applied math for engineers -...

Applied Math for Engineers

Ming Zhong

Lecture 13

March 12, 2018

Ming Zhong (JHU) AMS Spring 2018 1 / 24

Linear Algebra

Table of Contents

1 Linear Algebra

2 Differential Vector Calculus

3 Integral Vector Calculus

4 First Order ODEs

5 Second Order ODEs

6 High Order ODEs

7 First Order ODE Systems


Linear Algebra

Summary

We have discussed:

A + B, cA, set of m × n matrices form a Vector Space.

Linear Independence/Dependence, Basis, Dimension.

C = AB, non-communtative.

A = A>, A = −A>, A−1 = A>.

A~x = ~b has a solution iff A is non-singular iff A is invertible iffrank(A) = n iff det(A) 6= 0.

A~x = λ~x , det(A− λI ) = 0 and (A− λI )~x = ~0.

det(A) =∏n

j=1 λj .

If A has an eigen-basis, then A is diagonalizable.

If A has n different eigenvalues, then A has an eigenbasis.


Linear Algebra

Example

Example

Transform, 9x21 − 6x1x2 + 17x2

2 = 26, to the canonical form (to principal

axes). Express

[x1

x2

]in terms of the new variables

[y1

y2

].

A =

[9 −3−3 17

].

det(A− λI ) = λ2 − 26λ+ 144 = (x − 8)(x − 18).

λ = 8, A− 8I =

[1 −3−3 9

], ~x1 =

[3√101√10

].

λ = 18, A− 18I =

[−9 −3−3 −1

], ~x1 =

[1√10−3√

10

].


Linear Algebra

Example, Cont.

Example

Originally, we have B = ~x>A~x − 36 = 0

P =

[3√10

1√10

1√10

−3√10

], P−1 = P>.

A = PDP−1 = PDP>, so ~x>PDP>~x = (P>~x)>D(P>~x).

~y = P>~x , so ~x = P~y .

B = λ1y21 + λ2y

22 − 36 = 8y2

1 + 18y22 − 36 = 0.


Differential Vector Calculus

Table of Contents

1 Linear Algebra



4 First Order ODEs

5 Second Order ODEs

6 High Order ODEs




Summary

We have discussed:

~a · ~b =∣∣∣∣~a∣∣∣∣ · ∣∣∣∣~b∣∣∣∣ cos(γ),

∣∣∣∣~a× ~b∣∣∣∣ =

∣∣∣∣~a∣∣∣∣ · ∣∣∣∣~b∣∣∣∣ sin(γ).

~a · (~b × ~c) = (~a× ~b) · ~c , and∣∣~a · (~b × ~c)

∣∣.Given f (x , y , z), grad(f ) = ∇f =

∂f∂x∂f∂y∂f∂z

.

D~uf = ~u · grad(f ), only when ~u is a unit vector.

div(~f ) = ∇ · ~f = ∂f1∂x + ∂f2

∂y + ∂f3∂z .

curl(~f ) = ∇× ~f .

curl(grad(f )) = ~0 and div(curl(~f )) = 0.



Eample

Show that div(curl(~f )) = 0.

div(curl(~f )) = ∇ · curl(~f ) = ∇ · (∇× ~f ) = (∇×∇) · ~f = 0.


Integral Vector Calculus

Table of Contents

1 Linear Algebra



4 First Order ODEs

5 Second Order ODEs

6 High Order ODEs




Summary

We have discussed:∫C~f (~r) · d~r =

∫ ba~f (~r(t)) · d~r

dt dt.

~f is path independence if ~f = grad(f ).

Green’s theorem:∫∫

R(∂f2∂x −∂f1∂y ) dA =

∮C~f (~r) · d~r , R is a closed and

simply connected Region, and C is its boundary.∫∫S~f (~r)· d ~S =

∫∫R~f (~r(u, v)) · (~ru × ~rv )dA (check orientation).

Divergence theorem of Gauss:∫∫∫

V div(~f ) dV =∫∫

S~f (~r) · d ~S (the

surface normal ~n is pointing out and V is an enclosed region, S is itsboundary surface).

Stokes’s theorem:∫∫

S curl(~f (~r)) · d ~S =∮C~f (~r) · d~r (S is simply

connected).



