boundary value problems (sect. 10.1)....boundary value problems (sect. 10.1). i two-point bvp. i...

Boundary Value Problems (Sect. 10.1).

I Two-point BVP.

I Example from physics.

I Comparison: IVP vs BVP.

I Existence, uniqueness of solutions to BVP.

I Particular case of BVP: Eigenvalue-eigenfunction problem.

Two-point Boundary Value Problem.

DefinitionA two-point BVP is the following: Given functions p, q, g , andconstants x1 < x2, y1, y2, b1, b2, b̃1, b̃2,

find a function y solution of the differential equation

y ′′ + p(x) y ′ + q(x) y = g(x),

together with the extra, boundary conditions,

b1 y(x1) + b2 y ′(x1) = y1,

b̃1 y(x2) + b̃2 y ′(x2) = y2.

Remarks:I Both y and y ′ might appear in the boundary condition,

evaluated at the same point.

I In this notes we only study the case of constant coefficients,

y ′′ + a1 y ′ + a0 y = g(x).


Example

Examples of BVP.

Assume x1 6= x2.

(1) Find y solution of

y ′′ + a1 y ′ + a0 y = g(x), y(x1) = y1, y(x2) = y2.


y ′′ + a1 y ′ + a0 y = g(x), y ′(x1) = y1, y ′(x2) = y2.


y ′′ + a1 y ′ + a0 y = g(x), y(x1) = y1, y ′(x2) = y2.


Example

Examples of BVP. Assume x1 6= x2.


y ′′ + a1 y ′ + a0 y = g(x), y(x1) = y1, y(x2) = y2.


y ′′ + a1 y ′ + a0 y = g(x), y ′(x1) = y1, y ′(x2) = y2.


y ′′ + a1 y ′ + a0 y = g(x), y(x1) = y1, y ′(x2) = y2.


I Two-point BVP.





Example from physics.

Problem: The equilibrium (time independent) temperature of abar of length L with insulated horizontal sides and the bar verticalextremes kept at fixed temperatures T0, TL is the solution of theBVP:

T ′′(x) = 0, x ∈ (0, L), T (0) = T0, T (L) = TL,

y

x0

z insulation

insulation

T 0

T L

L x


I Two-point BVP.





Comparison: IVP vs BVP.

Review: IVP:Find the function values y(t) solutions of the differential equation

y ′′ + a1 y ′ + a0 y = g(t),

together with the initial conditions

y(t0) = y1, y ′(t0) = y2.

Remark: In physics:

I y(t): Position at time t.

I Initial conditions: Position and velocity at the initial time t0.

Comparison: IVP vs BVP.

Review: BVP:Find the function values y(x) solutions of the differential equation

y ′′ + a1 y ′ + a0 y = g(x),

together with the initial conditions

y(x1) = y1, y(x2) = y2.

Remark: In physics:

I y(x): A physical quantity (temperature) at a position x .

I Boundary conditions: Conditions at the boundary of theobject under study, where x1 6= x2.


I Two-point BVP.





Existence, uniqueness of solutions to BVP.

Review: The initial value problem.

Theorem (IVP)

Consider the homogeneous initial value problem:

y ′′ + a1 y ′ + a0 y = 0, y(t0) = y0, y ′(t0) = y1,

and let r± be the roots of the characteristic polynomial

p(r) = r2 + a1 r + a0.

If r+ 6= r−, real or complex, then for every choice of y0, y1, thereexists a unique solution y to the initial value problem above.

Summary: The IVP above always has a unique solution, no matterwhat y0 and y1 we choose.


Theorem (BVP)

Consider the homogeneous boundary value problem:

y ′′ + a1 y ′ + a0 y = 0, y(0) = y0, y(L) = y1,

and let r± be the roots of the characteristic polynomial

p(r) = r2 + a1 r + a0.

(A) If r+ 6= r−, real, then for every choice of L 6= 0 and y0, y1,there exists a unique solution y to the BVP above.

(B) If r± = α± iβ, with β 6= 0, and α, β ∈ R, then the solutionsto the BVP above belong to one of these possibilities:

(1) There exists a unique solution.(2) There exists no solution.(3) There exist infinitely many solutions.


Proof of IVP: We study the case r+ 6= r-.

The general solution is

y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.

The initial conditions determine c1 and c2 as follows:

y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0

Using matrix notation,[er- t0 er+ t0

r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].

The linear system above has a unique solution c1 and c2 for everyconstants y0 and y1 iff the det(Z ) 6= 0, where

Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].


Proof of IVP: We study the case r+ 6= r-. The general solution is

y(t) = c1 er- t + c2 er+ t ,

c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.

The initial conditions determine c1 and c2

as follows:

y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0)

= c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0)

= c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].

The linear system above has a unique solution c1 and c2 for everyconstants y0 and y1 iff

the det(Z ) 6= 0, where

Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].

The linear system above has a unique solution c1 and c2 for everyconstants y0 and y1 iff the det(Z ) 6= 0,

where

Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]

⇒ Z

[c1

c2

]=

[y0

y1

].



y(t) = c1 er- t + c2 er+ t , c1, c2 ∈ R.


y0 = y(t0) = c1 er- t0 + c2 er+ t0

y1 = y ′(t0) = c1r- er- t0 + c2r+ er+ t0


r- er-t0 r+ er+ t0

] [c1

c2

]=

[y0

y1

].


Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].


Proof of IVP:

Recall: Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].

A simple calculation shows

det(Z ) =(r+ − r-

)e(r++r-) t0 6= 0 ⇔ r+ 6= r-.

Since r+ 6= r-, the matrix Z is invertible and so[c1

c2

]= Z−1

[y0

y1

].

We conclude that for every choice of y0 and y1, there exist a uniquevalue of c1 and c2, so the IVP above has a unique solution.


Proof of IVP:

Recall: Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].


det(Z ) =(r+ − r-

)e(r++r-) t0

6= 0 ⇔ r+ 6= r-.


c2

]= Z−1

[y0

y1

].



Proof of IVP:

Recall: Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].


det(Z ) =(r+ − r-

)e(r++r-) t0 6= 0 ⇔ r+ 6= r-.


c2

]= Z−1

[y0

y1

].



Proof of IVP:

Recall: Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].


det(Z ) =(r+ − r-

)e(r++r-) t0 6= 0 ⇔ r+ 6= r-.

Since r+ 6= r-, the matrix Z is invertible

and so[c1

c2

]= Z−1

[y0

y1

].



Proof of IVP:

Recall: Z =

[er- t0 er+ t0

r- er- t0 r+ er+ t0

]⇒ Z

[c1

c2

]=

[y0

y1

].


det(Z ) =(r+ − r-

)e(r++r-) t0 6= 0 ⇔ r+ 6= r-.


c2

]= Z−1

[y0

y1

].



Proof of BVP: The general solution is

y(x) = c1 er- x + c2 er+ x ,

c1, c2 ∈ R.

The boundary conditions determine c1 and c2 as follows:

y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L

Using matrix notation,[1 1

er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.

The boundary conditions determine c1 and c2

as follows:

y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0)

= c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0) = c1 + c2.

y1 = y(L)

= c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].

The linear system above has a unique solution c1 and c2 for everyconstants y0 and y1 iff

the det(Z ) 6= 0, where

Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].

