HW

Chapter 4

1.17

a)

verify theorem 1 concerning the existen ce of a unique Green's function
Theorem 1:

"3: ((SuperscriptBox[d, 2] y)/(dx^2))=f(x), 0<x<(π/2), y(0)=0, y(π/2)=0"

"4: \!\(\*FormBox[
RowBox[{
RowBox[{
RowBox[{FractionBox[
RowBox[{SuperscriptBox["d", "2"], "y"}], SuperscriptBox["dx", "2"]], "-", 
RowBox[{SuperscriptBox["b", "2"], "y"}]}], "=", 
RowBox[{"f", "(", "x", ")"}]}], ",", " ", 
RowBox[{"0", "<", "x", "<", "1"}], ",", " ", 
RowBox[{
RowBox[{"y", "(", "0", ")"}], "=", "0"}], ",", " ", 
RowBox[{
RowBox[{"y", "(", "1", ")"}], "=", "0"}]}],
TraditionalForm]\)"

"5: \!\(\*FormBox[
RowBox[{
RowBox[{
RowBox[{FractionBox[
RowBox[{SuperscriptBox["d", "2"], "y"}], SuperscriptBox["dx", "2"]], "-", "y"}], "=", "x"}], ",", " ", 
RowBox[{"0", "<", "x", "<", "1"}], ",", " ", 
RowBox[{
RowBox[{"y", "(", "0", ")"}], "=", "0"}], ",", 
RowBox[{
RowBox[{"y", "(", "0", ")"}], "=", "1"}]}],
TraditionalForm]\)"

"6: ((SuperscriptBox[d, 2] y)/(dx^2))=f(x,y(x)), 0<x<1, y(0)=0, y(1)=0"

since 0 is a workable solution, they all have a unique solution.

b) i)

(((d^2) y)/(dx^2))=f[x], 0<x<1, y[0]=y'[1], y'[0]=y[1]
is of the form in eqn 4.1, with (Subscript[A, 1]) and (Subscript[A, 2]) being 0
the general form of the solution being y[x]=-(∫_a^b)g[x,t]f[t]dt
our solution u[x] will be in the form (Subscript[w, 1])[x](Subscript[v, 1])[x]+(Subscript[w, 2])[x](Subscript[v, 2])[x]
the derivative is u[x]=(Subscript[w, 1])[x](Subscript[v, 1])'[x]+(Subscript[w, 2])[x](Subscript[v, 2])'[x]
aaaaand from here on im confused.

b) ii)

(((d^2) y)/(dx^2))=f[x], 0<x<1, y[0]=y[1], y'[0]=y'[1]
looping region!  a horizonal line works, so any y=C is a valid solution
infinite number of solutions C

b) iii)

(((d^2)y)/(dx^2))+y=f[x], 0<x<π, y[0]=y[π]
since its over some distance π, i assume it relates to sines and cosines.
infinite number of solutions

1.22

a)

"Prove that the kernel K(x,t) = (x^3)(t^3) and L(x,t)=(x^2)(t^2) are orthogonal on {0≤x≤1, 0≤t≤1}"

K[x_,t_]=(x^3)(t^3);
L[x_,t_]=(x^2)(t^2);
Print["K[x,t]*L[x,t]:",Integrate[K[x,τ]*L[τ,t],{τ,0,1}]]
Print["K[x,t]*L[x,t]:",Integrate[L[x,τ]*K[τ,t],{τ,0,1}]]
Print["not orthogonal"]

Print["However, on a range of -1≤x≤1,"]
Print["K[x,t]*L[x,t]:",Integrate[K[x,τ]*L[τ,t],{τ,-1,1}]]
Print["K[x,t]*L[x,t]:",Integrate[L[x,τ]*K[τ,t],{τ,-1,1}]]
Print["orthogonal"]

b)

"Prove K(x,t)=sin(x-2t) is orthogonal to itself on {0≤x≤2π,0≤t≤2π}"

K[x_,t_]=Sin[x-2 t]
Print["K[x,t]*K[x,t]:",Integrate[K[x,τ]*K[τ,t],{τ,0,2π}]]
Print["orthogonal"]

Chapter 5

1.1

"solve and verify your solution to these fredholm equations"

a)

"u(x)=sin(x)-λ(∫_0^(π/2))sin(x)*cos(t)*u(t)\[DifferentialD]t"

