Why Check Existence and Uniqueness?
An initial-value problem specifies a differential equation and a value of its solution at one point. For example, \(y'=f(x,y)\) together with \(y(x_0)=y_0\) asks for a function whose slope at each point is prescribed by \(f\), and whose graph passes through \((x_0,y_0)\).
Finding a formula is not always the first or most important question. Before solving, it is useful to ask whether a solution is guaranteed to exist near the initial point and whether that solution is the only one. The local existence-and-uniqueness theorem answers these questions when the right-hand side meets suitable conditions.
The Local Existence-and-Uniqueness Theorem
A rectangle around \((x_0,y_0)\) is a set of the form \(|x-x_0|\leq a\), \(|y-y_0|\leq b\), where \(a,b>0\). The theorem below uses a rectangle on which the right-hand side is continuous and satisfies a Lipschitz condition in its second variable.
The inequality is called a Lipschitz condition in \(y\). It compares the values of \(f\) at two \(y\)-coordinates while keeping \(x\) fixed. The constant \(L\) must work throughout the rectangle, not just at the initial point. A function that satisfies such a condition on a neighborhood of \((x_0,y_0)\) is called locally Lipschitz in \(y\) there.
A common practical test is available when the partial derivative \(f_y\) exists and is continuous on a rectangle. On a smaller closed rectangle, continuity gives a bound \(|f_y(x,y)|\leq L\). The mean value theorem, applied as a function of \(y\) with \(x\) fixed, then gives
Thus, a continuous partial derivative with respect to \(y\) is a sufficient way to establish the Lipschitz condition. The theorem itself only requires the Lipschitz condition; it does not require that \(f_y\) exist.
Why the Conditions Give a Unique Local Solution
The main idea is to rewrite the differential equation as an integral equation. If \(y\) is a solution, the fundamental theorem of calculus gives
A function satisfying this equation is a fixed point of the operation that sends a proposed function \(u\) to \(y_0+\int_{x_0}^{x}f(t,u(t))\,dt\). Continuity and a bound on \(f\) keep this operation inside a small rectangle; the Lipschitz condition makes repeated application draw proposed functions closer together. The following argument supplies the local existence and uniqueness justification.
Take a rectangle \(|x-x_0|\leq a,\ |y-y_0|\leq b\) where the hypotheses hold. Continuity on this closed rectangle gives a bound \(|f(x,y)|\leq M\). Choose \(h>0\) with \(h\leq a\), \(Mh\leq b\), and \(Lh<1\). If \(L=0\), the last inequality holds for any \(h>0\).
Consider continuous functions \(u\) on \([x_0-h,x_0+h]\) satisfying \(|u(x)-y_0|\leq b\). Define \(Tu\) by \(Tu(x)=y_0+\int_{x_0}^{x}f(t,u(t))\,dt\). The bound on \(f\) gives \(|Tu(x)-y_0|\leq Mh\leq b\), so the operation stays within the chosen set of functions.
For any two such functions \(u,v\), the Lipschitz inequality gives \(|Tu(x)-Tv(x)|\leq Lh\sup_t|u(t)-v(t)|\). Thus the maximum distance between \(Tu\) and \(Tv\) is at most \(Lh\) times the maximum distance between \(u\) and \(v\), with \(Lh<1\). Starting with \(u_0(x)=y_0\) and repeatedly setting \(u_{n+1}=Tu_n\), successive differences are bounded by a geometric sequence. The iterates therefore converge uniformly to a continuous function \(y\).
The Lipschitz estimate also shows that \(Tu_n\) converges uniformly to \(Ty\); since \(u_{n+1}=Tu_n\) converges to \(y\), it follows that \(Ty=y\). Hence \(y(x)=y_0+\int_{x_0}^{x}f(t,y(t))\,dt\). The integrand is continuous, so differentiating gives \(y'=f(x,y)\), and \(y(x_0)=y_0\). If \(z\) is another solution in this interval and rectangle, it also satisfies \(Tz=z\). Therefore \(\sup|y-z|=\sup|Ty-Tz|\leq Lh\sup|y-z|\). Since \(Lh<1\), this forces \(\sup|y-z|=0\), so \(y=z\).
This argument explains the distinct roles of the hypotheses: continuity supplies a bounded, well-behaved integral operation, while the Lipschitz condition supplies the contraction that ensures uniqueness. The result is local because the construction only controls the solution within a suitably small interval and rectangle.
Applying the Theorem
Worked Example: A Smooth Right-Hand Side
Consider \(y'=x+y,\ y(0)=1\). Here \(f(x,y)=x+y\), which is continuous, and \(f_y(x,y)=1\). On the rectangle \(|x|\leq1,\ |y-1|\leq1\), the possible \(y\)-values satisfy \(0\leq y\leq2\). The mean value theorem in \(y\), or direct subtraction, gives
Thus the Lipschitz condition holds with \(L=1\). Also \(|f(x,y)|=|x+y|\leq |x|+|y|\leq3\), so \(M=3\) is a bound on the rectangle. For example, choose \(h=\frac14\). Then \(h\leq1\), \(Mh=\frac34\leq1\), and \(Lh=\frac14<1\). The theorem guarantees exactly one solution on a sufficiently small interval around \(0\); this choice of bounds supports the construction on \([-\frac14,\frac14]\). No explicit formula is needed to make that local guarantee.
