kappa_1 \frac{\epsilon}{\beta}e^{\beta x/\epsilon}+\kappa_2+\frac{1}{\beta}x\ .$$ Next, fix $\kappa_1$ and $\kappa_2$ by requiring that $u(0)=u(1)=0$ as $$\kappa_1\epsilon/\beta+\kappa_2=0 \quad\text{and}\quad \kappa_1(\epsilon/\beta)e^{\beta/\epsilon}+\kappa_2+1/\beta=0\ .$$

Consider the homogenous equation $$-\varepsilon y'' + \beta y' =0$$ its general solutions are of the form $$y=C_1 +C_2 e^{\frac{\varepsilon}{\beta} x}$$ thus a general solutions of your equation are $$u(x) =C_1 +\frac{x}{\beta} +C_2 e^{\frac{\varepsilon}{\beta} x}$$ since $u(0)=0 , u(1)=0.$ We get $C_1 +C_2 =0 \wedge C_1 + \frac{1}{\beta} + C_2 e^{\frac{\varepsilon}{\beta} } =0$ therefore $$C_1 =\frac{1}{\beta e^{\frac{\varepsilon}{\beta} } -\beta} , C_2 =\frac{1}{-\beta e^{\frac{\varepsilon}{\beta} } +\beta}$$ and the solution is equal to $$u(x)=\frac{1}{\beta e^{\frac{\varepsilon}{\beta} } -\beta} +\frac{x}{\beta} +\frac{1}{-\beta e^{\frac{\varepsilon}{\beta} } +\beta} e^{\frac{\varepsilon}{\beta} x}$$

This is a 2nd order inhomogeneous ODE with constant coefficients. Such an equation can always be solved with the method of variation of parameters. There 4 steps:

1. Find the general solution to the homogeneous equation $$-\epsilon u'' + \beta u' = 0$$ using the ansatz $u(x) = e^{kx}$. You will find two values $k_1,k_2$ for which $e^{kx}$ is a solution, so the general solution is then $u(x) = c_1 e^{k_1 x} + c_2 e^{k_2 x}$.

2. Reduce the inhomogeneous equation to a 1st order equation as follows: assume that $u(x) = c_1(x) e^{k_1 x} + c_2(x) e^{k_2 x}$, subject to the constraint $$c_1'(x) e^{k_1 x} + c_2'(x) e^{k_2 x} = 0.$$ Insert this expression for $u(x)$ back into the inhomogeneous equation, and using the fact that $e^{k_1 x}, e^{k_2 x}$ are solutions to the homogeneous equation, together with the constraint above, you will see that many terms cancel, and in the end you are left with $$-\epsilon( k_1 c_1'(x) e^{k_1 x} + k_2 c_2'(x) e^{k_2 x}) = 1.$$

3. Solve this 1st order equation. From the constraint, we have that $c_2'(x) = -c_1'(x) e^{(k_1-k_2)x}$. Plug this into the 1st order equation from (2), and rearrange to solve for $c_1'(x)$. You will get an equation of the form $$c_1'(x) = F(x),$$ so the solution is $c_1(x) = \int F(x) dx + a_1$, where $a_1$ is an integration constant. Now that you know $c_1(x)$, you can similarly solve for $c_2(x)$ (you will get another integration constant).

4. Use your boundary conditions $u(0) = 0, u(1) = 0$ to solve for the integration constants.

Since this smells of homework, I hesitate to give additional details. But it is easy to find many worked examples of variation of parameters, which should make the technique very easy to understand.

First of all make the substitution $t=u'$ so you'll have a first order equation: $$-\epsilon t'+\beta t=1$$ Then solve the homogenous equation: $$-\epsilon t_0'+\beta t_0=0 \iff t_0'-\frac{\beta}{\epsilon}t_0=0 \iff \frac{dt_0}{dx}=\frac{\beta}{\epsilon}t_0 \iff \frac{dt_0}{t_0}=\frac{\beta}{\epsilon}dx \iff$$ $$\iff \ln t_0=\frac{\beta}{\epsilon}x+k \iff t_0=ce^{\beta / \epsilon x}$$ Now looking at the RHS of the equation we know that the solutiono for the inhomogeneus equation differs from the solution of the homogenus by a zeroth degree polynomial: $$t=t_0+q$$ Substituting in the equation youll find that $q=\frac{1}{\beta}$ so: $$t=ce^{\beta / \epsilon x}+\frac{1}{\beta}$$ Now we can integrate $t$ to find $u$: $$u=\int t dx \iff u=c\frac{\epsilon }{\beta}e^{\beta / \epsilon x}+\frac{x}{\beta}+d$$ Now we can force the contour conditions to find $c$ and $d$.\left\{\begin{align*} u(0) &= c\frac{\epsilon }{\beta}+d &= 0\\ u(1) &= c\frac{\epsilon }{\beta}e^{\beta / \epsilon }+\frac{1}{\beta}+d &= 0 \end{align*}\right. then $$c=\frac{1}{\epsilon}\frac{1}{1-e^{\beta / \epsilon }}$$ $$d=-\frac{1}{\beta}\frac{1}{1-e^{\beta / \epsilon }}$$ and the final expression for $u$ is: $$\frac{1}{\beta}\frac{1}{1-e^{\beta / \epsilon }}\left ( e^{\beta / \epsilon x}+(1-e^{\beta / \epsilon })x-1\right )$$