01. Second Order Linear Boundary Value Problems

1.1 Basic Notation and Concepts

Second-Order Linear ODE

$$L[y(x)]=f(x)$$

where $f$ is a forcing function and $L$ is a linear differential operator that is

$$\begin{array}{rl}L[y(x)]& \equiv P_2(x)y^{\prime \prime }(x)+P_1(x)y^{\prime }(x)+P_0(x)y(x)\\ & \equiv P_2(x)\frac{d^{2}y}{d{x}^2}+P_1(x)\frac{dy}{dx}+P_0(x)y(x)\end{array}$$

for some given coefficients $P_2(x),P_1(x)$ and $P_0(x)$

$L$ is linear

$$L[{\alpha }_1{y}_1(x)+{\alpha }_2{y}_2(x)]\equiv a_{1}L[y_1(x)]+{\alpha }_{2}L(y_2(x))$$

for some constants ${\alpha }_i$ and functions $y_i(x)$

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Homogeneous and Inhomogeneous Versions of an ODE

Let $L$ be a linear differential operator then

Homogeneous

$$L[y]=0$$

Inhomogeneous

$$L[y]=f\not\equiv 0$$

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Boundary Value Problem of a Second-Order ODE

Generally to have two boundary conditions for a unique solution

Boundary Value Problem refers to the way in how the boundary conditions are imposed so
ODE is posed on an interval, say $a<x<b$ and boundary conditions involve

$$y\text{ and/or }{y}^{\prime }\text{ at both ends of domain }x=a\text{ and }x=b$$

provided coefficients $P_i(x)$ and $f(x)$ are sufficiently well behaved and $P_2(x)\ne 0$

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1.2 Space of Solutions

Space of Solutions to an ODE

Ignoring the Boundary Conditions

Let $(H)$ denote the Homogenous Equation ($Ly=0$)
Let $(N)$ denote the Inhomogeneous Equation ($Ly=f$)

The following properties of the solutions of $(H)$ and $(N)$ are

1. Solutions of $(H)$ form a vector space since
   If $L[y_1]+L[y_2]=0$ then

$$L[\alpha y_1+\beta y_2]=0$$

2. If $y_1$ and $y_2$ satisfy $(N)$ then $y_1-y_2$ satisfy $(H)$

3. It follows that the general solution of $(N)$ may be written in the form

$$y(x)=\underset{\text{any solution of }(N)}{\underbrace{y_{\text{PI}}(x)}}+\underset{\text{general solution of }(H)}{\underbrace{y_{\text{CF}}(x)}}$$

where $y_{\text{ PI }}(x)$ is called the particular integral and $y_{\text{CF}}(x)$ the complementary function

4. For a Second-Order ODE , then the vector space of solutions to $(H)$ has dimension $2$
   Complementary Function then takes form

$$y_{\text{CF}}(x)=c_1{y}_1(x)+c_2{y}_2(x)$$

where $c_1$ and $c_2$ are arbitrary constants, and $y_1,y_2$ are two linearly independent solutions to $(H)$

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1.3 Linear Independence / The Wronskian

Linear Independence

Let $y_1(x),y_2(x)$ be a pair of functions

$y_1(x),y_2(x)$ is linearly independent if

$$c_1{y}_1(x)+c_2{y}_2(x)\equiv 0⟺c_1=c_2=0$$

(that is there is no non-trivial linear combination that vanishes identically)

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Linear Dependence

Let $y_1(x),y_2(x)$ be a pair of functions then

If $c_i$ are not both zero then there exists combination such that

$$c_1{y}_1(x)+c_2{y}_2(x)\equiv 0$$

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Wronskian

Let $y_1(x),y_2(x)$ be a pair of functions then

$$W(x)=W[y_1,y_2]=\det \left(\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right)=y_1(x)y_2^{\prime }(x)-y_2(x)y_1^{\prime }(x)$$

