In mathematics, specifically the study of differential equations, the Picard–Lindelöf theorem gives a set of sufficient (but not necessary) conditions under which an initial value problem has a unique solution. It is also known as Picard's existence theorem, the Cauchy–Lipschitz theorem, or the existence and uniqueness theorem. The theorem is named after Émile Picard, Ernst Lindelöf, Rudolf Lipschitz and Augustin-Louis Cauchy.
Theorem Let D ⊆ R × R n {\displaystyle D\subseteq \mathbb {R} \times \mathbb {R} ^{n}} be a closed rectangle with ( t 0 , y 0 ) {\displaystyle (t_{0},y_{0})} a point in the interior of D . {\displaystyle D.} Let f : D → R n {\displaystyle f:D\to \mathbb {R} ^{n}} be a function that is continuous in t {\displaystyle t} and Lipschitz continuous in y {\displaystyle y} (with Lipschitz constant independent from t ) . {\displaystyle t).} Then there exists some ε > 0 {\displaystyle \varepsilon >0} such that the initial value problem
y ′ ( t ) = f ( t , y ( t ) ) , y ( t 0 ) = y 0 {\displaystyle y'(t)=f(t,y(t)),\qquad y(t_{0})=y_{0}}
has a unique solution y ( t ) {\displaystyle y(t)} on the interval [ t 0 − ε , t 0 + ε ] . {\displaystyle [t_{0}-\varepsilon ,t_{0}+\varepsilon ].}
Proof sketch A standard proof relies on transforming the differential equation into an integral equation, then applying the Banach fixed-point theorem to prove the existence and uniqueness of solutions. Integrating both sides of the differential equation y ′ ( t ) = f ( t , y ( t ) ) {\textstyle y'(t)=f(t,y(t))} shows that any solution to the differential equation must also satisfy the integral equation
y ( t ) − y ( t 0 ) = ∫ t 0 t f ( s , y ( s ) ) d s . {\displaystyle y(t)-y(t_{0})=\int _{t_{0}}^{t}f(s,y(s))\,ds.}
Given the hypotheses that f {\displaystyle f} is continuous in t {\displaystyle t} and Lipschitz continuous in y {\displaystyle y} , this integral operator is a contraction (See detailed proof below) and so the Banach fixed-point theorem proves that a solution can be obtained by fixed-point iteration of successive approximations. In this context, this fixed-point iteration method is known as Picard iteration. Set
φ 0 ( t ) = y 0 {\displaystyle \varphi _{0}(t)=y_{0}}
and
φ k + 1 ( t ) = y 0 + ∫ t 0 t f ( s , φ k ( s ) ) d s . {\displaystyle \varphi _{k+1}(t)=y_{0}+\int _{t_{0}}^{t}f(s,\varphi _{k}(s))\,ds.}
It follows from the Banach fixed-point theorem that the sequence of "Picard iterates" φ k {\textstyle \varphi _{k}} is convergent and that its limit is a solution to the original initial value problem:
lim k → ∞ φ k ( t ) = y ( t ) {\displaystyle \lim _{k\to \infty }\varphi _{k}(t)=y(t)} . Since the Banach fixed-point theorem states that the fixed-point is unique, the solution found through this iteration is the unique solution to the differential equation given an initial value.
