6 Existence and uniqueness
(All solutions of higher order complex linear autonomous ODEs) Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(k\ge1,\) \(a_{0},a_{1},\ldots,a_{k-1}\in\mathbb{C}.\) Consider the ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0.\tag{5.124}\] Assume that the characteristic polynomial \(p(r)=r^{k}+\sum_{l=0}^{k-1}a_{l}r^{l}\) has \(k'\) distinct roots \(r_{1},\ldots,r_{k'}\in\mathbb{C},\) each of multiplicity \(m_{1},m_{2},\ldots,m_{k'}\in\{1,2,\ldots\},\) where \(m_{1}+\ldots+m_{k'}=k.\) Let \[x_{1}(t),\ldots,x_{k}(t)\text{ denote the functions }e^{r_{1}t},te^{r_{1}t},\ldots,t^{m_{1}-1}e^{r_{1}t},e^{r_{2}t},te^{r_{2}t},\ldots,t^{m_{k'}-1}e^{r_{k'}t}.\]
a) A function \(x\) solves the ODE ([eq:ch5_ho_lin_aut_complex_all_sols_hom_ODE]) iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\tag{5.125}\] for some \(\alpha\in\mathbb{C}^{k}.\)
b) For any \(t_{*}\in I,\) the possibly complex \(k\times k\) matrix \(D\) with entries \[D_{i,l}=x_{i}^{(l)}(t_{*}),\] is non-degenerate.
c) Let \(t_{*}\in I\) and \(x_{*}\in\mathbb{C}^{k}.\) A function \(x\) solves the constrained homogeneous ODE \[x^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}x^{(l)}(t)=0,\quad\quad t\in I,\quad\quad x^{(l)}(t_{*})=x_{*,l+1},0\le l\le k-1,\tag{5.126}\] iff \[x(t)=\sum_{i=1}^{k}\alpha_{i}x_{i}(t),\] where \(\alpha\in\mathbb{C}^{k}\) is the unique solution to \[D\alpha=x_{*},\quad\quad\text{i.e. }\quad\quad\alpha=D^{-1}x_{*}.\]
d) If \(b\in\mathbb{C}\) and \(a_{0}\ne0\) a function \(y\) solves the non-homogeneous autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,\tag{5.127}\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] for an \(x(t)\) as in part a). If \(t_{*}\in I,y_{*}\in\mathbb{C}\) then a function \(y\) solves the constrained non-homogeneous second order autonomous ODE \[y^{(k)}(t)+\sum_{l=0}^{k-1}a_{l}y^{(l)}(t)=b,\quad\quad t\in I,y^{(l)}(t_{*})=y_{*,l+1},0\le l\le k-1,\] iff \[y(t)=x(t)+\frac{b}{a_{0}},\quad\quad t\in I,\] where \(x(t)\) is the unique function from part c) with \[\alpha=D^{-1}z_{*}\quad\quad\text{ for }\quad\quad z_{*}=y_{*}-\left(\begin{matrix}\frac{b}{a_{0}}\\
0\\
\ldots\\
0
\end{matrix}\right).\]
Let \(I\subset\mathbb{R}\) be an interval and let \(a,b:I\to\mathbb{R}\) be continuous and Lebesgue integrable on \(I.\)
A function \(y:I\to\mathbb{R}\) is a solution to the ODE \[\dot{y}(t)=a(t)y(t)+b(t),\quad\quad t\in I,\tag{3.60}\] iff \[y(t)=\alpha e^{\int_{t_{0}}^{t}a(s)ds}+\int_{t_{0}}^{t}e^{\int_{r}^{t}a(s)ds}b(r)dr,\quad\quad t\in I,\tag{3.61}\] for some \(\alpha\in\mathbb{R}.\)
6.1 Uniqueness - first order
We start with the standard uniqueness theorem for general first order ODEs. We have already seen an example of the failure of uniqueness in Section 3.7. In particular, we saw that the IVP \[y'=|y|^{1/2},\quad\quad y(0)=0,\tag{6.1}\] is solved by the functions \[y(t)=\begin{cases} 0 & \text{\,if }t\le s,\\ \frac{1}{4}\left(t-s\right)^{2} & \text{ for }t>s, \end{cases}\] for any \(s\ge0.\)
The uniqueness theorem essentially speaking says that the constrained first order ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0},\] has at most one solution if the function \(F(t,y)\) is sufficiently well-behaved. At most, because the theorem makes no claim about the existence of solutions, it only says that if one exists it is unique. We will treat conditions for existence later in the chapter. The conditions on \(F\) for uniqueness are that it is bounded and Lipschitz continuous in the \(y\) variable.
\(\text{ }\) Lipschitz Continuity Recall that if \(f:A\to B\) is a function for \(A,B\subset\mathbb{R}\) and \[\left|f(x)-f(y)\right|\le L\left|x-y\right|\text{ for all }x,y\in A,\tag{6.2}\] we say that \(f\) is Lipschitz continuous with Lipschitz constant \(L.\) For short, we say that \(f\) is \(L\)-Lipschitz. A note on this terminology: we do not require that the constant \(L\) is “optimal”; \(f\) is \(L\)-Lipschitz for any \(L\) such that ([eq:ch6_uniq_lipschitz_def]) holds, and if it is \(L\)-Lipschitz then it is also \(L'\)-Lipschitz for any \(L'\ge L.\) More generally, the same condition ([eq:ch6_uniq_lipschitz_def]) defines Lipschitz continuity for functions \(f:A\to B\) where \(A\) and \(B\) are any normed spaces: for us the cases where \(A\) and \(B\) are \(\mathbb{R}^{d}\) or \(\mathbb{C}^{d}\) for \(d\ge1\) will be relevant.
\(\)
\(f:\mathbb{R}\to\mathbb{R},f(x)=x\) is Lipschitz continuous with constant \(L=1.\)
For any \(a,b\in\mathbb{R}\) the function \(f:\mathbb{R}\to\mathbb{R},f(x)=ax+b\) is Lipschitz continuous with constant \(L=a.\)
\(f:\mathbb{R}\to\mathbb{R},f(x)=x^{2}\) is not Lipschitz continuous, since \(\left|x^{2}-y^{2}\right|=\left|x-y\right|\left|x+y\right|,\) where \(\left|x+y\right|\) can be arbitrarily large.
\(f:[-1,1]\to\mathbb{R},f(x)=x^{2}\) is Lipschitz with constant \(L=2,\) since \(\left|x^{2}-y^{2}\right|=\left|x-y\right|\left|x+y\right|\le2\left|x-y\right|\) for any \(x,y\in[-1,1].\)
If \(f:A\to\mathbb{R}\) for \(A\subset\mathbb{R}\) is differentiable and \(\sup_{x\in A}\left|f'(x)\right|<\infty,\) then \(f\) is Lipschitz with constant \(L=\sup_{x\in A}\left|f'(x)\right|,\) since by the fundamental theorem of calculus \[\left|f(x)-f(y)\right|=\left|\int_{y}^{x}f'(z)dz\right|\le\int_{y}^{x}\left|f'(z)\right|dz\le\int_{y}^{x}Ldz=\left|x-y\right|L.\]
If on the other hand \(f(x)\) is differentiable and \(\left|f'(x)\right|\) is not bounded on \(A,\) then \(f\) is not Lipschitz (the proof is part of exercise sheet 6).
\(f:[-1,1]\to\mathbb{R},f(x)=\left|x\right|^{1/2}\) is not Lipschitz, since \(\frac{\left|f(x)-f(0)\right|}{\left|x\right|}=\frac{1}{\left|x\right|^{1/2}}\to\infty\) as \(x\to0,\) so there is no constant \(L\) such that \(\left|f(x)-f(0)\right|\le L\left|x\right|\) for all \(x\in[-1,1].\) Alternatively, it’s not Lipschitz continuous by f), since on the the smaller set \((0,1)\) its differentiable but \(\left|f'(x)\right|\to\infty\) as \(x\to0.\)
For any \(\varepsilon>0\) the function \(f:[\varepsilon,\infty)\to\mathbb{R},f(x)=\left|x\right|^{1/2}\) is Lipschitz continuous by g), since \(f\) is differentiable and \(\left|f'(x)\right|=\frac{1}{2}\frac{1}{\left|x\right|^{1/2}}\le\frac{1}{2\varepsilon^{1/2}}<\infty\) for all \(x\ge\varepsilon.\)
In the example ([eq:ch6_uniq_fo_ex_failure]) where solutions fail to be unique we have \(F(t,y)=|y|^{1/2},\) which we can see from g) above is not Lipschitz in \(y\) on any interval containing zero.