The linear system above has a unique solution c1 and c2 for everyconstants y0 and y1 iff the det(Z ) 6= 0,

where

Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]

⇒ Z

[c1

c2

]=

[y0

y1

].



y(x) = c1 er- x + c2 er+ x , c1, c2 ∈ R.


y0 = y(0) = c1 + c2.

y1 = y(L) = c1 er- L + c2 er+ L


er- L er+ L

] [c1

c2

]=

[y0

y1

].


Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].


Proof of IVP: Recall: Z =

[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].


det(Z ) = er+ L − er- L 6= 0 ⇔ er+ L 6= er- L.

(A) If r+ 6= r- and real-valued, then det(Z ) 6= 0.

We conclude: For every choice of y0 and y1, there exist aunique value of c1 and c2, so the BVP in (A) above has aunique solution.

(B) If r± = α± iβ, with α, β ∈ R and β 6= 0, then

det(Z ) = eαL(e iβL − e−iβL

)⇒ det(Z ) = 2i eαL sin(βL).

Since det(Z ) = 0 iff βL = nπ, with n integer,

(1) If βL 6= nπ, then BVP has a unique solution.(2) If βL = nπ then BVP either has no solutions or it has infinitely

many solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].


det(Z ) = er+ L − er- L

6= 0 ⇔ er+ L 6= er- L.








many solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].










many solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].



(A) If r+ 6= r- and real-valued,

then det(Z ) 6= 0.







many solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].










many solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].





(B) If r± = α± iβ, with α, β ∈ R and β 6= 0,

then





many solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].







)

⇒ det(Z ) = 2i eαL sin(βL).



many solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].










many solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].









(1) If βL 6= nπ, then BVP has a unique solution.

(2) If βL = nπ then BVP either has no solutions or it has infinitelymany solutions.



[1 1

er- L er+ L

]⇒ Z

[c1

c2

]=

[y0

y1

].










many solutions.


Example

Find y solution of the BVP

y ′′ + y = 0, y(0) = 1, y(π) = −1.

Solution: The characteristic polynomial is

p(r) = r2 + 1 ⇒ r± = ±i .


y(x) = c1 cos(x) + c2 sin(x).

The boundary conditions are

1 = y(0) = c1, − 1 = y(π) = −c1 ⇒ c1 = 1, c2 free.

We conclude: y(x) = cos(x) + c2 sin(x), with c2 ∈ R.

The BVP has infinitely many solutions. C


Example


y ′′ + y = 0, y(0) = 1, y(π) = −1.


p(r) = r2 + 1

⇒ r± = ±i .




1 = y(0) = c1, − 1 = y(π) = −c1 ⇒ c1 = 1, c2 free.




Example


y ′′ + y = 0, y(0) = 1, y(π) = −1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, − 1 = y(π) = −c1 ⇒ c1 = 1, c2 free.




Example


y ′′ + y = 0, y(0) = 1, y(π) = −1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1,

− 1 = y(π) = −c1 ⇒ c1 = 1, c2 free.




Example


y ′′ + y = 0, y(0) = 1, y(π) = −1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, − 1 = y(π) = −c1

⇒ c1 = 1, c2 free.




Example


y ′′ + y = 0, y(0) = 1, y(π) = −1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, − 1 = y(π) = −c1 ⇒ c1 = 1, c2 free.




Example


y ′′ + y = 0, y(0) = 1, y(π) = 0.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, 0 = y(π) = −c1

The BVP has no solution. C


Example


y ′′ + y = 0, y(0) = 1, y(π) = 0.


p(r) = r2 + 1

⇒ r± = ±i .




1 = y(0) = c1, 0 = y(π) = −c1



Example


y ′′ + y = 0, y(0) = 1, y(π) = 0.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, 0 = y(π) = −c1



Example


y ′′ + y = 0, y(0) = 1, y(π) = 0.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1,

0 = y(π) = −c1



Example


y ′′ + y = 0, y(0) = 1, y(π) = 0.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, 0 = y(π) = −c1



Example


y ′′ + y = 0, y(0) = 1, y(π/2) = 1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, 1 = y(π/2) = c2 ⇒ c1 = c2 = 1.

We conclude: y(x) = cos(x) + sin(x).

The BVP has a unique solution. C


Example


y ′′ + y = 0, y(0) = 1, y(π/2) = 1.


p(r) = r2 + 1

⇒ r± = ±i .




1 = y(0) = c1, 1 = y(π/2) = c2 ⇒ c1 = c2 = 1.




Example


y ′′ + y = 0, y(0) = 1, y(π/2) = 1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, 1 = y(π/2) = c2 ⇒ c1 = c2 = 1.




Example


y ′′ + y = 0, y(0) = 1, y(π/2) = 1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1,

1 = y(π/2) = c2 ⇒ c1 = c2 = 1.




Example


y ′′ + y = 0, y(0) = 1, y(π/2) = 1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, 1 = y(π/2) = c2

⇒ c1 = c2 = 1.




Example


y ′′ + y = 0, y(0) = 1, y(π/2) = 1.


p(r) = r2 + 1 ⇒ r± = ±i .




1 = y(0) = c1, 1 = y(π/2) = c2 ⇒ c1 = c2 = 1.




I Two-point BVP.





Particular case of BVP: Eigenvalue-eigenfunction problem.

Problem:Find a number λ and a non-zero function y solutions to theboundary value problem

y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Remark: This problem is similar to the eigenvalue-eigenvectorproblem in Linear Algebra: Given an n × n matrix A, find λ and anon-zero n-vector v solutions of

Av− λ v = 0.

Differences:

I A −→

{computing a second derivative and

applying the boundary conditions.

}I v −→ {a function y}.



y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Remark: This problem is similar to the eigenvalue-eigenvectorproblem in Linear Algebra:

Given an n × n matrix A, find λ and anon-zero n-vector v solutions of

Av− λ v = 0.

Differences:

I A −→






y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


Av− λ v = 0.

Differences:

I A −→






y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


Av− λ v = 0.

Differences:

I A −→



}

I v −→ {a function y}.



y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


Av− λ v = 0.

Differences:

I A −→





Example

Find every λ ∈ R and non-zero functions y solutions of the BVP

y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Remarks: We will show that:

(1) If λ 6 0, then the BVP has no solution.

(2) If λ > 0, then there exist infinitely many eigenvalues λn andeigenfunctions yn, with n any positive integer, given by

λn =(nπ

L

)2, yn(x) = sin

(nπx

L

),

(3) Analogous results can be proven for the same equation butwith different types of boundary conditions. For example, fory(0) = 0, y ′(L) = 0; or for y ′(0) = 0, y ′(L) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.



(2) If λ > 0, then there exist infinitely many eigenvalues λn andeigenfunctions yn, with n any positive integer,

given by

λn =(nπ

L

)2, yn(x) = sin

(nπx

L

),



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.




λn =(nπ

L

)2, yn(x) = sin

(nπx

L

),



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.




λn =(nπ

L

)2, yn(x) = sin

(nπx

L

),

(3) Analogous results can be proven for the same equation butwith different types of boundary conditions.

For example, fory(0) = 0, y ′(L) = 0; or for y ′(0) = 0, y ′(L) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.