Start with: u[x] = Sin[x] - λ(∫_0^(π/2))Sin[x]*Cos[t]*u[t]dt
the integral separates, giving
u[x] = Sin[x] - λ*Sin[x](∫_0^(π/2))Cos[t]*u[t]dt
say  C = (∫_0^(π/2))Cos[t]*u[t]dt
leaving
u[x] = Sin[x] - λ*Sin[x]*C
plugging this u[x] back into C gives us
C = (∫_0^(π/2))Cos[t]*(Sin[t]-λ*Sin[t]*C)dt
applying algebra to solve for C
C = (∫_0^(π/2))Cos[t]*Sin[t]dt-λ*C*(∫_0^(π/2))Cos[t]*Sin[t]dt
C+λ*C*(∫_0^(π/2))Cos[t]*Sin[t]dt =  (∫_0^(π/2))Cos[t]*Sin[t]dt
C(1+λ*(∫_0^(π/2))Cos[t]*Sin[t]dt)=(∫_0^(π/2))Cos[t]*Sin[t]dt
C=((∫_0^(π/2))Cos[t]*Sin[t]dt)/(1+λ*(∫_0^(π/2))Cos[t]*Sin[t]dt)
C=(1/2)/(1+λ/2)
C = (1/(2+λ))
plugging this back into our u[x] equation, we get
u[x] =  Sin[x] - λ*Sin[x]*(1/(2+λ))
or after bashing it around with algebra,
u[x] = Sin[x]*(1-(λ/(2+λ)))
u[x] = Sin[x]*(((2+λ)/(2+λ))-(λ/(2+λ)))
u[x] = Sin[x]*((2/(2+λ)))

"prove the solution"

u[x] = Sin[x] - λ*Sin[x](∫_0^(π/2))Cos[t]*u[t]dt
with u[x] = Sin[x]*((2/(2+λ)))
Sin[x]*((2/(2+λ))) = Sin[x] - λ*Sin[x](∫_0^(π/2))Cos[t]*Sin[t]*((2/(2+λ)))dt
then add algebra to show the left side equals the right!
Sin[x]*((2/(2+λ))) = Sin[x] - λ*Sin[x]*((2/(2+λ)))*(∫_0^(π/2))Cos[t]*Sin[t]dt
Sin[x]*((2/(2+λ))) = Sin[x] - λ*Sin[x]*((2/(2+λ)))*(1/2)
Sin[x]*((2/(2+λ))) = Sin[x] (((2+λ)/(2+λ))- (λ/(2+λ)))
Sin[x]*((2/(2+λ))) = Sin[x] (((2+λ-λ)/(2+λ)))
Sin[x]*((2/(2+λ))) = Sin[x] ((2/(2+λ)))
so its good.

c)

"u[x] = 2x - π + 4(∫_0^π)Sin[x]^2*u[t]\[DifferentialD]t"

start with: u[x] = 2x - π + 4(∫_0^π)((Sin[x])^2)*u[t]dt
separate the kernel
u[x] = 2x - π + 4*((Sin[x])^2)(∫_0^π)u[t]dt
and set the integral as C
C = (∫_0^π)u[t]dt
with
u[x] = 2x - π + 4*((Sin[x])^2)*C
plugging this u[x] into our C equation gives us
C = (∫_0^π)2t - π + 4*((Sin[x])^2)*Cdt
more algebra thrashing to solve for C gives us
C = 2(∫_0^π)tdt - π(∫_0^π)dt + 4C(∫_0^π)((Sin[x])^2)dt
C = (π^2) - (π^2) + 2 C π
C = 2 C π
C(1-2π)=0
C=0
putting this result into our u[x] that has a C, 
u[x] = 2x - π

"prove the solution works"

u[x] = 2x - π + 4(∫_0^π)((Sin[x])^2)*u[t]dt
with u[x] = 2 x - π
2x-π = 2x - π + 4(∫_0^π)((Sin[x])^2)*(2t-π)dt
make the right side equal to the left!
2x-π = 2x - π + 4((Sin[x])^2)((∫_0^π)2tdt-(∫_0^π)πdt)
2x-π = 2x - π + 4((Sin[x])^2)(2(∫_0^π)tdt-π(∫_0^π)dt)
2x-π = 2x - π + 4((Sin[x])^2)((π^2)-(π^2))
2x-π = 2x - π
so, its good

(∫_0^π)((Sin[t])^2) \[DifferentialD]t

1.2

"Solve the homogeneous Fredholm equations by finding eigenvalues and eigenfunctions of the equations"

a)

"u[x]=λ(∫_0^(2  π))Sin[x]Sin[t]u[t]\[DifferentialD]t"

separate the kernel
u[x]=λSin[x](∫_0^(2π))Sin[t]u[t]dt
set the integral to equal C
C = (∫_0^(2π))Sin[t]u[t]dt
u[x]=λ C Sin[x]
plug the new u[x] into the equation for C
C = (∫_0^(2π))Sin[t]λ C Sin[t]dt
algebra to solve for C
C = λ C (∫_0^(2π))((Sin[t])^2)dt
C(1- λ(∫_0^(2π))((Sin[t])^2)dt)=0 
C=0  and 1-λπ = 0
we dont care about C = 0,
so λ = 1/π is our first eigenvalue.
now, our u[x,C] equation becomes
(Subscript[u, 1])[x]=((λ C Sin[x])/π)  is it's eigenfunction

b)

