TITLE: Help with this nonlinear PDE please!
QUESTION [2 upvotes]: I need help solving the following nonlinear PDE. I've been struggling for days!
$$
\frac{\partial^2V}{\partial^2 x}\frac{\partial V}{\partial t} = c\left(\frac{\partial V}{\partial x}\right)^2
$$
for $(x,t)\in [0,\infty)\times [0,T)$, with boundary condition
$$
V(T,x)=(x-H)^2
$$
for some known $H$ and $c$.

REPLY [1 votes]: Perhaps this will help. Let's try separation of variables. Doing so, we let $V(x,t) = X(x)T(t)$. Then the PDE becomes
$$X''(x)T(t)X(x)T'(t) = cX'(x)^2T(t)^2.$$
Isolating $x$ and $t$ gives
$$ \frac{1}{X'(x)^2}X''(x)X(x) = c\frac{T(t)}{T'(t)}$$
Note that 
$$\frac{T(t)}{T'(t)} = \frac{c}{(\log T(t))'}$$
and we can also gather terms on the left to get
$$\frac{X''(x)}{X'(x)}\frac{X(x)}{X'(x)} = (\log X'(x))'\frac{1}{(\log X(x))'}.$$
Since each side is a function of a different variable, we see that each side is constant. Thus
$$\frac{(\log X'(x))'}{(\log X(x))'} = D = \frac{c}{(\log T(t))'}$$
Thus,
$$(\log X'(x))' = D(\log X(x))' \Rightarrow \log X'(x) = D\log X(x) + E.$$
Do something similar with $T$.
This trick of noticing that $\frac{y'}{y} = (\log y)'$ is very useful when solving ODEs and PDEs as this problem no doubt shows!
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