Applications to Differential Equations

From an Initial-Value Problem to a Fixed Point

The contraction method for equations also applies to differential equations, but the unknown is now an entire function rather than a number. Consider an initial-value problem

A continuously differentiable solution must satisfy the corresponding integral equation

Conversely, if a continuous function satisfies this integral equation and \(f\) is continuous along its graph, the Fundamental Theorem of Calculus implies that the function is continuously differentiable and solves the initial-value problem. We can therefore define an operator on functions by

A fixed point of \(T\) is exactly a solution of the integral equation. The aim is to choose an interval and a set of functions so that \(T\) maps that set into itself and is a contraction. The Contraction Mapping Theorem then gives a unique fixed point and convergence of successive applications of \(T\).

The Uniform Metric on Continuous Functions

Let \(J=[t_0-h,t_0+h]\), where \(h>0\). Write \(C(J)\) for the set of continuous real-valued functions on \(J\), and equip it with the uniform metric

The supremum is finite because a continuous function on a closed bounded interval is bounded. The key fact is that this metric space is complete. We give the proof because completeness is what permits the contraction theorem to produce a solution.

Proof. Let \((u_n)\) be a Cauchy sequence in \(C(J)\). For each fixed \(t\in J\),

The Cauchy property in the uniform metric therefore makes \((u_n(t))\) a Cauchy sequence of real numbers. Since \(\mathbb{R}\) is complete, it has a limit; define \(u(t)=\lim_{n\to\infty}u_n(t)\).

We next show that \(u_n\) converges uniformly to \(u\). Given \(\varepsilon>0\), choose \(N\) such that \(d_\infty(u_n,u_m)<\varepsilon\) whenever \(n,m\geq N\). Fix \(n\geq N\) and \(t\in J\). Letting \(m\) tend to infinity in \(|u_n(t)-u_m(t)|<\varepsilon\) gives \(|u_n(t)-u(t)|\leq\varepsilon\). Since this holds for every \(t\in J\), \(d_\infty(u_n,u)\leq\varepsilon\) for every \(n\geq N\). Applying the argument with any smaller positive tolerance shows \(d_\infty(u_n,u)\to0\).

It remains to check that \(u\) is continuous. Fix \(t\in J\) and \(\varepsilon>0\). Choose \(n\) such that \(d_\infty(u_n,u)<\varepsilon/3\). By continuity of \(u_n\) at \(t\), there is a \(\delta>0\) such that \(|u_n(s)-u_n(t)|<\varepsilon/3\) whenever \(s\in J\) and \(|s-t|<\delta\). For such \(s\),

Thus \(u\) is continuous, so \(u\in C(J)\), and the Cauchy sequence converges to \(u\) in \(C(J)\). This proves completeness. \(\square\)

A closed ball in \(C(J)\) is also complete: a convergent sequence in the ball has its limit in the ball, since uniform convergence preserves the defining bound. We will use the ball centered at the constant function \(y_0\),

where \(b>0\). Its functions stay within distance \(b\) of the initial value, so a bound on \(f\) over that range can control the integral operator.

A Local Existence and Uniqueness Theorem

Assume \(f\) is continuous on the rectangle

where \(a,b>0\). Suppose there are finite constants \(M\geq0\) and \(L\geq0\) such that throughout this rectangle,

whenever \(y,z\in[y_0-b,y_0+b]\) and \(t\in[t_0-a,t_0+a]\). The second condition is a uniform Lipschitz condition in the dependent variable. Choose \(h>0\) so that

Such an \(h\) can be chosen because \(a,b>0\) and \(M,L\) are finite; if either \(M\) or \(L\) is zero, its corresponding restriction imposes no obstacle. On \(J=[t_0-h,t_0+h]\), define \(T\) by the integral formula above.

Proof. First, \(T\) is well defined on \(\mathcal{B}\). For \(u\in\mathcal{B}\), the point \((s,u(s))\) lies in \(R\) for each \(s\in J\), and continuity of \(f\) makes \(s\mapsto f(s,u(s))\) continuous. Consequently \(Tu\) is continuous. For every \(t\in J\), using the bound on \(f\),

The estimate holds whether \(t\) is to the left or the right of \(t_0\), since the absolute value of an oriented integral is bounded by the integral of the absolute value. Thus \(Tu\in\mathcal{B}\).

Next take \(u,v\in\mathcal{B}\). The Lipschitz condition gives, for every \(t\in J\),

Taking the supremum over \(t\) yields \(d_\infty(Tu,Tv)\leq Lh\,d_\infty(u,v)\). Since \(Lh<1\), \(T\) is a contraction on \(\mathcal{B}\). The completeness theorem for \(C(J)\) and closedness of \(\mathcal{B}\) show that \(\mathcal{B}\) is a nonempty complete metric space; it is nonempty because it contains the constant function \(y_0\). The Contraction Mapping Theorem now gives a unique fixed point \(y\in\mathcal{B}\), and iteration from any \(u_0\in\mathcal{B}\) converges uniformly to \(y\).

The fixed-point identity says \(y(t)=y_0+\int_{t_0}^t f(s,y(s))\,ds\). The integrand is continuous, so the Fundamental Theorem of Calculus gives \(y'(t)=f(t,y(t))\), and setting \(t=t_0\) gives \(y(t_0)=y_0\).

Finally, any other continuously differentiable solution on \(J\) whose values lie in \([y_0-b,y_0+b]\) satisfies the same integral equation, by the Fundamental Theorem of Calculus, and belongs to \(\mathcal{B}\). It is therefore another fixed point of \(T\), so it must equal \(y\). This proves the stated uniqueness. \(\square\)

The theorem is local: the interval length is selected using bounds in a rectangle around the initial data. It does not claim that a solution exists for all time. The successive functions \(u_{n+1}=Tu_n\) are called Picard iterates, or successive approximations. Their uniform convergence follows from the contraction theorem, not merely from a formal repeated substitution into the differential equation.

Worked Examples: Applying the Theorem

What the Bounds Do—and Do Not—Establish

The two estimates have different roles. The bound \(M\) controls how far a function can move from its initial value during one application of the integral operator. It ensures that the operator preserves the chosen ball. The Lipschitz constant \(L\) controls how much the operator changes when its input function changes. The strict inequality \(Lh<1\) ensures contraction. A bound on \(f\) alone does not establish uniqueness, and a Lipschitz estimate alone does not show that the operator maps the chosen ball into itself.

The interval must also be chosen consistently with both estimates. A smaller interval can make \(Lh<1\) hold and can keep the integral displacement within the allowed range \(b\). However, this argument only establishes the solution on the interval for which all the hypotheses have been verified. It should not be read as a global existence claim.

Once the hypotheses hold, the contraction theorem supplies more than existence: the solution is the unique fixed point in the ball, and Picard iteration from any function in that ball converges uniformly to it. This is the function-space version of using a contraction to solve an equation. Here the unknown is a trajectory, and the integral operator encodes both the differential equation and the initial value.