0

I am trying to show that the solution to: $$ \frac{\partial D}{\partial \tau} =\alpha - \beta D + \gamma D^2 $$ with terminal condition $$D(u,0)=0$$ and $\beta$ is a function of $u$, is $$ D(u,\tau)= ~(r_-)\frac{1-e^{-d\tau}}{1-ge^{-d\tau}} $$ Where $g=\frac{r_-}{r_+}$

So far, I have simplified so that: $$ \frac{\partial D}{\partial \tau} =\alpha - \beta D + \gamma D^2 = \gamma(D-r_+)(D-r_-) $$ Where $$ r_{\pm}=\frac{\beta \pm \sqrt{\beta^2 - 4 \alpha \gamma}}{2 \gamma} = \frac{\beta \pm d}{\sigma^2} $$ any hints as to how to proceed?

WeakLearner
  • 5,982
  • What is $u$?? Anyway, if you know what the solution should be, why don't you just verify it by substituting it into the equation, instead of trying to solve the equation from scratch? – Hans Lundmark Sep 28 '14 at 09:46
  • forgot to mention that $\beta$ is a function of u. Well I'm trying to learn the method of getting there? – WeakLearner Sep 28 '14 at 09:49
  • 1
    OK, that's a good reason! Here's a similar example which might be of help: http://math.stackexchange.com/questions/78560/how-do-you-solve-the-initial-value-probelm-dp-dt-10p1-p-p0-0-1/ – Hans Lundmark Sep 28 '14 at 10:24

1 Answers1

0

The question seems to be about a function $G$ of a single variable $\tau$, such that $G(0)=0$ and solving the differential equation $$G'(\tau)=\gamma(G(\tau)-r_+)(G(\tau)-r_-),$$ for some fixed constants $(\gamma,r_+,r_-)$. The values of $(\gamma,r_+,r_-)$ depend on $u$ but for each $u$, the triplet $(\gamma,r_+,r_-)$ is fixed and $G(\ )=D(u,\ )$ solves such an equation.

To be complete, note that the differential equation above implies that $$G'(\tau)\left(\frac1{G(\tau)-r_+}-\frac1{G(\tau)-r_-}\right)=\gamma(r_+-r_-),$$ hence $$\frac{G(\tau)-r_+}{G(\tau)-r_-}=C\,\mathrm e^{\gamma(r_+-r_-)\tau},$$ for some constant $C$. The initial condition $G(0)=0$ yields $C=r_+/r_-$ and, finally, $$G(\tau)=r_+r_-\frac{1-\mathrm e^{\gamma(r_+-r_-)\tau}}{r_--r_+\mathrm e^{\gamma(r_+-r_-)\tau}}.$$ The parameters $r_+$ and $r_-$ solve the quadratic equation $\gamma r^2-\beta r+\alpha=0$ hence the prefactor is $$r_+r_-=\alpha/\gamma,$$ and the exponent of the exponential is $$\gamma(r_+-r_-)=\sqrt{\beta^2-4\alpha\gamma}.$$ (The square root in the RHS might be your $d$ but then the mysterious $\eta^2$ in your denominator for $r_\pm$ should read $2\gamma$. In any case, a unique exponential function is involved, not two with different exponents, as your answer seems to imply.)

Did
  • 279,727