I have the following PDE (with small parameter $0<\epsilon\ll 1$):
$$
\begin{cases}
\dfrac{\partial u(x,t)}{\partial t}+\dfrac{\partial ^2u(x,t)}{\partial x^2}-\epsilon u(x,t)^2=0, & x\in (0,1), t>0\\
\\
u(x,0)=u_0(x), &x\in (0,1)\\
\\
u(0,t)=u(1,t)=0, &t>0
\end{cases}
$$
I now have to find the equations for the first three terms of the formal asymptotic expansion of the solution of this problem.
How would one go about doing this?
Edit: Substituting $u(x,t)$ by its asymptotic expansion $u_0+u_1\epsilon+u_2\epsilon^2+O(\epsilon^3)$ gives $\dfrac{\partial u_0}{\partial t}+\dfrac{\partial (u_1\epsilon)}{\partial t}+\dfrac{\partial (u_2\epsilon^2)}{\partial t}+\dfrac{\partial ^2u_0}{\partial x^2}+\dfrac{\partial ^2(u_1\epsilon)}{\partial x^2}+\dfrac{\partial ^2(u_2\epsilon^2)}{\partial x^2}-\epsilon u_0^2-2u_0u_1\epsilon^2+O(\epsilon^3)=0$
But I'm confused as to what to do with the first few terms.
Best regards,
Noah
By grouping the terms in order of powers of $\epsilon$, we have $$ \left({u_0}_t + {u_0}_{xx}\right) + \epsilon\left({u_1}_t + {u_1}_{xx} - u_0^2\right) + \epsilon^2\left({u_2}_t + {u_2}_{xx} - 2u_0u_1\right) + \mathcal{O}(\epsilon^3)=0\ \\ \implies \\ \begin{aligned} {u_0}_t + {u_0}_{xx} &=0 \\ {u_1}_t + {u_1}_{xx} &=u_0^2 \\ {u_2}_t + {u_2}_{xx} &=2u_0u_1. \end{aligned} $$ To obtain initial conditions, we set $u_0(x,0) + \epsilon u_1(x,0) + \epsilon^2 u_2(x,0) + \mathcal{O}(\epsilon^3)= u_{init}(x)$, which gives us $u_0(x,0) = u_{init}(x)$, and $u_1(x,0)=u_2(x,0)=0$. The boundary conditions are homogeneous Dirichlet for all equations. If they were nonhomogeneous, we would handle them the same way as we handled the initial condition.