Proof - Logic for it $y_1,y_2$ are linearly dependent then there exists $c_1,c_2$ such that

If

$$c_1{y}_1(x)+c_2{y}_2(x)\equiv 0$$

So also $c_1{y}_1^{\prime }(x)+c_2{y}_2^{\prime }(x)\equiv 0$ hence

$$\left(\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)\equiv 0$$

(assuming $y_1,y_2$ are differentiable)

Thus there exists non-trivial solutions $(c_1,c_2)$ if and only if determinant of matrix is zero
Which leads to the definition of the Wronskian

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Wronskian of Linearly Dependent Functions

If two functions are linearly dependent then Wronskian vanishes

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Wronskian is identically zero or never zero for Homogenous Second Order Linear ODEs

Let $y_1,y_2$ be two solutions to $(H)$

If $W=0$ at one point then $W\equiv 0$ everywhere
Conversely of $W\ne 0$ at one point then $W\ne 0$ everywhere

Proof

Let $y_1$ and $y_2$ be two solution to $(H)$ so

$$\begin{array}{rl}{P}_2{y}_1^{\prime \prime }+P_1{y}_1^{\prime }+P_0{y}_1& =0\\ P_2{y}_2^{\prime \prime }+P_1{y}_2^{\prime }+P_0{y}_2& =0\end{array}$$

Eliminating $P_0$ term then

$$P_2(y_1{y}_2^{\prime \prime }-y_2{y}_1^{\prime \prime })+P_1(y_1{y}_2^{\prime }-y_2{y}_1^{\prime })=0$$

So

$$P_2\frac{dW}{dx}+P_{1}W=0$$

Assuming that $P_2$ is not zero anywhere then

$$W(x)=\text{const}\cdot \exp \left(-\int \frac{P_1(x)}{P_2(x)}dx\right)$$

Since the exponential can’t vanish then result follows

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Relation between Wronskian and Linearly Dependence of Solution to a Homogenous Second-Order ODE

Let $(H)$ denote the Homogenous Equation ($Ly=0$)
Let $y_1(x),y_2(x)$ be two solutions of a given homogeneous second-order ODE $(H)$

$y_1$ and $y_2$ are linearly dependent if and only if their Wronskian is zero

Proof

Suppose $y_1$ and $y_2$ are two solutions of $(H)$

If they are linearly dependent then their Wronskian is zero

So instead suppose that their Wronskian is zero (everywhere)

If $y_1$ is the zero function then $y_1$ and $y_2$ are certainly linearly dependent
Otherwise assume there exists at least one value $x=a$ such that $y_1(a)\ne 0$

Let $u$ be such that $y_2(a)=\mu y_1(a)$ then

$$0=W(a)=y_1(a)y_2^{\prime }(a)-y_2(a)y_1^{\prime }(a)=y_1(a)(y_2^{\prime }(a)-\mu y_1^{\prime }(a))$$

Since $y_1(a)\ne 0$ by assumption then

$$y_2^{\prime }(a)=\mu y_1^{\prime }(a)$$

Define

$$y(x)=y_2(x)-\mu y_1(x)$$

By linearity then $y(x)$ is a solution of $(H)$
So $y(x)$ satisfies the initial conditions (by construction from above)

$$y(a)=0=y^{\prime }(a)$$

So by uniqueness of solution of $(H)$ (by Picard’s Theorem, assuming $P_2\ne 0$)
Thus

$$y(x)\equiv 0$$

as it satisfies the initial conditions and by uniqueness it must be the only solution

Hence $y_1$ and $y_2$ are linearly dependent

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1.4 A Basis of Solutions to $(H)$

Basis of Solutions to $(H)$

Let $(H)$ denote the Homogenous Equation ($Ly=0$)