λn =(nπ

L

)2, yn(x) = sin

(nπx

L

),

(3) Analogous results can be proven for the same equation butwith different types of boundary conditions. For example, fory(0) = 0, y ′(L) = 0;

or for y ′(0) = 0, y ′(L) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.




λn =(nπ

L

)2, yn(x) = sin

(nπx

L

),



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Case λ = 0. The equation is

y ′′ = 0 ⇒ y(x) = c1 + c2x .

The boundary conditions imply

0 = y(0) = c1, 0 = c1 + c2L ⇒ c1 = c2 = 0.

Since y = 0, there are NO non-zero solutions for λ = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Case λ = 0.

The equation is

y ′′ = 0 ⇒ y(x) = c1 + c2x .


0 = y(0) = c1, 0 = c1 + c2L ⇒ c1 = c2 = 0.



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


y ′′ = 0

⇒ y(x) = c1 + c2x .


0 = y(0) = c1, 0 = c1 + c2L ⇒ c1 = c2 = 0.



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


y ′′ = 0 ⇒ y(x) = c1 + c2x .


0 = y(0) = c1, 0 = c1 + c2L ⇒ c1 = c2 = 0.



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


y ′′ = 0 ⇒ y(x) = c1 + c2x .


0 = y(0)

= c1, 0 = c1 + c2L ⇒ c1 = c2 = 0.



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


y ′′ = 0 ⇒ y(x) = c1 + c2x .


0 = y(0) = c1,

0 = c1 + c2L ⇒ c1 = c2 = 0.



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


y ′′ = 0 ⇒ y(x) = c1 + c2x .


0 = y(0) = c1, 0 = c1 + c2L

⇒ c1 = c2 = 0.



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


y ′′ = 0 ⇒ y(x) = c1 + c2x .


0 = y(0) = c1, 0 = c1 + c2L ⇒ c1 = c2 = 0.



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Case λ < 0.

Introduce the notation λ = −µ2. Thecharacteristic equation is

p(r) = r2 − µ2 = 0 ⇒ r± = ±µ.


y(x) = c1 eµx + c2 e−µx .

The boundary condition are

0 = y(0) = c1 + c2,

0 = y(L) = c1 eµL + c2 e−µL.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Case λ < 0. Introduce the notation λ = −µ2.

Thecharacteristic equation is

p(r) = r2 − µ2 = 0 ⇒ r± = ±µ.




0 = y(0) = c1 + c2,

0 = y(L) = c1 eµL + c2 e−µL.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Case λ < 0. Introduce the notation λ = −µ2. Thecharacteristic equation is

p(r) = r2 − µ2 = 0

⇒ r± = ±µ.




0 = y(0) = c1 + c2,

0 = y(L) = c1 eµL + c2 e−µL.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 − µ2 = 0 ⇒ r± = ±µ.




0 = y(0) = c1 + c2,

0 = y(L) = c1 eµL + c2 e−µL.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 − µ2 = 0 ⇒ r± = ±µ.




0 = y(0)

= c1 + c2,

0 = y(L) = c1 eµL + c2 e−µL.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 − µ2 = 0 ⇒ r± = ±µ.




0 = y(0) = c1 + c2,

0 = y(L) = c1 eµL + c2 e−µL.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Recall: y(x) = c1 eµx + c2 eµx and

c1 + c2 = 0, c1 eµL + c2 e−µL = 0.

We need to solve the linear system[1 1

eµL e−µL

] [c1

c2

]=

[00

]⇔ Z

[c1

c2

]=

[00

], Z =

[1 1

eµL e−µL

]Since det(Z ) = e−µL− eµL 6= 0 for L 6= 0, matrix Z is invertible, sothe linear system above has a unique solution c1 = 0 and c2 = 0.

Since y = 0, there are NO non-zero solutions for λ < 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


c1 + c2 = 0, c1 eµL + c2 e−µL = 0.


eµL e−µL

] [c1

c2

]=

[00

]

⇔ Z

[c1

c2

]=

[00

], Z =

[1 1

eµL e−µL




Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


c1 + c2 = 0, c1 eµL + c2 e−µL = 0.


eµL e−µL

] [c1

c2

]=

[00

]⇔ Z

[c1

c2

]=

[00

],

Z =

[1 1

eµL e−µL




Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


c1 + c2 = 0, c1 eµL + c2 e−µL = 0.


eµL e−µL

] [c1

c2

]=

[00

]⇔ Z

[c1

c2

]=

[00

], Z =

[1 1

eµL e−µL

]

Since det(Z ) = e−µL− eµL 6= 0 for L 6= 0, matrix Z is invertible, sothe linear system above has a unique solution c1 = 0 and c2 = 0.



Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


c1 + c2 = 0, c1 eµL + c2 e−µL = 0.


eµL e−µL

] [c1

c2

]=

[00

]⇔ Z

[c1

c2

]=

[00

], Z =

[1 1

eµL e−µL




Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Case λ > 0.

Introduce the notation λ = µ2. Thecharacteristic equation is

p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .


y(x) = c1 cos(µx) + c2 sin(µx).


0 = y(0) = c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Case λ > 0. Introduce the notation λ = µ2.

Thecharacteristic equation is

p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .




0 = y(0) = c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Case λ > 0. Introduce the notation λ = µ2. Thecharacteristic equation is

p(r) = r2 + µ2 = 0

⇒ r± = ±µi .




0 = y(0) = c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .




0 = y(0) = c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .




0 = y(0)

= c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .




0 = y(0) = c1,

⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .




0 = y(0) = c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .




0 = y(0) = c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL),

c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .




0 = y(0) = c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0

⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


p(r) = r2 + µ2 = 0 ⇒ r± = ±µi .




0 = y(0) = c1, ⇒ y(x) = c2 sin(µx).

0 = y(L) = c2 sin(µL), c2 6= 0 ⇒ sin(µL) = 0.


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.

Solution: Recall: c1 = 0, c2 6= 0, and sin(µL) = 0.

The non-zero solution condition is the reason for c2 6= 0. Hence

sin(µL) = 0 ⇒ µnL = nπ ⇒ µn =nπ

L.

Recalling that λn = µ2n, and choosing c2 = 1, we conclude

λn =(nπ

L

)2, yn(x) = sin

(nπx

L

). C


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.


The non-zero solution condition is the reason for c2 6= 0.

Hence


L.


λn =(nπ

L

)2, yn(x) = sin

(nπx

L

). C


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.



sin(µL) = 0

⇒ µnL = nπ ⇒ µn =nπ

L.


λn =(nπ

L

)2, yn(x) = sin

(nπx

L

). C


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.



sin(µL) = 0 ⇒ µnL = nπ

⇒ µn =nπ

L.


λn =(nπ

L

)2, yn(x) = sin

(nπx

L

). C


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.




L.


λn =(nπ

L

)2, yn(x) = sin

(nπx

L

). C


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.




L.

Recalling that λn = µ2n, and choosing c2 = 1,

we conclude

λn =(nπ

L

)2, yn(x) = sin

(nπx

L

). C


Example


y ′′(x) + λ y(x) = 0, y(0) = 0, y(L) = 0, L > 0.




L.


