Consider Thiele's differential equation for $t\in[0,\infty)$ (all the other functions are continuous on $[0,\infty)$, too.)

$$ \begin{align} V'(t)&=\Big(\phi(t)+\lambda(t)\Big)V(t)+\pi(t)-\lambda(t)A(t)\\ V(0)&=0 \end{align} $$

I am reading a proof about the unique solution being

$$V(t)=\int_0^t \big(\pi(s)-\lambda(s)A(s)\big)\exp\Big(\int_s^t \big(\phi(u)+\lambda(u)\big)du\Big)ds$$

So the first thing happening in the proof is that the author solves the equation


and finding the solution by variation of the constant afterwards. The last equation is equivalent to


and therefore

$$\int_0^t\frac{V'(s)}{V(s)}ds=\int_0^t \big(\phi(s)+\lambda(s)\big)ds$$

Now he states something I do not understand:

$$\log V(t)=\int_0^t \big(\phi(s)+\lambda(s)\big)ds + c$$

In my opinion, using that $\log' V(t)= \frac{V'(t)}{V(t)}$, it should be

$$\log V(t) -\log V(0)=\int_0^t \big(\phi(s)+\lambda(s)\big)ds,$$

which seems to be not well defined, since $V(0)=0$. Is this some sort of method to solve this equation or is this just wrong? What is the procedure here?

Thanks in advance for any help!

  • $\begingroup$ The method is not wrong in itself, but dividing by $V(t)$ requires additional explanation. It is much better to multiply both sides of $V'(t)-(\phi(t)+\lambda(t))V(t)=\pi(t)-\lambda(t)A(t)$ by the integrating factor $\exp(-\int\limits_s^t(\phi(u)+\lambda(u))\,\mathrm{d}u)$ and notice that the LHS is just the derivative of $V(t)\exp(-\int\limits_s^t(\phi(u)+\lambda(u))\,\mathrm{d}u)$. Perhaps you could take another textbook? $\endgroup$ – user539887 May 1 at 6:40

You can't apply $V(0)=0$ because this boundary condition applies to the full equation, not the equation without the forcing term.

Instead solve this auxiliary equation with a random constant, use this solution to obtain a solution to the full equation and only then apply the boundary condition.


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