\(\text{ }\) Statement of uniqueness result We now formulate the uniqueness theorem for general first order ODEs.
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
\(\)
\(f:\mathbb{R}\to\mathbb{R},f(x)=x\) is Lipschitz continuous with constant \(L=1.\)
For any \(a,b\in\mathbb{R}\) the function \(f:\mathbb{R}\to\mathbb{R},f(x)=ax+b\) is Lipschitz continuous with constant \(L=a.\)
\(f:\mathbb{R}\to\mathbb{R},f(x)=x^{2}\) is not Lipschitz continuous, since \(\left|x^{2}-y^{2}\right|=\left|x-y\right|\left|x+y\right|,\) where \(\left|x+y\right|\) can be arbitrarily large.
\(f:[-1,1]\to\mathbb{R},f(x)=x^{2}\) is Lipschitz with constant \(L=2,\) since \(\left|x^{2}-y^{2}\right|=\left|x-y\right|\left|x+y\right|\le2\left|x-y\right|\) for any \(x,y\in[-1,1].\)
If \(f:A\to\mathbb{R}\) for \(A\subset\mathbb{R}\) is differentiable and \(\sup_{x\in A}\left|f'(x)\right|<\infty,\) then \(f\) is Lipschitz with constant \(L=\sup_{x\in A}\left|f'(x)\right|,\) since by the fundamental theorem of calculus \[\left|f(x)-f(y)\right|=\left|\int_{y}^{x}f'(z)dz\right|\le\int_{y}^{x}\left|f'(z)\right|dz\le\int_{y}^{x}Ldz=\left|x-y\right|L.\]
If on the other hand \(f(x)\) is differentiable and \(\left|f'(x)\right|\) is not bounded on \(A,\) then \(f\) is not Lipschitz (the proof is part of exercise sheet 6).
\(f:[-1,1]\to\mathbb{R},f(x)=\left|x\right|^{1/2}\) is not Lipschitz, since \(\frac{\left|f(x)-f(0)\right|}{\left|x\right|}=\frac{1}{\left|x\right|^{1/2}}\to\infty\) as \(x\to0,\) so there is no constant \(L\) such that \(\left|f(x)-f(0)\right|\le L\left|x\right|\) for all \(x\in[-1,1].\) Alternatively, it’s not Lipschitz continuous by f), since on the the smaller set \((0,1)\) its differentiable but \(\left|f'(x)\right|\to\infty\) as \(x\to0.\)
For any \(\varepsilon>0\) the function \(f:[\varepsilon,\infty)\to\mathbb{R},f(x)=\left|x\right|^{1/2}\) is Lipschitz continuous by g), since \(f\) is differentiable and \(\left|f'(x)\right|=\frac{1}{2}\frac{1}{\left|x\right|^{1/2}}\le\frac{1}{2\varepsilon^{1/2}}<\infty\) for all \(x\ge\varepsilon.\)
\(\)
\(f:\mathbb{R}\to\mathbb{R},f(x)=x\) is Lipschitz continuous with constant \(L=1.\)
For any \(a,b\in\mathbb{R}\) the function \(f:\mathbb{R}\to\mathbb{R},f(x)=ax+b\) is Lipschitz continuous with constant \(L=a.\)
\(f:\mathbb{R}\to\mathbb{R},f(x)=x^{2}\) is not Lipschitz continuous, since \(\left|x^{2}-y^{2}\right|=\left|x-y\right|\left|x+y\right|,\) where \(\left|x+y\right|\) can be arbitrarily large.
\(f:[-1,1]\to\mathbb{R},f(x)=x^{2}\) is Lipschitz with constant \(L=2,\) since \(\left|x^{2}-y^{2}\right|=\left|x-y\right|\left|x+y\right|\le2\left|x-y\right|\) for any \(x,y\in[-1,1].\)
If \(f:A\to\mathbb{R}\) for \(A\subset\mathbb{R}\) is differentiable and \(\sup_{x\in A}\left|f'(x)\right|<\infty,\) then \(f\) is Lipschitz with constant \(L=\sup_{x\in A}\left|f'(x)\right|,\) since by the fundamental theorem of calculus \[\left|f(x)-f(y)\right|=\left|\int_{y}^{x}f'(z)dz\right|\le\int_{y}^{x}\left|f'(z)\right|dz\le\int_{y}^{x}Ldz=\left|x-y\right|L.\]
If on the other hand \(f(x)\) is differentiable and \(\left|f'(x)\right|\) is not bounded on \(A,\) then \(f\) is not Lipschitz (the proof is part of exercise sheet 6).
\(f:[-1,1]\to\mathbb{R},f(x)=\left|x\right|^{1/2}\) is not Lipschitz, since \(\frac{\left|f(x)-f(0)\right|}{\left|x\right|}=\frac{1}{\left|x\right|^{1/2}}\to\infty\) as \(x\to0,\) so there is no constant \(L\) such that \(\left|f(x)-f(0)\right|\le L\left|x\right|\) for all \(x\in[-1,1].\) Alternatively, it’s not Lipschitz continuous by f), since on the the smaller set \((0,1)\) its differentiable but \(\left|f'(x)\right|\to\infty\) as \(x\to0.\)
For any \(\varepsilon>0\) the function \(f:[\varepsilon,\infty)\to\mathbb{R},f(x)=\left|x\right|^{1/2}\) is Lipschitz continuous by g), since \(f\) is differentiable and \(\left|f'(x)\right|=\frac{1}{2}\frac{1}{\left|x\right|^{1/2}}\le\frac{1}{2\varepsilon^{1/2}}<\infty\) for all \(x\ge\varepsilon.\)
\(\text{ }\) Grönwall’s lemma To prove the uniqueness theorem we use Grönwall’s lemma, which is an important result of real analysis. It can be described as a “solution” of the “first order differential inequality” \[u'(t)\le u(t)a(t).\tag{6.6}\] The lemma gives an upper bound on \(u(t)\) that holds whenever \(u(t)\) satisfies the inequality ([eq:ch6_gronwall_intro_diff_ass]) on an interval. More precisely, the bound is \[u(t)\le u(t_{0})\exp\left(\int_{t_{0}}^{t}a(s)ds\right),\tag{6.7}\] for \(t\ge t_{0}\). This is indeed a natural upper bound, since the inequality ([eq:ch6_gronwall_intro_diff_ass]) suggests that \(u\) grows slower than a function \(v\) which satisfies \[v'(t)=v(t)a(t),\quad\quad t\ge t_{0},\quad\quad v(t_{0})=u(t_{0}),\tag{6.8}\] i.e. ([eq:ch6_gronwall_intro_diff_ass]) but with equality. But ([eq:ch6_gronwall_intro_diff_ODE_v]) is just a first order IVP. In fact it is also linear and homogeneous, and in Section 3.5 we learned that its unique solution is \[v(t)=u(t_{0})\exp\left(\int_{t_{0}}^{t}a(s)ds\right).\] This is precisely the r.h.s of the upper bound ([eq:ch6_gronwall_intro_diff_claim]) that Grönwall’s lemma provides.
In our proof of uniqueness \(u\) will be the difference of two solutions of the same IVP ([eq:ch6_fo_uniquness_thm_ODE]) , so that \(u(t_{0})=0\) and we will use the lemma to deduce that \(u(t)\le0\) for \(t\ge0.\)
Grönwall’s lemma has two forms: the one discussed above is the differential form. The alternative integral form instead assumes the integral inequality \[u(t)\le\int_{t_{0}}^{t}a(s)u(s)ds\tag{6.9}\] and deduces an upper bound for \(u(t).\) Note that if we integrate the differential inequality ([eq:ch6_gronwall_intro_diff_ass]) we obtain the integral version ([eq:ch6_gronwall_intro_int_ass]). The integral form has the technical advantage that it applies even without the assumption that \(u\) is differentiable. This is convenient here, so we will use the integral form. We include the differential form for completeness, and because it is a bit simpler to state and prove. Interestingly, the proof of Grönwall’s lemma uses essentially the same argument as the proof of existence an uniqueness of linear first order ODEs we saw in Section 3.5.
