The following is our main result.
Theorem 3.1
Let
\([a,b]\)
be a compact interval, and let
\(Y\in {\mathcal{A}}_{n}\)
be a closed, connected and locally connected subset of
\({\mathbf{R}}^{n}\). Let
\(f:[a,b]\times{\mathbf{R}}^{n}\times{\mathbf{R}}^{n}\times Y\to {\mathbf{R}}\)
be a given function, \(\Sigma \subseteq Y\times Y\)
a countable set, dense in
\(Y\times Y\), \(D^{\prime}\)
and
\(D^{\prime\prime}\)
two dense subset of
Y. Assume that there exist 2n
sets
$$V_{1},V_{2},\ldots,V_{2n}\in{\mathcal{B}}({ \mathbf{R}}), $$
with
\(m_{1}(V_{i})=0\)
for all
\(i=1,\ldots,2n\), such that, if one puts
$$\begin{aligned}& \Omega :=\prod_{i=1}^{2n}[{\mathbf{R}} \setminus V_{i}], \\& f^{*}:[a,b]\times\Omega \times Y\to{\mathbf{R}},\qquad f^{*}:=f|_{[a,b]\times\Omega \times Y}, \end{aligned}$$
one has:
-
(i)
for all
\((y_{1},y_{2})\in\Sigma \), one has
$$\bigl\{ (t,x,z)\in[a,b]\times\Omega :f^{*}(t,x,z,y_{1})< 0< f^{*}(t,x,z,y_{2}) \bigr\} \in {\mathcal{L}} \bigl([a,b]\bigr)\otimes{\mathcal{B}}(\Omega ); $$
-
(ii)
for a.e. \(t\in[a,b]\), and for all
\(y\in D^{\prime}\), the function
\(f^{*}(t,\cdot ,\cdot ,y)\)
is l.s.c. over Ω;
-
(iii)
for a.e. \(t\in[a,b]\), and for all
\(y\in D^{\prime \prime}\), the function
\(f^{*}(t,\cdot ,\cdot ,y)\)
is u.s.c. over Ω;
-
(iv)
for a.e. \(t\in[a,b]\), and for all
\((x,z)\in\Omega \), the function
\(f(t,x,z,\cdot )\)
is continuous over
Y,
$$0\in\operatorname{int}_{\mathbf{R}}\bigl(f(t,x,z,Y)\bigr) $$
and
$$\operatorname{int}_{Y}\bigl(\bigl\{ y\in Y: f(t,x,z,y)=0\bigr\} \bigr)= \emptyset ; $$
-
(v)
there exist
\(p\in[1,+\infty]\)
and a positive function
\(\beta \in L^{p}([a,b])\)
such that, for a.e. \(t\in[a,b]\), and for all
\((x,z)\in\Omega \), one has
$$\bigl\{ y\in Y:f(t,x,z,y)=0\bigr\} \subseteq \overline{B}_{n} \bigl(0_{{\mathbf {R}}^{n}},\beta (t)\bigr). $$
Then, there exists
\(u\in W^{2,p}([a,b],{\mathbf{R}}^{n})\)
such that
$$\textstyle\begin{cases} f(t,u(t), u^{\prime}(t),u^{\prime\prime}(t))=0&\textit{for a.e. }t\in [a,b], \\ u(a)=u(b)=0_{{\mathbf{R}}^{n}}, \end{cases} $$
and
$$\bigl(u(t), u^{\prime}(t)\bigr)\in\Omega \quad \textit{for a.e. } t \in[a,b]. $$
Proof
From now on, we assume that assumptions (ii)-(v) are satisfied for all \(t\in[a,b]\) (it is routine matter to check that it is not restrictive to do this). First of all, we assume \(p<+\infty\). Fix \(t\in[a,b]\). Let
$$V_{t}:\Omega \to2^{Y}, \qquad E_{t}:\Omega \to2^{Y}, \qquad Q_{t}:\Omega \to2^{Y} $$
be the multifunctions defined by setting, for all \((x,z)\in\Omega \),
$$\begin{aligned}& V_{t}(x,z):=\bigl\{ y\in Y: f^{*}(t,x,z,y)=0\bigr\} , \\& E_{t}(x,z):=\bigl\{ y\in Y: y\mbox{ is a local extremum for } f^{*}(t,x,z,\cdot)\bigr\} , \\& Q_{t}(x,z):=V_{t}(x,z)\setminus E_{t}(x,z). \end{aligned}$$
By assumptions (ii), (iii), (iv), and Theorem 2.2 of [15], the multifunction \(Q_{t}:\Omega \to2^{Y}\) is lower semicontinuous in Ω with nonempty closed values (in Y, hence in \({\mathbf{R}}^{n}\) since Y is closed). Now, let
$$Q:[a,b]\times\Omega \to2^{Y}, \qquad Q(t,x,z):=Q_{t}(x,z). $$
By what precedes, the multifunction Q has nonempty closed values and, for each fixed \(t\in[a,b]\), the multifunction \(Q(t,\cdot ,\cdot )\) is lower semicontinuous in Ω.
