Download Linear Constant Coefficients Equations (§ 1.1) Linear Constant

Linear Constant Coefficients Equations (§ 1.1)
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Overview of Differential Equations
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Linear Ordinary Differential Equations
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Integrating a Differential Equation
Linear Constant Coefficients Equations
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The Integrating Factor Method
The Initial Value Problem
Linear Constant Coefficients Equations (§ 1.1)
I
Overview of Differential Equations
I
Linear Ordinary Differential Equations
I
Integrating a Differential Equation
Linear Constant Coefficients Equations
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I
I
The Integrating Factor Method
The Initial Value Problem
Overview of Differential Equations
Definition
A differential equation is an equation, the unknown is a function,
and both the function and its derivatives appear in the equation.
Example (Newton’s Second Law of Motion)
Mass × Acceleration = Force: m a = f .
The unknown is x(t), the particle position as function of time t
and the equation is
d2
1
x(t)
=
f (t, x(t)),
dt 2
m
with m the particle mass and f the force acting on the particle.
Remark: This is an Ordinary Differential Equations (ODE):
Derivatives with respect to only one variable appear in the
equation.
Overview of Differential Equations
Definition
A differential equation is an equation, the unknown is a function,
and both the function and its derivatives appear in the equation.
Example (Wave Equation)
Sound propagation in the air is described by the wave equation
2
∂2
2 ∂
u(t,
x)
=
v
u(t, x),
∂t 2
∂x 2
where function u represents air pressure and v the wave speed.
The unknown is u, which depends on two variables (t, x).
Remark: This is a Partial Differential Equations (PDE): Partial
derivatives of two or more variables appear in the equation.
Overview of Differential Equations
Remark: Differential equations are a central part in a physical
description of nature:
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Classical Mechanics:
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Fluid Dynamics:
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Einstein equation. (PDE)
Quantum Mechanics:
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Maxwell’s equations. (PDE)
General Relativity:
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Navier-Stokes equation. (PDE)
Electromagnetism:
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Newton second law of motion. (ODE)
Euler-Lagrange equations. (ODE)
Schrödinger’s equation. (PDE)
Standard Model of Particle Physics. (PDE)
Linear Constant Coefficients Equations (§ 1.1)
I
Overview of Differential Equations
I
Linear Ordinary Differential Equations
I
Integrating a Differential Equation
Linear Constant Coefficients Equations
I
I
I
The Integrating Factor Method
The Initial Value Problem
Linear Ordinary Differential Equations
Remark: We use the notation
y0 =
dy
,
dt
y 0 (t) =
dy
(t).
dt
Definition
An ODE determined by a functions f of two variables, with
unknown function y , is the equation
y 0 (t) = f (t, y (t)).
This equation is called linear iff there exist functions a, b such that
f (t, y ) = a(t) y + b(t). That is, f is linear on its argument y ,
hence a first order linear ODE is given by
y 0 (t) = a(t) y (t) + b(t).
Linear Ordinary Differential Equations
Example
A first order linear ODE is given by
y 0 (t) = −2 y (t) + 3.
Here a(t) = −2 and b(t) = 3. Since these functions do not depend
on t, the equation is called of constant coefficients or autonomous.
Example
A first order linear ODE is given by
2
y 0 (t) = − y (t) + 4t.
t
Here a(t) = −2/t and b(t) = 4t. Since these functions depend on
t, the equation is called of variable coefficients or nonautonomous.
Linear Constant Coefficients Equations (§ 1.1)
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Overview of Differential equations
I
Linear Ordinary Differential Equations
I
Integrating a Differential Equation
Linear Constant Coefficients Equations
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I
I
The Integrating Factor Method
The Initial Value Problem
Integrating a Differential Equation
Remarks:
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To integrate a differential equation y 0 = ay + b is to find all
functions y such that y 0 is equal to ay + b.
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Integrating on both sides of the equations fails.
For example, integrate on t on a constant coefficients equation:
Z
Z
Z
0
y (t) dt =
ay (t) + b dt ⇒ y (t) = a y (t) dt + bt + c
where c ∈ R is any integration constant.
Remarks:
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We have
R not solved the equation, because we do not know
what y dt actually is.
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We need a new idea to integrate the equation.
Linear Constant Coefficients Equations (§ 1.1)
I
Overview of Differential equations
I
Linear Ordinary Differential Equations
I
Integrating a Differential Equation
Linear Constant Coefficients Equations
I
I
I
The integrating Factor Method
The Initial Value Problem
The Integrating Factor Method
New idea: Transform the differential equation y 0 = ay + b into the
total derivative of a potential function ψ depending on (t, y ),
y 0 = ay + b
⇔
ψ 0 = 0.
The last equation is easy to integrate: ψ(t, y ) = c, for any c ∈ R.
How can the potential function ψ be obtained?
Multiplying the differential equation y 0 = ay + b by a nonzero
function, called the integrating factor.
Example
Find all functions y solution of the ODE y 0 = 2y + 3.
Solution: Write down the differential equation as y 0 − 2 y = 3.