Let \(t_{0},t_{1}\in\mathbb{R},t_{0}<t_{1}\) and \(I=[t_{0},t_{1}].\)
a) [Differential form] If \(a:I\to\mathbb{R}\) is continuous and \(u:I\to\mathbb{R}\) is a differentiable function such that \[u'(t)\le a(t)u(t)\quad\quad\forall t\in I,\tag{6.10}\] then \[u(t)\le u(t_{0})\exp\left(\int_{t_{0}}^{t}a(s)ds\right)\text{ for all }t\in I.\tag{6.11}\] b) [Integral form] If \(a,u:I\to\mathbb{R}\) are continuous and s.t. \[u(t)\le\int_{t_{0}}^{t}a(s)u(s)ds\quad\quad\forall t\in I,\tag{6.12}\] then \[u(t)\le0\quad\quad\forall t\in I.\tag{6.13}\]
Proof. a) Since \(a\) is continuous on the compact set \(I\) it is bounded, so the integral in ([eq:gronwall_claim_diff]) is well-defined. Now assume \(u:I\to\mathbb{R}\) is differentiable and satisfies ([eq:gronwall_assumption_diff]). Consider \[z(t)=\exp\left(-\int_{t_{0}}^{t}a(s)ds\right)u(t),\quad\quad t\in I.\tag{6.14}\] Because \(a\) is continuous the fundamental theorem of calculus implies that \(t\to\int_{t_{0}}^{t}a(s)ds\) is differentiable, with derivative \(a(t).\) Using the product and chain rules we therefore have that \(z\) is differentiable with \[z'(t)=-a(t)\exp\left(-\int_{t_{0}}^{t}a(s)ds\right)u(t)+\exp\left(-\int_{t_{0}}^{t}a(s)ds\right)u'(t).\] Combining this with the assumption ([eq:gronwall_assumption_diff]) gives \[z'(t)\le-a(t)\exp\left(-\int_{t_{0}}^{t}a(s)ds\right)u(t)+\exp\left(-\int_{t_{0}}^{t}a(s)ds\right)a(t)u(t)=0.\] So \(z'(t)\le0\) on \(I,\) which means that the function \(z:I\to\mathbb{R}\) is non-increasing. Furthermore \[z(t_{0})=\exp\left(-\int_{t_{0}}^{t_{0}}a(s)ds\right)u(t_{0})=u(t_{0}).\] Thus \[z(t)\le u(t_{0})\text{ for all }t\in I\tag{6.15}\] which by the definition ([eq:gronwall_z_def]) of \(z(t)\) implies ([eq:gronwall_claim_diff]).
b) Since \(a,u\) are continuous on the compact \(I\) set they are bounded, so the integrals in ([eq:gronwall_assumption_int]) and ([eq:gronwall_claim_int]) are well-defined. The proof consists in proving that the assumption ([eq:gronwall_assumption_int]) implies that \[\int_{t_{0}}^{t}a(s)u(s)ds\le0\text{ for all }t\in I.\tag{6.16}\] To this end define \[z(t)=\exp\left(-\int_{t_{0}}^{t}a(r)dr\right)\int_{t_{0}}^{t}a(s)u(s)ds.\tag{6.17}\] Since \(a,u\) are continuous the maps \(t\to\int_{t_{0}}^{t}a(r)dr\) and \(t\to\int_{t_{0}}^{t}a(s)u(s)ds\) are differentiable. Therefore also \(z\) is, and \[z'(t)=-a(t)\exp\left(-\int_{t_{0}}^{t}a(r)dr\right)\int_{t_{0}}^{t}a(s)u(s)ds+\exp\left(-\int_{t_{0}}^{t}a(r)dr\right)a(t)u(t).\] Using the assumption ([eq:gronwall_assumption_int]) shows that the r.h.s. is at most \(0,\) so that \[z'(t)\le0\quad\quad t\in I.\] The definition ([eq:gronwall_int_z_def]) of \(z\) implies that \(z(t_{0})=0,\) so we deduce that \[z(t)\le0\quad\quad t\in I.\tag{6.18}\] Since the first factor on the r.h.s. of ([eq:gronwall_int_z_def]) is positive this implies ([eq:gronwall_int_wts]).
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
Let \(t_{0}\in\mathbb{R}\) and let \(I=[t_{0},\infty),\) or \(I=[t_{0},t_{1})\) or \(I=[t_{0},t_{1}]\) for \(t_{1}>t_{0}.\) Let \(I_{y}\subset\mathbb{R}\) be an interval, and \(F:I\times I_{y}\to\mathbb{R}\) be bounded, and \(y_{0}\in I_{y}.\) Assume that \(L>0\) and that \(y\to F(t,y)\) is \(L\)-Lipschitz for all \(t\in I.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.19}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad t\in I.\tag{6.20}\]
Let \(t_{0}\in\mathbb{R}\) and let \(I=[t_{0},\infty),\) or \(I=[t_{0},t_{1})\) or \(I=[t_{0},t_{1}]\) for \(t_{1}>t_{0}.\) Let \(I_{y}\subset\mathbb{R}\) be an interval, and \(F:I\times I_{y}\to\mathbb{R}\) be bounded, and \(y_{0}\in I_{y}.\) Assume that \(L>0\) and that \(y\to F(t,y)\) is \(L\)-Lipschitz for all \(t\in I.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.19}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad t\in I.\tag{6.20}\]
Let \(t_{0},t_{1}\in\mathbb{R},t_{0}<t_{1}\) and \(I=[t_{0},t_{1}].\)
a) [Differential form] If \(a:I\to\mathbb{R}\) is continuous and \(u:I\to\mathbb{R}\) is a differentiable function such that \[u'(t)\le a(t)u(t)\quad\quad\forall t\in I,\tag{6.10}\] then \[u(t)\le u(t_{0})\exp\left(\int_{t_{0}}^{t}a(s)ds\right)\text{ for all }t\in I.\tag{6.11}\] b) [Integral form] If \(a,u:I\to\mathbb{R}\) are continuous and s.t. \[u(t)\le\int_{t_{0}}^{t}a(s)u(s)ds\quad\quad\forall t\in I,\tag{6.12}\] then \[u(t)\le0\quad\quad\forall t\in I.\tag{6.13}\]
Let \(t_{0}\in\mathbb{R}\) and let \(I=[t_{0},\infty),\) or \(I=[t_{0},t_{1})\) or \(I=[t_{0},t_{1}]\) for \(t_{1}>t_{0}.\) Let \(I_{y}\subset\mathbb{R}\) be an interval, and \(F:I\times I_{y}\to\mathbb{R}\) be bounded, and \(y_{0}\in I_{y}.\) Assume that \(L>0\) and that \(y\to F(t,y)\) is \(L\)-Lipschitz for all \(t\in I.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.19}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad t\in I.\tag{6.20}\]
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
Let \(t_{0}\in\mathbb{R}\) and let \(I=[t_{0},\infty),\) or \(I=[t_{0},t_{1})\) or \(I=[t_{0},t_{1}]\) for \(t_{1}>t_{0}.\) Let \(I_{y}\subset\mathbb{R}\) be an interval, and \(F:I\times I_{y}\to\mathbb{R}\) be bounded, and \(y_{0}\in I_{y}.\) Assume that \(L>0\) and that \(y\to F(t,y)\) is \(L\)-Lipschitz for all \(t\in I.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.19}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad t\in I.\tag{6.20}\]
Let \(t_{0}\in\mathbb{R}\) and let \(I=[t_{0},\infty),\) or \(I=[t_{0},t_{1})\) or \(I=[t_{0},t_{1}]\) for \(t_{1}>t_{0}.\) Let \(I_{y}\subset\mathbb{R}\) be an interval, and \(F:I\times I_{y}\to\mathbb{R}\) be bounded, and \(y_{0}\in I_{y}.\) Assume that \(L>0\) and that \(y\to F(t,y)\) is \(L\)-Lipschitz for all \(t\in I.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.19}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad t\in I.\tag{6.20}\]
We have proved a uniqueness result for general first order ODEs. What about higher order? In fact one can deduce uniqueness for higher order ODEs from the first order result, as we will see in the next section.