Worked Example: Local Uniqueness Does Not Mean Global Existence
Consider \(y'=y^2,\ y(0)=1\). Here \(f(x,y)=y^2\) is continuous, and \(f_y(x,y)=2y\) is continuous. On \(|y-1|\leq\frac12\), the \(y\)-values lie between \(\frac12\) and \(\frac32\), so \(|f_y|\leq3\). The theorem therefore guarantees a unique solution near \(x=0\).
In this case a formula can also be checked. For \(x<1\), let \(y(x)=\frac{1}{1-x}\). Differentiating and evaluating the right-hand side gives
Also \(y(0)=\frac{1}{1-0}=1\), so the formula satisfies both the equation and the initial condition. It becomes unbounded as \(x\) approaches \(1\) from below. The theorem's conclusion is a local guarantee, not a claim that a solution remains finite for every \(x\).
When Continuity Does Not Ensure Uniqueness
If the Lipschitz condition fails, the theorem does not settle the question of uniqueness. In particular, continuity alone cannot replace that condition in the uniqueness conclusion.
Worked Example: A Continuous Right-Hand Side With Multiple Solutions
Consider \(y'=\sqrt{|y|},\ y(0)=0\). The function \(f(x,y)=\sqrt{|y|}\) is continuous. However, it is not locally Lipschitz in \(y\) near \(y=0\): for \(y>0\),
and this ratio grows without bound as \(y\) approaches \(0\). The theorem's uniqueness hypothesis is not met. In fact, there are multiple solutions. One is \(y(x)=0\) for all \(x\). For any fixed \(a\geq0\), another is
For \(x<a\), its derivative is \(0=\sqrt{|y_a(x)|}\). For \(x>a\), its derivative is \(\frac{x-a}{2}\), while \(\sqrt{|y_a(x)|}=\sqrt{\frac{(x-a)^2}{4}}=\frac{x-a}{2}\). At \(x=a\), both the derivative and the right-hand side are \(0\). Thus the equation holds at every point, and \(y_a(0)=0\) when \(a\geq0\). Taking \(a=0\) gives a solution different from the constant zero solution for every \(x>0\).
What If a Hypothesis Fails?
The theorem is a sufficient condition: when its hypotheses hold, it guarantees local existence and uniqueness. If a hypothesis fails, it does not automatically follow that no solution exists or that uniqueness fails. The problem must be examined by other reasoning. The next example shows a case where discontinuity really does prevent a solution.
Worked Example: A Discontinuous Right-Hand Side
Define \(f(x,y)=0\) when \(x\ne0\), and \(f(0,y)=1\). Consider \(y'=f(x,y),\ y(0)=0\). The right-hand side is discontinuous at every point with \(x=0\), so the theorem does not apply. Suppose a differentiable solution existed on an interval around \(0\). For every \(x\ne0\), it would have \(y'(x)=0\). The mean value theorem then implies that \(y\) is constant on each side of \(0\). Continuity of \(y\) and \(y(0)=0\) force both constants to be \(0\). Consequently \(y'(0)=0\), but the equation at \(x=0\) requires \(y'(0)=f(0,0)=1\), a contradiction. Therefore this initial-value problem has no differentiable solution on an interval around \(0\).
Common Pitfalls
A frequent error is to treat continuity as sufficient for uniqueness. For example, one might see that \(\sqrt{|y|}\) is continuous and conclude that \(y'=\sqrt{|y|},\ y(0)=0\) has exactly one solution. That conclusion is wrong: the zero solution and the \(a=0\) solution in the preceding example are distinct solutions with the same initial value.
The correction is to check the specific Lipschitz condition in \(y\), or use a continuous partial derivative \(f_y\) to establish it. If neither check succeeds, the theorem gives no uniqueness conclusion. Also avoid treating a local theorem as a global one: it guarantees a unique solution only near the initial point, within a region where its assumptions hold.
How to Assess an Initial-Value Problem
For an initial-value problem \(y'=f(x,y),\ y(x_0)=y_0\), organize the assessment around the initial point:
- Identify the domain where \(f\) is defined and check that it contains a neighborhood of \((x_0,y_0)\).
- Check whether \(f\) is continuous on a rectangle around that point.
- Check for a Lipschitz bound in \(y\). If \(f_y\) exists and is continuous on a rectangle, bound \(|f_y|\) there to obtain such a bound.
- If both conditions hold, conclude that a unique solution exists on some interval around \(x_0\).
- If a condition fails, state only that this theorem does not apply; investigate existence or uniqueness separately if needed.
These checks clarify what can be concluded before solving an equation. They also explain why two equations that look similarly straightforward may behave differently when they share the same initial value.
Check Your Understanding
For each question, distinguish carefully between what the theorem guarantees and what its hypotheses do not establish.
- State the two main hypotheses on \(f\) in the local existence-and-uniqueness theorem.
- Why does a continuous \(f_y\) on a rectangle imply a Lipschitz condition in \(y\) on that rectangle?
- For \(y'=y^2,\ y(0)=1\), why does the theorem establish a local conclusion rather than a global one?
- What feature of \(f(x,y)=\sqrt{|y|}\) at \(y=0\) prevents the Lipschitz test used in the theorem from succeeding?
- If an initial-value problem fails one of the theorem's hypotheses, what can and cannot be concluded from that failure alone?