Let $y_1$ and $y_2$ be two particular solutions of $(H)$ satisfying initial conditions at some $x=a$

$$y_1(a)=1,y_1^{\prime }(a)=0,y_2(a)=0,y_2^{\prime }(a)=1$$

Then if $y(x)$ is a solution of $(H)$ then

$$y(x)\text{ is a linear combination of }{y}_1\text{ and }{y}_2$$

Proof

By Picard’s Theorem both $y_1(x)$ and $y_2(x)$ exist
Both are unique at least in a neighbourhood of $x=a$ provided $P_2(a)\ne 0$

Wronskian has $W=1$ at $x=a$ so is non-zero in the same neighbourhood of $x=a$
Hence they are linearly independent

Suppose $y(x)$ is any other solution of $(H)$ so set

$$Y(x)=y_1(x)y(a)+y_2(x)y^{\prime }(a)$$

So $Y(x)$ is a solution of $(H)$ and satisfies initial conditions

$$Y(a)=y(a),Y^{\prime }(a)=y^{\prime }(a)$$

By uniqueness then $Y(x)\equiv y(x)$ so $y(x)$ is a linear combination of $y_1$ and $y_2$
Hence $y_1$ and $y_2$ spans the vector space of solution $Y$, hence a basis

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Properties of Space of Solutions of $(H)$

1. Dimension of the space of solutions of $(H)$ is $2$

2. Any pair of solutions of $(H)$ with $W\ne 0$ is a basis

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1.5 Solution Methods for Homogenous Problems

1.5.1 Constant Coefficients

Example

Let $(H)$ denote the Homogenous Equation ($Ly=0$)
If $P_2,P_1,P_0$ are constant then $(H)$ has exponential solutions in form of

$$y=e^{mx}$$

where $m$ satisfies quadratic equation (also known as the auxiliary equation)

$$P_2{m}^2+P_{1}m+P_0=0$$

If roots are repeated or complex then care must be taken

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1.5.2 Cauchy-Euler Equation

Example

Coefficients are in the form of

$$P_2(x)=\alpha x^2,P_1(x)=\beta x,P_0(x)=\gamma$$

where $\alpha ,\beta ,\gamma$ are constants

So $(H)$ is in form

$$\alpha x^2\frac{d^{2}y}{d{x}^2}+\beta x\frac{dy}{dx}+\gamma y=0$$

Note that the powers of each for each term is the same

Hence solutions are in the form

$$y(x)=x^m$$

where $m$ satisfies quadratic equation

$$\alpha m(m-1)+\beta m+\gamma =0$$

If roots are repeated or complex then care must be taken

Alternate Substitution (Equivalent)

Substitute $x=e^t$ to get constant-coefficients equation

$$\alpha \frac{d^{2}y}{d{t}^2}+(\beta -\alpha )\frac{dy}{dt}+\gamma y=0$$

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1.5.3 Reduction of Order

Example

If one solution $y_1(x)$ is known then
General Solution can be found by solving an ODE of reduced order

Express solution to ODE $(H)$ in form of

$$y(x)=v(x)y_1(x)$$

Substituting into $(H)$ then using the fact that $y_1$ is a solution of $(H)$ obtain

$$P_2{y}_1{v}^{\prime \prime }+(2P_2{y}_1^{\prime }+P_1{y}_1)v^{\prime }=0$$

which is a separable first-order ODE for $v^{\prime }$ with solution

$$v^{\prime }(x)=\frac{\text{const}}{y_1(x{)}^2}\exp \left(-\int \frac{P_1(x)}{P_2(x)}dx\right)$$

Integrating once more gives $v(x)$ and hence general solution $y(x)=v(x)y_1(x)$

Alternative Derivation from Wronskian

Constructing the general solution from a single known solution can also be derived from Wronskian

$$\begin{array}{rl}W(x)& =y_1(x)y_2^{\prime }(x)-y_2(x)y_1^{\prime }(x)\\ & =y_1(x{)}^2\frac{d}{dx}\left(\frac{y_2(x)}{y_1(x)}\right)\\ & =\text{ const }\cdot \exp \left(-\int \frac{P_1(x)}{P_2(x)}dx\right)\end{array}$$

