A regularization of discontinuous
differential equations with application to
state-dependent DDEs
Giorgio Fusco & Nicola Guglielmi, L’Aquila
Outline
(1) A class of neutral state dependent problems.
(2) Existence termination of classical solutions.
(3) Regularization.
– p.1/20
A class of neutral problems
´ ³
´´
³
³

 ẏ(t) = f y(t), y α(t, y(t)) , ẏ α(t, y(t)) , t ∈ [t0 , tf ]

y(t) = ϕ(t), t ≤ t0
(NP)
– p.2/20
A class of neutral problems
´ ³
´´
³
³

 ẏ(t) = f y(t), y α(t, y(t)) , ẏ α(t, y(t)) , t ∈ [t0 , tf ]

y(t) = ϕ(t), t ≤ t0
(NP)
Assumptions
•
•
•
•
y, f ∈ Rd (d ≥ 1);
f and α smooth functions;
ϕ (piecewise) smooth vector function;
α(t, y(t)) ≤ t.
– p.2/20
Example
³
ẏ(t) = 1 − 2 ẏ y(t) − 1
´
with y(t) ≡ 0, t ≤ 0.
y+
1.0
1.0
y
ẏ
ẏ +
y−
ẏ −
.0
ξ=1
.0
ξ=1
At ξ = 1, where y(ξ) = 1, the equation is ẏ(ξ) = 1 − 2 ẏ(0).
The only possible solutions y + , y − don’t satisfy the equation.
– p.3/20
Classical solutions and possible regularizations
Due to the discontinuous right-hand side, the existence of a
solution beyond ξ is not assured for a state-dependent delay.
To define weak solutions we consider possible regularizations.
– p.4/20
Classical solutions and possible regularizations
Due to the discontinuous right-hand side, the existence of a
solution beyond ξ is not assured for a state-dependent delay.
To define weak solutions we consider possible regularizations.
(i) Replacing the jumping component by a time average.
Zt ³
´
1
Example. ẏ(t) = 1 − 2
ẏ y(s) − 1 ds.
ε
t−ε
– p.4/20
Classical solutions and possible regularizations
Due to the discontinuous right-hand side, the existence of a
solution beyond ξ is not assured for a state-dependent delay.
To define weak solutions we consider possible regularizations.
(i) Replacing the jumping component by a time average.
Zt ³
´
1
Example. ẏ(t) = 1 − 2
ẏ y(s) − 1 ds.
ε
t−ε
(ii) Augmenting the dimension of the system of differential
equations and introducing a singular perturbation.
(
ẏ(t) = z(t) ³
´
Example.
εż(t) = 1 − 2 z y(t) − 1 − z(t)
– p.4/20
Example in R2

³
´
 ẏ1 (t) = 1 − 2 ẏ1 y1 (t) − 1
³
´
 ẏ2 (t) = 1 + ẏ2 y1 (t) − 1
with zero initial data. For t ≤ 1 solution is y1 (t) = y2 (t) = t.
– p.5/20
Example in R2

³
´
 ẏ1 (t) = 1 − 2 ẏ1 y1 (t) − 1
³
´
 ẏ2 (t) = 1 + ẏ2 y1 (t) − 1
with zero initial data. For t ≤ 1 solution is y1 (t) = y2 (t) = t.
As long as y1 (s) ≤ 2, it holds ẏ(y1 (s) − 1) = H(y1 (s) − 1).
– p.5/20
Example in R2

³
´
 ẏ1 (t) = 1 − 2 ẏ1 y1 (t) − 1
³
´
 ẏ2 (t) = 1 + ẏ2 y1 (t) − 1
with zero initial data. For t ≤ 1 solution is y1 (t) = y2 (t) = t.
As long as y1 (s) ≤ 2, it holds ẏ(y1 (s) − 1) = H(y1 (s) − 1).
averaging.
Regularization by time
t
Z
³
´


2

ε
ε

(t)
=
1
−
H
y
ẏ

1 (s) − 1 ds
1

ε



1

ε

ẏ2 (t) = 1 +



ε
t−ε
Zt
t−ε
³
´
H y1ε (s) − 1 ds
– p.5/20
Suitable reformulation of the example
Equivalent formulation
(
p if h(y) > 0
ẏ(t) =
q if h(y) < 0
¶
µ
µ ¶
−1
1
,q=
with p =
1
2
and h(y) = 1 − y1 .
y1
q
p
h(y) < 0
h(y) > 0
y2
– p.6/20
Suitable reformulation of the example
Equivalent formulation
(
p if h(y) > 0
ẏ(t) =
q if h(y) < 0
¶
µ
µ ¶
−1
1
,q=
with p =
1
2
and h(y) = 1 − y1 .
y1
q
p
h(y) < 0
h(y) > 0
y2
Regularized problem.
Zt h ³
´
³
³
´´ i
1
H h (y ε (s)) p + 1 − H h (y ε (s))
ẏ ε (t) =
q ds
ε
t−ε
For constant vector fields the solution can be found explicitly.
– p.6/20
Solution of the regularized example
The solution y1ε , for t ≥ 1, is periodic of period 2ε and C 1