6.2 Uniqueness - systems of ODEs and higher order ODEs
The proof of uniqueness for higher order ODEs goes via uniqueness for systems of first order ODEs.
\(\text{ }\)Systems of ODEs. A system of ODEs are several ODEs that are coupled. For instance, consider \[\begin{array}{l} x'(t)=x(t)-y(t),\\ y'(t)=x(t)y(t). \end{array}\] The first equation specifies the derivative \(x'(t)\) in terms of the value \(x(t),\) as in a single first order ODE, but also in terms of the value \(y(t)\) of another function \(y(t).\) And the second equation specifies its derivative \(y'(t)\) in terms of both \(x(t)\) and \(y(t).\)
Systems of ODEs are important in their own right, for modelling a system whose state is described by more than one number. They are also an important tool for solving and proving theorems about the solutions of general higher order ODEs. A general system of \(d\ge1\) first order ODEs takes the form \[\begin{array}{ccc} x_{1}^{'} & = & F_{1}(t,x_{1},\ldots,x_{d}),\\ x_{2}^{'} & = & F_{2}(t,x_{1},\ldots,x_{d}),\\ \ldots & \ldots & \ldots\\ x_{d}^{'} & = & F_{d}(t,x_{1},\ldots,x_{d}), \end{array}\tag{6.30}\] for \(d\) real-valued functions \(F_{i}(t,x_{1},\ldots,x_{d})\) of \(d+1\) real variables. The system is solved by finding \(d\) real-valued functions \(x_{i}\) of the variable \(t\) that satisfy all the equations. If time permits we will return to systems of ODEs in later chapters.
\(\text{ }\)From higher order ODE to first order system. The importance of systems of ODEs for studying higher order ODEs is that every \(k\)-th order ODE is equivalent to a system of first order ODEs. For instance, consider the second order ODE \[y''-y'+y=t.\tag{6.31}\] For any solution \(y\) we can define \(x_{1}(t)=y(t)\) and \(x_{2}(t)=y'(t).\) Then by definition \[x_{1}^{'}(t)=x_{2}(t),\] and since \(x_{2}^{'}(t)=y''(t)\) it follows from ([eq:ch6_uniq_higher_order_so_example_ODE]) that \[x_{2}^{'}(t)-x_{2}(t)+x_{1}(t)=t.\] Thus \(x_{1},x_{2}\) satisfy \[\begin{array}{l} x_{1}^{'}(t)=x_{2}(t),\\ x_{2}^{'}(t)=t+x_{2}(t)-x_{1}(t), \end{array}\tag{6.32}\] which is a system of coupled first order ODEs. Conversely, if \(x_{1}\) and \(x_{2}\) are any differentiable functions that satisfy the system ([eq:ch6_uniq_higher_order_so_to_sys_example_sys]) and we let \(y(t)=x_{1}(t),\) then \(y\) is differentiable and \(y'(t)=x_{1}^{'}(t),\) but \(x_{1}^{'}(t)=x_{2}(t)\) and \(x_{2}\) is also differentiable, so \(y\) is in fact twice differentiable, and \[y''(t)-y'(t)+y(t)=x_{2}^{'}(t)-x_{2}(t)+x_{1}(t).\tag{6.33}\] By the second line of ([eq:ch6_uniq_higher_order_so_to_sys_example_sys]) the r.h.s. equals \(t,\) so \(y\) is a solution of ([eq:ch6_uniq_higher_order_so_example_ODE]). Thus there is a one-to-one correspondence between solutions of the system ([eq:ch6_uniq_higher_order_so_to_sys_example_sys]) and the solutions of the second order ODE ([eq:ch6_uniq_higher_order_so_example_ODE]).
In general, the second order ODE \[y''=F(t,y',y),t\in I\] is equivalent to the system of two first order ODEs \[\begin{array}{cl} x_{1}^{'}(t) & =x_{2}(t)\\ x_{2}^{'}(t) & =F(t,x_{2}(t),x_{1}(t)) \end{array}\] via the change of variables \(y(t)=x_{1}(t),y'(t)=x_{2}(t),\) so that \(y''(t)=x_{2}^{'}(t).\)
This trick works for any order: a \(k\)-th order ODE \[y^{(k)}=F(t,y^{(k-1)},y^{(k-2)},\ldots,y',y),\] is equivalent to the system of \(k\) first order ODEs \[\begin{array}{lcl} x_{1}^{'}(t) & = & x_{2}(t)\\ x_{2}^{'}(t) & = & x_{3}(t)\\ \ldots & \ldots & \ldots\\ x_{k-1}^{'}(t) & = & x_{k}(t)\\ x_{k}^{'}(t) & = & F(t,x_{k}(t),\ldots,x_{1}(t)) \end{array}\tag{6.34}\] via the change of variables \(y(t)=x_{1}(t),y'(t)=x_{2}(t),\ldots,y^{(k-1)}(t)=x_{k}(t),\) so that \(y^{(k)}(t)=x_{k}^{'}(t).\)
\(\text{ }\)System as single vector-valued ODE. A system of ODEs can also be viewed as a single ODE for a vector-valued function. Indeed let \(x(t)\) be an \(\mathbb{R}^{d}\)-valued function, so that \(x(t)\in\mathbb{R}^{d},\) and the coordinates \(x_{1}(t),\ldots,x_{d}(t)\) are real-valued functions such that \[x(t)=(x_{1}(t),\ldots,x_{d}(t)).\] Then the derivative in \(x'(t)\) is naturally defined as the coordinate-wise derivative \[x'(t)=(x_{1}^{'}(t),\ldots,x_{d}^{'}(t))\in\mathbb{R}^{d}.\tag{6.35}\] For any real-valued functions \(F_{1},\ldots,F_{d}\) as in the system ([eq:ch6_general_fo_system]) of \(d\) first order ODEs, we can define an \(\mathbb{R}^{d}\)-valued function \[F(t,x)=\left(F_{1}(t,x_{1},\ldots,x_{d}),\ldots,F_{d}(t,x_{1},\ldots,x_{d})\right)\] of the real \(t\in\mathbb{R}\) and vector \(x=(x_{1},\ldots,x_{d})\in\mathbb{R}^{d}.\) Then the system ([eq:ch6_general_fo_system]) can be written compactly as \[x'(t)=F(t,x),\quad\quad t\in I.\] Note that this notation is conveniently exactly the same as that for a single first order ODE!
\(\text{ }\)Higher order systems and higher order vector-valued ODEs A system of \(d\) second order ODEs takes the form \[\begin{array}{ccc} y_{1}^{''} & = & F_{1}(t,y_{1}^{'},\ldots,y_{d}^{'},y_{1},\ldots,y_{d}),\\ \ldots & \ldots & \ldots\\ y_{d}^{''} & = & F_{d}(t,y_{1}^{'},\ldots,y_{d}^{'},y_{1},\ldots,y_{d}). \end{array}\] It can similarly be written as a second order ODE for a single \(\mathbb{R}^{d}\)-valued function \(y(t)=(y_{1}(t),\ldots,y_{d}(t))\) in the form \[y''(t)=F(t,y'(t),y),\] where \(F(t,y_{1},y_{0})=(F_{1}(t,y_{1},y_{0}),\ldots,F_{d}(t,y_{1},y_{0}))\) is \(\mathbb{R}^{d}\)-valued a function of \(t\in\mathbb{R},y_{1}\in\mathbb{R}^{d},y_{2}\in\mathbb{R}^{d},\) and by the convention ([eq:ch6_derivative_of_vector_valued_func]) the second derivative \(y''(t)\) and indeed all higher derivatives are coordinate-wise: \[y^{(k)}(t)=(y_{1}^{(k)}(t),\ldots,y_{d}^{(k)}(t))\in\mathbb{R}^{d}.\]
A system of \(k\)-th order ODEs can of course also be transformed into a first order system in higher dimension using the construction ([eq:ch6_from_ho_ODE_to_fo_system]): if the original \(k\)-th order system is for an \(\mathbb{R}^{d}\)-valued function, then the new ODE is a \(\mathbb{R}^{k\times d}\)-valued first order ODE, or equivalently a system of \(k\times d\) first order ODEs.