"u[x] = λ (∫_0^(π/2))Sin[x]Cos[t]u[t]\[DifferentialD]t"

separate the kernel
u[x]=λ Sin[x](∫_0^(π/2))Cos[t]u[t]dt
set the integral as a constant
C=(∫_0^(π/2))Cos[t]u[t]dt
leaving us with
u[x]=λ Sin[x]C
which can be plugged into the C= equation
C=(∫_0^(π/2))Cos[t]λSin[t]Cdt
and with a little algebra, we find
C=λC(∫_0^(π/2))Cos[t]Sin[t]dt
C=λC/2
C(1-λ/2)=0
ignoring C = 0, our first eigenvalue is
λ=2
with this, we find our eigenfunction is
u[x]=2 Sin[x] C

Integrate[Cos[t]Sin[t],{t,0,π/2}]

1.3

a)

"discuss the existence of the solution of this integral equation:"
"u[x]=f[x]+λ(∫_0^(2  π))Sin[x+t]u[t]\[DifferentialD]t"
"for the cases of
	(i)  λ=1/(Sqrt[π]), f[x]=(x^2)
	(ii) λ=1/π, f[x]=Sin[3x]"

the eigenvalues of this equation are  ± 1/π, with eigenfunctions sin[x]+cos[x] and sin[x]-cos[x], from pg 222 in the book
(i)as 1/(Sqrt[π]) isn't an eigenvalue, there will exist one unique solution, no matter what f[x] is
(ii)as 1/π is an eigenvalue, we then test to see if it's f[x] is orthogonal to the kernel on the interval {0,2π}.
(∫_0^(2π))Sin[3x]*(Sin[x]-Cos[x])dt = 0
since it is orthogonal, it has an infinite number of solutions

b)

"find the solution of
	u[x] = (x^2)+(1/(Sqrt[π]))(∫_0^(2  π))Sin[x+t]u[t]\[DifferentialD]t"

start here,
u[x] = (x^2)+(1/(Sqrt[π]))(∫_0^(2π))Sin[x+t]u[t]dt
then separate the kernel
u[x] = (x^2)+(1/(Sqrt[π]))(∫_0^(2π))(Sin[x]Cos[t]+Cos[x]Sin[t])u[t]dt
power of algebra, i choose you! to split up the integral
u[x] = (x^2)+(1/(Sqrt[π]))((∫_0^(2π))Sin[x]Cos[t]u[t]dt+(∫_0^(2π))Cos[x]Sin[t]u[t]dt)
u[x] = (x^2)+(1/(Sqrt[π]))(Sin[x](∫_0^(2π))Cos[t]u[t]dt+Cos[x](∫_0^(2π))Sin[t]u[t]dt)
then replace each integral with a (Subscript[c, 1]) or (Subscript[c, 2])
(Subscript[c, 1])=(∫_0^(2π))Cos[t]u[t]dt
(Subscript[c, 2])=(∫_0^(2π))Sin[t]u[t]dt
leaving us with
u[x] = (x^2)+(1/(Sqrt[π]))(Sin[x](Subscript[c, 1])+Cos[x](Subscript[c, 1]))
now we make a matrix of each combination of (Subscript[c, 1]) and (Subscript[c, 2])
(Subscript[a, 11])=(∫_0^(2π))Cos[t]Sin[t]dt = 0
(Subscript[a, 12])=(∫_0^(2π))Cos[t]Cos[t]dt = π
(Subscript[a, 21])=(∫_0^(2π))Sin[t]Sin[t]dt = π
(Subscript[a, 22])=(∫_0^(2π))Sin[t]Cos[t]dt = 0
and a solution vector
(Subscript[f, 1])=(∫_0^(2π))Cos[t](t^2)dt = 4π
(Subscript[f, 2])=(∫_0^(2π))Sin[t](t^2)dt = -4π
with these, our matrix equation becomes
(I-λA)C=(({{1, 0}, {0, 1}}))(({{(Subscript[c, 1])}, {(Subscript[c, 2])}}))-λ(({{0, π}, {π, 0}}))(({{(Subscript[c, 1])}, {(Subscript[c, 2])}}))=(({{0}, {0}}))
(({{1, -λπ}, {-λπ, 1}}))(({{(Subscript[c, 1])}, {(Subscript[c, 2])}}))=(({{0}, {0}}))
we can then take the determinate of this matrix to see what the possible eigenvalues are
(Subscript[λ, 1])=1/π and (Subscript[λ, 2])=-1/π
to find the eigenfunctions we plug these back into our matrix to find the relationship between (Subscript[c, 1]) and (Subscript[c, 2]).
(Subscript[c, 1])=(Subscript[c, 2])
(Subscript[ϕ, 1])=(1/π)((Subscript[c, 1])Sin[x]+(Subscript[c, 1])Cos[x]) = ((Subscript[c, 1])/π)(Sin[x]+Cos[x])
and the other is 
(Subscript[ϕ, 2])=(1/π)((Subscript[c, 1])Sin[x]-(Subscript[c, 1])Cos[x]) = ((Subscript[c, 1])/π)(Sin[x]-Cos[x])
now we use all of this to find the unique solutions
u[x]=f[x]+λ((Subscript[c, 1])Sin[x]-(Subscript[c, 2])Cos[x])
so,
u[x]=(x^2)+(1/(Sqrt[π]))((Subscript[c, 1])Sin[x]-(Subscript[c, 2])Cos[x])
go go matrix powers!
C-λAC = F = 
(({{(Subscript[c, 1])}, {(Subscript[c, 2])}}))-3(({{0, π}, {π, 0}}))(({{(Subscript[c, 1])}, {(Subscript[c, 2])}}))=(({{4π}, {-4π}}))
(({{1, -3π}, {-3π, 1}}))(({{(Subscript[c, 1])}, {(Subscript[c, 2])}}))=(({{4π}, {-4π}}))
(Subscript[c, 1])=((4 π)/(1+3 π)) and (Subscript[c, 2])=-((4 π)/(1+3 π))
so, our unique solution is
u[x]=(x^2)+(1/(Sqrt[π]))(((4 π)/(1+3 π))Sin[x]+((4 π)/(1+3 π))Cos[x])

