6.A. Linear second order ODEs

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6.A. Linear second order ODEs


  1. Introduction
  2. Cauchy problem (aka IVP)
  3. BVP
  4. BVP.II

Introduction

This is not a required reading but at some moment you would like to see how problems we discuss here for PDEs are solved for ODEs (consider it as a toy-model)

We consider ODE \begin{equation} Ly:=y'' + a_1(x)y + a_2(x)y'=f(x). \label{eq-6.A.1} \end{equation} Let $\{y_1(x),y_2(x)\}$ be a fundamental system of solutions of the corresponding homogeneous equation \begin{equation} Ly:=y'' + a_1(x)y + a_2(x)y'=0). \label{eq-6.A.2} \end{equation} Recall that then Wronskian \begin{equation} W(y_1,y_2; x):= \left| \begin{matrix} y_1(x) & y_2(x)\\ y'_1(x) &y'_2(x)\end{matrix}\right| \label{eq-6.A.3} \end{equation} does not vanish.

Cauchy problem (aka IVP)

Consider equation (\ref{eq-6.A.1}) with the initial conditions \begin{equation} y(x_0)=b_1, \qquad y'(x_0)=b_2. \label{eq-6.A.4} \end{equation} Without any loss of the generality one can assume that \begin{equation} \begin{aligned} &y_1(x_0)=1, &&y'_1(x_0)=0,\\ &y_2(x_0)=0, &&y'_1(x_0)=1. \end{aligned} \label{eq-6.A.5} \end{equation} Indeed, replacing $\{y_1(x),y_2(x)\}$ by $\{z_1(x),z_2(x)\}$ with $z_j=\alpha_{j1}y_1+\alpha_{j2}y_2$ we reach (\ref{eq-6.A.5}) by solving the systems \begin{align*} &\alpha_{11}y_1(x_0)+\alpha_{12}y_2(x_0)=1, &&& &\alpha_{21}y_1(x_0)+\alpha_{22}y_2(x_0)=0\\ &\alpha_{11}y'_1(x_0)+\alpha_{12}y'_2(x_0)=0, &&& &\alpha_{21}y'_1(x_0)+\alpha_{22}y'_2(x_0)=1 \end{align*} which have unique solutions because $W(y_1,y_2;x_0)\ne 0$.

Then the general solution to (\ref{eq-6.A.2}) is $y=C_1y_1+C_2y_2$ with constants $C_1,C_2$. To find the general solution to (\ref{eq-6.A.1}) we apply method of variations of parameters; then \begin{equation} \begin{aligned} &C'_1y_1+C'_2y_2=0,\\ &C'_1y'_1+C'_2y'_2=f(x) \end{aligned} \label{eq-6.A.6} \end{equation} and then \begin{equation} C'_1= -\frac{1}{W} y_2f,\qquad C'_2= \frac{1}{W} y_12f \label{eq-6.A.7} \end{equation} and \begin{equation} \begin{aligned} &C_1(x)= -\int _{x_0}^x \frac{1}{W(x')} y_2(x')f(x')\,dx'+c_1,\\ &C_2(x)= \ \ \int _{x_0}^x \frac{1}{W(x')} y_1(x')f(x')\,dx'+c_2 \end{aligned} \label{eq-6.A.8} \end{equation} and \begin{equation} y(x)=\int _{x_0}^x G(x;x')f(x')\,dx'+b_1y_1(x)+b_2y_2(x) \label{eq-6.A.9} \end{equation} with \begin{equation} G(x;x')=\frac{1}{W(x')} \bigl(y_2(x)y_1(x')-y_1(x)y_2(x')\bigr) \label{eq-6.A.10} \end{equation} and $c_1=b_1$, $c_2=b_2$ found from initial data.

Definition 1. $G(x,x')$ is a Green function (called in the case of IVP also Cauchy function).

This formula (\ref{eq-6.A.9}) could be rewritten as \begin{equation} y(x)=\int _{x_0}^x G(x;x')f(x')\,dx'+G'_x(x;x_0)b_1 +G(x;x_0)b_2. \label{eq-6.A.11} \end{equation}

BVP

Consider equation (\ref{eq-6.A.1}) with the boundary conditions \begin{equation} y(x_1)=b_1,\qquad y(x_2)=b_2 \label{eq-6.A.12} \end{equation} where $x_1< x_2$ are the ends of the segment $[x_1,x_2]$.