We call any ODE of the form \[\dot{x}(t)=H(t,x(t)),\quad\quad t\in I,\tag{3.40}\] for a function \(H:\mathbb{R}^{2}\to\mathbb{R}\) and interval \(I\subset\mathbb{R}\) an explicit first order ODE.
We call any ODE of the form \[\dot{x}(t)=H(x(t)),\quad\quad t\in I,\tag{3.41}\] for a function \(H:\mathbb{R}\to\mathbb{R}\) and interval \(I\subset\mathbb{R}\) an explicit first order autonomous ODE.
We call ([eq:def_ch3_fo_explicit]) or ([eq:def_ch3_fo_aut]) together with the constraint \[x(t_{0})=x_{0},\tag{3.42}\] where \(t_{0}\) is the left-endpoint of \(I\) and \(x_{0}\in\mathbb{R}\) an explicit first order Initial Value Problem (IVP).
Let \(d\ge1\) and \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C}.\)
[ODE]Let \(I\subset\mathbb{R}\) be an non-empty interval, let \(d\ge1\) and \(I_{y}\subset\mathbb{F}^{d}\) and \[F:I\times I_{y}\to\mathbb{F}^{d}\] be a function. We call the expression \[y'(t)=F(t,y),\quad\quad t\in I,\tag{6.37}\] an \(\mathbb{F}^{d}\)-valued first order ODE (i.e. \(\mathbb{R}^{d}\)-valued or \(\mathbb{C}^{d}\)-valued), or equivalently a system of \(d\) first order \(\mathbb{F}\)-valued ODEs.
[IVP]For any \(t_{0}\in I\) and \(y_{0}\in I_{y}\) we call the constrained ODE \[y'(t)=F(t,y)\quad\quad t\ge t_{0}\quad\quad y(t_{0})=y_{0},\tag{6.38}\] a \(\mathbb{F}^{d}\)-valued IVP.
[Solution]A solution to ([eq:ch6_def_vector_valued_ODE]) or ([eq:ch6_def_vector_valued_IVP]) is any function \(x:I\to I_{y}\) which is differentiable and such that ([eq:ch6_def_vector_valued_ODE]) or ([eq:ch6_def_vector_valued_IVP]) is satisfied for \(y=x.\)
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C},\) \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(I_{y}\subset V\) a non-empty set, and \(F:I\times I_{y}\to V\) a function. Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad\forall t\in I.\]
Let \(t_{0}\in\mathbb{R}\) and let \(I=[t_{0},\infty),\) or \(I=[t_{0},t_{1})\) or \(I=[t_{0},t_{1}]\) for \(t_{1}>t_{0}.\) Let \(I_{y}\subset\mathbb{R}\) be an interval, and \(F:I\times I_{y}\to\mathbb{R}\) be bounded, and \(y_{0}\in I_{y}.\) Assume that \(L>0\) and that \(y\to F(t,y)\) is \(L\)-Lipschitz for all \(t\in I.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.19}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad t\in I.\tag{6.20}\]
Let \(t_{0}\in\mathbb{R}\) and let \(I=[t_{0},\infty),\) or \(I=[t_{0},t_{1})\) or \(I=[t_{0},t_{1}]\) for \(t_{1}>t_{0}.\) Let \(I_{y}\subset\mathbb{R}\) be an interval, and \(F:I\times I_{y}\to\mathbb{R}\) be bounded, and \(y_{0}\in I_{y}.\) Assume that \(L>0\) and that \(y\to F(t,y)\) is \(L\)-Lipschitz for all \(t\in I.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.19}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad t\in I.\tag{6.20}\]
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
Let \(I,I_{y}\subset\mathbb{R}\) be intervals and \(F:I\times I_{y}\to\mathbb{R}.\) Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad t\in I,\quad\quad y(t_{0})=y_{0}\tag{6.4}\] are in fact equal: \[y_{1}(t)=y_{2}(t)\quad\quad\forall t\in I.\]
Check line-by-line that indeed every step of the two proofs works for \(V=\mathbb{R}^{d}\) and \(V=\mathbb{C}^{d}.\)
Note that for a \(\mathbb{F}^{d}\)-valued ODE, one generically needs one \(\mathbb{F}^{d}\)-valued constraint \(y(t_{0})=y_{0}\in\mathbb{F}^{d}\) to uniquely specify a solution.
Thanks to the correspondence between systems and higher order ODEs, we can also derive from this a uniqueness result for higher order ODEs. Let us record the correspondence formally here.
Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C}\) and \(I_{y,0},\ldots,I_{y,k-1}\subset\mathbb{F}\) be non-empty sets and \[F:I\times I_{y,k-1}\times I_{y,k-2}\times\ldots\times I_{y,1}\times I_{y,0}\to\mathbb{F}\] be a function.
A function \(y\) is a solution to the \(k\)-th order ODE \[y^{(k)}(t)=F(t,y^{(k-1)}(t),\ldots,y'(t),y(t))\quad\quad t\in I\] iff \(x(t)=(y(t),\ldots,y^{(k-1)}(t))\) is a solution to the first order \(\mathbb{F}^{k}\)-valued ODE \[\dot{x}=G(t,x)\quad\quad t\in I\] where \[G(t,x)=(x_{2},x_{3},\ldots,x_{k},F(t,x_{k-1},\ldots,x_{1})).\]
Let \(t_{0}\in I\) and \(y_{0,l}\in I_{y,l},l=0,\ldots,k-1.\) A function \(y\) is a solution to a \(k\)-th order constrained ODE \[y^{(k)}(t)=F(t,y^{(k-1)}(t),\ldots,y'(t),y(t)),\quad\quad y\in I,\quad\quad y^{(l)}(t_{0})=y_{0,l},l=0,\ldots,k-1\] iff \(x(t)=(y(t),\ldots,y^{(k-1)}(t))\) is a solution to the first order \(\mathbb{F}^{k}\)-valued constrained ODE \[\dot{x}=G(t,x),\quad\quad t\in I,\quad\quad x(t_{0})=x_{0},\] where \(x_{0}=(y_{0,0},\ldots,y_{0,l}).\)
Proof. Exercise - argument is exactly as described around ([eq:ch6_uniq_higher_order_so_example_ODE])-([eq:ch6_uniq_higher_order_so_example_ODE_last]).
We can now state the uniqueness result for \(k\)-th order ODEs.
Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C}\) and \(I_{y,0},\ldots,I_{y,k-1}\subset\mathbb{F}\) be non-emtpy sets and \[F:I\times I_{y,k-1}\times I_{y,k-2}\times\ldots\times I_{y,1}\times I_{y,0}\to\mathbb{F}\] a function. Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y),y\in I_{y,k-1}\times\ldots\times I_{y,0}\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0,l}\in I_{y,l},l=0,\ldots,k-1.\) Then any \(k\) times differentiable functions \(y_{i}:I\to I_{y,0},i=1,2,\) such that \(y_{i}^{(l)}(t)\in I_{y,l},l=0,\ldots,k-1\) for all \(t\in I,\) and which solve the constrained ODE \[y^{(k)}(t)=F(t,y^{(k-1)}(t),y^{(k-2)}(t),\ldots,y'(t),y(t)),\quad t\in I,\quad y^{(l)}(t_{0})=y_{0,l},l=0,\ldots,k-1,\tag{6.41}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad\forall t\in I.\]
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C},\) \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(I_{y}\subset V\) a non-empty set, and \(F:I\times I_{y}\to V\) a function. Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad\forall t\in I.\]
Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C}\) and \(I_{y,0},\ldots,I_{y,k-1}\subset\mathbb{F}\) be non-emtpy sets and \[F:I\times I_{y,k-1}\times I_{y,k-2}\times\ldots\times I_{y,1}\times I_{y,0}\to\mathbb{F}\] a function. Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y),y\in I_{y,k-1}\times\ldots\times I_{y,0}\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0,l}\in I_{y,l},l=0,\ldots,k-1.\) Then any \(k\) times differentiable functions \(y_{i}:I\to I_{y,0},i=1,2,\) such that \(y_{i}^{(l)}(t)\in I_{y,l},l=0,\ldots,k-1\) for all \(t\in I,\) and which solve the constrained ODE \[y^{(k)}(t)=F(t,y^{(k-1)}(t),y^{(k-2)}(t),\ldots,y'(t),y(t)),\quad t\in I,\quad y^{(l)}(t_{0})=y_{0,l},l=0,\ldots,k-1,\tag{6.41}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad\forall t\in I.\]
An important aspect of this result concerns the number of conditions needed to specify a solution uniquely. In Chapter 5 we saw that for linear autonomous \(k\)-th order equations, \(k\) condition on the \(k\) derivatives \(y^{(l)},l=0,\ldots,k-1\) ensured the uniqueness of solutions. Since ([eq:ch6_ho_uniquness_IVP]) has the exact same \(k\) conditions, we see that this is the case generically for \(k\)-th order ODEs.