2.3

"consider this fredholm equation:
u[x]=Cos[2x]+2(∫_0^(π/2))K[x,t]u[t]\[DifferentialD]t, 
where K[x,t]={GridBox[{{RowBox[{RowBox[{RowBox[{Sin, [,x,]}], RowBox[{Cos, [,t,]}]}], , }], RowBox[{0, ≤,x,≤,t}]}, {RowBox[{RowBox[{RowBox[{Sin, [,t,]}], RowBox[{Cos, [,x,]}]}], , }], RowBox[{t, ≤,x,≤,FractionBox["π", "2"]}]}}]}"

a)

symmetric kernel means K[x,t]=K[t,x],

K[x_,t_]:=Piecewise[{
{Sin[x]Cos[t],0≤x≤t},
{Sin[t]Cos[x],t≤x≤π/2}}]
{Plot3D[K[x,t],{x,0,3},{t,0,8},AxesLabel->{"x","t","K"}],Plot3D[K[t,x],{x,0,8},{t,0,3},AxesLabel->{"t","x","K"}]}

they are the same.  the slight differences in the graphs is that the x axis is reversed in one of them.
to test if it is square integrable, we evaluate the following integral and make sure it is less than ∞
(∫_a^b)(∫_a^b)((K[x,t])^2)dxdt = (∫_0^(π/2))(∫_0^(π/2))((K[x,t])^2)dxdt= (1/16) (-8+(π^2))

b)

"reduce u[x]=λ(∫_0^(π/2))K[x,t]u[t]\[DifferentialD]t to a differential equation to obtain the eigenvalues and eigenfunctions, with
where K[x,t]={GridBox[{{RowBox[{RowBox[{RowBox[{Sin, [,x,]}], RowBox[{Cos, [,t,]}]}], , }], RowBox[{0, ≤,x,≤,t}]}, {RowBox[{RowBox[{RowBox[{Sin, [,t,]}], RowBox[{Cos, [,x,]}]}], , }], RowBox[{t, ≤,x,≤,FractionBox["π", "2"]}]}}]}"

