I am trying to establish uniqueness for a solution to a bigger problem, and it boils down to whether or not the following differential equation has a unique solution:


Clearly, one solution to this differential equation is $f(t)=t+K$. Are there other solutions to this differential equation?


Let $g(t)=f(t)-t$ then $f$ is a solution if and only if $g$ solves the autonomous differential equation $$g'(t)=A(g(t)),\qquad A:x\mapsto(K-x)/x,$$ for every $x\ne0$. Studying the sign of $A$, one sees that every initial condition $g(0)=x_0$ yields:

  • a decreasing solution $g$ with limit $-\infty$ if $x_0\lt0$,
  • an increasing solution $g$ with limit $K$ if $0\lt x_0\lt K$,
  • the constant solution $g=K$ if $x_0=K$,
  • a decreasing solution $g$ with limit $K$ if $x_0\gt K$.

Thus, there are lots of solutions... Solving the ODE in $g$ yields, for every $t$, $$(f(t)-t-K)\cdot\exp(f(t)/K)=(f(0)-K)\cdot\exp(f(0)/K).$$

  • $\begingroup$ Thanks. This is super helpful. It looks like the solution is monotonic regardless of the initial condition. If I insist on the solution (to the transformed problem of solving for g(t)) being bounded, and t is allowed to range from -∞ to ∞, is the constant solution the only solution? (i.e., is the constant solution the only bounded solution if the domain is R?) $\endgroup$ – Mike Sep 10 '14 at 22:26
  • $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$ – Did Sep 10 '14 at 22:43

Short answer; Yes. Long answer;

Let $$v(t) := f(t)-t, v'(t)= f'(t)-1$$

$$(v'(t)+1) v(t) = k \iff v'(t) = \frac{k-v(t)}{v(t)} \iff \frac{v'(t) v(t)}{k-v(t)} = 1 \iff \int \frac{v'(t) v(t)}{k-v(t)} \mathrm{d}t = \int 1 \mathrm{d}t \iff -(k \log(-k+v(t)))-v(t) = t+c_1 \iff v(t) = k \left (W \left (\frac{1}{k e^{\frac{t+c_1}{k}}}\right )+1 \right ) \iff f(t) = k+t+k W\left (\frac{1}{k e^{\frac{t+c_1}{k}}} \right )$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.