2
τ

 1+τ −
for 0 ≤ τ ≤ ε
ε
ε
y1 (1 + τ ) =
2
(τ
−
ε)

 1 − (τ − ε) +
for ε ≤ τ ≤ 2ε
ε
The solution y2ε is the sum of a periodic and a linear function
1 ε
3
ε
(1 + τ ) − y1 (1 + τ )
y2 (1 + τ ) =
for τ ≥ 0.
2
2
– p.7/20
Solution of the regularized example
The solution y1ε , for t ≥ 1, is periodic of period 2ε and C 1

2
τ

 1+τ −
for 0 ≤ τ ≤ ε
ε
ε
y1 (1 + τ ) =
2
(τ
−
ε)

 1 − (τ − ε) +
for ε ≤ τ ≤ 2ε
ε
The solution y2ε is the sum of a periodic and a linear function
1 ε
3
ε
(1 + τ ) − y1 (1 + τ )
y2 (1 + τ ) =
for τ ≥ 0.
2
2
Note: for all t ≥ 1 the solution remains ε-close to the manifold
M = {y : h(y) = 0} = {y : y1 = 1}.
– p.7/20
Weak solution
1.003
y1ε
.997
.99
Solution y1ε (t ≥ 1)
y1ε = 1 + O(ε).
ε = 10−2
1.10
– p.8/20
Weak solution
1.003
Solution y1ε (t ≥ 1)
y1ε
y1ε = 1 + O(ε).
ε = 10−2
.997
.99
1.10
We have that y1 and y2 converge as ε → 0 in the C 0 -topology,
lim y1ε (1 + τ ) =: y10 (τ ) = 1
ε→0
lim y2ε (1
ε→0
+ τ ) =:
y20 (τ )
3
=1+ τ
2
for
τ ≥ 0.
– p.8/20
Limit problem
The weak solution solves the problem
µ
¶
p q
0
ẏ(t) = + =
,
3/2
2 2
t ≥ 1.
– p.9/20
Limit problem
The weak solution solves the problem
µ
¶
p q
0
ẏ(t) = + =
, t ≥ 1.
3/2
2 2
The r.h.s. is the unique convex combination of p and q which
lies on the manifold.
y1
y1
q
p
h(y) < 0
h(y) > 0
M
p+q
2
y2
This agrees with Filippov’s generalized solution.
y2
– p.9/20
General case
ẏ(t) =
(
f (y(t)) if h(y(t)) > 0
g(y(t)) if h(y(t)) < 0
The manifold M = {y : h(y) = 0} of codimension 1
separates the space into two regions.
Let n be the normal to M at ȳ and let f and g the smooth
vector fields in the regions S + (identified by n) and S − .
– p.10/20
General case
ẏ(t) =
(
f (y(t)) if h(y(t)) > 0
g(y(t)) if h(y(t)) < 0
The manifold M = {y : h(y) = 0} of codimension 1
separates the space into two regions.
Let n be the normal to M at ȳ and let f and g the smooth
vector fields in the regions S + (identified by n) and S − .
S+
The solution terminates
to exist at ȳ ∈ M if
(
f
hf (ȳ), ni < 0
n
M
hg(ȳ), ni > 0
g
S−
– p.10/20
Regularized problem (RP)
Zt h ³
´
³
´
i
1
ε
ε
ε
ε
ε
H h(y (s) f (y (s))+H −h(y (s) g (y (s)) ds
ẏ (t) =
ε
t−ε
with y ε (t) = Φ(t) ∀t ≤ 0.
– p.11/20
Regularized problem (RP)
Zt h ³
´
³
´
i
1
ε
ε
ε
ε
ε
H h(y (s) f (y (s))+H −h(y (s) g (y (s)) ds
ẏ (t) =
ε
t−ε
with y ε (t) = Φ(t) ∀t ≤ 0.
yε
Illustrative case
• Φ is a C 1 function;
• h (Φ(t)) 6= 0 ∀t < 0;
M
Φ
• h (Φ(0)) = 0.
– p.11/20
Existence of a solution ε-close to M
Theorem
Let f, g be smooth. There exist T > 0, ε0 > 0 s.t. problem
(RP) has a C 1 -solution y ε : [0, T ] → Rd ∀ε ∈ (0, ε0 ) with
•
|h (y ε (t)) | ≤ Cε ∀t ∈ [0, T ].
– p.12/20
Existence of a solution ε-close to M
Theorem
Let f, g be smooth. There exist T > 0, ε0 > 0 s.t. problem
(RP) has a C 1 -solution y ε : [0, T ] → Rd ∀ε ∈ (0, ε0 ) with
•
|h (y ε (t)) | ≤ Cε ∀t ∈ [0, T ].
Moreover there is a C 1 function y 0 such that
¡ 0 ¢
• h y (t) ≡ 0;
•
•
lim ky ε − y 0 kC 0 [0,T ] = 0;
ε→0
³
´
³
´
ẏ 0 (t) = λ(t)f y 0 (t) + (1 − λ(t))g y 0 (t) ∈ T M