λn =(nπ

L

)2, yn(x) = sin

(nπx

L

). C

Overview of Fourier Series (Sect. 10.2).

I Origins of the Fourier Series.

I Periodic functions.

I Orthogonality of Sines and Cosines.

I Main result on Fourier Series.

Origins of the Fourier Series.

Summary:Daniel Bernoulli (∼ 1750) foundsolutions to the equation thatdescribes waves propagating on avibrating string.

x

u(t,x)

x

y

0

The function u, measuring the vertical displacement of the string,is the solution to the wave equation,

∂2t u(t, x) = v2 ∂2

xu(t, x), v ∈ R, x ∈ [0, L], t ∈ [0,∞),

with initial conditions,

u(0, x) = f (x), ∂tu(0, x) = 0,

and boundary conditions,

u(t, 0) = 0, u(t, L) = 0.



x

u(t,x)

x

y

0

The function u, measuring the vertical displacement of the string,

is the solution to the wave equation,

∂2t u(t, x) = v2 ∂2

xu(t, x), v ∈ R, x ∈ [0, L], t ∈ [0,∞),


u(0, x) = f (x), ∂tu(0, x) = 0,


u(t, 0) = 0, u(t, L) = 0.



x

u(t,x)

x

y

0

The function u, measuring the vertical displacement of the string,is the solution to the wave equation,

∂2t u(t, x) = v2 ∂2

xu(t, x), v ∈ R, x ∈ [0, L], t ∈ [0,∞),


u(0, x) = f (x), ∂tu(0, x) = 0,


u(t, 0) = 0, u(t, L) = 0.


Summary:Bernoulli found particular solutions to the wave equation.

If the initial condition is fn(x) = sin(nπx

L

),

then the solution is un(t, x) = sin(nπx

L

)cos

(vnπt

L

).

Bernoulli also realized that

UN(t, x) =N∑

n=1

an sin(nπx

L

)cos

(vnπt

L

), an ∈ R

is also solution of the wave equation with initial condition

FN(x) =N∑

n=1

an sin(nπx

L

).

Remark: The wave equation and its solutions provide amathematical description of music.


Remarks:I Bernoulli claimed he had obtained all solutions to the problem

above for the wave equation.

I However, he did not prove that claim.

I A proof is: Given a function F with F (0) = F (L) = 0, butotherwise arbitrary, find N and the coefficients an such that Fis approximated by an expansion FN given in the previous slide.

I Joseph Fourier (∼ 1800) provided such formula for thecoefficients an, while studying a different problem:The heat transport in a solid material.

I Find the temperature function u solution of the heat equation

∂tu(t, x) = k ∂2xu(t, x), k > 0, x ∈ [0, L], t ∈ [0,∞),

I.C. u(0, x) = f (x),

B.C. u(t, 0) = 0, u(t, L) = 0.






I Joseph Fourier (∼ 1800) provided such formula for thecoefficients an, while studying a different problem:

The heat transport in a solid material.


∂tu(t, x) = k ∂2xu(t, x), k > 0, x ∈ [0, L], t ∈ [0,∞),

I.C. u(0, x) = f (x),

B.C. u(t, 0) = 0, u(t, L) = 0.








∂tu(t, x) = k ∂2xu(t, x), k > 0, x ∈ [0, L], t ∈ [0,∞),

I.C. u(0, x) = f (x),

B.C. u(t, 0) = 0, u(t, L) = 0.








∂tu(t, x) = k ∂2xu(t, x),

k > 0, x ∈ [0, L], t ∈ [0,∞),

I.C. u(0, x) = f (x),

B.C. u(t, 0) = 0, u(t, L) = 0.








∂tu(t, x) = k ∂2xu(t, x), k > 0,

x ∈ [0, L], t ∈ [0,∞),

I.C. u(0, x) = f (x),

B.C. u(t, 0) = 0, u(t, L) = 0.








∂tu(t, x) = k ∂2xu(t, x), k > 0, x ∈ [0, L],

t ∈ [0,∞),

I.C. u(0, x) = f (x),

B.C. u(t, 0) = 0, u(t, L) = 0.








∂tu(t, x) = k ∂2xu(t, x), k > 0, x ∈ [0, L], t ∈ [0,∞),

I.C. u(0, x) = f (x),

B.C. u(t, 0) = 0, u(t, L) = 0.


Remarks:Fourier found particular solutions to the heat equation.

If the initial condition is fn(x) = sin(nπx

L

),

then the solution is un(t, x) = sin(nπx

L

)e−k( nπ

L)2t .

Fourier also realized that

UN(t, x) =N∑

n=1

an sin(nπx

L

)e−k( nπ

L)2t , an ∈ R

is also solution of the heat equation with initial condition

FN(x) =N∑

n=1

an sin(nπx

L

).

Remark: The heat equation and its solutions provide amathematical description of heat transport in a solid material.


Remarks:

I However, Fourier went farther than Bernoulli.

Fourier found aformula for the coefficients an in terms of the function F .

I Given an initial data function F , satisfying F (0) = F (L) = 0,but otherwise arbitrary, Fourier proved that one can constructan expansion FN as follows,

FN(x) =N∑

n=1

an sin(nπx

L

),

for N any positive integer, where the an are given by

an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .

I To find all solutions to the heat equation problem above onemust prove one more thing: That FN approximates F for largeenough N. That is, limN→∞ FN = F . Fourier didn’t show this.


Remarks:

I However, Fourier went farther than Bernoulli. Fourier found aformula for the coefficients an in terms of the function F .


FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .



Remarks:


I Given an initial data function F , satisfying F (0) = F (L) = 0,but otherwise arbitrary,

Fourier proved that one can constructan expansion FN as follows,

FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .



Remarks:


I Given an initial data function F , satisfying F (0) = F (L) = 0,but otherwise arbitrary, Fourier proved that one can constructan expansion FN

as follows,

FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .



Remarks:



FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .



Remarks:



FN(x) =N∑

n=1

an sin(nπx

L

),

for N any positive integer,

where the an are given by

an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .



Remarks:



FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .



Remarks:



FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .

I To find all solutions to the heat equation problem above onemust prove one more thing:

That FN approximates F for largeenough N. That is, limN→∞ FN = F . Fourier didn’t show this.


Remarks:



FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .

I To find all solutions to the heat equation problem above onemust prove one more thing: That FN approximates F for largeenough N.

That is, limN→∞ FN = F . Fourier didn’t show this.


Remarks:



FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .

I To find all solutions to the heat equation problem above onemust prove one more thing: That FN approximates F for largeenough N. That is, limN→∞ FN = F .

Fourier didn’t show this.


Remarks:



FN(x) =N∑

n=1

an sin(nπx

L

),


an =2

L

∫ L

0F (x) sin

(nπx

L

)dx .



Remarks:

I Based on Bernoulli and Fourier works, people have been ableto prove that.

Every continuous, τ -periodic function can beexpressed as an infinite linear combination of sine and cosinefunctions.

I More precisely: Every continuous, τ -periodic function F , thereexist constants a0, an, bn, for n = 1, 2, · · · such that

FN(x) =a0

2+

N∑n=1

[an cos

(2nπx

τ

)+ bn sin

(2nπx

τ

)],

satisfies limN→∞

FN(x) = F (x) for every x ∈ R.

