We can figure out that for the ODE $u'(t)=f(u(t))$ the solution is global if $\int_{M}^{\infty}\frac{du}{f(u)}=\infty$. $(M>0)$
- I cannot find out why this condition is sufficient for blow up of above ODE.
- Use the ODE above to conclude that the solution for $u_t-\Delta u=f(u)$ is global iff $\int_{M}^{\infty}\frac{du}{f(u)}=\infty$. $(M>0)$.
Can anyone suggest me good references where I can find out the theory for blow up of Equations by comparing with ODEs.
Regards
I will assume $f(u)>0$ for $u\ge M$.
The solution with initial condition $u(0)=M$ is given implicitely by $$ \int_M^u\frac{ds}{f(s)}=t. $$ Let $F(u)=\int_M^u\frac{ds}{f(s)}$. $F$ is strictly increasing on $[M,\infty)$ and $\lim_{u\to\infty}F(u)=\infty$. It has an inverse $F^{-1}\colon[0,\infty)\to[M,\infty)$ and $$ u(t)=F^{-1}(t). $$
Any solution of $u_t-\Delta u=f(u)$ with initial value $u_0$ such that $0\le u_0(x)\le M$ is bounded by the solution of $u'=f(u)$ with $u(0)=M$. The iff part is wrong. Blow up may depend both on $f$ and on the size of $u_0$. For instance, $u_t-\Delta u=u^p$ has global solutions for sufficiently small $u_0$ if $p>1+2/n$.