The method

Suppose we have an ODE (with initial conditions) x ⁣(t)=f ⁣(x ⁣(t),t),x ⁣(t0)=C,x'\!\left(t\right)=f\!\left(x\!\left(t\right),t\right), \quad x\!\left(t_0\right)=C,(1)\p{1} where xx is the unknown function, and ff is Lipschitz continuous in its first argument and continuous in its second argument. By Picard–Lindelöf theorem, we can seek the unique solution within t[t0ε,t0+ε]t\in\left[t_0-\varepsilon,t_0+\varepsilon\right].

Here, I propose the following method: we can write out a sequence of functions defined by x0 ⁣(t)C,x_0\!\left(t\right)\coloneqq C,(2)\p{2} xn+1 ⁣(t)t0tf ⁣(xn ⁣(s),s)ds+Cx_{n+1}\!\left(t\right)\coloneqq\int_{t_0}^tf\!\left(x_n\!\left(s\right),s\right)\,\mathrm ds+C(3)\p{3} (the properties of ff guarantee that the integral is well-defined). Then, if the sequence (xn)\left(x_n'\right) converges uniformly on [tε,t+ε]\left[t-\varepsilon,t+\varepsilon\right] (question: can this condition actually be proved?), then the sequence (xn)\left(x_n\right) converges uniformly to a function xx on [tε,t+ε]\left[t-\varepsilon,t+\varepsilon\right], which is the unique solution to Equation 1.


The proof is easy. Note from Equation 3 that xn+1 ⁣(t)=f ⁣(xn ⁣(t),t).x_{n+1}'\!\left(t\right)=f\!\left(x_n\!\left(t\right),t\right). Then, take the limit nn\to\infty. By the uniform convergence, we recovers Equation 1.

An example

Suppose f ⁣(x,t)xf\!\left(x,t\right)\coloneqq x and t00t_0\coloneqq0, then xn ⁣(t)=Cj=0n1tjj!.x_n\!\left(t\right)=C\sum_{j=0}^{n-1}\frac{t^j}{j!}. This is because 0txn ⁣(s)ds+C=Cj=0n1tj+1j!(j+1)+C=xn+1 ⁣(t).\int_0^tx_n\!\left(s\right)\,\mathrm ds+C=C\sum_{j=0}^{n-1}\frac{t^{j+1}}{j!\left(j+1\right)}+C=x_{n+1}\!\left(t\right). By taking the limit nn\to\infty, we get x ⁣(t)=Cetx\!\left(t\right)=C\mathrm e^t, which means that the unique solution to x ⁣(t)=x ⁣(t)x'\!\left(t\right)=x\!\left(t\right) with initial condition x ⁣(0)=Cx\!\left(0\right)=C is x ⁣(t)=Cetx\!\left(t\right)=C\mathrm e^t, expectedly.


This method is good because integration is sometimes much easier than solving ODE. We can use integration to get functions that are close to the exact solution. A natural question to ask is how close xnx_n is to the exact solution xx.

If Equation 1 is defined on Rn\mathbb R^n (where we may define differences and infinitesimals), it is clear that xnxx_n-x is an infinitesimal of higher order than (tt0)n\left(t-t_0\right)^n.