Can somebody help me with this question,I don't have a clue really.
Q: I The spread of an impulse from one end of a one-dimensional nerve fibre is modelled by the partial differential equation $$\frac{\partial u}{\partial t}=\nu \frac{\partial^2 u}{\partial x^2}-\frac{3 \gamma}{2} (u-k)^2, x \geq0,$$
where $u(x,t)$ is the strength of the impulse and $\nu, \gamma$ and $k$ are positive parameters.
(a) Write down the ordinary differential equation for stationary solutions $u(x,t)=w(x)$, and show that the quantity $$C=\left(\frac{dw}{dx}\right)^2-\frac{\gamma}{\nu}(w-k)^3$$ is constant for these solutions
(b) By solving the above equation for $C=0$, or otherwise, show that there is a solution of the form $$w(x)=k+\frac{A}{(x+x_0)^2},$$ where $A$ is a constant to be found, and $x_0$ is another constant. Sketch this solution and comment on any restrictions that should be made on the parameter $x_0$, so that the solution is bounded on the nerve fibre.
A: The only thing I kind of get is that if $u(x,t)=w(x)$, then $\frac{ \partial u}{\partial t}=0$, and $\frac{ \partial u}{\partial x}=\frac{dw}{dx}$, so obviously $\frac{\partial^2 u}{\partial x^2}=\frac{d^2w}{dx^2}$. Giving us: $$\nu \frac{d^2w}{dx^2}-\frac{3 \gamma}{2}(w-k)^2=0.$$ What do I do from here?
Is the following valid $$\nu \frac{d^2w}{dx^2}-\frac{3 \gamma}{2}(w-k)^2=0 \Leftrightarrow \frac{d^2w}{dx^2}-\frac{3 \gamma}{2\nu}(w-k)^2=0$$ Then get $$\frac{d^2w}{dx^2}=\frac{3 \gamma}{2\nu}(w-k)^2$$ After integrating this we end up $$\left(\frac{dw}{dx}\right)^2=\frac{\gamma}{2\nu}(w-k)^3+C_1$$. Then after rescaling coefficients, we get our desired result?
– user152911 May 26 '14 at 12:57