Now we prove that the multifunction Q is \({\mathcal{L}}([a,b])\times {\mathcal{B}} (\Omega )\)-measurable. To this aim, let \(A\subseteq Y\) be a nonempty open connected set, such that \(Q^{-}(A)\ne\emptyset \). We claim that
$$ Q^{-}(A)=\bigcup_{(y_{1},y_{2})\in(A\times A)\cap\Sigma } \bigl\{ (t,x,z)\in[a,b]\times \Omega :f(t,x,z,y_{1})< 0< f(t,x,z,y_{2}) \bigr\} . $$
(8)
To this aim, fix any \((t_{0},x_{0},z_{0})\in Q^{-}(A)\). Therefore, \((t_{0},x_{0},z_{0})\in[a,b]\times\Omega \) and there exists \(y_{0}\in A\cap Q(t_{0},x_{0},z_{0})\). That is, \(y_{0}\in A\), \(f(t_{0},x_{0},z_{0},y_{0})=0\) and \(y_{0}\) is not a local extremum for the function \(f(t_{0},x_{0},z_{0},\cdot )\). This implies that there exist two points \(y^{\prime},y^{\prime\prime }\in A\) such that
$$f\bigl(t_{0},x_{0},z_{0},y^{\prime} \bigr)< 0 \quad \mbox{and}\quad f\bigl(t_{0},x_{0},z_{0},y^{\prime\prime} \bigr)>0. $$
By the continuity of the function \(f(t_{0},x_{0},z_{0},\cdot )\), there exist two open sets \(B_{1},B_{2}\subseteq Y\) such that \(y^{\prime}\in B_{1}\), \(y^{\prime \prime}\in B_{2}\), and
$$\begin{aligned}& f(t_{0},x_{0},z_{0},y)< 0\quad \mbox{for all } y \in B_{1}, \\& f(t_{0},x_{0},z_{0},y)>0 \quad \mbox{for all } y\in B_{2}. \end{aligned}$$
Put \(A_{1}:=B_{1}\cap A\), \(A_{2}:=B_{2}\cap A\). Of course, both \(A_{1}\) and \(A_{2}\) are open in Y and nonempty (since they contain \(y^{\prime}\) and \(y^{\prime\prime}\), respectively).
Let \((y_{1},y_{2})\in(A_{1}\times A_{2})\cap\Sigma \) (this last set is nonempty since Σ is dense in \(Y\times Y\)). We see that
$$(t_{0},x_{0},z_{0})\in \bigl\{ (t,x,z)\in[a,b] \times\Omega :f(t,x,z,y_{1})< 0< f(t,x,z,y_{2}) \bigr\} , $$
hence \((t_{0},x_{0},z_{0})\) belongs to the right-hand side of (8). Conversely, let \((t^{*},x^{*},z^{*})\) belong to the right-hand side of (8). Thus, there exists \((y_{1},y_{2})\in(A\times A)\cap\Sigma \) such that
$$ f\bigl(t^{*},x^{*},z^{*},y_{1}\bigr)< 0< f\bigl(t^{*},x^{*},z^{*},y_{2} \bigr). $$
(9)
Since A is connected and \(f(t^{*},x^{*},z^{*},\cdot )\) is continuous in Y, there exists \(y_{3}\in A\) such that \(f(t^{*},x^{*},z^{*},y_{3})=0\). We now distinguish two cases.
-
(1)
If \(y_{3}\) is not a local extremum for the function \(f(t^{*},x^{*},z^{*},\cdot )\), then we get \(y_{3}\in Q(t^{*},x^{*},z^{*})\cap A\), hence \((t^{*},x^{*},z^{*})\in Q^{-}(A)\), as desired.
-
(2)
If \(y_{3}\) is a local extremum for the function \(f(t^{*},x^{*},z^{*},\cdot )\) (not absolute by assumption (iv)), then \(y_{3}\) is also a local extremum for the function \(f(t^{*},x^{*},z^{*},\cdot )|_{A}\) (not absolute by (9)). Moreover, since A is open in Y, by assumption (v) we get
$$\operatorname{int}_{A}\bigl(\bigl\{ y\in A: f\bigl(t^{*},x^{*},z^{*},y \bigr)=0\bigr\} \bigr)=\emptyset . $$
Consequently, by Lemma 2.1 of [15], there exists a point \(y^{*}\in A\) such that \(f(t^{*},x^{*},z^{*},y^{*})=0\) and \(y^{*}\) is not a local extremum for the function \(f(t^{*},x^{*},z^{*},\cdot )|_{A}\). This easily implies that \(y^{*}\) is not a local extremum for the function \(f(t^{*},x^{*},z^{*},\cdot )\) (considered over the whole set Y). Therefore, we have \(y^{*}\in Q(t^{*},x^{*},z^{*})\cap A\), hence \((t^{*},x^{*},z^{*})\in Q^{-}(A)\), as desired.
Thus, the equality (8) is proved, and therefore, by assumption (i), the set \(Q^{-}(A)\) is \({\mathcal{L}}([a,b])\otimes{\mathcal{B}}(\Omega )\)-measurable.
Our assumptions on Y imply that it has a countable base of connected open sets. Therefore, it follows that the multifunction Q is \({\mathcal{L}}([a,b])\times{\mathcal {B}}(\Omega )\)-weakly measurable. By Theorem 3.5 of [9], the multifunction Q is also \({\mathcal{L}}([a,b])\otimes{\mathcal{B}}(\Omega )\)-measurable.
