$$\frac{\partial^2 u}{\partial x^2} + \frac{1}{L} \frac{\partial u}{\partial x} = 0$$

The solution is given as $$u = \exp(-\frac{1}{L}x) + 0.2 ,$$ with far field boundary condition as $u_\infty=0.2$.

Can anyone tell how does one end up with that kind of solution ?

  • 1
    $\begingroup$ Try solving the characteristic equation $\endgroup$ – Dylan Dec 19 '18 at 15:39

First, we can integrate with respect to $x$: $$u' +\frac u L = c_0, \quad c_0\text{ is a constant}$$

This is easy to integrate (hint: look at the derivative $\left(\exp\left(\frac xL\right)u\right)'$), too, it gives $$\exp\left(\frac xL\right)u=Lc_0\exp\left(\frac xL\right) +c_1,$$or $$u=Lc_0 +c_1\exp\left(-\frac xL\right)$$ Plug your boundary conditions for $x\to+\infty$, it would give you $Lc_0=0.2$. You will need some other condition to conclude that $c_1=1$.

edit How to solve a differential equation for constant $a$ and $b$ $$u' + au = b$$ The standard trick is to multiply by a well-chosen exponent in order to obtain "a derivative of a product" on the left-hand side. We know that $(\exp(ct))' = c\exp(ct)$, so what happens if we multiply both sides by $\exp(at)$: $$u'(t)\exp(at) + a\exp(at)u(t) = \left(u(t)\exp(at)\right)' = b\exp(at).$$ Now we have a derivative on the left-hand side and something that does not depend on $u$ on the right-hand side, so we can happily integrate: $$u(t) \exp(at) = \frac{b}{a}\exp(at) + d.$$ Multiply by $\exp(-at)$ to arrive to the final result. This approach has multiple generalizations, but the idea remains the same - multiply the equation to recover a derivative of a product in order to reduce the number of terms containing the derivative of the unknown function.

  • $\begingroup$ Thanks. Can you briefly explain the second line..how do you get that without looking at the solution? $\endgroup$ – newstudent Dec 19 '18 at 15:26
  • 1
    $\begingroup$ @newstudent see edit $\endgroup$ – TZakrevskiy Dec 19 '18 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.