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C},\) \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(I_{y}\subset V\) a non-empty set, and \(F:I\times I_{y}\to V\) a function. Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad\forall t\in I.\]
6.3 Existence - first order
We now turn to proving existence of solutions to ODEs. As for uniqueness, we can use vector-valued ODEs to treat ODEs of any order and systems of ODEs in a unified manner.
We have already seen an example of the failure of global existence in Section 3.7. There we saw that the longest interval on which \[y'=y^{2},\quad\quad y(0)=y_{0},\tag{6.42}\] has a solution is \([0,\frac{1}{y_{0}}),\) when \(y_{0}>0.\) It is impossible for a solution to exist on a longer interval. Note however, that whatever value you pick for \(y_{0},\) there does exist a solution on some non-empty interval - the interval may be very short if \(y_{0}\) is large, but its length is never zero.
It turns out that if the function \(F\) in the general first order ODE \(y'=F(t,y)\) satisfies some natural conditions, we are always guaranteed the existence of a solution to an IVP for some non-empty interval - possibly very small, but always of positive length. The conditions for this is Lipschitz continuity of \(F(t,y)\) in \(y,\) which was also a condition of the uniqueness result, together with continuity of \(F(t,y)\) in both variables. We now state the theorem for vector-valued real or complex ODEs.
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C},\) and \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(t_{*}\in I,\) and \(I_{y}\subset V\) be a set and \(y_{*}\in I_{y}\) a point such for some \(r>0\) the ball \(\left\{ y\in V:\left|y-y_{*}\right|\le r\right\}\) is contained in \(I_{y}.\) For \(\varepsilon>0\) let \[I_{\varepsilon}=[t_{*}-\varepsilon,t_{*}+\varepsilon]\cap I.\] Assume that \(F:I\times I_{y}\to V\) is continuous on \(I\times I_{y}\) and \(L\)-Lipschitz in \(y\) for all \(t\in I,\) for some \(L<\infty.\) Then there exists an \(\varepsilon>0\) such the constrained ODE \[y'=F(t,y),\quad\quad t\in I_{\varepsilon},\quad\quad y(t_{*})=y_{*},\tag{6.43}\] has a solution \(y:I_{\varepsilon}\to\mathbb{R}.\)
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C},\) \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(I_{y}\subset V\) a non-empty set, and \(F:I\times I_{y}\to V\) a function. Assume that \(F\) is bounded, and that for some \(L>0\) the maps \(y\to F(t,y)\) are \(L\)-Lipschitz for all \(t\in I.\) Let \(t_{0}\in I\) and \(y_{0}\in I_{y}.\) Then any two solutions \(y_{i}:I\to I_{y},i=1,2,\) of the constrained ODE \[y'=F(t,y),\quad\quad t\in I,\quad\quad y(t_{0})=y_{0}\] are in fact equal: \[y_{1}(t)=y_{2}(t),\quad\quad\forall t\in I.\]
\(\text{ }\)Fixed-point iteration Consider an equation for a real number \(y\in\mathbb{R}\) of the form \[y=G(y),\tag{6.44}\] for some \(G:\mathbb{R}\to\mathbb{R}.\) Such an equation can be called a fixed point equation, since a solution \(y\) is a fixed point of the function \(G\): applying \(G\) to \(y\) returns the very same number \(y.\) There is a simple algorithm to solve such equations, which is to start with some \(y_{0}\in\mathbb{R}\) and then compute the sequence \[y_{n+1}=G(y_{n}),n\ge1\tag{6.45}\] iteratively. In AD 60 the ancient Greek mathematician Heron described a method for computing the square root of a number \(s\) with such an iteration. He used \(G(y)=\frac{1}{2}\left(y+\frac{s}{y}\right),\) i.e. \[y_{n+1}=\frac{1}{2}\left(y_{n}+\frac{s}{y_{n}}\right).\] The square root \(\sqrt{s}\) is a fixed point of this \(G,\) since \(G(s)=\frac{1}{2}\left(\sqrt{s}+\frac{s}{\sqrt{s}}\right)=s.\) Applying the iteration for \(s=2\) and \(y_{0}=2\) we obtain the sequence of approximations \[y_{0}=2\quad\quad y_{1}=\frac{3}{2}=1.5\quad\quad y_{2}=\frac{17}{12}\approx1.4167\quad\quad y_{3}=\frac{577}{408}\approx1.41422,\] which do in fact rapidly converge to the fixed point \(\sqrt{2}=1.41421....\) Indeed for general continuous \(G,\) if the sequence \(y_{n}\) converges to some \(y_{\infty},\) that is \[\lim_{n\to\infty}y_{n}=y_{\infty},\] then by the continuity of \(G\) \[\lim_{n\to\infty}G(y_{n})=G(y_{\infty}).\] Thus both sides of ([eq:ch6_exist_fixed_point_it_first_example]) converge, and since the sequences corresponding to each side are equal for all \(n,\) the limits are also equal: then \[y_{\infty}=G(y_{\infty})\] and \(y_{\infty}\) is a solution to the equation ([eq:ch6_exist_fixed_point_eq_real])!
An attractive feature of this algorithm is that it is by no means limited to equations for a real number \(y\in\mathbb{R}.\) It can be used just as well for solving equations for vectors \(y\in\mathbb{R}^{d},\) whereby \(G:\mathbb{R}^{d}\to\mathbb{R}^{d},\) or indeed in principle for an equation for \(y\in\mathcal{Y}\) in any space where the concepts of convergence and continuity are defined (that is, for any topological space \(\mathcal{Y}\)).
The success of the algorithm depends on the sequence \(y_{n}\) converging, which is by no means guaranteed - but when it does, it is a remarkably simple algorithm for solving what can be quite a difficult problem.
\(\text{ }\)Picard iteration for ODEs The proof of the existence theorem for ODEs is based on such a fixed-point iteration. To prove the existence of a solution to a general first order constrained ODE of the form \[y'=F(t,y),\quad\quad y(t_{*})=y_{*},\tag{6.46}\] it starts with the simple function \(y_{0}(t)=y_{*}\) (this initial \(y_{0}(t)\) at least satisfies the constraint) and applies the iteration defined by \[y_{n+1}^{'}(t)=F(t,y_{n}(t)),\quad\quad y_{n+1}(t_{*})=y_{*}.\] Note that this is a trivial ODE, one that can be solved simply by integrating both sides to obtain \[y_{n+1}(t)=y_{*}+\int_{t_{*}}^{t}F(t,y_{n}(t))dt.\tag{6.47}\] This is a fixed-point iteration on the space of functions on \(I.\) The key step in the proof of the existence theorem is to prove that the iterates \(y_{n}\) converge to a function \(y_{\infty},\) which is a fixed point of the iteration, i.e. \[y_{\infty}(t)=y_{*}+\int_{t_{*}}^{t}F(t,y_{\infty}(t))dt.\] This relation is implies \[y_{\infty}^{'}(t)=F(t,y_{\infty}^{'}(t)),\] i.e. implies that \(y_{\infty}\) is a solution of the ODE ([eq:ch6_picard_overview_ODE]). In this way we can prove that a solution exists, even if we don’t have a formula for the solution. The method is named after the French mathematician Émile Picard, who introduced it at the end of the 1800s.
