I am trying to solve an HJB equation in a very simple case that put me in trouble.
For a function $v(t,x)$ regular enough defined on $[0,T]$ with condition $v(T,x) = 0$ , I want to find a solution of
$$ \partial_{t}v(t,x) -\frac{1}{2}\left(\partial_{x}v(t,x)\right)^2 + 2x\partial_{x}v(t,x) + \frac{1}{2}x^2 = 0 $$
I have a small background in ode but I heard that such equations are simple to solve.
My attempt is the following :
I start with a function of two variables of the form $v(t,x)=\phi(t)x^2$ that can be differentiate easily :
$$ \partial_{t}v(t,x) = \phi’(t)x^2\quad\text{and}\quad\partial_{x}v(t,x) = 2\phi(t)x $$
The pde becomes
$$ \phi’(t)x^2 + \frac{1}{2}x^2 + 4\phi(t)x^2 -2(\phi(t)x)^2 = 0 $$
Then I should solve
$$ \phi’(t)+ \frac{1}{2}+ 4\phi(t)-2\phi(t)^2 = 0 $$
With terminal condition $\phi(T) = 0$.
However I don’t know how to solve this ode. It seems to be closely related to the Riccati equation but the constant term does not allow me to do an interesting change of variable if I am right.
If you have some hints on how to pursue in the direction I have chosen, or another directions to propose for solving this ode let me know please.
Using the hint given by Gonçalo we have for $\phi(t) = -\frac{v’(t)}{2v(t)}$
$$ v’’+ 4v’-v = 0 $$
$\Delta = 20$ hence there is two roots $r_1 = -2-\sqrt(5)$ and $r_2=-2+\sqrt(5)$. Hence
$$ v(t)=\alpha e^{r_1 t} + \beta e^{r_2 t} $$
Since $\phi(T)=0$ we have $v’(T)=0$ which implies $v(T)=c$. From this we get $\alpha=\frac{c-\beta e^{r_2 T}}{e^{r_1 T}}$. Hence
$$ v(t) = \frac{c-\beta e^{r_2 T}}{e^{r_1 T}}e^{r_1 t} + \beta e^{r_2 t} $$
It follows that
$$ \phi(t) = \frac{-r_{1}(c-\beta e^{r_{2} T})e^{r_{1}(t-T)} - r_{2}\beta e^{r_{2}t}}{2\left[\left(c-\beta e^{r_{2} T}\right)e^{r_{1}(t-T)} + \beta e^{r_{2}t}\right]} $$
Which can be simplified to
$$ \phi(t) = \frac{-r_{1}}{2\left[1+ \frac{\beta e^{r_{2} t}}{2\left(c-\beta e^{r_{2} T}\right)e^{r_{1}(t-T)}}\right]} - \frac{-r_{2}}{2\left[\frac{\left(c-\beta e^{r_{2} T}\right)e^{r_{1}(t-T)}}{\beta e^{r_{2} T}} + 1\right]} $$
I did not find the value of $\beta$ while I have used, I think, all the conditions on my dynamic ?
Hint. The differential equation $\phi'(t)+ \frac{1}{2}+ 4\phi(t)-2\phi(t)^2 = 0$ is separable. The solution that satisfies the condition $\phi(T)=0$ is given by $$ \int_{\phi}^0\frac{du}{2u^2-4u-\frac{1}{2}}=\int_t^Tds=T-t. $$
Addendum
The differential equation $\phi'(t)+ \frac{1}{2}+ 4\phi(t)-2\phi(t)^2 = 0$ is a Riccati equation. It can be transformed into a linear second order ODE with the substitution $\phi=-\frac{v'}{2v}$: $$ -\frac{v''}{2v}+\frac{(v')^2}{2v^2}+\frac{1}{2}-\frac{2v'}{v}-\frac{(v')^2}{2v^2}=0 \implies v''+4v'-v=0. $$