Notation: F (x) =a0

2+

∞∑n=1

[an cos

(2nπx

τ

)+ bn sin

(2nπx

τ

)].


Remarks:

I Based on Bernoulli and Fourier works, people have been ableto prove that. Every continuous, τ -periodic function can beexpressed as an infinite linear combination of sine and cosinefunctions.

I More precisely: Every continuous, τ -periodic function F , thereexist constants a0, an, bn, for n = 1, 2, · · · such that

FN(x) =a0

2+

N∑n=1

[an cos

(2nπx

τ

)+ bn sin

(2nπx

τ

)],

satisfies limN→∞

FN(x) = F (x) for every x ∈ R.

Notation: F (x) =a0

2+

∞∑n=1

[an cos

(2nπx

τ

)+ bn sin

(2nπx

τ

)].


The main problem in our class:Given a continuous, τ -periodic function f , find the formulas for an

and bn such that

f (x) =a0

2+

∞∑n=1

[an cos

(2nπx

τ

)+ bn sin

(2nπx

τ

)].

Remarks: We need to review two main concepts:

I The notion of periodic functions.

I The notion of orthogonal functions, in particular theorthogonality of Sines and Cosines.

Fourier Series (Sect. 10.2).





Periodic functions.

DefinitionA function f : R → R is called periodic iff there exists τ > 0 suchthat for all x ∈ R holds

f (x + τ) = f (x).

Remark: f is invariant under translations by τ .

DefinitionA period T of a periodic function f is the smallest value of τ suchthat f (x + τ) = f (x) holds.

Notation:A periodic function with period T is also called T -periodic.

Periodic functions.

Example

The following functions are periodic, with period T ,

f (x) = sin(x), T = 2π.

f (x) = cos(x), T = 2π.

f (x) = tan(x), T = π.

f (x) = sin(ax), T =2π

a.

The proof of the latter statement is the following:

f(x +

2π

a

)= sin

(ax + a

2π

a

)= sin(ax + 2π) = sin(ax) = f (x).

C

Periodic functions.

Example


f (x) = sin(x), T = 2π.

f (x) = cos(x), T = 2π.

f (x) = tan(x), T = π.


a.


f(x +

2π

a

)

= sin(ax + a

2π

a

)= sin(ax + 2π) = sin(ax) = f (x).

C

Periodic functions.

Example


f (x) = sin(x), T = 2π.

f (x) = cos(x), T = 2π.

f (x) = tan(x), T = π.


a.


f(x +

2π

a

)= sin

(ax + a

2π

a

)

= sin(ax + 2π) = sin(ax) = f (x).

C

Periodic functions.

Example


f (x) = sin(x), T = 2π.

f (x) = cos(x), T = 2π.

f (x) = tan(x), T = π.


a.


f(x +

2π

a

)= sin

(ax + a

2π

a

)= sin(ax + 2π)

= sin(ax) = f (x).

C

Periodic functions.

Example


f (x) = sin(x), T = 2π.

f (x) = cos(x), T = 2π.

f (x) = tan(x), T = π.


a.


f(x +

2π

a

)= sin

(ax + a

2π

a

)= sin(ax + 2π) = sin(ax)

= f (x).

C

Periodic functions.

Example


f (x) = sin(x), T = 2π.

f (x) = cos(x), T = 2π.

f (x) = tan(x), T = π.


a.


f(x +

2π

a

)= sin

(ax + a

2π

a

)= sin(ax + 2π) = sin(ax) = f (x).

C

Periodic functions.

Example

Show that the function below is periodic, and find its period,

f (x) = ex , x ∈ [0, 2), f (x − 2) = f (x).

Solution: We just graph the function,

y = f(x)y

0 1 x

So the function is periodic with period T = 2. C

Periodic functions.

TheoremA linear combination of T -periodic functions is also T-periodic.

Proof: If f (x + T ) = f (x) and g(x + T ) = g(x), then

af (x + T ) + bg(x + T ) = af (x) + bg(x),

so (af + bg) is also T -periodic.

Example

f (x) = 2 sin(3x) + 7 cos(3x) is periodic with period T = 2π/3. C

Remark: The functions below are periodic with period T =τ

n,

f (x) = cos(2πnx

τ

), g(x) = sin

(2πnx

τ

),

Since f and g are invariant under translations by τ/n, they are alsoinvariant under translations by τ .

Periodic functions.

Corollary

Any function f given by

f (x) =a0

2+

∞∑n=1

[an cos

(2nπx

τ

)+ bn sin

(2nπx

τ

)]is periodic with period τ .

Remark: We will show that the converse statement is true.

TheoremA function f is τ -periodic iff holds

f (x) =a0

2+

∞∑n=1

[an cos

(2nπx

τ

)+ bn sin

(2nπx

τ

)].

Fourier Series (Sect. 10.2).





Orthogonality of Sines and Cosines.

Remark:From now on we work on the following domain: [−L, L].

L x

y

T = 2 L

cos ( pi x / L )

sin ( pi x / L )

−L


Theorem (Orthogonality)

The following relations hold for all n, m ∈ N,∫ L

−Lcos

(nπx

L

)cos

(mπx

L

)dx =

0 n 6= m,

L n = m 6= 0,

2L n = m = 0,∫ L

−Lsin

(nπx

L

)sin

(mπx

L

)dx =

{0 n 6= m,

L n = m,∫ L

−Lcos

(nπx

L

)sin

(mπx

L

)dx = 0.

Remark:

I The operation f · g =

∫ L

−Lf (x) g(x) dx is an inner product in

the vector space of functions. Like the dot product is in R2.

I Two functions f , g , are orthogonal iff f · g = 0.




−Lcos

(nπx

L

)cos

(mπx

L

)dx =

0 n 6= m,

L n = m 6= 0,

2L n = m = 0,∫ L

−Lsin

(nπx

L

)sin

(mπx

L

)dx =

{0 n 6= m,

L n = m,∫ L

−Lcos

(nπx

L

)sin

(mπx

L

)dx = 0.

Remark:


∫ L


the vector space of functions.

Like the dot product is in R2.





−Lcos

(nπx

L

)cos

(mπx

L

)dx =

0 n 6= m,

L n = m 6= 0,

2L n = m = 0,∫ L

−Lsin

(nπx

L

)sin

(mπx

L

)dx =

{0 n 6= m,

L n = m,∫ L

−Lcos

(nπx

L

)sin

(mπx

L

)dx = 0.

Remark:


∫ L


the vector space of functions. Like the dot product is in R2.



Recall: cos(θ) cos(φ) =1

2

[cos(θ + φ) + cos(θ − φ)

];

sin(θ) sin(φ) =1

2

[cos(θ − φ)− cos(θ + φ)

];

sin(θ) cos(φ) =1

2

[sin(θ + φ) + sin(θ − φ)

].

Proof: First formula: If n = m = 0, it is simple to see that∫ L

−Lcos

(nπx

L

)cos

(mπx

L

)dx =

∫ L

−Ldx = 2L.

In the case where one of n or m is non-zero, use the relation∫ L

−Lcos

(nπx

L

)cos

(mπx

L

)dx =

1

2

∫ L

−Lcos

[(n + m)πx

L

]dx

+1

2

∫ L

−Lcos

[(n −m)πx

L

]dx .