Example

Example

Evaluate the integral,∫∫

S~f (~r) · d ~S , directly or if possible by the divergence

theorem. Show details:

~f =

111

, S : x2 + y2 + 4z2 = 4, z ≥ 0.

S does not enclose any area, so no divergence.

S : x2 + y2 + 4z2 = 4⇒ f (x , y) =√

1− x2

4 −y2

4 , Domain:

R : x2 + y2 ≤ 4.

~r(x , y) =

xy√

1− x2

4 −y2

4

.



Example, Cont.

Example

~rx =

10−x4√

1− x2

4− y2

4

and ~ry =

01−y

4√1− x2

4− y2

4

.

~rx × ~ry =x4√

1− x2

4− y2

4

~i +y4√

1− x2

4− y2

4

~j + ~k = −fx~i +−fy~j + ~k.

~f · (~rx × ~ry ) =x+y

4√1− x2

4− y2

4

+ 1.

Use Polar: x = r cos(θ) and y = r sin(θ), 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π.

Any other way?


First Order ODEs

Table of Contents

1 Linear Algebra



4 First Order ODEs

5 Second Order ODEs

6 High Order ODEs



First Order ODEs

Summary

We have discussed:

F (x , y , y ′) = 0 (implicit form) and y ′ = f (x , y) (explicit form).

Separable ODEs: g(y)dy = h(x)dx , integrate, G (y) = H(x) + C .

y ′ = f (x , y) = f ( yx ), let u = yx .

Exact ODEs: M(x , y)dx + N(x , y)dy = 0 when ∂M∂y = ∂N

∂x .

Integrate M w.r.t. x , obtain u(x , y), then differentiate u w.r.t. y .

P(x , y)dx + Q(x , y)dy is not exact, then use integrating factor.

Linear ODEs: y ′+ p(x)y = r(x), integrating factor: F (x) = e∫p(x) dx .

Bernoulli ODEs: y ′ + p(x)y = g(x)yα, when α 6= 0 and αneq1, thenu = y1−α.


First Order ODEs

Example

Example

Solve the IVP. Indicate the method used. Show the details of your work,

3 sec(y)dx +1

3sec(x)dy = 0, y(0) = 0.

It is separable: cos(y)dy = −9 cos(x)dx .

sin(y) = −9 sin(x) + C , y(0) = 0, 0 = −9 · 0 + C , C = 0.

sin(y) = −9 sin(x).


Second Order ODEs

Table of Contents

1 Linear Algebra



4 First Order ODEs

5 Second Order ODEs

6 High Order ODEs



Second Order ODEs

Summary

We have discussed:

y ′′ = f (x , y , y ′) (explicit form).Linear Second Order ODEs: y ′′ + p(x)y ′ + q(x)y = r(x).Linear Second Order Homogeneous ODEs: y ′′ + p(x)y ′ + q(x)y = 0.superposition principle: yh = c1y1(x) + c2y2(x); and reduction of order: giveny1, y2 = uy1; Wronskian: W (y1, y2) = y ′1y2 − y2y

′1 6= 0.

Linear Second Order Homogeneous ODEs with Constant Coefficients:y ′′ + ay ′ + by = 0. Assume that y = eλx , obtain the characteristic equation:λ2 + aλ+ b = 0. Three cases: a2 − 4b > 0, = 0, or < 0.Linear Second Order non-Homogeneous ODEs with Constant Coefficients:y ′′ + ay ′ + by = r(x). Method of Undetermined Coefficients (elementaryfunctions), and y = yh + yp.Linear Second Order non-Homogeneous ODEs: y ′′ + p(x)y ′ + q(x)y = r(x);Method of Variation of Parameters: y = yh + yp, andyp = −y1

∫ y2rW dx + y2

∫ y1rW dx .

Euler-Cauchy ODEs: x2y ′′ + xay ′ + by = 0. Assume y = xm, we havem(m − 1) + am + b = 0. Three cases.