our equation starts as u[x]=λ(∫_0^(π/2))K[x,t]u[t]dt
which we then write with the kernel
u[x]=λ(∫_0^x)Sin[t]Cos[x]u[t]dt+λ(∫_x^1)Sin[x]Cos[t]u[t]dt
separate it out
u[x]=λ Cos[x](∫_0^x)Sin[t]u[t]dt+λ Sin[x](∫_x^1)Cos[t]u[t]dt
applying the leibnitz rule to each,
(d/(d x))(∫_(α[x])^(β[x]))F[x,t]dt=(∫_(α[x])^(β[x]))((d F[x,t])/(d x))dt+F[x,β[x]]((d β[x])/(d x))-F[x,α[x]]((d α[x])/(d x))
where for the first integral, F[x,t]=λCos[x]Sin[t]u[t], α=0, β=x
and the second, F[x,t]=λSin[x]Cos[t]u[t], α=x, β=π/2
u'[x]=-λ(∫_0^x)Sin[x]Sin[t]u[t]dt+λ Cos[x]Sin[x]u[x]+λ(∫_x^1)Cos[x]Sin[t]u[t]dt-λSin[x]Cos[x]u[x]
u'[x]=-λ(∫_0^x)Sin[x]Sin[t]u[t]dt+λ(∫_x^1)Cos[x]Cos[t]u[t]dt
and again
(Subscript[F, 1])=-λ Sin[x]Sin[t]u[t], (Subscript[α, 1])=0, (Subscript[β, 1])=x
(Subscript[F, 2])=   λ Cos[x]Cos[t]u[t], (Subscript[α, 2])=x, (Subscript[β, 2])=π/2
u''[x]=(∫_((Subscript[α, 1])[x])^((Subscript[β, 1])[x]))((d (Subscript[F, 1])[x,t])/(d x))dt+(Subscript[F, 1])[x,(Subscript[β, 1])[x]]((d (Subscript[β, 1])[x])/(d x))-(Subscript[F, 2])[x,(Subscript[α, 1])[x]]((d (Subscript[α, 1])[x])/(d x))+(∫_((Subscript[α, 2])[x])^((Subscript[β, 2])[x]))((d (Subscript[F, 2])[x,t])/(d x))dt+(Subscript[F, 2])[x,(Subscript[β, 2])[x]]((d (Subscript[β, 2])[x])/(d x))-(Subscript[F, 2])[x,(Subscript[α, 2])[x]]((d (Subscript[α, 2])[x])/(d x))
u''[x]=(∫_0^x)((d (Subscript[F, 1])[x,t])/(d x))dt+(Subscript[F, 1])[x,x]((d x)/(d x))-(Subscript[F, 1])[x,0]((d 0)/(d x))+(∫_x^1)((d (Subscript[F, 2])[x,t])/(d x))dt+(Subscript[F, 2])[x,1]((d 1)/(d x))-(Subscript[F, 2])[x,x]((d x)/(d x))
u''[x]=(∫_0^x)-λCos[x]Sin[t]u[t]dt-((λSin[x])^2)u[x]+(∫_x^1)-λSin[x]Cos[t]u[t]dt-((λCos[x])^2)u[x]
u''[x]=-λ(∫_x^(π/2))Sin[x]Cos[t]u[t]dt-λ(∫_0^x)Cos[x]Sin[t]u[t]dt-λ u[x] (((Sin[x])^2)+((Cos[x])^2))
u''[x]=-λ(∫_x^(π/2))Sin[x]Cos[t]u[t]dt-λ(∫_0^x)Cos[x]Sin[t]u[t]dt-λ u[x]
u''[x]=-λ(∫_0^(π/2))K[x,t]u[t]dt-λ u[x]
u''[x]=-u[x]-λ u[x]
u''[x]=-u[x](1+λ)
u''[x]+u[x](1+λ)=0
from this differential equation, we get a solution of u[x]=A*Cos[u[x]=A Cos[(Sqrt[1+λ])x]+B Sin[(Sqrt[1+λ])x], 
with boundary conditions u[0]=u[π/2]=0
solving with these conditions gives us
u[x]=B Sin[(Sqrt[1+λ])x]
which only works if Sin[blah] ≠0.  to get this, then (Sqrt[1+λ])(π/2) =n π 
or,
(Sqrt[1+λ])=2n
1+λ=4(n^2)
λ=4(n^2)-1, our eigenvalues.
the eigenfunctions are just our original solution with these values for λ plugged in, giving us B Sin[2nx].
then we normalize it to find B
1=(∫_0^(π/2))(B^2)((Sin[2 n x])^2)dx
(1/(B^2))=(∫_0^(π/2))((Sin[2 n x])^2)dx=π/4, so B = Sqrt[4/π]
so, finally, our eigenfunctions are
Eig=(2/(Sqrt[π])) Sin[2 n x], with eigenvaules of λ=4(n^2)-1

c)

u[x]=Cos[2x]+2(∫_0^(π/2))K[x,t]u[t]dt
our f[x]=Cos[2x], and our λ is 2
we can use the resolvent kernel, from eqns 5.55 and 5.46 to construct our solution
u[x]=f[x]+λ(∫_a^b)Γ[x,t;λ]f[t]dt, where
Γ[x,t;λ]=(∑_(n=1)^∞)((Eig[x]*Eig[t])/((Subscript[λ, n])-λ))
so,
u[x]=Cos[2x]+2(∫_0^(π/2))(∑_(n=1)^∞)(((4/π)Sin[2n x]Sin[2n t])/((4(n^2)-1)-2))Cos[2t]dt

d)

since Cos[2x] is orthogonal to all the eigenfunctions(Sin[2 n x]), there exists only one unique solution

3.7

a)