Key idea: The left hand side above is a total derivative if we
multiply it by an exponential, e −2t (y 0 − 2 y ) = 3e −2t .
The Integrating Factor Method
Example
Find all functions y solution of the ODE y 0 = 2y + 3.
Solution: Recall: e −2t (y 0 − 2 y ) = 3e −2t .
0
e −2t y 0 − 2 e −2t y = 3 e −2t ⇔ e −2t y 0 + e −2t y = 3 e −2t .
0
The derivative of a product implies e −2t y = 3 e −2t , hence
h
−2t 0 h 3 −2t i0
3 −2t i0
e
y = − e
⇔
y+
e
= 0.
2
2
So e −2t is an Integrating Factor. A Potential Function is
3 −2t
and the equation y 0 = 2y + 3 is: ψ 0 = 0.
ψ= y+
e
2
3 −2t
3
Integrating, ψ = c ⇒ y + e
= c ⇒ y (t) = c e 2t − . C
2
2
The Integrating Factor Method
Example
Find all functions y solution of the ODE y 0 = 2y + 3.
Solution:
We concluded that the ODE has
infinitely many solutions, given by
y
c>0
3
y (t) = c e 2t − ,
2
c ∈ R.
Since we did one integration, it is
reasonable that the solution
contains a constant of
integration, c ∈ R.
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Verification: y 0 = 2c e 2t , but we know that 2y + 3 = 2c e 2t ,
therefore we conclude that y satisfies the ODE y 0 = 2y + 3.
t
c=0
c<0
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The Integrating Factor Method
Remark: Solutions to first order linear ODE can be obtained with
the Integrating Factor Method.
Theorem (Constant Coefficients)
If the constants a, b ∈ R satisfy a 6= 0, then the linear equation
y 0 (t) = a y (t) + b
has infinitely many solutions, one for each value of c ∈ R, given by
b
y (t) = c e at − .
a
Remarks:
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This expression for y is called the General Solution.
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Here we present the main idea of the proof using an
exponential integrating factor.
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In the Lecture Notes we show that this is essentially the only
integrating factor.
The Integrating Factor Method
Main ideas of the Proof: Write down the differential equation as
y 0 − a y = b.
Key idea: The left-hand side above is a total derivative if we
multiply it by an exponential e −at (y 0 − a y ) = b e −at .
0
e −at y 0 − a e −at y = b e −at ⇔ e −at y 0 + e −at y = b e −at .
0
The derivative of a product implies e −at y = b e −at , hence
h
−at 0 h b −at i0
b −at i0
e
y = − e
⇔
y+
e
= 0.
a
a
b −at
Introducing a potential function ψ = y +
e , the equation
a
y 0 = ay + b is now ψ 0 = 0. We can now integrate the equation,
b −at
b
ψ=c ⇔ y+
e
= c ⇔ y (t) = c e at − .
a
a
Linear Constant Coefficients Equations (§ 1.1)
I
Overview of Differential Equations
I
Linear Ordinary Differential Equations
I
Integrating a Differential Equation
Linear Constant Coefficients Equations
I
I
I
The Integrating Factor Method
The Initial Value Problem
The Initial Value Problem
Definition
The Initial Value Problem (IVP) for a linear ODE is the following:
Given functions a, b and a constant y0 ∈ R, find a solution y of
y 0 = a(t) y + b(t),
y (0) = y0 .
Remark: The initial condition selects one solution of the ODE.
Theorem (IVP Constant Coefficients)
Given constants a, b, y0 ∈ R, with a 6= 0, the initial value problem
y 0 = a y + b,
y (0) = y0
has the unique solution
b at b
y (t) = y0 +
e − .
a
a
The Initial Value Problem
Example
Find the solution to the initial value problem
y 0 = 2y + 3,
y (0) = 1.
Solution: Every solution of the ODE above is given by
3
y (t) = c e 2t − ,
2
c ∈ R.
The initial condition y (0) = 1 selects only one solution:
1 = y (0) = c −
We conclude that y (t) =
3
2
⇒
5
c= .
2
5 2t 3
e − .
2
2
The Initial Value Problem
Example
Find the solution y to the IVP y 0 = −3y + 1, y (0) = 1.
Solution: Write down the differential equation as y 0 + 3 y = 1.
Key idea: The left-hand side above is a total derivative if we
multiply it by the exponential e 3t . Indeed,
0
e 3t y 0 + 3 e 3t y = e 3t ⇔ e 3t y 0 + e 3t y = e 3t .
This is the key idea, because the derivative of a product implies
3t 0
e y = e 3t .
The exponential e 3t is called an integrating factor. Integrating,
1
1
e 3t y = e 3t + c ⇔ y (t) = c e −3t + .
3
3
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The Initial Value Problem
Example
Find the solution y to the IVP y 0 = −3y + 1, y (0) = 1.
Solution: Every solution of the ODE above is given by
1
y (t) = c e −3t + ,
3
c ∈ R.
The initial condition y (0) = 1 selects only one solution:
1 = y (0) = c +
We conclude that y (t) =
1
3
2 −3t 1
e
+ .
3
3
⇒
2
c= .
3
C