By Corollary 6.6.7 of [10], the set Ω is a Souslin set since it belongs to \({\mathcal{B}}({\mathbf{R}}^{2n})\). By Theorem 2.1 (where the spaces \([a,b]\) and R are considered with the usual one-dimensional Lebesgue measure \(m_{1}\) over their Borel families), there exist \(Q_{1},\ldots,Q_{2n}\in{\mathcal{B}}({\mathbf{R}})\), with \(m_{1}(Q_{1})=0\) for all \(i=1,\ldots,2n\), a set \(K_{0}\in{\mathcal{L}}([a,b])\), with \(m_{1}(K_{0})=0\), and a function \(\phi:[a,b]\times\Omega \to 2^{{\mathbf{R}}^{n}}\), such that:
-
(a)
\(\phi(t,x,z)\in Q(t,x,z)\) for all \((t,x,z)\in[a,b]\times \Omega \) (hence, in particular, the function ϕ takes its values in Y);
-
(b)
for all \((x,z)\in\Omega \setminus[\bigcup_{i=1}^{2n} P_{2n,i}^{-1}(Q_{i})]\), the function \(\phi( \cdot ,x,z)\) is \({\mathcal{L}}([a,b])\)-measurable;
-
(c)
for all \(t\in[a,b]\setminus K_{0}\), one has
$$\bigl\{ (x,z)\in\Omega :\phi(t,\cdot ,\cdot ) \mbox{ is discontinuous at }(x,z) \bigr\} \subseteq \Omega \cap \Biggl[\bigcup_{i=1}^{2n}P_{2n,i}^{-1}(Q_{i}) \Biggr]. $$
Now, let \(\psi:[a,b]\times{\mathbf{R}}^{n}\times{\mathbf{R}}^{n}\to {\mathbf{R}}^{n}\) be defined by
$$\psi(t,x,z)= \textstyle\begin{cases} \phi(t,x,z)&\mbox{if }(x,z)\in\Omega , \\ 0_{{\mathbf{R}}^{n}}&\mbox{if }(x,z)\notin\Omega . \end{cases} $$
Observe that \({\mathbf{R}}^{2n}\setminus\Omega =\bigcup_{i=1}^{2n}P_{2n,i}^{-1}(V_{i})\), and also
$$\Omega \Bigm\backslash \Biggl[\bigcup_{i=1}^{2n}P_{2n,i}^{-1}(Q_{i}) \Biggr]= {\mathbf{R}}^{2n}\Bigm\backslash \Biggl[\bigcup _{i=1}^{2n}P_{2n,i}^{-1}(V_{i} \cup Q_{i}) \Biggr] =\prod_{i=1}^{2n} \bigl[{\mathbf{R}}\setminus(V_{i}\cup Q_{i}) \bigr]. $$
Let D be a countable subset of \(\Omega \setminus[\bigcup_{i=1}^{2n}P_{2n,i}^{-1}(Q_{i})]\), dense in \({\mathbf{R}}^{2n}\). Of course, such a set D exists since
$$m_{2n}\Biggl(\bigcup_{i=1}^{2n}P_{2n,i}^{-1}(V_{i} \cup Q_{i})\Biggr)=0. $$
Now we want to apply Proposition 2.6 to the function ψ, choosing \(E={\mathbf{R}}^{2n}\setminus\Omega \). To this aim, observe that:
-
(1)
for all \(t\in[a,b]\), and all \((x,z)\in{\mathbf{R}}^{n}\times {\mathbf{R}}^{n}\), we have \(\psi(t,x,z)\in\overline{B}_{n}(0_{{\mathbf{R}}^{n}},\beta (t))\) (this follows by the construction of ϕ and Q and by assumption (v)), hence for all \(t\in[a,b]\) the function \(\psi(t,\cdot ,\cdot )\) is bounded;
-
(2)
for all \((x,z)\in D\), the function \(\psi( \cdot ,x,z)=\phi( \cdot ,x,z)\) is \({\mathcal{L}} ([a,b])\)-measurable.