Consider first homogeneous equation (\ref{eq-6.A.2}); then $y=c_1y_1+c_2y_2$ and (\ref{eq-6.A.12}) becomes \begin{equation*} \begin{aligned} &c_1y_1(x_1)+c_2y_2(x_1)=b_1,\\ &c_1y_1(x_2)+c_2y_2(x_2)=b_2 \end{aligned} \end{equation*} and this system is solvable for any $b_1,b_2$ and this solution is unique if and only if determinant is not $0$: \begin{equation} \left|\begin{matrix} y_1(x_1) & y_2(x_1)\\y_1(x_2) & y_2(x_2)\end{matrix}\right|\ne 0. \label{eq-6.A.13} \end{equation} Assume that this condition is fulfilled. Then without any loss of the generality one can assume that \begin{equation} y_1(x_1)=1,\quad y_1(x_2)=0, \quad y_2(x_1)=0,\quad y_2(x_2)=1; \label{eq-6.A.14} \end{equation} otherwise as before we can replace them by their linear combinations. Consider inhomogeneous equation. Solving it by method of variations of parameters we have again (\ref{eq-6.A.7}) but its solution we write in a form slightly different from (\ref{eq-6.A.8}) \begin{equation} \begin{aligned} &C_1(x)= -\int _{x_1}^x \frac{1}{W(x')} y_2(x')f(x')\,dx'+c_1,\\ &C_2(x)= -\int _x^{x_2} \frac{1}{W(x')} y_1(x')f(x')\,dx'+c_2. \end{aligned} \label{eq-6.A.15} \end{equation} Then \begin{equation} y(x)=\int _{x_1}^{x_2} G(x;x')f(x')\,dx'+c_1y_1(x)+c_2y_2(x) \label{eq-6.A.16} \end{equation} where \begin{equation} G(x;x')=-\frac{1}{W(x')} \left\{\begin{aligned} &y_2(x')y_1(x) && x_1<x'<x,\\ &y_1(x')y_2(x)&& x<x'<x_2. \end{aligned}\right. \label{eq-6.A.17} \end{equation} From boundary conditions one can check easily that $c_1=b_1$, $c_2=b_2$. One can also $y_1(x)= -G'_{x'}(x;x')|_{x'=x_1}$, $y_2(x)=-G'_{x'}(x;x')|_{x'=x_2}$ and therefore \begin{multline} y(x)=\int _{x_1}^{x_2} G(x;x')f(x')\,dx'\\ -G'_{x'}(x;x')|_{x'=x_1} b_1+ G'_{x'}(x;x')|_{x'=x_2} b_2.\qquad\qquad \label{eq-6.A.18} \end{multline}

Definition 2. $G(x,x')$ is a Green function.

BVP. II

Assume now that (\ref{eq-6.A.13}) is violated. Then we cannot expect that the problem is uniquely solvable but let us salvage what we can. Without any loss of the generality we can assume now that \begin{equation} y_2(x_1)= y_2(x_2)=0; \label{eq-6.A.19} \end{equation} Using for a solution the same formulae (\ref{eq-6.A.8}), (\ref{eq-6.A.10}) but with $x_0$ replaced by $x_1$, plugging into boundary conditions and using (\ref{eq-6.A.19}) we have \begin{equation*} c_1y_1(x_1)=b_1, \qquad\bigl(c_1-\int_{x_1}^{x_2}\frac{1}{W(x')}y_2(x')\,dx'\bigr)y_1(x_2)=b_2 \end{equation*} which could be satisfied if and only iff \begin{equation} \int_{x_1}^{x_2}\frac{1}{W(x')}y_2(x')\,dx'-\frac{b_1}{y_1(x_1)}+\frac{b_2}{y_1(x_2)}=0 \label{eq-6.A.20} \end{equation} but solution is not unique: it is defined modulo $c_2 y_2(x)$.

Remark 1. More general boundary conditions \begin{equation} \alpha_1 y'(x_1)+\beta_1y(x_1)=b_1,\qquad \alpha_2 y'(x_2)+\beta_2y(x_2)=b_2 \label{eq-6.A.21} \end{equation} could be analyzed in a similar way.


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