Existence and Uniqueness of Differential Equations
Existence and Uniqueness of Differential Equations
Theorem
1 Existence
Theorem
(Formal):Suppose
Supposeff(t,
function
in ainrectangle
Theorem
1 Existence
Theorem
(Formal):
(t, y)
y)isisa acontinuous
continuous
function
a rectangle
Uniqueness
the {(t,
formy){(t,
|ta<<b,t <c b,
in the
ty-plane.IfIf(t(t00,,yy00)) is
is aa point
then
of the of
form
| a y)
<
< cy <<yd<} din} the
ty-plane.
pointininthis
thisrectangle,
rectangle,
then
there
exists
an
ε
>
0
and
a
function
y(t)
defined
for
t
−
ε
<
t
<
t
+
ε
that
solves
the
initial-value
Uniqueness
0
0
there exists an ! > 0 and a function y(t) defined for t0 − ! < t < t0 + ! that solves the initial-value
problem
r the initial value
problem
problem
dy problem
Consider the initial value
= f (t, y), y(t0 ) = y0 .
dy
2/3
dt = 0.
= f (t, y) = 3y , y(0)
dy
dt
= is
f (t,
y) positive
= 3y 2/3value
, y(0)
= 0. ε can take on so
(The statement ”there exists an ε > 0” means thatdt
there
some
the variable
that the statement becomes true. The theorem
does not tell us how large that value is. )
Uniqueness
.
existsvalue
an ! problem.
> 0” means that there is some positive value the variable ! can
that y(t) = t3The
is astatement
solution of”there
the initial
Consider
initial
value
problem
take
on sothe
that
the
statement
becomes
The of
theorem
doesvalue
not tell
us how large that value is.
Show
that
y(t)
= t3 is atrue.
solution
the initial
problem.
Consider1.the
initial
value
problem
dy
= f (t, y) = 3y 2/3 , y(0) = 0.
dt
.
1.
Show that y(t) = t3 is a solution of the initial value problem
1. Show that y(t) = t3 is a solution of the initial value problem.
that y(t) = 0 is also
solution
the= initial
problem.
2. a Show
thatofy(t)
0 is alsovalue
a solution
of the initial value problem.
2. Show that y(t) = 0 is also a solution of the initial value problem.
3. Is there a unique solution to the initial value problem
2. Show that y(t) = 0 is also a solution
of the initial value problem.
re a unique solution to the initial value problem dy
y(0) = 0?
dt = f (t, y),
3. Is there a unique solution to the initial value problem
dy
dt
= f (t, y),
y(0) = 0?
4. 2/3
Is continuous?
the function f(t, y) = 3y2/3 continuous?
function f (t, y) = 3y
4. Is the function f (t, y) = 3y 2/3 continuous?
dy
3. Is the
there
a unique
solution
to the
initial
value
(t, y), value
y(0)problem?
= 0?
oes this say about
existence
of
a say
solution
of
initial
value
problem?
dt
What
does this
about
thethe
existence
of
aproblem
solution
of =
thefinitial
What does this say about the existence of a solution of the initial value problem?
late the first partial derivative of the function f (t, y) = 3y 2/3
5. Calculate the first partial derivative of the function f (t, y) = 3y 2/3
5. Calculate the first partial
derivative of the function f(t, y) = 3y2/3
2/3 continuous?
4. Is the function
f
(t,
y)
=
3y
∂f
∂y =
What does this say about the existence of a solution of the initial value problem?
artial derivative function
continuous?
Is this partial
derivative function continuous?
5. Calculate the
first
partialderivative
derivativefunction
of the function
f (t, y) = 3y 2/3
Is this partial
continuous?
∂f
∂y
=
1
Is this partial derivative function continuous?
2
Existence and Uniqueness of Differential Equations
Theorem 1 Existence Theorem (Formal): Suppose f (t, y) is a continuous function in a rectangle
continuous
function inthen
a
of the Theorem
form {(t, 2y)Uniqueness
| a < t < b,Theorem
c < y < (Formal):
d } in theSuppose
ty-plane.f(t,Ify)(tand
) is aare
point
in this rectangle,
0 , y0∂f/∂y
rectangle
of
the
form
{(t,
y)
|
a
<
t
<
b,
c
<
y
<
d
}
in
the
ty-plane.
If
(t
,
y
)
is
a
point
in
this
0 solves the initial-value
there exists an ! > 0 and a function y(t) defined for t0 − ! < t < t0 + ! 0that
rectangle and if y1(t) and y2(t) are two functions that solve the t solves the initial-value problem
problem
dy
= f (t, y), y(t0 ) = y0 .
dt
for all t in the interval for t0 − ε < t < t0 + ε (where ε is some positive number) then
y1(t) = y2(t)
for
t
∈
(t
−
ε
,
t
+
ε
).
0
The statement0 ”there
exists an ! > 0” means that there is some positive value the variable ! can
take on so that the statement becomes true. The theorem does not tell us how large that value is.
This theorem says that the solution of the initial value problem is unique.
Implications of Uniqueness Solution curves in the ty-plane cannot touch or cross.
Why?
1