Crank-Nicolson for quadratic PDE

Question

I would like to solve the equation $$\partial_t u = C(t)\cdot u-u^2$$ using the Crank-Nicolson approach. That resulted in the equations $$\begin{align} \frac{u_1-u_0}{\Delta t}&=0.5\left((C_1u_1-u_1^2)+(C_0u_0-u_0^2)\right)\\ u_1\left(1-\frac{\Delta t}{2}C_1+\frac{\Delta t}{2}u_1\right)&=u_0\left(1+\frac{\Delta t}{2}C_0-\frac{\Delta t}{2}u_0\right) \end{align}$$ Without the square part the solution would be easy, but I am lost with the $u_1$ in the left bracket. How can I fix that? Or is the Crank-Nicolson-Approach not usable for that problem?

Answer 1

This is not an answer to the question asking for Crank-Nicolson approach. This a comment but too long to be edited in the comments section. The exact analytic solution below may be used to compare with the approximative results from numerical method.

Comment :

Only one variable $t$ appears in the equation : $$\partial_t u = C(t)\cdot u-u^2$$ Thus this is a PDE reduced to an ODE : $$\frac{du}{dt}=C(t)u(t)-\big(u(t)\big)^2$$ To solve this Riccati ODE the change of function is : $$u(t)=\frac{1}{y(t)}\frac{dy}{dt}$$ $$u'=\frac{y''}{y}-\frac{(y')^2}{y^2}=C(t)\frac{y'}{y}-\left(\frac{y'}{y}\right)^2$$ $$y''=C(t)y'\quad\implies\quad \frac{y''}{y'}=C(t)$$ $$y'=c_1e^{\int C(t)dt}$$ $$y=c_1\int\left(e^{\int C(t)dt}\right)dt+c_2$$ $$u(t)=\frac{c_1e^{\int C(t)dt}}{c_1\int\left(e^{\int C(t)dt}\right)dt+c_2}$$ The exact solution is $$\boxed{u(t)=\frac{e^{\int C(t)dt}}{\int\left(e^{\int C(t)dt}\right)dt+c}}$$ $c=\frac{c_2}{c_1}=$constant (to be determined according to some initial condition not specified in the wording of the question).