with λ(t) ∈ [0, 1] ∀t ∈ [0, T ] (according to Filippov
theory).
– p.12/20
Local problem (RL)
By a local diffeomorfism we may flatten the manifold. From
this and the rescaling y(t) = εx(t/ε), we obtain the following
problem (RL):
ẋ(t) =
Zt ·
t−1
³
´
³
H(xd (s)) p + εP (x(s), ε) + 1 − H(xd (s)
´³
q + εQ(x(s), ε)
´¸
– p.13/20
ds
Local problem (RL)
By a local diffeomorfism we may flatten the manifold. From
this and the rescaling y(t) = εx(t/ε), we obtain the following
problem (RL):
ẋ(t) =
Zt ·
t−1
³
´
³
H(xd (s)) p + εP (x(s), ε) + 1 − H(xd (s)
´³
q + εQ(x(s), ε)
´¸
S+
Here p, q are constant vector
fields and P, Q smooth vector
fields s.t. P (x, 0) = Q(x, 0) = 0 M
and M = {x : xd = 0}.
p + εP (x, ε)
n
q + εQ(x, ε)
S−
– p.13/20
ds
The case of constant vector-fields
.5
ϕ
σ η
ξ
σ̂ η
0
−.5
ξ becomes periodic of period 2
Assume ϕ(t) : (−1, 0] → Rd of
class C 1 , ϕ(t) having a unique
non-degenerate zero in [−1, 0).
˙
= 0).
Let σ > 0 (σ = 0 ⇒ ξ(0)
Then ξ can be found explicitly.
.7
σ̂
.0
∗
.0 σ
σ
η
.7
– p.14/20
Idea of the proof
Looking for solutions of the form xε (t) = ξ(t) + εz(t), we
apply a nested iteration. The external iteration proceeds on
intervals given by two consecutive zeros t0 , t̃0 (see the figure);
the internal iteration is concerned with a step-by-step
integration which covers every single interval.
– p.15/20
Idea of the proof
Looking for solutions of the form xε (t) = ξ(t) + εz(t), we
apply a nested iteration. The external iteration proceeds on
intervals given by two consecutive zeros t0 , t̃0 (see the figure);
the internal iteration is concerned with a step-by-step
integration which covers every single interval.
.5
xε (t)
0
t0
ξ(t)
t̃0
ϕ
−.5
Let ξ(t) solution of the constant case. At each step xε (t) is
determined by contraction mapping arguments.
– p.15/20
Sketch of the proof (for simplicity 1−d case)
³
Z t Zτ
´
1
A z (t)−z(tj ) =
ε
tj
|
+
Z t Zτ
tj τ −1
+
|
Z t Zτ
tj τ −1
|
τ −1
h
i
1Iτtj (s) H (xε (s)) − H (ξ(s)) (p − q) ds dτ
{z
}
I1
h
i
1Iτtj (s) H (xε (s)) P (xε (s)) − H (ξ(s)) P (ξ(s)) ds dτ
{z
I2
h
i
}
1Iτtj (s) H (−xε (s)) Q(xε (s)) − H (−ξ(s)) Q(ξ(s)) ds dτ
{z
I3
}
– p.16/20
Sketch of the proof (for simplicity 1−d case)
³
Z t Zτ
´
1
A z (t)−z(tj ) =
ε
tj
|
+
Z t Zτ
tj τ −1
+
|
Z t Zτ
tj τ −1
|
τ −1
h
i
1Iτtj (s) H (xε (s)) − H (ξ(s)) (p − q) ds dτ
{z
}
I1
h
i
1Iτtj (s) H (xε (s)) P (xε (s)) − H (ξ(s)) P (ξ(s)) ds dτ
{z
I2
h
i
}
1Iτtj (s) H (−xε (s)) Q(xε (s)) − H (−ξ(s)) Q(ξ(s)) ds dτ
{z
I3
}
We show that the map A is a contraction on a complete metric
space Y .
– p.16/20
Sketch of the proof (ctd)
.5
ξ(t) + εz(t)
ξ(t)
0
t0
tj r
−.5
– p.17/20
Sketch of the proof (ctd)
.5
ξ(t) + εz(t)
ξ(t)
0
t0
tj r
−.5
With r < 1, |z(tj )| ≤ jK and K suitably chosen, set
©
ª
0
Y = z ∈ C [tj , tj + r] : kz − z(tj )kC 0 ([tj ,tj +r]) ≤ K .
– p.17/20
Sketch of the proof (ctd)
.5
ξ(t) + εz(t)
ξ(t)
0
t0
tj r
−.5
With r < 1, |z(tj )| ≤ jK and K suitably chosen, set
©
ª
0
Y = z ∈ C [tj , tj + r] : kz − z(tj )kC 0 ([tj ,tj +r]) ≤ K .