"collocation method on u[x]=x+(∫_(-1)^1)x*t*u[t]\[DifferentialD]t"
"find an approx solution in terms of (Subscript[ϕ, 1])=1, (Subscript[ϕ, 2])=Sin[x], and (Subscript[ϕ, 3])=Cos[x]"

so our solution becomes a linear combination of these ϕ's
(Subscript[u, s])[x]=(∑_(n=1)^3)(Subscript[c, n])(Subscript[ϕ, n])=(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Cos[x]
and plugging this into our original equation gives us
(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Cos[x] = x+(∫_(-1)^1)x t ((Subscript[c, 1])+(Subscript[c, 2])Sin[t]+(Subscript[c, 3])Cos[t])dt+ϵ(x,(Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3]))
or
(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Cos[x] = x+x(∫_(-1)^1) t ((Subscript[c, 1])+(Subscript[c, 2])Sin[t]+(Subscript[c, 3])Cos[t])dt+ϵ(x,(Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3]))
then executing the integral
(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Cos[x] = x+x((∫_(-1)^1) (Subscript[c, 1])tdt+(Subscript[c, 2])(∫_(-1)^1)Sin[t]dt+(Subscript[c, 3])(∫_(-1)^1)Cos[t]dt)+ϵ(x,(Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3]))
(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Cos[x] = x+x 2(Subscript[c, 3])Sin[1]+ϵ(x,(Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3]))
we'll say that the error vanishes at (Subscript[x, 1])=0, (Subscript[x, 2])=π/2, and  (Subscript[x, 3])=π
so, we get the following system of equations:
for (Subscript[x, 1]),  (Subscript[c, 1])+(Subscript[c, 2])Sin[0]+(Subscript[c, 3])Cos[0] = (0)+(0) 2(Subscript[c, 3])Sin[1]
for (Subscript[x, 2]),  (Subscript[c, 1])+(Subscript[c, 2])Sin[π/2]+(Subscript[c, 3])Cos[π/2] = (π/2)+(π/2) 2(Subscript[c, 3])Sin[1]
for (Subscript[x, 3]),  (Subscript[c, 1])+(Subscript[c, 2])Sin[π]+(Subscript[c, 3])Cos[π] = (π)+(π) 2(Subscript[c, 3])Sin[1]
or simplified,
for (Subscript[x, 1]),  (Subscript[c, 1])+(Subscript[c, 3]) = 0   ->  (Subscript[c, 1])=-(Subscript[c, 3])
for (Subscript[x, 2]),  (Subscript[c, 1])+(Subscript[c, 2]) = (π/2)+π (Subscript[c, 3])Sin[1]
for (Subscript[x, 3]),  (Subscript[c, 1])-(Subscript[c, 3]) = π+2π (Subscript[c, 3])Sin[1]
and then a system of equations gives us (Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3])
(Subscript[c, 1])=(π/(2+3 π Sin[1])), (Subscript[c, 2])=((-2 π+(π^2) Sin[1])/(2 (2+3 π Sin[1]))),   (Subscript[c, 3])= -(π/(2+3 π Sin[1]))
which makes our solution
(Subscript[u, s])[x]=(π/(2+3 π Sin[1]))+((-2 π+(π^2) Sin[1])/(2 (2+3 π Sin[1])))Sin[x]+-(π/(2+3 π Sin[1]))Cos[x]

"and in mathematica";
varlist=Table[(Subscript[c, i]),{i,3}]
zeroptlist=Table[i*π/2,{i,0,2}]
soleqn[x_]=(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Cos[x] 
eqn[x_]:={soleqn[x]==(∫_(-1)^1)x t soleqn[t]\[DifferentialD]t}
eqsnlist=Table[eqn[zeroptlist[[i]]][[1]],{i,1,3}]
Solve[eqsnlist,varlist]

"find an approx solution in terms of ϕ=1,Sin[x],Sin[2x],Sin[3x],Sin[4x],Cos[x],Cos[2x],Cos[3x]"

so our solution becomes a linear combination of these ϕ's.
(Subscript[u, s])[x]=(∑_(n=1)^8)(Subscript[c, n])(Subscript[ϕ, n])=(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Sin[2x]+(Subscript[c, 4])Sin[3x]+(Subscript[c, 5])Sin[4x]+(Subscript[c, 6])Cos[x]+(Subscript[c, 7])Cos[2x]+(Subscript[c, 8])Cos[3x]
so our integral equation becomes
(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Sin[2x]+(Subscript[c, 4])Sin[3x]+(Subscript[c, 5])Sin[4x]+(Subscript[c, 6])Cos[x]+(Subscript[c, 7])Cos[2x]+(Subscript[c, 8])Cos[3x]=(∫_(-1)^1)x*t*((Subscript[c, 1])+(Subscript[c, 2])Sin[t]+(Subscript[c, 3])Sin[2t]+(Subscript[c, 4])Sin[3t]+(Subscript[c, 5])Sin[4t]+(Subscript[c, 6])Cos[t]+(Subscript[c, 7])Cos[2t]+(Subscript[c, 8])Cos[3t])dt+ϵ(x,(Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3]),(Subscript[c, 4]),(Subscript[c, 5]),(Subscript[c, 6]),(Subscript[c, 7]),(Subscript[c, 8]))
executing the integral gives us
(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Sin[2x]+(Subscript[c, 4])Sin[3x]+(Subscript[c, 5])Sin[4x]+(Subscript[c, 6])Cos[x]+(Subscript[c, 7])Cos[2x]+(Subscript[c, 8])Cos[3x]=(-2 x (Cos[1]-Sin[1]) (Subscript[c, 2])+(1/2) x (-2 Cos[2]+Sin[2]) (Subscript[c, 3])+(2/9) x (-3 Cos[3]+Sin[3]) (Subscript[c, 4])+(1/8) x (-4 Cos[4]+Sin[4]) (Subscript[c, 5]))+ϵ(x,(Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3]),(Subscript[c, 4]),(Subscript[c, 5]),(Subscript[c, 6]),(Subscript[c, 7]),(Subscript[c, 8]))
for kicks and giggles, we'll say that the error is zero at a bunch of points that turn our sine's and cosine's to zeros
then, feeding it to mathematica,

