Theory
Power Series Solutions to 2nd order ODE’s
D^2y/dx^2 + a1(x) dy/dx + a2(x)*y =0
Y”+a1(x)y’+a2(x)y=0
Find y(x)
The coefficients a1(x),a2(x)
- clue you to the specific form of power series
- Assurance of existence
Theorem:
A1(x)] Nice, Well behaved functions
A2(x)] Not so nice functions
Nice: typical function that does not have asymptotes
Not nice: have asymptotes
If both a1(x) and a2(x) are nice
Then: the ODE has 2 linearly independent solutions of the form:
Y(x) = sum(An*x^n)
If a1(x) and/or a2(x) is not nice
a. If a1(x)*x and/or a2(x)*x^2 are nice
i. At least one solution exists with the form: y(x)=sum[An*x^(n+s)]
ii. A second one (Linearly independent) may or may not exist
b. If a1(x)*x and/or a2(x)*x^2 are not nice
i. No Assurances
Y”+y=0
A1(x)=0]  y(x) =C1*sum(An*x^n)+C2*sum(Bn*x^n)
A2(x)=1]
Y”+(1/x^2)*y=0
A1(x)=0
]
A2(x)=(1/x^2)] a2(x)*x^2=1 
y1(x)=sum[An*x^(n+s)], and possibly y2(x)=sum[Bn*x^(n+s)]
y”+(1/x^3)*y=0
a1(x)=0
a2(x)=(1/x^3)  a2(x)*x^2=1/x, no assurances
RELATED TO HW:
Y”+y=0
Y(x)=C1*sin(x)+C2*cos(x)
A1(x)=0]  y(x) =C1*sum(An*x^n)+C2*sum(Bn*x^n)
A2(x)=1]
Try y(x)=Ao+A1x+a2x^2+a3x^3+a4x^4+a5x^5+…
Y’(x)=A1+2A2x+3A2x^2+4A4x^3+5A5x^4+…
Y”(x)=2A2+6A3x+12A4x^2+20A5x^3+…
(2A2+6A3x+12A4x^2+20A5x^3+…) +(Ao+A1x+a2x^2+a3x^3+a4x^4+a5x^5+…)=0
Collect common terms
(2A2+ Ao)+( 6A3x+ A1x)+( 12A4x^2+ a2x^2)+( 20A5x^3+ a3x^3)+…=0
X is arbitrary (independent variable), all terms on left hand side need to equal 0
Hence: 2A2+ Ao=0, 6A3+ A1=0, 12A4+ a2=0, 20A5+ a3=0
A2=-Ao/2, A3=-A1/6, A4=-A2/12=Ao/24, A5=-A3/20=A1/120…
Since: y(x)=Ao+A1x+a2x^2+a3x^3+a4x^4+a5x^5+…, substitute  terms of Ao and A1
= Ao(1-1/2x^2+1/24x^4-…)+A1(x-1/6x^3+1/120x^5-…)
Ao=cos(x), A1=sin(x)
HW:
2x^2*y”-x*y’+(1+x)*y=0
Div by 2x^2: y”-{1/2x)y’+(1+x)/(2x^2)y=0
A1(x)=-1/2x
]  not nice
A2(x)=(1+x)/(2x^2) ]
X*a1(x)=-1/2
]  nice!!!, expect at least solution of the typical form
X^2*a2(x)=(1/x)/2 ]
Try y1(x)=sum[An*x^(n+s)]
Y’1(x)=sum[(n+s)*An*x^(x+s-1)]
Y”1(x)=sum[(n+s-1)*(n+s)*An*x&(n+s-2)]
2x^2y”=sum[2*(n+s-1)*(n+s)*An*x^(n+s)]
-xy’=-sum[(n+s)An*x^(n+s)]
(1+x)y=y+xy=sum[An*x^(n+s)]+sum[An*x^(n+s+1)]
Sum[2*(n+s+1)*(n+s)*An*x^(n+s)] –
sum[(n+s)*An*x^(n+s)]+sum[An*x^(n+s)]+sum[An*x^(n+s+1)]=0
Note: n is dummy index, replace n by n-1
sum[An*x^(n+s+1)]= sum[A(n-1)*x^(n+s)]
Sum[2*(n+s-1)*(n+s)*An*x^(n+s)] –
sum[(n+s)*An*x^(n+s)]+sum[An*x^(n+s)]+sum[An*x^(n+s)]=0
But sum[An*x^(n+s)]=Ao*x^s+sumn=1[An*x(n+s)]
Sum[2*(n+s-1)*(n+s)*An*x^(n+s)] = 2*(s-1)*(s)*Ao*x^(s)+ Sum[2*(n+s1)*(n+s)*An*x^(n+s)]
… collect all terms
Inditial equation: setting common terms equal to 0 to solve for s
S=1,.5