Consequently, if \(G:[a,b]\times{\mathbf{R}}^{n}\times{\mathbf {R}}^{n}\to2^{{\mathbf{R}}^{n}}\) is the multifunction defined by setting, for all \((t,x,z)\in[a,b]\times{\mathbf{R}}^{n}\times{\mathbf{R}}^{n}\),
$$\begin{aligned} G(t,x,z)&=\bigcap_{m\in{\mathbf{N}}}\overline{ \operatorname{conv}} \overline { \biggl(\bigcup_{\substack{{(v,w)\in D}\\{\|(v,w)-(x,z)\|_{2n}\le{1\over m}}}} \bigl\{ \psi(t,v,w)\bigr\} \biggr)} \\ &=\bigcap_{m\in{\mathbf{N}}}\overline{\operatorname{conv}} \overline { \biggl(\bigcup_{\substack{{(v,w)\in D}\\{\|(v,w)-(x,z)\|_{2n}\le{1\over m}}}}\bigl\{ \phi(t,v,w) \bigr\} \biggr) }, \end{aligned}$$
by Proposition 2.6 we see that:
- (a)′:
-
G has nonempty closed convex values;
- (b)′:
-
for all \((x,z)\in{\mathbf{R}}^{n}\times {\mathbf{R}}^{n}\), the multifunction \(G( \cdot ,x,z)\) is \({\mathcal{L}}([a,b])\)-measurable;
- (c)′:
-
for all \(t\in[a,b]\), the multifunction \(G( t, \cdot ,\cdot )\) has closed graph;
- (d)′:
-
if \(t\in[a,b]\), and the function \(\psi(t,\cdot ,\cdot )|_{\Omega}=\phi(t,\cdot ,\cdot )\) is continuous at \((x,z)\in\Omega \), then one has
$$G(t,x,z)=\bigl\{ \psi(t,x,z)\bigr\} =\bigl\{ \phi(t,x,z)\bigr\} . $$
Moreover, observe that by the above construction we see that
$$ G(t,x,z)\subseteq \overline{B}_{n}\bigl(0_{{\mathbf{R}}^{n}},\beta (t) \bigr)\cap \overline{\operatorname{conv}}(Y)\quad \mbox{for all } (t,x,z)\in[a,b] \times{\mathbf{R}}^{n}\times{\mathbf{R}}^{n}. $$
(10)
Consequently, by Theorem 3 of [16], there exist \(u\in W^{2,p}([a,b],{\mathbf{R}}^{n})\) and a set \(K_{1}\in{\mathcal {L}}([a,b])\), with \(m_{1}(K_{1})=0\), such that
$$ \textstyle\begin{cases} u^{\prime\prime}(t)\in G(t,u(t), u^{\prime}(t))&\mbox{for all } t\in[a,b]\setminus K_{1}, \\ u(a)=u(b)=0_{{\mathbf{R}}^{n}}. \end{cases} $$
(11)
In particular, by (10) we get
$$ u^{\prime\prime}(t)\in{\overline{\operatorname{conv}}}(Y) \quad \mbox{for all } t\in[a,b]\setminus K_{1}. $$
(12)
Now, fix \(i\in\{1,\ldots,n\}\). Of course, (12) implies that
$$u_{i}^{\prime\prime}(t)\in \bigl[\inf P_{n,i}\bigl({ \overline{\operatorname{conv}}}(Y)\bigr), \sup P_{n,i}\bigl({\overline{ \operatorname{conv}}}(Y)\bigr) \bigr] \quad \mbox{for all } t\in[a,b]\setminus K_{1} $$
(of course, we are denoting by \(u_{i}(t)\) the ith component of \(u(t)\)). In particular, taking into account that \(Y\in{\mathcal{A}}_{n}\), this implies that \(u_{i}^{\prime\prime}(t)\) has constant sign for all \(t\in[a,b]\setminus K_{1}\). Assume that
$$u_{i}^{\prime\prime}(t)>0 \quad \mbox{for all } t\in[a,b]\setminus K_{1}. $$
Since \(u_{i}^{\prime}(t)\) is absolutely continuous, we have
$$u_{i}^{\prime}(t)=u_{i}^{\prime}(a)+ \int_{a}^{t} u_{i}^{\prime\prime}(s) \, ds \quad \mbox{for all } t\in[a,b], $$
hence \(u_{i}^{\prime}(t)\) is strictly increasing in \([a,b]\). Consequently, by Theorem 2 of [17], the function
$$\bigl(u^{\prime}_{i}\bigr)^{-1}:u^{\prime}_{i} \bigl([a,b]\bigr)\to[a,b] $$
is absolutely continuous. Moreover, since \(u_{i}(a)=u_{i}(b)=0\), there exists \(c_{i}\in\, ]a,b[\) such that
$$u^{\prime}_{i}(t)< 0 \quad \mbox{for all } t \in[a,c_{i}[ $$
and
$$u^{\prime}_{i}(t)>0 \quad \mbox{for all } t\in\, ]c_{i},b]. $$
Again by Theorem 2 of [17], the functions
$$(u_{i}|_{[a,c_{i}]})^{-1}:u_{i} \bigl([a,c_{i}]\bigr)\to[a,c_{i}] $$
and
$$(u_{i}|_{[c_{i},b]})^{-1}:u_{i} \bigl([c_{i},b]\bigr)\to[c_{i},b] $$
are absolutely continuous. Consequently, by Theorem 18.25 of [18], for each Lebesgue measurable \(U\subseteq {\mathbf{R}}\), with \(m_{1}(U)=0\), the sets
$$\bigl(u^{\prime}_{i}\bigr)^{-1}(U):=\bigl\{ t \in[a,b]:u^{\prime}_{i}(t)\in U\bigr\} $$
and
$$\begin{aligned} (u_{i})^{-1}(U)&:=\bigl\{ t\in[a,b]:u_{i}(t)\in U \bigr\} \\ &=\bigl\{ t\in[a,c_{i}]:u_{i}(t)\in U\bigr\} \cup\bigl\{ t \in[c_{i},b]:u_{i}(t)\in U\bigr\} \end{aligned}$$
have null Lebesgue measure.
If, conversely, one has
$$u_{i}^{\prime\prime}(t)< 0 \quad \mbox{for all } t\in[a,b]\setminus K_{1}, $$
by an analogous argument we can get the same conclusion.
