I have a PDE system that I am trying to solve at steady state. When I make the appropriate substitutions, I get an equation of the following form:
$$\frac{1+M}{M}\frac{d^2 M}{d x^2}=1$$ Is there a closed form expression for this? Mathematica finds a nasty solution involving integrals after a lot of thinking. Is this perhaps something that requires nice boundary conditions to solve?
For those that are interested, the system that this originated from is below
$$\frac{\partial M}{\partial t}=D\frac{\partial^2 M}{\partial x^2}-k_{on}R_{tot}M(1-R)+k_{off}R$$ $$\frac{\partial R}{\partial t}=k_{on}R_{tot}M(1-R)-k_{off}R$$
With $$\frac{\partial M}{\partial t}(0,t)=\frac{\nu}{R_{tot}}-k_{on}R_{tot}M(1-R)+k_{off}R$$ $$M(L,t)=R(L,t)= M(x,0)=R(x,0)=0$$
I want to solve this for the steady state, i.e. when $\frac{\partial M}{\partial t}=\frac{\partial R}{\partial t}=0$
I don't think you will get a nice expression. First, we write the differential equation as $$ M''=\frac{M}{M+1}. $$ Multiplying with $M'$ we find that this is $$ \frac{1}{2}\bigl((M')^2\bigr)'=\frac{M}{M+1}M' $$ Integrating, we find that $$ \frac{1}{2}(M')^2=M-\log|1+M|+C. $$ Now you have a separable differential equation, and you will have to integrate it, $$ \pm\frac{M'}{\sqrt{2(M-\log|1+M|+C)}}=1, $$ so $$ \pm \int\frac{1}{\sqrt{2(M-\log|1+M|+C)}}\,dM=x+\widehat{C}. $$ I think this is not possible to solve in elementary functions$\dots$