Example borrowed from the book of H. S. Bear, p. 145.
Note furthermore that in this case the Picard iteration essentially amounts to a series solution of the ODE, as we saw in Chapter 4. This is because for \(F(t,y)=2ty\) any iterate which is a polynomial in \(t\) yields a next iterate which is also a polynomial in \(t.\) It is thus a very special property of this particular ODE, for general \(F\) the iterates will not take such a simple explicit form.
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C},\) and \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(t_{*}\in I,\) and \(I_{y}\subset V\) be a set and \(y_{*}\in I_{y}\) a point such for some \(r>0\) the ball \(\left\{ y\in V:\left|y-y_{*}\right|\le r\right\}\) is contained in \(I_{y}.\) For \(\varepsilon>0\) let \[I_{\varepsilon}=[t_{*}-\varepsilon,t_{*}+\varepsilon]\cap I.\] Assume that \(F:I\times I_{y}\to V\) is continuous on \(I\times I_{y}\) and \(L\)-Lipschitz in \(y\) for all \(t\in I,\) for some \(L<\infty.\) Then there exists an \(\varepsilon>0\) such the constrained ODE \[y'=F(t,y),\quad\quad t\in I_{\varepsilon},\quad\quad y(t_{*})=y_{*},\tag{6.43}\] has a solution \(y:I_{\varepsilon}\to\mathbb{R}.\)
Restricting to a smaller interval \(I_{\varepsilon}\) alleviates this problem, since in a small interval of “time” between \(t_{*}\) and \(t\) the next iterate \(y_{n+1}(t)\) cannot move very far from \(y_{*},\) and if it does not move too far it must stay in \(I_{y}\) (here we use that \(y_{*}\) is contained in a ball which lies inside \(I_{y}\)). Indeed for \(t\in I_{\varepsilon}=[t_{*}-\varepsilon,t_{*}+\varepsilon]\cap I\) we have \[\left|y_{n+1}(t)-y_{*}\right|=\left|\int_{t_{*}}^{t}F(t,y_{n}(t))dt\right|\le\int_{t_{*}}^{t}\left|F(t,y_{n}(t))\right|dt\le M\left|t-t_{*}\right|,\] where \(M\) is the maximum of \(\left|F(t,y)\right|.\) Thus if pick \(\varepsilon=\frac{\delta}{M}\) and make \(\delta\) small enough so that \([y_{*}-\delta,y_{*}+\delta]\subset I_{y}\) then we always have that \[y_{n+1}(t)\in[y_{*}-M\varepsilon,y_{*}+M\varepsilon]\subset[y_{*}-\delta,y_{*}+\delta]\subset I_{y}\text{ for all }t\in I_{\varepsilon},\] so all the iterates \[y_{n}:I_{\varepsilon}\to I_{y},n=0,1,2,\ldots\] are well defined.
\(\text{ }\)Proving convergence. In addition to the above, the most crucial points of the argument is proving that
the iterates \(y_{n}\) do indeed converge to a limiting function \(y_{\infty}:I_{\varepsilon}\times I_{y}\to\mathbb{R},\)
that the limiting function \(y_{\infty}\) is differentiable,
the limiting function \(y_{\infty}\) and its derivative \(y_{\infty}^{'}\) satisfy the ODE.
To prove convergence of \(y_{n}\) we use the standard result from real analysis which states that an absolutely convergent sequence is convergent:
If \(a_{n}\in\mathbb{R},n\ge0,\) is a sequence and \[\sum_{n=0}^{\infty}\left|a_{n}\right|<\infty,\] then the infinite sum \(\sum_{n\ge0}a_{n}\) is convergent, i.e. there exists an \(a\in\mathbb{R}\) such that \[\lim_{m\to\infty}\sum_{n=0}^{m}a_{n}=a.\]
We will use this result for the telescoping sequence of differences \[a_{n}(t)=y_{n+1}(t)-y_{n}(t),\] for which \[y_{n}(t)=y_{0}(t)+\sum_{m=0}^{n-1}a_{m}(t).\] Similarly to in the uniqueness proof we can bound the difference \(a_{n}(t)=y_{n+1}(t)-y_{n}(t)\) by rewriting it in terms of \(F\) and the previous iterates \[a_{n}(t)=\int_{t_{*}}^{t}F(t,y_{n}(s))ds-\int_{t_{*}}^{t}F(t,y_{n-1}(s))ds,\] and using the Lipschitz property of \(y\to F(t,y)\): \[\begin{array}{ccl} \left|a_{n}(t)\right| & = & \left|\int_{t_{*}}^{t}F(t,y_{n}(s))ds-\int_{t_{*}}^{t}F(t,y_{n-1}(s))ds\right|\\ & \le & \int_{t_{*}}^{t}\left|F(t,y_{n}(s))-F(t,y_{n-1}(s))\right|ds\\ & \le & \int_{t_{*}}^{t}L\left|y_{n}(s)-y_{n-1}(s)\right|ds\\ & = & L\left|t-t_{*}\right|\sup_{t\in I_{\varepsilon}}\left|a_{n-1}(t)\right|\\ & \le & L\varepsilon\sup_{t\in I_{\varepsilon}}\left|a_{n-1}(t)\right|. \end{array}\] Iterating this inequality shows that the supremum of \(a_{n}\) on \(I_{\varepsilon}\) satisfies the bound \[\sup_{t\in I_{\varepsilon}}\left|a_{n}(t)\right|\le\left(L\varepsilon\right)^{n}\sup_{t\in I_{\varepsilon}}\left|a_{0}(t)\right|.\] This gives a very rapidly decaying bound on \(a_{n}\) if \(L\varepsilon<1\) - for instance if \(L\varepsilon\le\frac{1}{2}\) then \[\sup_{t\in I_{\varepsilon}}\left|a_{n}(t)\right|\le2^{-n}\sup_{t\in I_{\varepsilon}}\left|a_{0}(t)\right|.\] Thus by possibly shrinking \(\varepsilon,\) we have with \(C=\sup_{t\in I_{\varepsilon}}\left|a_{0}(t)\right|<\infty\) that \[\sup_{t\in I_{\varepsilon}}\left|a_{n}(t)\right|\le C\sum_{n=0}^{\infty}2^{-n}<\infty.\] Thus for each \(t\in I_{\varepsilon}\) the infinite sum \(\sum_{n=0}^{\infty}a_{n}(t)\) is convergent, so we can define \[y_{\infty}(t)=y_{0}(t)+\sum_{n=0}^{\infty}a_{n}(t),t\in I_{\varepsilon},\] and this \(y_{\infty}\) is the limit of \(y_{n}\): \[y_{\infty}(t)=\lim_{n\to\infty}y_{n}(t),\quad\quad t\in I_{\varepsilon}.\] The convergence is in fact uniform in \(t\), since \[\sup_{t\in I_{\varepsilon}}\left|y_{\infty}(t)-y_{n}(t)\right|\le\sum_{m\ge n}\sup_{t\in I_{\varepsilon}}\left|a_{m}(t)\right|\le C\sum_{m\ge n}2^{-m}=2C\cdot2^{-n}\overset{\text{ as }n\to\infty}{\to}0.\]
It then remains to show that this \(y_{\infty}\) solves the ODE. Recall that by construction \[y_{n+1}(t)=y_{*}+\int_{t_{*}}^{t}F(s,y_{n}(s))ds\quad\quad\forall t\in I_{\varepsilon},n\ge0.\] The r.h.s. converges to \(y_{\infty}(t)\) for all \(t,\) and using the uniform converge and continuity of \(F\) one can show that the r.h.s. converges to \(\int_{t_{*}}^{t}F(s,y_{\infty}(s))ds.\) From this we will conclude that \[y_{\infty}(t)=y_{*}+\int_{t_{*}}^{t}F(s,y_{\infty}(s))ds\quad\quad\forall t\in I_{\varepsilon},\] implying that \(y_{\infty}\) is differentiable and satisfies \[y_{\infty}^{'}(t)=F(t,y_{\infty}(t))\quad\quad\forall t\in I_{\varepsilon}.\]
