Linear and non-linear equations - University of Notre Dametaylor/Math20580/Lectures/Lb2... ·...

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Linear and non-linear equations LRT 04/08/2019 1 / 10

Transcript of Linear and non-linear equations - University of Notre Dametaylor/Math20580/Lectures/Lb2... ·...

Linear and non-linear equations

LRT04/08/2019

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First order linear ODE’s

A first order linear ODE will always be written as

y′ + p(t)y = g(t)

We understand these equations very well.

Given any initial value y(t0) = y0 and any interval I such that t0 ∈ Iand both p(t) and q(t) are continuous on I, then there is a unique φ(t)which is differentiable for all t ∈ I; φ(t) satisfies the differentialequation; and φ(t0) = y0.

The above result is not the last word but it is more than enough formost applications.

For an example not covered by the theorem let G(x) = x2 sin

(1

x

),

x 6= 0, G(0) = 0. You can check that G′(x) = 2x sin

(1

x

)− cos

(1

x

),

x 6= 0, G′(0) = 0. This G(x) is a “simple” example of a functionwhich is differentiable on all of R but the derivative is not continuous.Hence if g(x) = G′(x), y′ = g(x) has a solution even though g(x) isnot continuous.

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Non-linear first order ODE

Non-linear equations are significantly more difficult. We will studyfirst order ODE’s of the form y′ = f(x, y). An existence anduniqueness theorem covering most examples is the following.

Let I and J be open intervals in R with values x0 ∈ I and y0 ∈ J . LetR ⊂ R2 be the rectangle of all points (x, y) with x ∈ I and y ∈ J .

Suppose that f(x, y) is continuous on R and that∂f

∂yis defined and

continuous on R. Then there exists a unique function φ(x) satisfyingthe following conditions.

1. φ(x) is differentiable on some interval I0 ⊂ I with x0 ∈ I0.

2. φ(x0) = y0.

3. φ′(x) = f(x, φ(x)

)for all x ∈ I0.

There are many ways in which this existence and uniqueness is lesssatisfactory than the result for linear ODE’s. First of all it is harderto determine if a function of two variables is continuous than decidingif a one variable function is continuous. In practice this is not somuch of a problem since you have theorems from calc III.

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A much more significant problem is that the interval I0 over whichthe solution is defined is not described by the result - only that somesuch interval exists. The only reliable way to figure out the interval isto solve the equation and this can be difficult. At this point the onlynon-linear examples we have discussed are the separable equations.We will do a few more later this week.

The book seems to think you also know the following improvement tothe existence and uniqueness theorem. Suppose I = (a−, a+) andI0 = (α−, α+. If I0 is the largest interval over which φ is defined andsatisfies the equation, then for any [r, s] ⊂ J φ(x) ∈ J − [r, s] for all xsufficiently near α+ or sufficiently near α−.

One reason we stress existence and uniqueness theorems is thefollowing remark. Suppose y′ = f(x, y) has unique solutions for allpoints in an open region R. Suppose φ1(x) and φ2(x) are twosolutions of the equation. If the graphs of these two solutionsintersect at a point in R then they are equal on any interval whichcontains the intersection point in the intersection of the two domains.

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Example of the issueRecall from section 2.2 where we solved y′ − x2y2 = x2. The generalexistence and uniqueness result holds for this equation and we foundthe solution to be

y = tan

(x3 − x30

3+ arctan(y0)

)with interval of definition

3

√x30 − 3 arctan(y0)− 3π

2< x <

3

√x30 − 3 arctan(y0) +

2

Notice how the interval of defini-tion depends on (x0, y0). Eachsolution has vertical asymptotesat the end points of its intervalof definition and the width of theinterval of definition gets smallerand smaller as you move awayfrom the origin.

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Peano’s improvement

Peano proved that solutions exist under the assumption only thatf(x, y) is continuous. Uniqueness can fail badly in this case. Consider

y′ = |y 13 |. Then f(x, y) is continuous but

∂f

∂yis undefined at y = 0.

Check that for any C ∈ R, if y+ =

(2

3(x+ C)

) 32

then

y′+ =3

2

(2

3(x+ C)

) 12 2

3=

(2

3(x+ C)

) 12

so this is a solution and

y+ > 0.

Similarly, if y− = −(−2

3(x+ C)

) 32

then

y′− = −3

2

(−2

3(x+ C)

) 12 −2

3=

(−2

3(x+ C)

) 12

so this is a solution

and y− 6 0.Also y = 0 is a solution.

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If y0 > 0 then there is a solution to our equation going through(x0, y0) for any x0. The function is y+ for some value of C:

y0 =

(2

3(x0 + C)

) 32

,2

3(x0 + C) = y

230 , x0 + C =

3

2y

230 , C =

3

2y

230 − x0

so a solution is y+ =

(2

3(x− x0) + y

230

) 32

Hence y+ is defined on the

interval

[x0 −

3

2y

230 ,∞

).

A similar calculation yields that a solution passing through (x0, y0)

for y0 < 0 is is y− = −(−2

3(x− x0) + y

230

) 32

Hence y− is defined on

the interval

(−∞, x0 +

3

2y

230

].

When x = −C the one-sided derivative from the right of y+ is 0 andthe one-sided derivative from the right of y− is 0.

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The green curves are the graphs of y+ and the blues curves are thegraphs of y−.

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Consider solutions to the initial value problem passing through thegreen dot. Our existence and uniqueness theorem says the solution isunique on the interval (1,∞).But there are infinitely many solutions with interval (−∞,∞). One ofthem is to come up the blue curve which ends at (1, 0) and then go upthe green curve. The one sided derivative remarks show this functionis differentiable at 0 and so is a solution for x ∈ (−∞,∞).A second solution follows the blue curve up to (0, 0), goes along thex-axis to (1, 0) and then goes up the green curve. There are infinitelymany variations on this idea.And of course you can come along the x-axis from −∞ to 1 and thenhead up the green curve.

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Consider solutions to the initial value problem passing through theblue dot. Our existence and uniqueness theorem says the solution isunique on the interval (−∞, 0) but an argument similar to the one onthe last slide shows there are infinitely many solution with interval(−∞,∞). Indeed there are infinitely many solutions on any intervalof the form (−∞, a) for a > 0.

Peano guarantees the existence of solutions passing through (x0, 0) forany x0 but there is no open interval on which such a solution isunique.

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