list=Table[i*π/4,{i,1,16,2}]
eqn[x_]={(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Sin[2x]+(Subscript[c, 4])Sin[3x]+(Subscript[c, 5])Sin[4x]+(Subscript[c, 6])Cos[x]+(Subscript[c, 7])Cos[2x]+(Subscript[c, 8])Cos[3x]==(-2 x (Cos[1]-Sin[1]) (Subscript[c, 2])+(1/2) x (-2 Cos[2]+Sin[2]) (Subscript[c, 3])+(2/9) x (-3 Cos[3]+Sin[3]) (Subscript[c, 4])+(1/8) x (-4 Cos[4]+Sin[4]) (Subscript[c, 5]))}
eqnlist=Table[eqn[list[[i]]][[1]],{i,1,8}];
varlist=Table[(Subscript[c, i]),{i,1,8}];
varsol=Solve[eqnlist,varlist][[1]];
Sol=(Subscript[c, 1])+(Subscript[c, 2])Sin[x]+(Subscript[c, 3])Sin[2x]+(Subscript[c, 4])Sin[3x]+(Subscript[c, 5])Sin[4x]+(Subscript[c, 6])Cos[x]+(Subscript[c, 7])Cos[2x]+(Subscript[c, 8])Cos[3x]
N[Sol/.varsol]

"well, it sorta solves it, but i can't figure out whats wrong.  i understand the method though, i think.  inserting that error function, then saying it is zero at a few different points gives us some extra information thats needed to solve for the coefficients."

3.9

"Galerkin method to find the approx soln to:"
"u[x]=x+(∫_(-1)^1)x t u[t]\[DifferentialD]t"

a)

ϕ[x_]={1,Sin[x],Cos[x]};
n=Length[ϕ[x]];
coeff=Table[(Subscript[c, i]),{i,1,n}];
(Subscript[u, s])[x_]=Sum[ϕ[x][[i]]*coeff[[i]],{i,1,n}]
errorfn[x_]=(Subscript[u, s])[x]-x+(∫_(-1)^1)x t (Subscript[u, s])[t]\[DifferentialD]t;  Print["ϵ(x,(Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3])) is:  ",errorfn[x]]
"error must be orthogonal to n linearly independent functions, (Subscript[ψ, 1]),(Subscript[ψ, 2])...(Subscript[ψ, n])";
ψlist=Table[Integrate[ϕ[x][[i]]*errorfn[x],{x,-1,1}]==Integrate[ϕ[x][[i]]*x,{x,-1,1}],{i,1,n}];Print["ψ equations are:  ",TableForm[ψlist]]
Print["i can use these ψ's to solve for our (Subscript[c, i])'s"]
sol=Solve[ψlist,coeff][[1]]
solution=(Subscript[u, s])[x]/.sol;
Print["the solution then becomes:  (Subscript[u, s])[x]=",solution]
gr1=Plot[solution,{x,0,4}]

"again for 8 functions"
ϕ[x_]={1,Sin[x],Cos[x],Sin[2x],Cos[2x],Sin[3x],Cos[3x],Sin[4x]}
n=Length[ϕ[x]];
coeff=Table[(Subscript[c, i]),{i,1,n}];
(Subscript[u, s])[x_]=Sum[ϕ[x][[i]]*coeff[[i]],{i,1,n}]
errorfn[x_]=(Subscript[u, s])[x]-x+(∫_(-1)^1)x t (Subscript[u, s])[t]\[DifferentialD]t;  Print["ϵ(x,(Subscript[c, 1]),(Subscript[c, 2]),(Subscript[c, 3])) is:  ",errorfn[x]]
"error must be orthogonal to n linearly independent functions, (Subscript[ψ, 1]),(Subscript[ψ, 2])...(Subscript[ψ, n])";
ψlist=Table[Integrate[ϕ[x][[i]]*errorfn[x],{x,-1,1}]==Integrate[ϕ[x][[i]]*x,{x,-1,1}],{i,1,n}];Print["ψ equations are:  ",TableForm[ψlist]]
Print["i can use these ψ's to solve for our (Subscript[c, i])'s"]
sol=NSolve[ψlist,coeff][[1]]
solution2=(Subscript[u, s])[x]/.sol;
Print["the solution then becomes:  (Subscript[u, s])[x]=",solution2]

gr2=Plot[solution2,{x,0,4}];

