Definition of ' blow up time' in the context of PDE 
$$ \partial_t u - x \partial_x u = -u^2$$
$$ u(x,0) = u_0 (x)$$
(i) Find the blow up time $t_*$ for the cauchy data $u_0(x) = cosx $ and determine the position $x(t_*)$ which the solution develops poles.

Found a solution for this PDE using method of characteristics, but never heard
the term ' blow up time ' ever for the context of PDE. Seems like it has
something to do with solution going infinity but not sure. Any help how to solve this problem??
 A: $$\frac{\partial u}{\partial t}-x\frac{\partial u}{\partial x}=-u^2$$
On the characteristic curves, the differential equations are :
$$\frac{dt}{1}=\frac{dx}{-x}=\frac{du}{-u^2}$$
A first characteristic curve comes from $\quad dt=-\frac{dx}{x} \quad\to\quad xe^t=c_1$
A second characteristic curve comes from $\quad dt=-\frac{du}{u^2}\quad\to\quad \frac{1}{u}-t=c_2$ 
The general solution of the PDE can be expressed on the form of implicit equation $\Phi(c_1,c_2)=0$ where $\Phi$ is any differentiable function of two variables :
$$\Phi\left(xe^t,\frac{1}{u}-t\right)=0$$
Or, on an equivalent explicit form, with $F$ any differentiable function :
$$\frac{1}{u}-t=F(xe^t)\quad\to\quad u(x,t)=\frac{1}{t+F(xe^t)}$$
Taking account of the initial condition :
$$u(x,0)=u_0(x)=\frac{1}{0+F(xe^0)} \quad\to\quad F(x)=\frac{1}{u_0(x)}$$
With $F(xe^t)=\frac{1}{u_0(xe^t)}$ , the particular solution in agreement with the initial condition is :
$$u(x,t)=\frac{1}{t+\frac{1}{u_0(xe^t)} }=\frac{u_0(xe^t) }{1+t\:u_0(xe^t)}$$
In case of $\quad u_0(x)=\cos(x)$
$$u(x,t)=\frac{\cos(xe^t) }{1+t\:\cos(xe^t)}$$
$|u|\to\infty$ , i.e.: "blow-up", when $\quad 1+t\:\cos(xe^t)=0$
$$
x_\text{blow-up}=e^{-t}\cos^{-1}\left(-\frac{1}{t}\right)\quad\text{with } t\geq 1 $$
No blow-up while $t<1$ .
A: In the course of applying the method of characteristics to this problem you will encounter the related Riccati ODE
$$
\begin{cases}
\dot{z}(t) = - (z(t))^2\\
z(0)=z_0
\end{cases}
$$
We solve this by noticing that if a solution exists, then
$$
\frac{d}{dt} \left(\frac{1}{z(t)} \right) = -\frac{\dot{z}(t)}{(z(t))^2} = 1,
$$
and so we can integrate to find
$$
\frac{1}{z(t)} - \frac{1}{z_0} = t.
$$
Hence
$$
z(t) = \frac{z_0}{1 + t z_0},
$$
and we can verify by direct computation this this is actually the solution.  Notice, though, that if $z_0 < 0$ then actually the solution becomes singular when $t = -1/z_0 >0$.  In fact, we have that 
$$
\lim_{t \to (-1/z_0)^-} z(t) = \infty,
$$
and this is typically referred to as the "blow up" of the solution at the "blow up time" $t = -1/z_0$.  Something similar happens if $z_0>0$, but in this case the singularity occurs for $t = -1/z_0 < 0$, so if we're only interested in $t\ge 0$ then no blow-up occurs.
The question you've been asked wants you to mimic the above analysis for your PDE.  This makes sense here, as you solve the problem using ODE methods, and so you can follow the above argument.