Two solutions are assured if two solutions are distinct and not an integer from eachother
of the following form
Y1(x)=sum[An*x^(n+s1)]
Y2(x)=sum[Bn*x^(n+s2)]
To find the coefficients An, need the recurrence relations
Collect common factors of x^(n+s)
Sum{[2*(n+s-1)*(n+s)+(n+s)+1]*An+A n=1}x^(n+s)=0
Since x is arbitrary (IV) then { }=0 for all n
An=( A n=1 )/ [2*(n+s-1)*(n+s)+(n+s)+1]*
Recurrence relation
… same for Bn
Y(x)= C1*sum[x^(n+s1)]+C2*sum[x^(n+s2)]
BVP and characteristic functions
Y”+k^2y=0, y(0)=0;y(1)=0, dirichlet type conditions
General solution y(x)=C1sin(kx) +C2cos(kx)
But cos(k0)=1C2=0 since (y(0)=0)
Since y(1)=0  C1sin(k)=0
C1~=0, sin(k)=0
K=pi,2pi,…
Kn=npi are the eigenvalues or characteristic values associated with the BVP
Sin(k*n*x)=sin(n*pi*x) are the eigen functions or characteristic functions associated
with the BVP, key property of eigenfunctions ORTHOGONALITY
Theorem
Let Qn(x) be a set of eigenfunctions associated with a particular BVP
Then, it is true that:
Integral[Qn(x)*Qm(x)W(x)dx]=0 when n~=m, n ~=0 when n=m
Here W(x) is called the weighing function (in the case of the BVP is the coefficient of
k^2y (i.e. W(x)=1))
Integral[Qn^2(x)W(x)dx]=L/2, norm of Qn(x)
Property of orthogonality of eigenfunctions
Representing arbitrary functions by means of linear combinations of eigenfunctions
Let Qn(x) be eigenfunctions
Let f(x) be arbitrary functions
F(x)=sum[An*Qn(x)]  Fourier
An=easily determined (orthogonality)
F(x)=sum[An*Qn(x)]
= Ao*Qo(x)+A1*Q1(x)+…
1. multiply LHS and RHS by Qm(x)*W(x)
Qm(x)*W(x)*f(x)=Qm(x)*W(x)*sum[An*Qn(x)]
2. Integrate LHS and RHS from 0 to L=1
Integral[Qm(x) *W(x)*f(x)*dx]= Integral{Qm(x) *W(x)*sum[An*Qn(x)]*dx}
3. Exchange the order of integration and summation on RHS
Integral[Qm(x) *W(x)*f(x)*dx]= Sum{An* Integral[Qm(x) *W(x)* dx]}
Integral[Qm(x) *W(x)*f(x)*dx]=Am*Integral[Qm(x)* Qm(x) *W(x)* dx]
Am= {Integral[Qm(x) *W(x)*f(x)*dx]}/ Integral[Qm^2(x) *W(x)* dx]
F(x)=sum[Am*Qm(x)]
Let f(x)=(1/20)*(x-x^5)
Let Qn(x) be eigenfunctions associated with BVP y”+k^2y=0, y(0)=0, y(1)=0
i.e. Qn(x) =sin(n*pi*x)
represent f(x) as a linear combination of Qn(x), find fourier coefficients AN such that
f(x)=sum[An*Qn(x)]
1. multiple LHS and RHS times sin(m*pi*x)
sin(m*pi*x)*f(x)= sin(m*pi*x)* sum[An*sin(n*pi*x)]
2. Integrate LHS and RHS from x=0 to x=1
integral[sin(m*pi*x)*f(x)*dx]=integral[sin(m*pi*x)* sum[An*sin(n*pi*x)]*dx]
3. Exchange integral and sum on RHS
integral[sin(m*pi*x)*f(x)*dx]=sum[An* integral[sin(m*pi*x)* sin(n*pi*x)*dx]
4. Orthogonality
integral[sin(m*pi*x)*f(x)*dx]=Am* integral[sin^2(m*pi*x)*dx]
Am={ integral[sin(m*pi*x)*f(x)*dx]}/{ integral[sin^2(m*pi*x)*dx]}