3
$\begingroup$

A (non-math) paper I'm working through presents the following differential equation (and solution) in a casual way:

$$ 0 = \frac{(1-\gamma)^{2}}{\gamma} e^{-\rho t} \bigg[ \frac{e^{\rho t} J_{x}}{\beta} \bigg]^{\frac{\gamma}{\gamma-1}} + J_{t} + [(1-\gamma) \eta / \beta + r x] J_{x} - \frac{J_{x}^{2}}{J_{xx}} \frac{(\alpha - r)^{2}}{\alpha \sigma^{2}}, $$ subject to $J(x,T)=0$.

A solution is $$ J(x,t) = \delta \beta^{-\gamma} e^{-\rho t} \bigg[ \frac{\delta(1-e^{-\left( \frac{\rho - \gamma \nu}{\delta}\right)(T-t)})}{\rho - \gamma \nu} \bigg]^{\delta} \bigg[ \frac{x}{\delta} + \frac{\eta}{\beta r}(1-e^{-r(T-t)}) \bigg]^{\gamma}, $$ where $\delta \equiv 1 - \gamma$ and $\nu \equiv r + (\alpha - r)^{2}/2 \delta \sigma^{2}$.

It seemed rather matter-of-fact, but I would have had no idea how to solve (what appears to be) such a complicated PDE. I've been looking through Polyanin's Handbook of Nonlinear Partial Differential Equations, but I haven't (yet) been able to find anything.

Any ideas on how they did it? (Or recommendations for relevant references?)

$\endgroup$
0

1 Answer 1

2
$\begingroup$

The second derivative $J_{xx}$ in the denominator is unusual, but maybe some progress is possible. The explicit time exponentials can be removed by substituting $$ J(x,t) = e^{(\gamma-1)\rho t}Y(x,t). $$ I won't keep track of the many constants, but you get an equation of the form $$ c_1 Y_{x}^{\gamma/(\gamma-1)} +c_2Y+c_3Y_t +(c_4+rx)Y_x-c_5\frac{Y_x^2}{Y_{xx}} = 0. $$ It is still a strange equation. But maybe it suggests powers of something plus $x$. So try $$ Y(x,t) = (h(t)+x)^\gamma. $$ We get $$ c_1\gamma^{\gamma/(\gamma-1)}(h+x)^\gamma + c_2(h+x)^\gamma+c_3\gamma(h+x)^{\gamma-1}h_t $$ $$ + (c_4+rx)\gamma(h+x)^{\gamma-1} -c_5\frac{(\gamma(h+x)^{\gamma-1})^2}{\gamma(\gamma-1)(h+x)^{\gamma-2}} = 0. $$ Factor out $(h+x)^\gamma$ to get $$ c_1\gamma^{\gamma/(\gamma-1)} + c_2+\gamma\frac{c_3 h_t+c_4+rx}{h+x} -c_5\frac{\gamma}{\gamma-1} = 0. $$ It is easy to find a function $h(t)$ satisfying $$ c_3h_t+c_4 = rh. $$ Then you are left with the question whether the constants $$ c_1\gamma^{\gamma/(\gamma-1)} + c_2+\gamma r -c_5\frac{\gamma}{\gamma-1} = 0? $$ If so then it seems that you have a solution. I have not tried to see whether it is the same as the one proposed in the paper, or even check the intial value.

Of course, the equation is so unusual that we probably don't know whether the intial value problem has a unique solution.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .