Let $f\in C^1(\Bbb R)$ s.t. $f(0)=0$ and $c\in\Bbb R$. Then is $u_t=u_{xx}+f(u)+c$ a well-posed PDE?

Question

For $u:\Bbb R\times [0,T)\to\Bbb R$ or $u:\Bbb T^1\times [0,T)\to\Bbb R$, and with $f\in C^1(\Bbb R)$ such that $f(0)=0$ and $c=0$, then I wish to consider the PDE $$u_t=u_{xx}+f(u)+c$$ with some sufficiently nice initial conditions $u\vert_{t=0}=u_0$. Now, it follow by contraction mapping arguments that if $c=0$, then this PDE is locally well-posed – the condition $f(0)=0$ comes in when we let $f(x) = x g(x)$ for $g\in C(\Bbb R)$ – however this argument does not adapt easily to the case where $c\ne 0$, and I see no way of using solutions to $$u_t=u_{xx}+f(u)$$ to obtain solutions to $$u_t=u_{xx}+f(u)+c.$$ How should I proceed?

Answer 1

Let $e^{t\Delta}$ the linear propagator, namely the flow of $u_t=u_{xx}$. Define the operator $$\Phi_{u^0}(u):=e^{t\Delta}u^0+\int_0^te^{(t-s)\Delta}(f(u(s))+c)ds,$$ it is easy to show that such an operator is a contraction on the Banach space $$X:=\{u(t,x)\in C^0(0,T),C^0((\mathbb{R})):\|u(t,x)\|_{C^0[(0,T),C^0(R)]}\leq2\|u^{0}\|_{C^0(\mathbb{R})}\}$$ if $T$ is chosen in such a way that $$T(\|f(u)\|_{C^0[(0,T),C^0(R)]}+c)< \|u^{0}\|_{C^0(\mathbb{R})}.$$ At least it is immediate to show that actually $\Phi$ maps $X$ in $X$, then the contraction argument does not see the constant $c$.