{Plot[3x,{x,0,4}],gr1,gr2}

4.3

a)

"Show that   f[x]=(∫_a^b)K[x,t]u[t]\[DifferentialD]t  with kernel K[x,t]=(∑_(k = 1)^n)(Subscript[a, k])[x](Subscript[b, k])[t]"
"only has a solution if f[x] is in the space of (Subscript[a, k])[x]"

by theorem 7, there exists a unique solution to that equation only if (∑_(n=1)^∞)|(Subscript[λ, n])(Subscript[a, n])((|)^2) converges to infinity, where (Subscript[λ, n]) are the eigenvalues of the kernel, and (Subscript[a, n]) are the fourier coefficients of the function f[x].  the act of coverting something into it into a linear combination of (Subscript[a, k]) requires the converted thing to bein that space to begin with, otherwise information will be lost and the linear combination will be incomplete.  an analogy would be to describe a 3D sphere in 2d space.  the linear combination of x and y cannot be combined to form the third "missing" dimension, z.  yay for linear independence!

b)

not necessarily.  continous-ness indicates whether something is stable or not.  the continuous-ness of the input has no effect on the continuous-ness of the output.
last paragraph on page 281: "the fourier coefficients of the increasing factor (Subscript[λ, n]) which clearly will cause δu(x), is magnified by the multiplicative increasing factor (Subscript[λ, n]) which clearly will cause δu[x] to be a larger change compared to the given small change δf[x].  Hence, there seems to be no continuous dependence of the solution u[x] on the given data f[x] in the eqn (5.97).

c)

as there exists a solution, then eqn 5.101 (∑_(n=1)^∞)|(Subscript[λ, n])(Subscript[a, n])((|)^2)  must converge.  in order for this to be the case, each (Subscript[a, n+1]) must be smaller than it's preceding term (Subscript[a, n]).  this indicates that if the (Subscript[λ, n]) terms grow, then the (Subscript[a, n]) terms must shrink faster so that each term in the summation is smaller than the last.

4.4

in theory, our solution can only be made up of eigenfunctions of the kernel in order for it to work, i.e. our solution exists in the eigenfunction space dictated by the kernel, not outside it.  thus the solution is restricted.
we want to show that the integral equation f[x]=(∫_a^b)K[x,t]u[t]dt becomes solvable only for a restricted class of functions regardless of u[x]
start with the equation f[x]=(∫_a^b)K[x,t]u[t]dt
and make the substitution  K[x,t]=(∑_(i=1)^n)(Subscript[c, i])[t](Subscript[ϕ, i])[t]
giving us f[x]=(∫_a^b)(∑_(i=1)^n)(Subscript[c, i])[t](Subscript[ϕ, i])[t]u[t]dt
rearrainging that gets us f[x]=(∑_(i=1)^n)(∫_a^b)(Subscript[c, i])[t](Subscript[ϕ, i])[t]u[t]dt
after executing the integral, we get f[x]=(∫_a^b)(∑_(i=1)^n)(Subscript[c, i])[t](Subscript[ϕ, i])[t]u[t]dt  hrm...   not sure what the hint meant.

(∑_(i=1)^n)(∫_a^b)((Subscript[c, i])[t](Subscript[ϕ, i])[t]u[t])\[DifferentialD]t

4.5

"f[x]=(∫_0^(2  π))Sin[x+t]u[t]\[DifferentialD]t"

a) i)

"find the eigenvalues and eigenfunctions of the kernel Sin[x+t]"

from example 4 on page 221, we know that the eigenfunctions of this probaem are Sin[x]+Cos[x] and Sin[x]-Cos[x], with eigenvalues 1/π and -1/π

a) ii)

first, we plug in the kernel, and separate it out into  a function of x multiplied by a function of t
f[x]=(∫_a^b)K[x,t]u[t]dt
f[x]=(∫_a^b)Sin[x+t]u[t]dt
f[x]=(∫_a^b)(Sin[x]Cos[t]+Cos[x]Sin[t])u[t]dt
f[x]=(∫_a^b)Sin[x]Cos[t]u[t]dt+(∫_a^b)Cos[x]Sin[t]u[t]dt
f[x]=Sin[x](∫_a^b)Cos[t]u[t]dt+Cos[x](∫_a^b)Sin[t]u[t]dtin this final equation,both integrals are now only a function of t,which when integrated,is replaced by constants.so,each integral amounts to a constant multiplied by a sine or a cosine,so our final f[x] must be a combination of sines and cosines.

b)

since sines and cosines are orthogonal, they can be used to describe any other function.  so when an additional function is added to our fredholm equation, all this amounts to is altering the constants on our final solution of  A sin[x] + B cos[x]

c)

when the function f[x] cannot be described in the space of the eigenfunctions of the kernel.