Now, for each \(i=1,\ldots, n\), let us put
$$S_{i}:=u_{i}^{-1}(V_{i}\cup Q_{i}),\qquad W_{i}:=\bigl(u^{\prime}_{i} \bigr)^{-1}(V_{n+i}\cup Q_{n+i}), $$
and
$$S:=K_{0}\cup K_{1}\cup \Biggl[ \bigcup _{i=1}^{n}(S_{i}\cup W_{i}) \Biggr]. $$
For what precedes, we have \(m_{1}(S)=0\). Now, fix \(t\in[a,b]\setminus S\). Since \(t\notin \bigcup_{i=1}^{n}(S_{i}\cup W_{i})\), for all \(i=1,\ldots, n\) we get
$$u_{i}(t)\notin V_{i}\cup Q_{i},\qquad u^{\prime}_{i}(t)\notin V_{n+i}\cup Q_{n+i}, $$
hence
$$\bigl(u(t),u^{\prime}(t)\bigr)\in \prod_{i=1}^{2n} \bigl[{\mathbf{R}}\setminus(V_{i}\cup Q_{i})\bigr]= \Omega \Bigm\backslash \Biggl[\bigcup_{i=1}^{2n}P_{2n,i}^{-1}(Q_{i}) \Biggr]. $$
Since \(t\notin K_{0}\), this implies that the function \(\phi(t,\cdot ,\cdot ):\Omega \to{\mathbf{R}}^{n}\) is continuous at \((u(t),u^{\prime}(t))\). Hence, by the property (d)′ and the above construction, taking into account that \(t\notin K_{1}\) and (11), we get
$$ u^{\prime\prime}(t)\in G\bigl(t,u(t), u^{\prime}(t)\bigr)= \bigl\{ \phi \bigl(t,u(t), u^{\prime}(t)\bigr)\bigr\} \subseteq Q\bigl(t,u(t), u^{\prime}(t)\bigr)\subseteq V_{t} \bigl(u(t), u^{\prime}(t)\bigr). $$
Therefore, we get \(u^{\prime\prime}(t)\in Y\) and
$$f\bigl(t, u(t),u^{\prime}(t),u^{\prime\prime}(t)\bigr)=0. $$
Thus, our conclusion is proved in the case \(p<+\infty\).
Now, assume \(p=+\infty\). Fix any \(q\in[1,+\infty[ \). Since \(\beta \in L^{q}([a,b])\), by the first part of the proof there exist a function \(u\in W^{2,q}([a,b],{\mathbf{R}}^{n})\) and a set \(U\in {\mathcal{L}}([a,b])\), with \(m_{1}(U)=0\), such that
$$\textstyle\begin{cases} f(t,u(t), u^{\prime}(t),u^{\prime\prime}(t))=0&\mbox{for all }t\in [a,b]\setminus U, \\ u(a)=u(b)=0_{{\mathbf{R}}^{n}}, \end{cases} $$
and
$$\bigl(u(t), u^{\prime}(t)\bigr)\in\Omega \quad \mbox{for all } t\in [a,b] \setminus U. $$
By (v) we get
$$\bigl\{ y\in Y: f\bigl(t,u(t),u^{\prime}(t),y\bigr)=0\bigr\} \subseteq \overline{B}_{n} \bigl(0_{{\mathbf{R}}^{n}},\beta (t)\bigr) \quad \mbox{for all } t\in [a,b]\setminus U, $$
hence
$$\bigl\Vert u^{\prime\prime}(t)\bigr\Vert _{n}\le\beta (t)\quad \mbox{for all } t\in[a,b]\setminus U. $$
This implies \(u^{\prime\prime}\in L^{\infty}([a,b],{\mathbf{R}}^{n})\), hence \(u\in W^{2,\infty}([a,b],{\mathbf{R}}^{n})\). The proof is now complete. □
Remark
Theorem 3.1 can be put in the following equivalent form.
Corollary 3.2
Let
\([a,b]\), Y, f, Σ, \(D^{\prime}\)
and
\(D^{\prime\prime}\)
be as in the statement of Theorem
3.1. Assume that there exists a set
\(F\in{\mathcal{F}}_{2n}\)
such that, if one puts
\(\Omega :={\mathbf{R}}^{2n}\setminus F\)
and
\(f^{*}:=f|_{[a,b]\times \Omega \times Y}\), then assumptions (i)-(v) of Theorem
3.1
are satisfied.
Then the same conclusion of Theorem
3.1
holds.
Proof
Let \(F_{1},F_{2},\ldots,F_{2n}\subseteq {\mathbf{R}}^{2n}\), with \(m_{1}(P_{2n,i}(F_{i}))=0\) for all \(i=1,\ldots,2n\), be such that \(F=\bigcup_{i=1}^{2n} F_{i}\). For each \(i=1,\ldots,2n\), let \(V_{i}\in{\mathcal{B}}({\mathbf{R}})\) be such that \(P_{2n,i}(F_{i})\subseteq V_{i}\) and \(m_{1}(V_{i})=0\). If one puts \(\Omega ^{\prime}:=\prod_{i=1}^{2n}({\mathbf{R}}\setminus V_{i})\), then \(\Omega ^{\prime}\subseteq \Omega \), hence all the assumptions of Theorem 3.1 are satisfied. The conclusion follows at once. □
Now we give a simple example of application of Theorem 3.1.