Second Order ODEs

Example

Example

Find a general solution. Show the details of your calculation.

y ′′ + 16y = 17ex , y(0) = 6, y ′(0) = −2.

y ′′ + 16y = 10, λ2 + 16λ = 0, λ1 = 0, and λ2 = −16.

yh = c1 + c2e−16x .

yp = Kex , y ′p = Kex , y ′′ = Kex , Kex + 16Kex = 16ex , K = 1.

y = c1 + c2e−16x + ex , y(0) = c1 + c2 + 1 = 6,

y ′(0) = −16c2 + 1 = −2.

c2 = 316 and c1 = 5− 3

16 .


High Order ODEs

Table of Contents

1 Linear Algebra



4 First Order ODEs

5 Second Order ODEs

6 High Order ODEs



High Order ODEs

Summary

We have discussed:

y (n) = f (x , y , y ′, · · · , y (n−1)) (Explicit form).Linear nth Order ODEs:y (n) + pn−1(x)y (n−1) + · · ·+ p1(x)y ′ + p0(x)y = r(x).Linear nth Order Homogeneous ODEs:y (n) + pn−1(x)y (n−1) + · · ·+ p1(x)y ′ + p0(x)y = 0. Superpositionprinciple, Wronskian: W (y1, y2, · · · , yn). Reduction of Order.Linear nth Order Homogeneous ODEs with constant coefficients:y (n) + pn−1y

(n−1) + · · ·+ p1y′ + p0y = 0. Assume that y = eλx ,

characteristic equation: λn + pn−1λn−1 + · · ·+ p1λ+ p0 = 0, it has n

roots. Simple (real or complex) roots, multiple (real or complex) roots.Linear nth Order non-Homogeneous ODEs with constant coefficients:y (n) + pn−1y

(n−1) + · · ·+ p1y′ + p0y = r(x). Method of undetermined

coefficients if r(x) is some elementary functions for yp, Theny = yh + yp.


High Order ODEs

Example

Linear nth Order non-Homogeneous ODEs:y (n) + pn−1(x)y (n−1) + · · ·+ p1(x)y ′ + p0(x)y = r(x). Method of

variation of parameters. yp =∑n

j=1 yj(x)∫ Wj (x)

W (x) r(x) dx . Wj is obtained

by replacing the j th column of W by[0, · · · , 0, 1

]>.

Example

Solve the given ODE. Show the details of your work,

y ′′′ + 4y ′′ + 13y ′ = 0.

λ3 + 4λ2 + 13λ = 0, λ(λ2 + 4λ+ 13) = 0.λ1 = 0, ∆ = 16− 52 = −36 = (6i)2 < 0.λ2,3 = −4±6i

2 = −2± 3i .yh = c1 + e−2x(c2 cos(3x) + c3 sin(3x)).


First Order ODE Systems

Table of Contents

1 Linear Algebra



4 First Order ODEs

5 Second Order ODEs

6 High Order ODEs




Summary

We have discussed:

~y ′ = ~f (t, ~y) (Explicit form). Caution: t or x , it does not really matter.

Linear Systems of ODEs: ~y ′ = A(t)~y + ~g(t), where A is a matrix.

Linear Homogeneous Systems of ODEs: ~y ′ = A(t)~y . Superpositionprinciple. Fundamental matrix Y =

[~y1, · · · , ~yn

], Wronskian,

W (~y1, · · · , ~yn) = det([~y1, · · · , ~yn

]) = det(Y ), ~yj ’s are the basis

solutions.

Linear Homogeneous Systems of ODEs with constant coefficients:~y ′ = A~y . A has an eigen-basis, ~xj ’s are n linearly independenteigen-vectors of A, then ~yj = ~xje

λjx .

Categorizing the critical point yc =

[00

]. We consider λ1 + λ2, λ1λ2

and (λ1 − λ2)2.



Example

Find a general solution. Determine the kind and stability of the criticalpoint.

y ′1 = 3y1 + 4y2

y ′2 = 3y1 + 2y2

A =

[3 43 2

], det(A− λI ) = λ2 + 5λ− 6 = 0.

λ1 = −6 and λ2 = 1.

~x1 =

[43

]and ~x2 =

[1−1

].

~yh = c1~x1e−6t + c2~x2e

t .

p = λ1 + λ2 = −5 < 0, q = λ1λ2 = −6 < 0,∆ = (λ1 − λ2)2 = 49 > 0, real, opposite signs, Saddle point.


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