Ordinary Differential Equations 8 | Existence and Uniqueness?
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For x' = x^2 with x(0)=1, the explicit solution x(t)=1/(1−t) blows up at t=1, so the maximal existence interval is t<1 rather than all real numbers.
Briefing
Existence and uniqueness for initial value problems in ordinary differential equations can fail in two different ways: solutions may not extend to all times, and even when solutions exist they may not be unique. Using concrete examples, the discussion shows how the behavior of the vector field V(x) determines both the maximal time interval of a solution and whether multiple trajectories can satisfy the same initial condition.
The first example starts with the initial value problem x' = x^2 and x(0) = 1. Solving gives x(t) = 1/(1 − t). While this function satisfies the differential equation and the initial condition, it cannot be defined for all real t because it blows up at t = 1. The maximal domain of definition is therefore t < 1, not the whole real line. Geometrically, the vector field points to the right for positive x and grows in magnitude as x increases, so a trajectory starting at x = 1 accelerates toward infinity. The key takeaway is that “finite-time blow-up” limits the time interval of existence, but it does not contradict the existence of a solution on its maximal interval.
The second example keeps existence but breaks uniqueness. The vector field is defined piecewise using square roots: for x ≥ 0, V(x) = √x, and for x < 0, V(x) = −√|x| (equivalently, the sign of x multiplies the square root of |x|). This V(x) is continuous everywhere, including at x = 0, so solutions exist. However, differentiability fails at x = 0 because the slope becomes unbounded there. With the initial condition x(0) = 0, one solution is the constant trajectory x(t) = 0, since the vector field vanishes at the origin. A second, distinct solution can “leave” the origin after any waiting time; the transcript gives one explicit choice: x(t) = (t^2)/4 for t > 0 (with x(t) = 0 for t ≤ 0). Both satisfy the differential equation and the initial value, yet they are different, meaning the same initial condition admits multiple solutions.
These two counterexamples motivate the general theory: existence corresponds to whether every point in the phase space lies on at least one orbit (so every initial condition can be followed along the direction field). Uniqueness corresponds to whether orbits fail to cross—crossing would imply that the same initial value problem can branch into multiple trajectories. The remedy is to impose a regularity condition on the vector field, specifically a Lipschitz condition, which prevents the kind of branching seen at x = 0 in the second example while also supporting the desired well-posedness of initial value problems. The next step is to formalize that Lipschitz criterion.
Cornell Notes
Initial value problems for ODEs can fail in two ways: solutions may exist only on a limited time interval, and solutions may not be unique. For x' = x^2 with x(0)=1, the solution x(t)=1/(1−t) blows up at t=1, so the maximal interval is t<1 rather than all real times. For a different vector field V(x)=sign(x)√|x|, continuity ensures existence, but non-differentiability at x=0 allows multiple solutions from x(0)=0. One solution stays at x(t)=0, while another can remain at 0 for a while and then follow x(t)=t^2/4 for t>0. Uniqueness is tied to whether orbits cross, and Lipschitz regularity is the standard condition that rules out this branching.
Why does the solution to x' = x^2 with x(0)=1 fail to exist for all t?
What geometric intuition explains the finite-time blow-up in the first example?
How can a continuous vector field still produce non-unique solutions?
How do two different solutions arise from the same initial condition x(0)=0 in the second example?
What orbit-based criterion links uniqueness to the behavior of trajectories?
Review Questions
- In the example x' = x^2, how do you determine the maximal interval of existence from the closed-form solution?
- What specific property of the vector field at x=0 allows multiple solutions to start from x(0)=0 in the second example?
- How does the Lipschitz condition relate to preventing orbit crossing and restoring uniqueness?
Key Points
- 1
For x' = x^2 with x(0)=1, the explicit solution x(t)=1/(1−t) blows up at t=1, so the maximal existence interval is t<1 rather than all real numbers.
- 2
Finite-time blow-up limits the time domain of a solution but does not negate existence on the maximal interval where the solution is defined.
- 3
A continuous vector field can still fail uniqueness if it is not regular enough at a point (non-differentiability at x=0 in the second example).
- 4
With V(x)=sign(x)√|x| and x(0)=0, both the constant solution x(t)=0 and a “delayed departure” solution x(t)=t^2/4 for t>0 satisfy the same initial value problem.
- 5
Uniqueness is best understood through orbit behavior: if orbits cross or branch, the same initial condition can generate multiple solutions.
- 6
Existence corresponds to every point lying on at least one orbit of the direction field, ensuring each initial value problem can be followed.
- 7
A Lipschitz condition on the vector field is the standard tool to guarantee both existence (in the needed sense) and uniqueness by preventing branching.