A basic lemma
There exist r > 0, ε0 > 0 s.t. for all ε ∈ (0, ε0 ) problem (RL)
has a unique C 1 -solution xε = ξ + εz : [tj , tj + r] → Rd and
•
kzkC 0 ([tj ,tj +r]) ≤ (j + 1)K
with r > const/j
– p.17/20
Proof of the lemma
We show that there is at most 1 zero of ξ in the interval
[t − 1, t] for t ∈ [tj , tj + r]. Moreover the zero is simple
and the slope of ξ is −|p|.
– p.18/20
Proof of the lemma
We show that there is at most 1 zero of ξ in the interval
[t − 1, t] for t ∈ [tj , tj + r]. Moreover the zero is simple
and the slope
of
ξ
is
−|p|.
n
o
Set Ωτ = s ∈ [τ − 1, τ ] : sign (xε (s)) 6= sign (ξ(s)) .
– p.18/20
Proof of the lemma
We show that there is at most 1 zero of ξ in the interval
[t − 1, t] for t ∈ [tj , tj + r]. Moreover the zero is simple
and the slope
of
ξ
is
−|p|.
n
o
Set Ωτ = s ∈ [τ − 1, τ ] : sign (xε (s)) 6= sign (ξ(s)) .
First prove the bound |Ω| ≤ D(K)ε for a certain constant
D(K) > 0 using the assumption on non-degenerate zeros
of ξ. Then get the estimates (K0 , K1 , K2 suitable constants)
|I1 | ≤ (1/ε)|Ωτ ||p − q|r ≤ K0 r
|I2 | ≤ |Ωτ | (kP k + kQk) r ≤ εK1 r
|I3 | ≤ εK2 r
– p.18/20
Proof of the lemma (ctd)
We use an induction argument. More precisely, assuming
|z(tj−1 )| ≤ jK, we get
|(Az)(t) − z(tj−1 )| ≤ rK3 jK< K
Moreover we get the estimate
|A(z − ẑ)(t)| ≤ rK4 kz − ẑk
– p.19/20
Proof of the lemma (ctd)
We use an induction argument. More precisely, assuming
|z(tj−1 )| ≤ jK, we get
|(Az)(t) − z(tj−1 )| ≤ rK3 jK< K
Moreover we get the estimate
|A(z − ẑ)(t)| ≤ rK4 kz − ẑk
1
The first implies that we have to choose r = rj = const .
j
This means that we cannot use a constant r but anyway we
can go up to the following zero by a finite number of steps.
This gives a contraction on Y .
– p.19/20
Proof of the lemma (ctd)
We use an induction argument. More precisely, assuming
|z(tj−1 )| ≤ jK, we get
|(Az)(t) − z(tj−1 )| ≤ rK3 jK< K
Moreover we get the estimate
|A(z − ẑ)(t)| ≤ rK4 kz − ẑk
1
The first implies that we have to choose r = rj = const .
j
This means that we cannot use a constant r but anyway we
can go up to the following zero by a finite number of steps.
This gives a contraction on Y .
Then we can repeat the procedure a number of times which is
proportional to 1/ε. This completes the proof of the theorem.
– p.19/20
Conclusion
Example (Castleton & Grimm and Enright & Hayashi)
´ 1
³
ẏ(t) = cos (t) 1 + y(ty(t)2 ) + y(t) ẏ(ty(t)2 )
2
The derivative is not smooth. The solution terminates.
Regularization
Solution and retarded argument of regularized equation.
– p.20/20
Conclusion
Example (Castleton & Grimm and Enright & Hayashi)
´ 1
³
ẏ(t) = cos (t) 1 + y(ty(t)2 ) + y(t) ẏ(ty(t)2 )
2
The derivative is not smooth. The solution terminates.
Regularization
Solution and retarded argument of regularized equation.
Software
Release 2.1 of the code RADAR5 is available at
http://univaq.it/∼guglielm
http://www.unige.ch/∼hairer/software.html
(where the first released version is actually available).
– p.20/20