Example 1
Let \(n=1\), \(Y=[1,+\infty[\), \(p\in[1,+\infty]\), and let \(\alpha \in L^{p}([0,1])\), with \(\alpha (t)>0\) for all \(t\in[0,1]\). Let
$$E:=\bigl\{ (x,z)\in{\mathbf{R}}^{2}: x\in{\mathbf{Q}} \mbox{ or } z\in{ \mathbf{Q}} \bigr\} = ({\mathbf{Q}}\times{\mathbf{R}})\cup({\mathbf{R}}\times{ \mathbf{Q}}), $$
and let \(f:[0,1]\times{\mathbf{R}}\times{\mathbf{R}}\times Y\to{\mathbf {R}}\) be defined by putting
$$f(t,x,z,y)=\textstyle\begin{cases}2y-\alpha (t)-10&\mbox{if }(x,z)\in E, \\ 3\cos(y+{\pi\over 2}-1)+y(2+|\cos x|)-|\cos(x+z)|-\alpha (t)-3 &\mbox{if }(x,z)\notin E. \end{cases} $$
It is immediate to check that for each fixed \((t,y)\in[0,1]\times Y\) one has
$$f(t,x,z,y)\ge2y-\alpha (t)-7\quad \mbox{for all } (x,z)\in {\mathbf{R}} ^{2}\setminus E. $$
Consequently, for each fixed \((t,y)\in[0,1]\times Y\), the function \(f(t,\cdot ,\cdot ,y)\)
is discontinuous at all points
\((x,z)\in{\mathbf{R}}^{2}\).
Now, let us observe that such a function f satisfies the assumptions of Theorem 3.1. To this aim, choose \(V_{1}=V_{2}={\mathbf{Q}}\), \(D^{\prime}=D^{\prime\prime}=Y\), and let Σ be any countable dense subset of \(Y\times Y\). In this case, we have
$$\Omega =({\mathbf{R}}\setminus{\mathbf{Q}})\times({\mathbf {R}}\setminus{ \mathbf{Q}}), $$
hence in particular we get \(\Omega ={\mathbf{R}}^{2}\setminus E\) and
$$f^{*}(t,x,z,y)=3\cos\biggl(y+{\pi\over 2}-1\biggr)+y\bigl(2+\vert \cos x\vert \bigr)-\bigl\vert \cos (x+z)\bigr\vert -\alpha (t)-3 $$
for all \((t,x,z,y)\in[0,1]\times\Omega \times Y\). At this point, observe what follows.
(a) For all fixed \(y\in Y\), the function \(f^{*}(\cdot ,\cdot ,\cdot ,y)\) is \({\mathcal{L}}([0,1])\otimes{\mathcal{B}}(\Omega )\)-measurable. Therefore, assumption (i) of Theorem 3.1 is satisfied.
(b) For all \((t,y)\in[0,1]\times Y\), the function \(f^{*}(t,\cdot ,\cdot ,y)\) is continuous over Ω.
(c) Let \(t\in[0,1]\) and \((x,z)\in\Omega \) be fixed. We see that \(f(t,x,z,\cdot )=f^{*}(t,x,z,\cdot )\) is continuous in Y, and also
$$f(t,x,z,1)\le-\alpha (t)< 0. $$
Since
$$\lim_{y\to+\infty}f(t,x,z,y)=+\infty, $$
we get \(0\in\operatorname{int}_{\mathbf{R}}(f(t,x,z,Y))\). Moreover, if we put \(s(\cdot):=f(t,x,z,\cdot )\), we get
$$s^{\prime}(y)= -3\sin\biggl(y+{\pi\over 2}-1\biggr)+2+|\cos x| ,\quad \forall y\in Y. $$
Consequently, the set \(\{y\in Y: s^{\prime}(y)=0\}\) is countable, hence \(s^{\prime}\) is never identically equal to 0 in any interval \(I\subseteq Y\). This implies that there exists no interval \(I\subseteq Y\) such that \(s(\cdot)\) is constant on I. Therefore, assumption (iv) of Theorem 3.1 is also satisfied. We also remark that if \(x\ne k\pi\) (with \(k\in{\mathbf{Z}}\)), then \(s^{\prime}\) takes both positive and negative values, hence \(f(t,x,z,\cdot )\) is not monotone.
(d) Let \(t\in[0,1]\) and \((x,z)\in\Omega \) be fixed, and let \(y\in Y\) be such that \(f(t,x,z,y)=0\). By the definition of Y and f we get
$$1\le y\le{1\over 2}\bigl(7+\alpha (t)\bigr). $$
Since the function
$$\beta (t)= {1\over 2}\bigl(7+\alpha (t)\bigr) $$
belongs to \(L^{p}([0,1])\), assumption (v) of Theorem 3.1 is also satisfied.
Therefore, all the assumptions of Theorem 3.1 are fulfilled. Consequently, there exists \(u\in W^{2,p}([0,1])\) such that
$$ \textstyle\begin{cases} f(t,u(t), u^{\prime}(t),u^{\prime\prime}(t))=0&\mbox{for a.e. }t\in [0,1], \\ u(0)=u(1)=0 \end{cases} $$
(13)
and also
$$\bigl(u(t), u^{\prime}(t)\bigr)\in({\mathbf{R}}\setminus{\mathbf{Q}}) \times ({\mathbf{R}}\setminus{\mathbf{Q}} ) \quad \mbox{for a.e. } t\in[0,1]. $$
Finally, we observe that problem (13) does not admit the trivial solution \(u(t)\equiv0\).