Hence we can construct $y_2(x)$ given $y_1(x)$

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1.6 Variation of Parameters

Variation of Parameters

Let $(N)$ denote the Inhomogeneous Equation ($Ly=f$)

Using Variation of Parameters then

$$y(x)=-{\int }^x\frac{f(\xi )y_2(\xi )y_1(x)}{P_2(\xi )W(\xi )}d\xi +{\int }^x\frac{f(\xi )y_1(\xi )y_2(x)}{P_2(\xi )W(\xi )}d\xi$$

Proof

General way to find the general solution for the inhomogeneous version $(N)$

Suppose $(H)$ is solved by

$$y(x)=c_1{y}_1(x)+c_2{y}_2(x)$$

where $y_1,y_2$ are linearly independent

Seeking a solution to $(N)$ in the form

$$y(x)=c_1(x)y_1(x)+c_2(x)y_2(x)$$

(that is we vary the parameter $c_1$ and $c_2$)

Differentiating

$$y^{\prime }=c_1{y}_1^{\prime }+c_2{y}_2^{\prime }+c_1^{\prime }{y}_1+c_2^{\prime }{y}_2$$

Eliminating the highest derivative of $c_i$, impose condition

$$c_1^{\prime }{y}_1+c_2^{\prime }{y}_2=0$$

on $c_1$ and $c_2$

Since two functions $c_1$ and $c_2$ define $y$, should be enough freedom to satisfy constraint

Under assumption then $y^{\prime }$ simplifies to

$$y^{\prime }=c_1{y}_1^{\prime }+c_2{y}_2^{\prime }$$

Differentiating and substituting into $(H)$ then

$$\begin{array}{rl}L[y]& =P_2(c_1{y}_1^{\prime \prime }+c_2{y}_2^{\prime \prime }+c_1^{\prime }{y}_1^{\prime }+c_2^{\prime }{y}_2^{\prime })+P_1(c_1{y}_1^{\prime }+c_2{y}_2^{\prime })+P_0(c_1{y}_1+c_2{y}_2)\\ & =c_{1}L[y_1]+c_{2}L[y_2]+P_2(c_1^{\prime }{y}_1^{\prime }+c_2^{\prime }{y}_2^{\prime })\end{array}$$

Since $y_i$ satisfy $(H)$ then inhomogeneous ODE $(N)$ becomes

$$L[y]=P_2(c_1^{\prime }{y}_1^{\prime }+c_2^{\prime }{y}_2^{\prime })=f$$

So then

$$\left(\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right)\left(\begin{array}{c}{c}_1^{\prime }\\ c_2^{\prime }\end{array}\right)=\left(\begin{array}{c}0\\ \frac{f}{P_2}\end{array}\right)$$

Determinant of matrix on left is Wronskian $W$ so it is non-zero (by linear independence)
By inverting then

$$\left(\begin{array}{c}{c}_1^{\prime }\\ c_2^{\prime }\end{array}\right)=\frac{1}{W}\left(\begin{array}{cc}{y}_2^{\prime }& -y_2\\ -y_1^{\prime }& y_1\end{array}\right)\left(\begin{array}{c}0\\ \frac{f}{P_2}\end{array}\right)=\frac{f}{P_{2}W}\left(\begin{array}{c}-y_2\\ y_1\end{array}\right)$$

Hence by integrating

$$c_1(x)=-{\int }^x\frac{f(\xi )y_2(\xi )}{P_2(\xi )W(\xi )}d\xi ,c_2(x)={\int }^x\frac{f(\xi )y_1(\xi )}{P_2(\xi )W(\xi )}d\xi$$

Hence by substitution

$$y(x)=-{\int }^x\frac{f(\xi )y_2(\xi )y_1(x)}{P_2(\xi )W(\xi )}d\xi +{\int }^x\frac{f(\xi )y_1(\xi )y_2(x)}{P_2(\xi )W(\xi )}d\xi$$

