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Exact equations (Sect. 2.6).
I (^) Exact differential equations.
I (^) The Poincar´e Lemma.
I (^) Implicit solutions and the potential function.
I (^) Generalization: The integrating factor method.
Exact differential equations.
Definition
Given an open rectangle R = (t 1 , t 2 ) × (u 1 , u 2 ) ⊂ R^2 and continuously differentiable functions M, N : R → R, denoted as
(t, u) 7 → M(t, u) and (t, u) 7 → N(t, u), the differential equation in the unknown function y : (t 1 , t 2 ) → R given by
N(t, y (t)) y ′ (t) + M(t, y (t)) = 0
is called exact iff for every point (t, u) ∈ R holds
∂t N(t, u) = ∂u M(t, u)
Recall: we use the notation: ∂t N =
∂N
∂t
, and ∂u M =
∂M
∂u
Exact differential equations.
Example
Show whether the differential equation below is exact,
2 ty (t) y ′ (t) + 2t + y 2 (t) = 0.
Solution: We first identify the functions N and M,
[
2 ty (t)
]
y ′ (t) +
[
2 t + y 2 (t)
]
N(t, u) = 2tu,
M(t, u) = 2t + u 2 .
The equation is exact iff ∂t N = ∂u M. Since
N(t, u) = 2tu ⇒ ∂t N(t, u) = 2u,
M(t, u) = 2t + u 2 ⇒ ∂u M(t, u) = 2u.
We conclude: ∂t N(t, u) = ∂u M(t, u). C
Remark: The ODE above is not separable and non-linear.
Exact differential equations.
Example
Show whether the differential equation below is exact,
sin(t)y ′ (y ) + t 2 e y (t) y ′ (t) − y ′ (t) = −y (t) cos(t) − 2 te y (t) .
Solution: We first identify the functions N and M, if we write
[ sin(t) + t 2 e y (t) − 1
]
y ′ (t) +
[
y (t) cos(t) + 2te y (t)
]
we can see that
N(t, u) = sin(t) + t 2 e u − 1 ⇒ ∂t N(t, u) = cos(t) + 2te u ,
M(t, u) = u cos(t) + 2te u ⇒ ∂u M(t, u) = cos(t) + 2te u .
The equation is exact, since ∂t N(t, u) = ∂u M(t, u). C
The Poincar´e Lemma.
Remark: The coefficients N and M of an exact equations are the
derivatives of a potential function ψ.
Lemma (Poincar´e)
Given an open rectangle R = (t 1 , t 2 ) × (u 1 , u 2 ) ⊂ R^2 , the
continuously differentiable functions M, N : R → R satisfy the equation ∂t N(t, u) = ∂u M(t, u)
iff there exists a twice continuously differentiable function ψ : R → R, called potential function, such that for all (t, u) ∈ R holds ∂u ψ(t, u) = N(t, u), ∂t ψ(t, u) = M(t, u).
Proof: (⇐) Simple:
∂t N = ∂t ∂u ψ,
∂u M = ∂u ∂t ψ,
⇒ ∂t N = ∂u M.
(⇒) Difficult: Poincar´e, 1880.
The Poincar´e Lemma.
Example
Show that the function ψ(t, u) = t 2
- tu 2 is the potential function for the exact differential equation
2 ty (t) y ′ (t) + 2t + y 2 (t) = 0.
Solution: We already saw that the differential equation above is exact, since the functions M and N,
N(t, u) = 2tu,
M(t, u) = 2t + u 2
⇒ ∂t N = 2u = ∂u M.
The potential function is ψ(t, u) = t 2
∂t ψ = 2t + u 2 = M, ∂u ψ = 2tu = N. C
Remark: The Poincar´e Lemma only states necessary and sufficient
conditions on N and M for the existence of ψ.
Exact equations (Sect. 2.6).
I (^) Exact differential equations.
I (^) The Poincar´e Lemma.
I (^) Implicit solutions and the potential function.
I (^) Generalization: The integrating factor method.
Implicit solutions and the potential function.
Theorem (Exact differential equations)
Let M, N : R → R be continuously differentiable functions on an
open rectangle R = (t 1 , t 2 ) × (u 1 , u 2 ) ⊂ R 2
. If the differential equation N(t, y (t)) y ′ (t) + M(t, y (t)) = 0 (1)
is exact, then every solution y : (t 1 , t 2 ) → R must satisfy the
algebraic equation ψ(t, y (t)) = c,
where c ∈ R and ψ : R → R is a potential function for Eq. (1).
Proof: 0 = N(t, y ) y
′
dy
dt
d
dt
ψ(t, y (t)) ⇔ ψ(t, y (t)) = c.
Exact equations (Sect. 2.6).
I (^) Exact differential equations.
I (^) The Poincar´e Lemma.
I (^) Implicit solutions and the potential function.
I (^) Generalization: The integrating factor method.
Remark:
Sometimes a non-exact equation can we transformed into an exact equation multiplying the equation by an integrating factor. Just like in the case of linear differential equations.
Generalization: The integrating factor method.
Theorem (Integrating factor)
Let M, N : R → R be continuously differentiable functions on R = (t 1 , t 2 ) × (u 1 , u 2 ) ⊂ R^2 , with N 6 = 0. If the equation
N(t, y (t)) y ′ (t) + M(t, y (t)) = 0
is not exact, that is, ∂t N(t, u) 6 = ∂u M(t, u), and if the function
N(t, u)
[
∂u M(t, u) − ∂t N(t, u)
]
does not depend on the variable u, then the equation
μ(t)
[
N(t, y (t)) y ′ (t) + M(t, y (t))
]
is exact, where
μ ′ (t)
μ(t)
N(t, u)
[
∂u M(t, u) − ∂t N(t, u)
]
Generalization: The integrating factor method.
Example
Find all solutions y to the differential equation [ t 2
]
y ′ (t) +
[
3 t y (t) + y 2 (t)
]
Solution: The equation is not exact:
N(t, u) = t 2
- tu ⇒ ∂t N(t, u) = 2t + u,
M(t, u) = 3tu + u 2 ⇒ ∂u M(t, u) = 3t + 2u,
hence ∂t N 6 = ∂u M. We now verify whether the extra condition in Theorem above holds: [ ∂u M(t, u) − ∂t N(t, u)
]
N(t, u)
(t^2 + tu)
[
(3t + 2u) − (2t + u)
]
[
∂u M(t, u) − ∂t N(t, u)
]
N(t, u)
t(t + u)
(t + u) =
t
Generalization: The integrating factor method.
Example
Find all solutions y to the differential equation [ t
2
]
y
′ (t) +
[
3 t y (t) + y
2 (t)
]
Solution:
[
∂u M(t, u) − ∂t N(t, u)
]
N(t, u)
t
We find a function μ solution of
μ′
μ
[
∂u M − ∂t N
]
N
, that is
μ ′ (t)
μ(t)
t
⇒ ln(μ(t)) = ln(t) ⇒ μ(t) = t.
Therefore, the equation below is exact:
[ t 3
]
y ′ (t) +
[
3 t 2 y (t) + t y 2 (t)
]