The following result is an immediate consequence of Theorem 3.1.
Theorem 3.3
Let
\([a,b]\)
be a compact interval, and let
\(Y\in {\mathcal{A}}_{n}\)
be a closed, connected and locally connected subset of
\({\mathbf{R}}^{n}\). Let
\(\psi:[a,b]\times Y\to{\mathbf{R}}\)
and
\(g:[a,b]\times{\mathbf {R}}^{n}\times{\mathbf{R}} ^{n}\to{\mathbf{R}}\)
be two given functions, \(D\subseteq Y\)
a countable set, dense in
Y. Assume that for all
\(y\in D\), the function
\(\psi(\cdot ,y)\)
is
\({\mathcal{L}} ([a,b])\)-measurable, and for a.e. \(t\in[a,b]\), the function
\(\psi(t,\cdot )\)
is continuous in
Y. Moreover, assume that there exist 2n
sets
$$V_{1},V_{2},\ldots,V_{2n}\in{\mathcal{B}}({ \mathbf{R}}), $$
with
\(m_{1}(V_{i})=0\)
for all
\(i=1,\ldots,2n\), such that, if one puts
$$\Omega :=\prod_{i=1}^{2n} [{\mathbf{R}} \setminus V_{i} ], $$
one has:
-
(i)
for all
\(t\in[a,b]\), the function
\(g( t,\cdot ,\cdot )|_{\Omega}\)
is continuous over Ω;
-
(ii)
for all
\((x,z)\in\Omega \), the function
\(g( \cdot ,x,z)\)
is
\({\mathcal{L}}([a,b])\)-measurable;
-
(iii)
for a.e. \(t\in[a,b]\), one has
\(g(t,\Omega )\subseteq \operatorname{int} _{\mathbf{R}} (\psi(t,Y))\);
-
(iv)
for a.e. \(t\in[a,b]\)
and for all
\(v\in\operatorname {int}_{\mathbf{R}} (\psi(t,Y))\), one has
\(\operatorname{int}_{Y}(\{y\in Y: \psi(t,y)=v\} )=\emptyset \);
-
(v)
there exist
\(p\in[1,+\infty]\)
and a positive function
\(\beta \in L^{p}([a,b])\)
such that, for a.e. \(t\in[a,b]\), and for all
\((x,z)\in\Omega \), one has
$$\bigl\{ y\in Y:\psi(t,y)=g(t,x,z)\bigr\} \subseteq \overline{B}_{n} \bigl(0_{{\mathbf {R}}^{n}},\beta (t)\bigr). $$
Then, there exists
\(u\in W^{2,p}([a,b],{\mathbf{R}}^{n})\)
such that
$$\textstyle\begin{cases} \psi(t,u^{\prime\prime}(t))= g(t,u(t),u^{\prime}(t))&\textit{for a.e. }t\in[a,b], \\ u(a)=u(b)=0_{{\mathbf{R}}^{n}}, \end{cases} $$
and
$$\bigl(u(t), u^{\prime}(t)\bigr)\in\Omega \quad \textit{for a.e. } t\in[a,b]. $$
Proof
Let us apply Theorem 3.1, with \(f(t,x,z,y)=\psi(t,y)-g(t,x,z)\), \(D^{\prime}=D^{\prime\prime}=Y\) and \(\Sigma =D\times D\). Observe that by assumptions (i) and (ii) and by the lemma at p.198 of [12], the function \(g|_{[a,b]\times\Omega }\) is \({\mathcal{L}}([a,b])\times{\mathcal {B}}(\Omega )\)-measurable. Consequently, assumption (i) of Theorem 3.1 is satisfied. It is a matter of routine to check that all the other assumptions of Theorem 3.1 are satisfied. Therefore, the conclusion follows at once. □
Arguing exactly as for Theorem 3.1, it can be easily seen that Theorem 3.3 can be put in the following equivalent form.
Corollary 3.4
Let
\([a,b]\), Y, ψ, g
and
D
be as in the statement of Theorem
3.3. Assume that there exists a set
\(F\in {\mathcal{F}}_{2n}\)
such that, if one puts
\(\Omega :={\mathbf{R}}^{2n}\setminus F\), then assumptions (i)-(v) of Theorem
3.3
are satisfied.
Then, the same conclusion of Theorem
3.3
holds.
Remark
Theorem 3.3 (or, equivalently, Corollary 3.4) should be compared with Theorem 2.2 of [6] (valid for the case where ψ does not depend on t and the set Y is bounded), where the function g is assumed to be continuous in all variables. We now give an example of an application of Theorem 3.3 in the vector case \(n=2\).
