ordinary differential equations-morphogen gradient

Question

I am reading a paper by Merkin and Sleeman (2005)

Find the approximation solution of

$(u')^2=\frac{2}{k}(u-\frac{1}{k}\ln(1+ku)); ~~u(0)=1$ for $k$ sufficiently small.

they gave the following approximate solution for $k$ small:

$u=e^{-x}+\frac{k}{3}(e^{-x}-e^{-2x})+\cdots$

but I have no ideia how they got this solution. I ask for help

Answer 1

That's not the complete solution: note that if $u(x)$ is a solution then so is $u(-x)$. So there's another solution

$$ u = e^x + \dfrac{k}{3} \left(e^{x} - e^{2x}\right) + \ldots$$

The expansion of the right side of your differential equation in powers of $k$ is $$ u^2 - \dfrac{2u^3}{3} k + \ldots $$ To 0'th order, the d.e. $(u')^2 = u^2$, $u(0)=1$ has solutions $u(x) = e^x$ and $u(x) = e^{-x}$. If we then choose $u(x) = e^{-x} + k u_1(x) + \ldots$, we find we need $$ u_1' = - u_1 + \dfrac{1}{3} e^{-2x},\ u_1(0)=0$$ so that $u_1(x) = (e^{-x} - e^{-2x})/3$, and thus $$ u(x) = e^{-x} + \dfrac{k}{3} (e^{-x} - e^{-2x}) + \ldots$$ This can be repeated to include as many powers of $k$ as you wish.

Answer 2

You can try expanding $\ln(1+kx)$ into Taylor series of $kx$. Then assume the asymptotics over $k$ $$u(k,x) = \sum_{i=0}^\infty k^iu_i(x)$$ where $u_i(x)$ is a smooth function of $x$ only. Then obtain the ordinary differential equations of $u_i(x)$ by matching the coefficient of $k^i$ of the left and right hand side. Solving these ODE's recursively, you should obtain the desired solution.

In addition you need to add the boundary condition at $x=\infty$ that $\lim\limits_{x\to\infty}u(k,x)=0$ to obtain the solution you list in your question. Otherwise you have another but explosive solution.

I already see the procedure pops out the zero'th order term $u_0=e^{-x}$.