2


];

sin(θ) sin(φ) =1

2


];

sin(θ) cos(φ) =1

2


].

Proof: First formula:

If n = m = 0, it is simple to see that∫ L

−Lcos

(nπx

L

)cos

(mπx

L

)dx =

∫ L

−Ldx = 2L.


−Lcos

(nπx

L

)cos

(mπx

L

)dx =

1

2

∫ L

−Lcos

[(n + m)πx

L

]dx

+1

2

∫ L

−Lcos

[(n −m)πx

L

]dx .



2


];

sin(θ) sin(φ) =1

2


];

sin(θ) cos(φ) =1

2


].

Proof: First formula: If n = m = 0, it is simple to see that∫ L

−Lcos

(nπx

L

)cos

(mπx

L

)dx =

∫ L

−Ldx = 2L.


−Lcos

(nπx

L

)cos

(mπx

L

)dx =

1

2

∫ L

−Lcos

[(n + m)πx

L

]dx

+1

2

∫ L

−Lcos

[(n −m)πx

L

]dx .


Proof: Since one of n or m is non-zero,

holds

1

2

∫ L

−Lcos

[(n + m)πx

L

]dx =

L

2(n + m)πsin

[(n + m)πx

L

]∣∣∣L−L

= 0.

We obtain that∫ L

−Lcos

(nπx

L

)cos

(mπx

L

)dx =

1

2

∫ L

−Lcos

[(n −m)πx

L

]dx .

If we further restrict n 6= m, then

1

2

∫ L

−Lcos

[(n −m)πx

L

]dx =

L

2(n −m)πsin

[(n −m)πx

L

]∣∣∣L−L

= 0.

If n = m 6= 0, we have that

1

2

∫ L

−Lcos

[(n −m)πx

L

]dx =

1

2

∫ L

−Ldx = L.

This establishes the first equation in the Theorem. The remainingequations are proven in a similar way.


Proof: Since one of n or m is non-zero, holds

1

2

∫ L

−Lcos

[(n + m)πx

L

]dx =

L

2(n + m)πsin

[(n + m)πx

L

]∣∣∣L−L

= 0.

We obtain that∫ L

−Lcos

(nπx

L

)cos

(mπx

L

)dx =

1

2

∫ L

−Lcos

[(n −m)πx

L

]dx .

If we further restrict n 6= m, then

1

2

∫ L

−Lcos

[(n −m)πx

L

]dx =

L

2(n −m)πsin

[(n −m)πx

L

]∣∣∣L−L

= 0.

If n = m 6= 0, we have that

1

2

∫ L

−Lcos

[(n −m)πx

L

]dx =

1

2

∫ L

−Ldx = L.

This establishes the first equation in the Theorem. The remainingequations are proven in a similar way.

Overview of Fourier Series (Sect. 10.2).





Main result on Fourier Series.

Theorem (Fourier Series)

If the function f : [−L, L] ⊂ R → R is continuous, then f can beexpressed as an infinite series

f (x) =a0

2+

∞∑n=1

[an cos

(nπx

L

)+ bn sin

(nπx

L

)](1)

with the constants an and bn given by

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , n > 0,

bn =1

L

∫ L

−Lf (x) sin

(nπx

L

)dx , n > 1.

Furthermore, the Fourier series in Eq. (1) provides a 2L-periodicextension of f from the domain [−L, L] ⊂ R to R.

Examples of the Fourier Theorem (Sect. 10.3).

I The Fourier Theorem: Continuous case.

I Example: Using the Fourier Theorem.

I The Fourier Theorem: Piecewise continuous case.


The Fourier Theorem: Continuous case.


If the function f : [−L, L] ⊂ R → R is continuous, then f can beexpressed as an infinite series

f (x) =a0

2+

∞∑n=1

[an cos

(nπx

L

)+ bn sin

(nπx

L

)](2)

with the constants an and bn given by

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , n > 0,

bn =1

L

∫ L

−Lf (x) sin

(nπx

L

)dx , n > 1.

Furthermore, the Fourier series in Eq. (2) provides a 2L-periodicextension of function f from the domain [−L, L] ⊂ R to R.


Sketch of the Proof:

I Define the partial sum functions

fN(x) =a0

2+

N∑n=1

[an cos

(nπx

L

)+ bn sin

(nπx

L

)]

with an and bn given by

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , n > 0,

bn =1

L

∫ L

−Lf (x) sin

(nπx

L

)dx , n > 1.

I Express fN as a convolution of Sine, Cosine, functions and theoriginal function f .

I Use the convolution properties to show that

limN→∞

fN(x) = f (x), x ∈ [−L, L].


Sketch of the Proof:

I Define the partial sum functions

fN(x) =a0

2+

N∑n=1

[an cos

(nπx

L

)+ bn sin

(nπx

L

)]with an and bn given by

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , n > 0,

bn =1

L

∫ L

−Lf (x) sin

(nπx

L

)dx , n > 1.

I Express fN as a convolution of Sine, Cosine, functions and theoriginal function f .

I Use the convolution properties to show that

limN→∞

fN(x) = f (x), x ∈ [−L, L].

Example: Using the Fourier Theorem.Example

Find the Fourier series expansion of the function

f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].

Solution: In this case L = 1. The Fourier series expansion is

f (x) =a0

2+

∞∑n=1

[an cos(nπx) + bn sin(nπx)

],

where the an, bn are given in the Theorem. We start with a0,

a0 =

∫ 1

−1f (x) dx =

∫ 0

−1(1 + x) dx +

∫ 1

0(1− x) dx .

a0 =(x +

x2

2

)∣∣∣0−1

+(x − x2

2

)∣∣∣10

=(1− 1

2

)+

(1− 1

2

)We obtain: a0 = 1.



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].

Solution: In this case L = 1.

The Fourier series expansion is

f (x) =a0

2+

∞∑n=1


],


a0 =

∫ 1

−1f (x) dx =

∫ 0

−1(1 + x) dx +

∫ 1

0(1− x) dx .

a0 =(x +

x2

2

)∣∣∣0−1

+(x − x2

2

)∣∣∣10

=(1− 1

2

)+

(1− 1

2

)We obtain: a0 = 1.



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


f (x) =a0

2+

∞∑n=1


],


a0 =

∫ 1

−1f (x) dx =

∫ 0

−1(1 + x) dx +

∫ 1

0(1− x) dx .

a0 =(x +

x2

2

)∣∣∣0−1

+(x − x2

2

)∣∣∣10

=(1− 1

2

)+

(1− 1

2

)We obtain: a0 = 1.



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


f (x) =a0

2+

∞∑n=1


],

where the an, bn are given in the Theorem.

We start with a0,

a0 =

∫ 1

−1f (x) dx =

∫ 0

−1(1 + x) dx +

∫ 1

0(1− x) dx .

a0 =(x +

x2

2

)∣∣∣0−1

+(x − x2

2

)∣∣∣10

=(1− 1

2

)+

(1− 1

2

)We obtain: a0 = 1.



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


f (x) =a0

2+

∞∑n=1


],


a0 =

∫ 1

−1f (x) dx

=

∫ 0

−1(1 + x) dx +

∫ 1

0(1− x) dx .

a0 =(x +

x2

2

)∣∣∣0−1

+(x − x2

2

)∣∣∣10

=(1− 1

2

)+

(1− 1

2

)We obtain: a0 = 1.



