$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ##6.A. Linear second order ODEs ----------- > 1. [Introduction](#sect-6.A.1) > 2. [Cauchy problem (aka IVP)](#sect-6.A.2) > 3. [BVP](#sect-6.A.3) > 4. [BVP.II](#sect-6.A.4) ###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\_1Definition 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. --------------- [$\Uparrow$](../contents.html)  [$\uparrow$](./S6.1.html)  [$\Rightarrow$](./S6.2.html)