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C},\) and \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(t_{*}\in I,\) and \(I_{y}\subset V\) be a set and \(y_{*}\in I_{y}\) a point such for some \(r>0\) the ball \(\left\{ y\in V:\left|y-y_{*}\right|\le r\right\}\) is contained in \(I_{y}.\) For \(\varepsilon>0\) let \[I_{\varepsilon}=[t_{*}-\varepsilon,t_{*}+\varepsilon]\cap I.\] Assume that \(F:I\times I_{y}\to V\) is continuous on \(I\times I_{y}\) and \(L\)-Lipschitz in \(y\) for all \(t\in I,\) for some \(L<\infty.\) Then there exists an \(\varepsilon>0\) such the constrained ODE \[y'=F(t,y),\quad\quad t\in I_{\varepsilon},\quad\quad y(t_{*})=y_{*},\tag{6.43}\] has a solution \(y:I_{\varepsilon}\to\mathbb{R}.\)
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C},\) and \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(t_{*}\in I,\) and \(I_{y}\subset V\) be a set and \(y_{*}\in I_{y}\) a point such for some \(r>0\) the ball \(\left\{ y\in V:\left|y-y_{*}\right|\le r\right\}\) is contained in \(I_{y}.\) For \(\varepsilon>0\) let \[I_{\varepsilon}=[t_{*}-\varepsilon,t_{*}+\varepsilon]\cap I.\] Assume that \(F:I\times I_{y}\to V\) is continuous on \(I\times I_{y}\) and \(L\)-Lipschitz in \(y\) for all \(t\in I,\) for some \(L<\infty.\) Then there exists an \(\varepsilon>0\) such the constrained ODE \[y'=F(t,y),\quad\quad t\in I_{\varepsilon},\quad\quad y(t_{*})=y_{*},\tag{6.43}\] has a solution \(y:I_{\varepsilon}\to\mathbb{R}.\)
If \(a_{n}\in\mathbb{R},n\ge0,\) is a sequence and \[\sum_{n=0}^{\infty}\left|a_{n}\right|<\infty,\] then the infinite sum \(\sum_{n\ge0}a_{n}\) is convergent, i.e. there exists an \(a\in\mathbb{R}\) such that \[\lim_{m\to\infty}\sum_{n=0}^{m}a_{n}=a.\]
Recall that by construction the identity ([eq:ch6_exist_proof_def_iteraton]) holds for all \(n.\) As \(n\to\infty\) its l.h.s. converges to \(y_{\infty}(t),\) for any \(t\in I_{\varepsilon}.\) For the r.h.s. we have \[\begin{array}{ccl} \left|\int_{t_{*}}^{t}F(s,y_{\infty}(s))ds-\int_{t_{*}}^{t}F(s,y_{n}(s))ds\right| & \le & L\int_{t_{*}}^{t}\left|y_{\infty}(s)-y_{n}(s)\right|ds\\ & \le & L\varepsilon\sup_{t\in I_{\varepsilon}}\left|y_{\infty}(t)-y_{n}(t)\right|\to0, \end{array}\] for any \(t\in I_{\varepsilon},\) so it converges to \(y_{*}+\int_{t_{*}}^{t}F(s,y_{\infty}(s))ds.\) Thus in fact \[y_{\infty}(t)=y_{*}+\int_{t_{*}}^{t}F(t,y_{\infty}(t)),\quad\quad t\in I_{\varepsilon}.\] Since the uniform limit of continuous functions are continuous we know that \(y_{\infty}(t)\) is continuous, and since also \(F\) is continuous the fundamental theorem of calculus implies that \(t\to\int_{t_{*}}^{t}F(t,y_{\infty}(t))\) is differentiable. Thus it follows that \(y_{\infty}(t)\) is differentiable and \[y_{\infty}^{'}(t)=F(t,y_{\infty}(t)),\quad\quad t\in I_{\varepsilon}.\] Thus \(y_{\infty}\) is a solution to the constrained ODE ([eq:ch6_fo_system_ODE_exist_ODE]) on the interval \(I_{\varepsilon},\) proving the claim.
6.4 Existence - higher order ODEs
As a direct consequence we obtain an existence theorem for higher order ODEs.
(Existence for \(k\)-th order ODE). Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C}\) and \(k\ge1.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(t_{*}\in I,\) and \(I_{y,0},\ldots I_{y,k-1}\subset\mathbb{F}\) be sets and \(y_{*,l}\in I_{y,l},l=0,\ldots,k-1\) be points such for some \(r>0\) the ball \(\left\{ y\in\mathbb{F}:\left|y-y_{*,l}\right|\le r\right\}\) is contained in \(I_{y,l}\) for each \(l.\) For \(\varepsilon>0\) let \[I_{\varepsilon}=[t_{*}-\varepsilon,t_{*}+\varepsilon]\cap I.\] Assume that \(F:I\times I_{y,k-1}\times\ldots\times I_{y,0}\to\mathbb{F}\) and that \(F(t,y)\) is continuous in \(t\) and \(y,\) and \(L\)-Lipschitz in \(y\) for all \(t\in I,\) for some \(L<\infty.\) Then there exists an \(\varepsilon>0\) such that the \(k\)-th order ODE \[y^{(k)}(t)=F(t,y^{(k-1)}(t),\ldots,y^{'}(t),y(t)),\quad\quad t\in I_{\varepsilon},\quad\quad y^{(l)}(t_{*})=y_{*,l},l=0,\ldots k-1\] has a solution \(y:I_{\varepsilon}\to\mathbb{R}.\)
Let \(I\subset\mathbb{R}\) be a non-empty interval. Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C}\) and \(I_{y,0},\ldots,I_{y,k-1}\subset\mathbb{F}\) be non-empty sets and \[F:I\times I_{y,k-1}\times I_{y,k-2}\times\ldots\times I_{y,1}\times I_{y,0}\to\mathbb{F}\] be a function.
A function \(y\) is a solution to the \(k\)-th order ODE \[y^{(k)}(t)=F(t,y^{(k-1)}(t),\ldots,y'(t),y(t))\quad\quad t\in I\] iff \(x(t)=(y(t),\ldots,y^{(k-1)}(t))\) is a solution to the first order \(\mathbb{F}^{k}\)-valued ODE \[\dot{x}=G(t,x)\quad\quad t\in I\] where \[G(t,x)=(x_{2},x_{3},\ldots,x_{k},F(t,x_{k-1},\ldots,x_{1})).\]
Let \(t_{0}\in I\) and \(y_{0,l}\in I_{y,l},l=0,\ldots,k-1.\) A function \(y\) is a solution to a \(k\)-th order constrained ODE \[y^{(k)}(t)=F(t,y^{(k-1)}(t),\ldots,y'(t),y(t)),\quad\quad y\in I,\quad\quad y^{(l)}(t_{0})=y_{0,l},l=0,\ldots,k-1\] iff \(x(t)=(y(t),\ldots,y^{(k-1)}(t))\) is a solution to the first order \(\mathbb{F}^{k}\)-valued constrained ODE \[\dot{x}=G(t,x),\quad\quad t\in I,\quad\quad x(t_{0})=x_{0},\] where \(x_{0}=(y_{0,0},\ldots,y_{0,l}).\)
Let \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C},\) and \(d\ge1\) and \(V=\mathbb{F}^{d}.\) Let \(I\subset\mathbb{R}\) be a non-empty interval and \(t_{*}\in I,\) and \(I_{y}\subset V\) be a set and \(y_{*}\in I_{y}\) a point such for some \(r>0\) the ball \(\left\{ y\in V:\left|y-y_{*}\right|\le r\right\}\) is contained in \(I_{y}.\) For \(\varepsilon>0\) let \[I_{\varepsilon}=[t_{*}-\varepsilon,t_{*}+\varepsilon]\cap I.\] Assume that \(F:I\times I_{y}\to V\) is continuous on \(I\times I_{y}\) and \(L\)-Lipschitz in \(y\) for all \(t\in I,\) for some \(L<\infty.\) Then there exists an \(\varepsilon>0\) such the constrained ODE \[y'=F(t,y),\quad\quad t\in I_{\varepsilon},\quad\quad y(t_{*})=y_{*},\tag{6.43}\] has a solution \(y:I_{\varepsilon}\to\mathbb{R}.\)