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


f (x) =a0

2+

∞∑n=1


],


a0 =

∫ 1

−1f (x) dx =

∫ 0

−1(1 + x) dx +

∫ 1

0(1− x) dx .

a0 =(x +

x2

2

)∣∣∣0−1

+(x − x2

2

)∣∣∣10

=(1− 1

2

)+

(1− 1

2

)We obtain: a0 = 1.



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


f (x) =a0

2+

∞∑n=1


],


a0 =

∫ 1

−1f (x) dx =

∫ 0

−1(1 + x) dx +

∫ 1

0(1− x) dx .

a0 =(x +

x2

2

)∣∣∣0−1

+(x − x2

2

)∣∣∣10

=(1− 1

2

)+

(1− 1

2

)

We obtain: a0 = 1.



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


f (x) =a0

2+

∞∑n=1


],


a0 =

∫ 1

−1f (x) dx =

∫ 0

−1(1 + x) dx +

∫ 1

0(1− x) dx .

a0 =(x +

x2

2

)∣∣∣0−1

+(x − x2

2

)∣∣∣10

=(1− 1

2

)+

(1− 1

2

)We obtain: a0 = 1.



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].

Solution: Recall: a0 = 1.

Similarly, the rest of the an are given by,

an =

∫ 1

−1f (x) cos(nπx) dx

an =

∫ 0

−1(1 + x) cos(nπx) dx +

∫ 1

0(1− x) cos(nπx) dx .

Recall the integrals

∫cos(nπx) dx =

1

nπsin(nπx), and∫

x cos(nπx) dx =x

nπsin(nπx) +

1

n2π2cos(nπx).



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].

Solution: Recall: a0 = 1. Similarly, the rest of the an are given by,

an =

∫ 1


an =

∫ 0

−1(1 + x) cos(nπx) dx +

∫ 1



∫cos(nπx) dx =

1


x cos(nπx) dx =x

nπsin(nπx) +

1

n2π2cos(nπx).



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


an =

∫ 1


an =

∫ 0

−1(1 + x) cos(nπx) dx +

∫ 1



∫cos(nπx) dx =

1

nπsin(nπx),

and∫x cos(nπx) dx =

x

nπsin(nπx) +

1

n2π2cos(nπx).



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


an =

∫ 1


an =

∫ 0

−1(1 + x) cos(nπx) dx +

∫ 1



∫cos(nπx) dx =

1


x cos(nπx) dx =x

nπsin(nπx) +

1

n2π2cos(nπx).

Example: Using the Fourier Theorem.

Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].

Solution: It is not difficult to see that

an =1

nπsin(nπx)

∣∣∣0−1

+[ x

nπsin(nπx) +

1

n2π2cos(nπx)

]∣∣∣0−1

+1

nπsin(nπx)

∣∣∣10−

[ x

nπsin(nπx) +

1

n2π2cos(nπx)

]∣∣∣10

an =[ 1

n2π2− 1

n2π2cos(−nπ)

]−

[ 1

n2π2cos(nπ)− 1

n2π2

].

We then conclude that an =2

n2π2

[1− cos(nπ)

].


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].

Solution: Recall: a0 = 1, and an =2

n2π2

[1− cos(nπ)

].

Finally, we must find the coefficients bn.

A similar calculation shows that bn = 0.

Then, the Fourier series of f is given by

f (x) =1

2+

∞∑n=1

2

n2π2

[1− cos(nπ)

]cos(nπx). C



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].

Solution: Recall: f (x) =1

2+

∞∑n=1

2

n2π2

[1− cos(nπ)

]cos(nπx).

We can obtain a simpler expression for the Fourier coefficients an.

Recall the relations cos(nπ) = (−1)n, then

f (x) =1

2+

∞∑n=1

2

n2π2

[1− (−1)n

]cos(nπx).

f (x) =1

2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1− cos(nπ)

]cos(nπx).


Recall the relations cos(nπ) = (−1)n,

then

f (x) =1

2+

∞∑n=1

2

n2π2

[1− (−1)n

]cos(nπx).

f (x) =1

2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).



f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1− cos(nπ)

]cos(nπx).


Recall the relations cos(nπ) = (−1)n, then

f (x) =1

2+

∞∑n=1

2

n2π2

[1− (−1)n

]cos(nπx).

f (x) =1

2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).

If n = 2k, so n is even, so n + 1 = 2k + 1 is odd, then

a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.

If n = 2k − 1, so n is odd, so n + 1 = 2k is even, then

a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).

If n = 2k,

so n is even, so n + 1 = 2k + 1 is odd, then

a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.


a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).

If n = 2k, so n is even,

so n + 1 = 2k + 1 is odd, then

a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.


a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).

If n = 2k, so n is even, so n + 1 = 2k + 1 is odd,

then

a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.


a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).


a2k =2

(2k)2π2(1− 1)

⇒ a2k = 0.


a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).


a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.


a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).


a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.

If n = 2k − 1,

so n is odd, so n + 1 = 2k is even, then

a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).


a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.

If n = 2k − 1, so n is odd,

so n + 1 = 2k is even, then

a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).


a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.

If n = 2k − 1, so n is odd, so n + 1 = 2k is even,

then

a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).


a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.


a2k−1 =2

(2k − 1)2π2(1 + 1)

⇒ a2k−1 =4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].


2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx).


a2k =2

(2k)2π2(1− 1) ⇒ a2k = 0.


a2k−1 =2

(2k − 1)2π2(1 + 1) ⇒ a2k−1 =

4

(2k − 1)2π2.


Example


f (x) =

{1 + x x ∈ [−1, 0),

1− x x ∈ [0, 1].

Solution:

Recall: f (x) =1

2+

∞∑n=1

2

n2π2

[1 + (−1)n+1

]cos(nπx), and

a2k = 0, a2k−1 =4

(2k − 1)2π2.

We conclude: f (x) =1

2+

∞∑k=1

4

(2k − 1)2π2cos((2k − 1)πx). C

The Fourier Theorem: Piecewise continuous case.

Recall:

DefinitionA function f : [a, b] → R is called piecewise continuous iff holds,

(a) [a, b] can be partitioned in a finite number of sub-intervalssuch that f is continuous on the interior of these sub-intervals.

(b) f has finite limits at the endpoints of all sub-intervals.

The Fourier Theorem: Piecewise continuous case.


If f : [−L, L] ⊂ R → R is piecewise continuous, then the function

fF (x) =a0

2+

∞∑n=1

[an cos

(nπx

L

)+ bn sin

(nπx

L

)]where an and bn given by

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , n > 0,

bn =1

L

∫ L

−Lf (x) sin

(nπx

L

)dx , n > 1.

satisfies that:

(a) fF (x) = f (x) for all x where f is continuous;

(b) fF (x0) =1

2

[lim

x→x+0

f (x) + limx→x−0

f (x)]

for all x0 where f is

discontinuous.


Example

Find the Fourier series of f (x) =

{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).

and periodic with period T = 2.