Example of Picard iteration
Let y ( t ) = tan ( t ) , {\displaystyle y(t)=\tan(t),} the solution to the equation y ′ ( t ) = 1 + y ( t ) 2 {\displaystyle y'(t)=1+y(t)^{2}} with initial condition y ( t 0 ) = y 0 = 0 , t 0 = 0. {\displaystyle y(t_{0})=y_{0}=0,t_{0}=0.} Starting with φ 0 ( t ) = 0 , {\displaystyle \varphi _{0}(t)=0,} we iterate
φ k + 1 ( t ) = ∫ 0 t ( 1 + ( φ k ( s ) ) 2 ) d s {\displaystyle \varphi _{k+1}(t)=\int _{0}^{t}(1+(\varphi _{k}(s))^{2})\,ds}
so that φ n ( t ) → y ( t ) {\displaystyle \varphi _{n}(t)\to y(t)} :
φ 1 ( t ) = ∫ 0 t ( 1 + 0 2 ) d s = t {\displaystyle \varphi _{1}(t)=\int _{0}^{t}(1+0^{2})\,ds=t}
φ 2 ( t ) = ∫ 0 t ( 1 + s 2 ) d s = t + t 3 3 {\displaystyle \varphi _{2}(t)=\int _{0}^{t}(1+s^{2})\,ds=t+{\frac {t^{3}}{3}}}
φ 3 ( t ) = ∫ 0 t ( 1 + ( s + s 3 3 ) 2 ) d s = t + t 3 3 + 2 t 5 15 + t 7 63 {\displaystyle \varphi _{3}(t)=\int _{0}^{t}\left(1+\left(s+{\frac {s^{3}}{3}}\right)^{2}\right)\,ds=t+{\frac {t^{3}}{3}}+{\frac {2t^{5}}{15}}+{\frac {t^{7}}{63}}}
and so on. Evidently, the functions are computing the Taylor series expansion of our known solution y = tan ( t ) . {\displaystyle y=\tan(t).} Since tan {\displaystyle \tan } has poles at ± π 2 , {\displaystyle \pm {\tfrac {\pi }{2}},} it is not Lipschitz-continuous in the neighborhood of those points, and the iteration converges toward a local solution for | t | < π 2 {\displaystyle |t|<{\tfrac {\pi }{2}}} only that is not valid over all of R {\displaystyle \mathbb {R} } .
Example of non-uniqueness To understand uniqueness of solutions, contrast the following two examples of first order ordinary differential equations for y(t). Both differential equations will possess a single stationary point y = 0. First, the homogeneous linear equation dy/dt = ay ( a < 0 {\displaystyle a<0} ), a stationary solution is y(t) = 0, which is obtained for the initial condition y(0) = 0. Beginning with any other initial condition y(0) = y0 ≠ 0, the solution y ( t ) = y 0 e a t {\displaystyle y(t)=y_{0}e^{at}} tends toward the stationary point y = 0, but it only approaches it in the limit of infinite time, so the uniqueness of solutions over all finite times is guaranteed. By contrast for an equation in which the stationary point can be reached after a finite time, uniqueness of solutions does not hold. Consider the homogeneous nonlinear equation dy/dt = ay 2/3, which has at least these two solutions corresponding to the initial condition y(0) = 0: y(t) = 0 and
y ( t ) = { ( a t 3 ) 3 t < 0 0 t ≥ 0 , {\displaystyle y(t)={\begin{cases}\left({\tfrac {at}{3}}\right)^{3}&t<0\\\ \ \ \ 0&t\geq 0,\end{cases}}}
so the previous state of the system is not uniquely determined by its state at or after t = 0. The uniqueness theorem does not apply because the derivative of the function f (y) = y 2/3 is not bounded in the neighborhood of y = 0 and therefore it is not Lipschitz continuous, violating the hypothesis of the theorem.
Detailed proof Let L {\displaystyle L} be the Lipschitz constant of ( t , y ) ↦ f ( t , y ) {\displaystyle (t,y)\mapsto f(t,y)} with respect to y . {\displaystyle y.}
