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You were right to question it, as I see an error. The algebra is a little messy and I will leave the details to correcting it to the OP.But basically, writing H(t,s) instead of H(t,a) and differentiating w.r.t "s" yieldsH' = mu H (1 - H), where the prime indicates s differentiation. The terminal condition is H(t,t) = f(t), and solving this is routine.

Alan is horribly correct!I tried a little bit before finding change of variable u=1/G to solve it.sorry for the bad notation.my true problem would be a little bit more complicated and it is H(t,a)=f(t-s)exp(...) and it's a true branching problem.I see it can be solved in the same way just more work to do and you have a more complicated equation to solve.great always bring the wilmott forum with you before going out in the forest:-)

QuoteOriginally posted by: AlanH' = mu H (1 - H), where the prime indicates s differentiation. The terminal condition is H(t,t) = f(t), and solving this is routine.Out of interest, how will one solve this initial value problem? A closed soluition would be ideal, but probably not possible in the general case. Will you use some ODE solver?At first glance, it looks like a doable integral equation (as well).

dH/(H*(1-H))=mu*dtdH/H+dH/(1-H)=mu*dtand then you integrate (you have to be prudent with the signs).Anyway I still didn't get the OP original equation.

QuoteOriginally posted by: frenchXdH/(H*(1-H))=mu*dtdH/H+dH/(1-H)=mu*dtand then you integrate (you have to be prudent with the signs).Anyway I still didn't get the OP original equation.I don't get this. 'dt' is being used as a discrete quantity but an ODE is the limit dt -> 0?

This is the "not so mathematically clean way used by physicists to solve ODE". H'=dH/dt and then you separate the infinitessimals such as f(H)*dH=g(t)*dt and then you integrate the two sides of the equality.

QuoteOriginally posted by: frenchXThis is the "not so mathematically clean way used by physicists to solve ODE". H'=dH/dt and then you separate the infinitessimals such as f(H)*dH=g(t)*dt and then you integrate the two sides of the equality.Ok The original equation was an integral equation so I am just wondering how/why we arrived at an ODE.

No problem I'll explainH(t,s)=f(t)exp(-mu*int(1-H(u),u=0..s))You differentiate both side with respect to 's' H'=-mu*(1-H)*f(t)*f(t)exp(-mu*int(1-H(u),u=0..s))and using the original integral equation you arrive atH'=-mu*(1-H)*HThe problem is that I'm really not sure that we are solving the original OP problem.

I'm with frenchX, we haven't yet had a clear presentation of the problem from the OP.P

QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: AlanH' = mu H (1 - H), where the prime indicates s differentiation. The terminal condition is H(t,t) = f(t), and solving this is routine.Out of interest, how will one solve this initial value problem? A closed soluition would be ideal, but probably not possible in the general case. Will you use some ODE solver?At first glance, it looks like a doable integral equation (as well).It looks like 'how to solve' has been answered by frenchX. If you carry it through, it will be closed-form solution.

We did have the statement from the OP that "a" (now "s") is a constant. That rather spoils a lot of this if it's true.We could try to reverse engineer the equation, to figure out what it is trying to solve, and then we'd know what was going on. For example, if "a" is a variable then it's relatively uncommon to have two time-like variables in a problem. P

The ODE I wrote comes up all the time in solutions of the Heston (CIR/square-root model/affine, etc) models. Two times are very common; I would think of "t" as a fixed terminal time, better "T". Then, "a" -> "s" or even better, now, re-label to "t" is simply the running time.

Yes, calibration problems are places I most often see this.But there are no squares or square roots so I would guess not vol calibration.P

OK, here is a forensic analysis. Suppose I said:Solve for h(t,x;T) where(*) -h_t = mu (x h_xx - x h_x)with the terminal condition h(T,x;T) = exp(f(T) x). [Here f( ) is the OP's f( )]Then, you could make the ansatz h(t,x;T) = exp( H(T,t) x) and H would solve theODE I wrote (with the new notation "t old" -> "T" and "a" -> t), and with H(T,T) = f(T).So that's a PDE problem associated with the solution.Now, underlying that PDE is the SDEdX = -mu X dt + sqrt(2 mu X) dW This is Feller's branching processIt's probably not going to represent an interest rate or a volatility because X(t) hits the origin and gets stuck. It's going to be a model of something (a population in the original apps) that can get extinguished at a finite time. Since riccardo24 indeed mentioned 'branching', this is probably on the right track.

I'm impressed:-)even if the process is not that one this shows a strict link. Now I would want to show the link in more depth.I'm studying proof of proposition 4.4 one passage is still not all clear to me but all is very interesting.Soon I should be able to show something I hope