Solution: We start computing the Fourier coefficients bn;

bn =1

L

∫ L

−Lf (x) sin

(nπx

L

)dx , L = 1,

bn =

∫ 0

−1(−1) sin

(nπx

)dx +

∫ 1

0(1) sin

(nπx

)dx ,

bn =(−1)

nπ

[− cos(nπx)

∣∣∣0−1

]+

1

nπ

[− cos(nπx)

∣∣∣10

],

bn =(−1)

nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



bn =1

L

∫ L

−Lf (x) sin

(nπx

L

)dx ,

L = 1,

bn =

∫ 0

−1(−1) sin

(nπx

)dx +

∫ 1

0(1) sin

(nπx

)dx ,

bn =(−1)

nπ

[− cos(nπx)

∣∣∣0−1

]+

1

nπ

[− cos(nπx)

∣∣∣10

],

bn =(−1)

nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



bn =1

L

∫ L

−Lf (x) sin

(nπx

L

)dx , L = 1,

bn =

∫ 0

−1(−1) sin

(nπx

)dx +

∫ 1

0(1) sin

(nπx

)dx ,

bn =(−1)

nπ

[− cos(nπx)

∣∣∣0−1

]+

1

nπ

[− cos(nπx)

∣∣∣10

],

bn =(−1)

nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).


Solution: bn =(−1)

nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].

bn =1

nπ

[1− cos(−nπ)− cos(nπ) + 1

]=

2

nπ

[1− cos(nπ)

],

We obtain: bn =2

nπ

[1− (−1)n

].

If n = 2k, then b2k =2

2kπ

[1− (−1)2k

], hence b2k = 0.

If n = 2k − 1, then b2k−1 =2

(2k − 1)π

[1− (−1)2k−1

],

hence b2k =4

(2k − 1)π.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].

bn =1

nπ


]

=2

nπ

[1− cos(nπ)

],

We obtain: bn =2

nπ

[1− (−1)n

].


2kπ

[1− (−1)2k

], hence b2k = 0.

If n = 2k − 1, then b2k−1 =2

(2k − 1)π

[1− (−1)2k−1

],

hence b2k =4

(2k − 1)π.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].

bn =1

nπ


]=

2

nπ

[1− cos(nπ)

],

We obtain: bn =2

nπ

[1− (−1)n

].


2kπ

[1− (−1)2k

], hence b2k = 0.

If n = 2k − 1, then b2k−1 =2

(2k − 1)π

[1− (−1)2k−1

],

hence b2k =4

(2k − 1)π.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].

bn =1

nπ


]=

2

nπ

[1− cos(nπ)

],

We obtain: bn =2

nπ

[1− (−1)n

].

If n = 2k,

then b2k =2

2kπ

[1− (−1)2k

], hence b2k = 0.

If n = 2k − 1, then b2k−1 =2

(2k − 1)π

[1− (−1)2k−1

],

hence b2k =4

(2k − 1)π.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].

bn =1

nπ


]=

2

nπ

[1− cos(nπ)

],

We obtain: bn =2

nπ

[1− (−1)n

].


2kπ

[1− (−1)2k

],

hence b2k = 0.

If n = 2k − 1, then b2k−1 =2

(2k − 1)π

[1− (−1)2k−1

],

hence b2k =4

(2k − 1)π.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].

bn =1

nπ


]=

2

nπ

[1− cos(nπ)

],

We obtain: bn =2

nπ

[1− (−1)n

].


2kπ

[1− (−1)2k

], hence b2k = 0.

If n = 2k − 1, then b2k−1 =2

(2k − 1)π

[1− (−1)2k−1

],

hence b2k =4

(2k − 1)π.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].

bn =1

nπ


]=

2

nπ

[1− cos(nπ)

],

We obtain: bn =2

nπ

[1− (−1)n

].


2kπ

[1− (−1)2k

], hence b2k = 0.

If n = 2k − 1,

then b2k−1 =2

(2k − 1)π

[1− (−1)2k−1

],

hence b2k =4

(2k − 1)π.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



nπ

[−1 + cos(−nπ)

]+

1

nπ

[− cos(nπ) + 1

].

bn =1

nπ


]=

2

nπ

[1− cos(nπ)

],

We obtain: bn =2

nπ

[1− (−1)n

].


2kπ

[1− (−1)2k

], hence b2k = 0.

If n = 2k − 1, then b2k−1 =2

(2k − 1)π

[1− (−1)2k−1

],

hence b2k =4

(2k − 1)π.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).


Solution: Recall: b2k = 0, and b2k =4

(2k − 1)π.

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , L = 1,

an =

∫ 0

−1(−1) cos

(nπx

)dx +

∫ 1

0(1) cos

(nπx

)dx ,

an =(−1)

nπ

[sin(nπx)

∣∣∣0−1

]+

1

nπ

[sin(nπx)

∣∣∣10

],

an =(−1)

nπ

[0− sin(−nπ)

]+

1

nπ

[sin(nπ)− 0

]⇒ an = 0.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



(2k − 1)π.

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx ,

L = 1,

an =

∫ 0

−1(−1) cos

(nπx

)dx +

∫ 1

0(1) cos

(nπx

)dx ,

an =(−1)

nπ

[sin(nπx)

∣∣∣0−1

]+

1

nπ

[sin(nπx)

∣∣∣10

],

an =(−1)

nπ

[0− sin(−nπ)

]+

1

nπ

[sin(nπ)− 0

]⇒ an = 0.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



(2k − 1)π.

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , L = 1,

an =

∫ 0

−1(−1) cos

(nπx

)dx +

∫ 1

0(1) cos

(nπx

)dx ,

an =(−1)

nπ

[sin(nπx)

∣∣∣0−1

]+

1

nπ

[sin(nπx)

∣∣∣10

],

an =(−1)

nπ

[0− sin(−nπ)

]+

1

nπ

[sin(nπ)− 0

]⇒ an = 0.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



(2k − 1)π.

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , L = 1,

an =

∫ 0

−1(−1) cos

(nπx

)dx +

∫ 1

0(1) cos

(nπx

)dx ,

an =(−1)

nπ

[sin(nπx)

∣∣∣0−1

]+

1

nπ

[sin(nπx)

∣∣∣10

],

an =(−1)

nπ

[0− sin(−nπ)

]+

1

nπ

[sin(nπ)− 0

]

⇒ an = 0.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).



(2k − 1)π.

an =1

L

∫ L

−Lf (x) cos

(nπx

L

)dx , L = 1,

an =

∫ 0

−1(−1) cos

(nπx

)dx +

∫ 1

0(1) cos

(nπx

)dx ,

an =(−1)

nπ

[sin(nπx)

∣∣∣0−1

]+

1

nπ

[sin(nπx)

∣∣∣10

],

an =(−1)

nπ

[0− sin(−nπ)

]+

1

nπ

[sin(nπ)− 0

]⇒ an = 0.


Example


{− 1 x ∈ [−1, 0),

1 x ∈ [0, 1).


Solution: Recall: b2k = 0, b2k =4

(2k − 1)π, and an = 0.

Therefore, we conclude that

fF (x) =4

π

∞∑k=1

1

(2k − 1)sin

((2k − 1)π x

).

C

boundary value problems (sect. 10.1)....boundary value problems (sect. 10.1). i two-point bvp. i...

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