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Variation of Parameters - $c_i$ definition

$$c_1(x)=-{\int }^x\frac{f(\xi )y_2(\xi )}{P_2(\xi )W(\xi )}d\xi ,c_2(x)={\int }^x\frac{f(\xi )y_1(\xi )}{P_2(\xi )W(\xi )}d\xi$$

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1.7 Fitting Boundary Conditions

Green's Function

Let $y_1$ and $y_2$ be linearly independent solutions to the Homogenous ODE $Ly=0$ where

$$Ly\equiv P_2{y}^{\prime \prime }+P_1{y}^{\prime }+P_{0}y=fa<x<b$$

with $y_1$ and $y_2$ satisfying one boundary condition each ($y_1(a)=0=y_2(b)$)

Let Green’s Function $g(x,\xi )$ be defined as

$$g(x,\xi )=\{\begin{array}{ll}\frac{y_1(\xi )y_2(x)}{P_2(\xi )W(\xi )}& a<\xi <x<b\\ \frac{y_2(\xi )y_1(x)}{P_2(\xi )W(\xi )}& a<x<\xi <b\end{array}$$

Note that it provides a kind of inverse to $Ly=f$

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Boundary Conditions

Attempting Variation of Parameters to solve $(N)$ with homogenous boundary conditions

Consider BVP

$$P_2(x)y^{\prime \prime }(x)+P_1(x)y^{\prime }(x)+P_0(x)y(x)=f(x)a<x<b$$

with boundary data

$$y(a)=0,y(b)=0$$

Then

$$y(x)={\int }_a^x\frac{f(\xi )y_1(\xi )y_2(x)}{P_2(\xi )W(\xi )}d\xi +{\int }_x^b\frac{f(\xi )y_2(\xi )y_1(x)}{P_2(\xi )W(\xi )}d\xi$$

which can be concisely written as

$$y(x)={\int }_a^{b}g(x,\xi )f(\xi )d\xi$$

where $g(x,\xi )$ is Green’s Function

Proof

Use Variation of Parameters but with condition

$$y_1(a)=0\text{ and }{y}_2(b)=0$$

where $y_1$ and $y_2$ are the two basis solutions

Assuming they are linearly independent so that

$$W[y_1,y_2]\ne 0$$

Hence

$$y_1(b)\ne 0\text{ and }{y}_2(a)\ne 0$$

Thus solution takes form

$$y(x)=c_1(x)y_1(x)+c_2(x)y_2(x)$$

where $c_1(x)$ and $c_2(x)$ come from above

Hence

$$\begin{array}{rl}y(a)& =c_1(a)y_1(a)+c_2(a)y_2(a)=c_2(a)y_2(a)=0\\ y(b)& =c_1(b)y_1(b)+c_2(b)y_2(b)=c_1(b)y_1(b)=0\end{array}$$

Thus

$$c_2(a)=0=c_1(b)$$

Imposing conditions on the formulae for $c_i$ to get explicit unique forms

$$c_1(x)={\int }_x^b\frac{f(\xi )y_2(\xi )}{P_2(\xi )W(\xi )}d\xi ,c_2(x)={\int }_a^x\frac{f(\xi )y_1(\xi )}{P_2(\xi )W(\xi )}d\xi$$

Note that the limits in the integral of $c_1$ are switched

Hence the solution to the BVP can be written as

$$y(x)={\int }_a^x\frac{f(\xi )y_1(\xi )y_2(x)}{P_2(\xi )W(\xi )}d\xi +{\int }_x^b\frac{f(\xi )y_2(\xi )y_1(x)}{P_2(\xi )W(\xi )}d\xi$$

which can be concisely written as

$$y(x)={\int }_a^{b}g(x,\xi )f(\xi )d\xi$$

where $g(x,\xi )$ is Green’s Function

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