The function f {\displaystyle f} is continuous as a function of ( t , y ) {\displaystyle (t,y)} . In particular, since t ↦ f ( t , y ) {\displaystyle t\mapsto f(t,y)} is a continuous function of t {\displaystyle t} , we have that for any point ( t 0 , y 0 ) {\displaystyle (t_{0},y_{0})} and ϵ > 0 {\displaystyle \epsilon >0} there exist δ > 0 {\displaystyle \delta >0}
such that | f ( t , y 0 ) − f ( t 0 , y 0 ) | < ϵ / 2 {\displaystyle |f(t,y_{0})-f(t_{0},y_{0})|<\epsilon /2} when | t − t 0 | < δ {\displaystyle |t-t_{0}|<\delta } . We have
| f ( t , y ) − f ( t 0 , y 0 ) | ≤ | f ( t , y ) − f ( t , y 0 ) | + | f ( t , y 0 ) − f ( t 0 , y 0 ) | < ϵ , {\displaystyle |f(t,y)-f(t_{0},y_{0})|\leq |f(t,y)-f(t,y_{0})|+|f(t,y_{0})-f(t_{0},y_{0})|<\epsilon ,}
provided | t − t 0 | < δ {\displaystyle |t-t_{0}|<\delta } and | y − y 0 | < ϵ / 2 L {\displaystyle |y-y_{0}|<\epsilon /2L} , which shows that f {\displaystyle f} is continuous at ( t 0 , y 0 ) {\displaystyle (t_{0},y_{0})} . Let a := 1 / 2 L {\displaystyle a:=1/2L} and take any b > 0 {\displaystyle b>0} such that
C a , b = I a ( t 0 ) × B b ( y 0 ) {\displaystyle C_{a,b}=I_{a}(t_{0})\times B_{b}(y_{0})}
is a subset of D , {\displaystyle D,} where
I a ( t 0 ) = [ t 0 − a , t 0 + a ] B b ( y 0 ) = [ y 0 − b , y 0 + b ] . {\displaystyle {\begin{aligned}I_{a}(t_{0})&=[t_{0}-a,t_{0}+a]\\B_{b}(y_{0})&=[y_{0}-b,y_{0}+b].\end{aligned}}}
Such a set exists because ( t 0 , y 0 ) {\displaystyle (t_{0},y_{0})} is in the interior of D , {\displaystyle D,} by assumption. Let
M = sup ( t , y ) ∈ C a , b ‖ f ( t , y ) ‖ , {\displaystyle M=\sup _{(t,y)\in C_{a,b}}\|f(t,y)\|,}
which is the supremum of (the absolute values of) the slopes of the function. The function f {\displaystyle f} attains a maximum on C a , b {\displaystyle C_{a,b}} because f {\displaystyle f} is continuous and C a , b {\displaystyle C_{a,b}} is compact. For a later step in the proof, we need that a < b / M , {\displaystyle a<b/M,} so if a ≥ b / M , {\displaystyle a\geq b/M,} then change a {\displaystyle a} to a := 1 2 min { 1 / L , b / M } , {\displaystyle a:={\tfrac {1}{2}}\min\{1/L,\ b/M\},} and update I a ( t 0 ) , {\displaystyle I_{a}(t_{0}),} B b ( y 0 ) , {\displaystyle B_{b}(y_{0}),} C a , b , {\displaystyle C_{a,b},} and M {\displaystyle M} accordingly (this update will be needed at most once since M {\displaystyle M} cannot increase as a result of restricting C a , b {\displaystyle C_{a,b}} ). Consider C ( I a ( t 0 ) , B b ( y 0 ) ) {\displaystyle {\mathcal {C}}(I_{a}(t_{0}),B_{b}(y_{0}))} , the function space of continuous functions I a ( t 0 ) → B b ( y 0 ) . {\displaystyle I_{a}(t_{0})\to B_{b}(y_{0}).}
We will proceed by applying the Banach fixed-point theorem using the metric on C ( I a ( t 0 ) , B b ( y 0 ) ) {\displaystyle {\mathcal {C}}(I_{a}(t_{0}),B_{b}(y_{0}))} induced by the uniform norm. Namely, for each continuous function φ : I a ( t 0 ) → B b ( y 0 ) , {\displaystyle \varphi :I_{a}(t_{0})\to B_{b}(y_{0}),} the norm of φ {\displaystyle \varphi } is
