# ordinary differential equations-morphogen gradient

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

• If you really need help, you should provide all the information that helps others to help you. For example, you should give details of the reference. Who knows what the title of the paper is, what journal it is published on. You should provide the link to the paper if it is available online. – Hans Jul 2 '15 at 21:48
• thank you! the reference is: J.H. Merkin, B.D.Sleeman, on the spread of morphogen, J.Math.Biol. 51,1-17(2005); – AMERICO Jul 2 '15 at 21:59

## 2 Answers

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.

• Thank you everyone for helping! Now I have an idea how to proceed ! – AMERICO Jul 2 '15 at 22:29

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 $u_0=e^{-x}$.