Example 2
Let \(n=2\). We now denote vectors of \({\mathbf{R}}^{2}\) by
$$x=(x_{1},x_{2}), \qquad z=(z_{1},z_{2}), \qquad y=(y_{1},y_{2}). $$
Let \(p\in[1,+\infty]\) and \(\alpha \in L^{p}([0,1])\) be fixed, with \(\alpha (t)\ge3\) for all \(t\in[0,1]\). Let
$$E:= \bigl\{ (x,z)\in{\mathbf{R}}^{4}: \mbox{at least one of } x_{1},x_{2},z_{1},z_{2} \mbox{ is rational} \bigr\} , $$
and let \(h:[0,1]\times{\mathbf{R}}^{2}\to{\mathbf{R}}\) and \(g:[0,1]\times{\mathbf{R}}^{2}\times{\mathbf{R}}^{2}\to{\mathbf{R}}\) be defined by
$$\begin{aligned}& h(t,y_{1},y_{2})=\sin\bigl[t(y_{1}-1) \bigr]+y_{1}+y_{2}, \\& g(t,x_{1},x_{2},z_{1},z_{2})= \textstyle\begin{cases} \alpha (t)+\cos^{2}(t\ x_{1}\ x_{2}\ z_{1}\ z_{2}) &\mbox{if }(x,z)\notin E, \\ \alpha (t)-2&\mbox{if }(x,z)\in E. \end{cases}\displaystyle \end{aligned}$$
Of course, for all \(t\in[0,1]\) the function \(g(t,\cdot ,\cdot )\)
is discontinuous at all points
\((x,z)\in{\mathbf{R}}^{4}\). However, Theorem 3.3 can be easily applied, by choosing
$$V_{1}=V_{2}=V_{3}=V_{4}={\mathbf{Q}}, \qquad Y:=[1,+\infty[\,\times[1,+\infty [,\qquad \psi:=h|_{[0,1]\times Y}, $$
and taking as D any countable dense subset of Y.
In particular, we observe that in this case we have \(\Omega ={\mathbf{R}} ^{4}\setminus E\) and
$$g(t,x_{1},x_{2},z_{1},z_{2})=\alpha (t)+\cos^{2}(t\ x_{1}\ x_{2}\ z_{1}\ z_{2}) \quad \mbox{for all } (x,z)\in\Omega . $$
Consequently, assumptions (i) and (ii) of Theorem 3.3 are satisfied. If we fix \(t\in[0,1]\), then \(\psi(t,1,1)=2\) and \(\psi(t,\cdot ,\cdot )\) is upperly unbounded, hence
$$g(t,\Omega )\subseteq [3,+\infty[\,\subseteq\, ] 2,+\infty [\,\subseteq \psi(t,Y). $$
Hence, assumption (iii) of Theorem 3.3 is also satisfied. We now prove that assumption (iv) is also satisfied (although this fact is quite intuitive, we provide a direct proof).
To this aim, fix \(t\in[0,1]\) and \(v^{*}\in\operatorname{int}(\psi (t,Y))\). Let \(y^{*}:=(y^{*}_{1},y^{*}_{2})\in Y\) be such that \(\psi(t, y^{*}_{1},y^{*}_{2})=v^{*}\), and let Σ be any open set in Y such that \((y^{*}_{1},y^{*}_{2})\in\Sigma \). This implies that there exists \(\delta >0\) such that
$$\bigl( \bigl]y^{*}_{1}-\delta ,y^{*}_{1}+\delta \bigr[\,\times\, \bigl]y^{*}_{2}-\delta ,y^{*}_{2}+\delta \bigr[\bigr)\cap Y\subseteq \Sigma . $$
If we choose any \(y_{2}\in\,]y_{2}^{*},y^{*}_{2}+\delta [\), we get \((y_{1}^{*},y_{2})\in\Sigma \) and
$$\psi\bigl(t,y_{1}^{*},y_{2}\bigr)>\psi\bigl(t,y_{1}^{*},y^{*}_{2} \bigr)=v^{*}. $$
Hence, the set \(\{y\in Y:\psi(t,y_{1},y_{2})=v^{*}\}\) has empty interior in Y, as desired.
Finally, let \(t\in[0,1]\) and \((x,z)\in\Omega \). Take \(y\in Y\) such that \(\psi(t,y)=g(t,x,z)\). By the definitions of Y, ψ and g we get
$$\|y\|_{2}\le\alpha (t)+2, $$
hence assumption (v) of Theorem 3.3 is also satisfied. Consequently, there exists \(u\in W^{2,p} ([0,1],{\mathbf{R}}^{2})\) such that
$$ \textstyle\begin{cases} h(t,u^{\prime\prime}(t))= g(t,u(t),u^{\prime}(t))&\mbox{for a.e. } t\in[0,1], \\ u(a)=u(b)=0_{{\mathbf{R}}^{2}}, \end{cases} $$
(14)
and also
$$u(t)\in({\mathbf{R}}\setminus{\mathbf{Q}})\times({\mathbf {R}}\setminus{ \mathbf{Q}}),\qquad u^{\prime}(t)\in({\mathbf{R}}\setminus{\mathbf{Q}}) \times ({\mathbf{R}}\setminus{\mathbf{Q}}), \qquad u^{\prime\prime}(t)\in[1,+ \infty[\,\times[1,+\infty[ $$
for a.e. \(t\in[0,1]\). We also observe that problem (14) does not admit the trivial solution \(u(t)\equiv0_{{\mathbf{R}}^{2}}\).