‖ φ ‖ ∞ = sup t ∈ I a ‖ φ ( t ) ‖ . {\displaystyle \|\varphi \|_{\infty }=\sup _{t\in I_{a}}\|\varphi (t)\|.}
The Picard operator Γ : C ( I a ( t 0 ) , B b ( y 0 ) ) → C ( I a ( t 0 ) , B b ( y 0 ) ) {\displaystyle \Gamma :{\mathcal {C}}{\big (}I_{a}(t_{0}),B_{b}(y_{0}){\big )}\to {\mathcal {C}}{\big (}I_{a}(t_{0}),B_{b}(y_{0}){\big )}} is defined for each φ ∈ C ( I a ( t 0 ) , B b ( y 0 ) ) {\displaystyle \varphi \in {\mathcal {C}}(I_{a}(t_{0}),B_{b}(y_{0}))} by Γ φ ∈ C ( I a ( t 0 ) , B b ( y 0 ) ) {\displaystyle \Gamma \varphi \in {\mathcal {C}}(I_{a}(t_{0}),B_{b}(y_{0}))} given by
Γ φ ( t ) = y 0 + ∫ t 0 t f ( s , φ ( s ) ) d s ∀ t ∈ I a ( t 0 ) . {\displaystyle \Gamma \varphi (t)=y_{0}+\int _{t_{0}}^{t}f(s,\varphi (s))\,ds\quad \forall t\in I_{a}(t_{0}).}
To apply the Banach fixed-point theorem, we must show that Γ {\displaystyle \Gamma } maps a complete non-empty metric space X into itself and also is a contraction mapping. We first show that Γ {\displaystyle \Gamma } takes B b ( y 0 ) {\displaystyle B_{b}(y_{0})} into itself in the space of continuous functions with the uniform norm. Here, B b ( y 0 ) {\displaystyle B_{b}(y_{0})} is a closed ball in the space of continuous (and bounded) functions "centered" at the constant function y 0 {\displaystyle y_{0}} . Hence we need to show that
‖ φ − y 0 ‖ ∞ ≤ b {\displaystyle \|\varphi -y_{0}\|_{\infty }\leq b}
implies
‖ Γ φ ( t ) − y 0 ‖ = ‖ ∫ t 0 t f ( s , φ ( s ) ) d s ‖ ≤ ∫ t 0 t ′ ‖ f ( s , φ ( s ) ) ‖ d s ≤ ∫ t 0 t ′ M d s = M | t ′ − t 0 | ≤ M a ≤ b {\displaystyle \left\|\Gamma \varphi (t)-y_{0}\right\|=\left\|\int _{t_{0}}^{t}f(s,\varphi (s))\,ds\right\|\leq \int _{t_{0}}^{t'}\left\|f(s,\varphi (s))\right\|ds\leq \int _{t_{0}}^{t'}M\,ds=M\left|t'-t_{0}\right|\leq Ma\leq b}
where t ′ {\displaystyle t'} is some number in [ t 0 − a , t 0 + a ] {\displaystyle [t_{0}-a,t_{0}+a]} where the maximum is achieved. The last inequality in the chain is true since a < b / M . {\displaystyle a<b/M.}
Now let us prove that Γ {\displaystyle \Gamma } is a contraction mapping as required to apply the Banach fixed-point theorem. In particular, we want to show that there exists 0 ≤ q < 1 , {\displaystyle 0\leq q<1,} such that
‖ Γ φ 1 − Γ φ 2 ‖ ∞ ≤ q ‖ φ 1 − φ 2 ‖ ∞ {\displaystyle \left\|\Gamma \varphi _{1}-\Gamma \varphi _{2}\right\|_{\infty }\leq q\left\|\varphi _{1}-\varphi _{2}\right\|_{\infty }}
for all φ 1 , φ 2 ∈ C ( I a ( t 0 ) , B b ( y 0 ) ) . {\displaystyle \varphi _{1},\varphi _{2}\in {\mathcal {C}}(I_{a}(t_{0}),B_{b}(y_{0})).}
Let q = a L {\displaystyle q=aL} and take any φ 1 , φ 2 ∈ C ( I a ( t 0 ) , B b ( y 0 ) ) . {\displaystyle \varphi _{1},\varphi _{2}\in {\mathcal {C}}(I_{a}(t_{0}),B_{b}(y_{0})).}
Take t {\displaystyle t} such that
‖ Γ φ 1 − Γ φ 2 ‖ ∞ = ‖ ( Γ φ 1 − Γ φ 2 ) ( t ) ‖ . {\displaystyle \|\Gamma \varphi _{1}-\Gamma \varphi _{2}\|_{\infty }=\left\|\left(\Gamma \varphi _{1}-\Gamma \varphi _{2}\right)(t)\right\|.}
Then, using the definition of Γ {\displaystyle \Gamma } ,
‖ ( Γ φ 1 − Γ φ 2 ) ( t ) ‖ = ‖ ∫ t 0 t ( f ( s , φ 1 ( s ) ) − f ( s , φ 2 ( s ) ) ) d s ‖