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4. Differential equations

4.1 Linear differential equations

4.1.1 First order linear DE

The general solution of a linear differential equation is given by

$y_{\rm A}=y_{\rm H}+y_{\rm P}$ , where

$y_{\rm H}$ is the solution of the homogeneous equation and

$y_{\rm P}$ is a particular solution.

A first order differential equation is given by: $y'(x)+a(x)y(x)=b(x)$ . Its homogeneous equation is $y'(x)+a(x)y(x)=0$ .

The solution of the homogeneous equation is given by $y_{\rm H}=k\exp\left(\int a(x)dx\right)$ Suppose that $a(x)=a=$ constant.

Substitution of $\exp(\lambda x)$ in the homogeneous equation leads to the characteristic equation $\lambda+a=0\Rightarrow\lambda=-a$ .

Suppose $b(x)=\alpha\exp(\mu x)$ . Than one can distinguish two cases:

$\lambda\neq\mu$ : a particular solution is: $y_{\rm P}=\exp(\mu x)$
$\lambda=\mu$ : a particular solution is: $y_{\rm P}=x\exp(\mu x)$

When a DE is solved by variation of parameters one writes: $y_{\rm P}(x)=y_{\rm H}(x)f(x)$ , and than one solves $f(x)$ from this.

4.1.2 Second order linear DE

A differential equation of the second order with constant coefficients is given by:

$y''(x)+ay'(x)+by(x)=c(x)$ . If

$c(x)=c=$ constant there exists a particular solution

$y_{\rm P}=c/b$ .

Substitution of $y=\exp(\lambda x)$ leads to the characteristic equation $\lambda^2+a\lambda+b=0$ .

There are now 2 possibilities:

$\lambda_1\neq\lambda_2$ : than $y_{\rm H}=\alpha\exp(\lambda_1 x)+\beta\exp(\lambda_2 x)$ .
$\lambda_1=\lambda_2=\lambda$ : than $y_{\rm H}=(\alpha +\beta x)\exp(\lambda x)$ .

If $c(x)=p(x)\exp(\mu x)$ where $p(x)$ is a polynomial there are 3 possibilities:

$\lambda_1,\lambda_2\neq\mu$ : $y_{\rm P}=q(x)\exp(\mu x)$ .
$\lambda_1=\mu,\lambda_2\neq\mu$ : $y_{\rm P}=xq(x)\exp(\mu x)$ .
$\lambda_1=\lambda_2=\mu$ : $y_{\rm P}=x^2q(x)\exp(\mu x)$ .

where

$q(x)$ is a polynomial of the same order as

$p(x)$ .

When: $y''(x)+\omega^2y(x)=\omega f(x)$ and $y(0)=y'(0)=0$ follows: $y(x)=\int\limits_0^xf(x)\sin(\omega(x-t))dt$ .

4.1.3 The Wronskian

We start with the LDE

$y''(x)+p(x)y'(x)+q(x)y(x)=0$ and the two initial conditions

$y(x_0)=K_0$ and

$y'(x_0)=K_1$ . When

$p(x)$ and

$q(x)$ are continuous on the open interval

$I$ there exists a unique solution

$y(x)$ on this interval.

The general solution can than be written as $y(x)=c_1y_1(x)+c_2y_2(x)$ and $y_1$ and $y_2$ are linear independent. These are also all solutions of the LDE.

The Wronskian is defined by: $W(y_1,y_2)= \left|\begin{array}{cc} y_1&y_2\\ y'_1&y'_2 \end{array}\right|=y_1y'_2-y_2y'_1$

$y_1$ and $y_2$ are linear independent if and only if on the interval $I$ when $\exists x_0\in I$ so that holds: $W(y_1(x_0),y_2(x_0))=0$ .

4.1.4 Power series substitution

When a series

$y=\sum a_nx^n$ is substituted in the LDE with constant coefficients

$y''(x)+py'(x)+qy(x)=0$ this leads to:

$\sum_n\left[n(n-1)a_nx^{n-2}+pna_nx^{n-1}+qa_nx^n\right]=0$ Setting coefficients for equal powers of

$x$ equal gives:

$(n+2)(n+1)a_{n+2}+p(n+1)a_{n+1}+qa_n=0$ This gives a general relation between the coefficients. Special cases are

$n=0,1,2$ .

4.2.1 Frobenius' method

Given the LDE

$\frac{d^2y(x)}{dx^2}+\frac{b(x)}{x}\frac{dy(x)}{dx}+\frac{c(x)}{x^2}y(x)=0$ with

$b(x)$ and

$c(x)$ analytical at

$x=0$ . This LDE has at least one solution of the form

$y_i(x)=x^{r_i}\sum_{n=0}^\infty a_nx^n~~~\mbox{with}~~i=1,2$ with

$r$ real or complex and chosen so that

$a_0\neq0$ . When one expands

$b(x)$ and

$c(x)$ as

$b(x)=b_0+b_1x+b_2x^2+...$ and

$c(x)=c_0+c_1x+c_2x^2+...$ , it follows for

$r$ :

$r^2+(b_0-1)r+c_0=0$ There are now 3 possibilities:

$r_1=r_2$ : than $y(x)=y_1(x)\ln|x|+y_2(x)$ .
$r_1-r_2\in I\hspace{-1mm}N$ : than $y(x)=ky_1(x)\ln|x|+y_2(x)$ .
$r_1-r_2\neq Z\hspace{-1ex}Z$ : than $y(x)=y_1(x)+y_2(x)$ .

4.2.2 Euler

Given the LDE

$x^2\frac{d^2y(x)}{dx^2}+ax\frac{dy(x)}{dx}+by(x)=0$ Substitution of

$y(x)=x^r$ gives an equation for

$r$ :

$r^2+(a-1)r+b=0$ . From this one gets two solutions

$r_1$ and

$r_2$ . There are now 2 possibilities:

$r_1\neq r_2$ : than $y(x)=C_1x^{r1}+C_2x^{r_2}$ .
$r_1=r_2=r$ : than $y(x)=(C_1\ln(x)+C_2)x^r$ .

4.2.3 Legendre's DE

Given the LDE

$(1-x^2)\frac{d^2y(x)}{dx^2}-2x\frac{dy(x)}{dx}+n(n-1)y(x)=0$ The solutions of this equation are given by

$y(x)=aP_n(x)+by_2(x)$ where the Legendre polynomials

$P(x)$ are defined by:

$P_n(x)=\frac{d^n}{dx^n}\left(\frac{(1-x^2)^n}{2^n n!}\right)$ For these holds:

$\|P_n\|^2=2/(2n+1)$ .

4.2.4 The associated Legendre equation

This equation follows from the

$\theta$ -dependent part of the wave equation

$\nabla^2\Psi=0$ by substitution of

$\xi=\cos(\theta)$ . Than follows:

$(1-\xi^2)\frac{d}{d\xi}\left((1-\xi^2)\frac{dP(\xi)}{d\xi}\right)+ [C(1-\xi^2)-m^2]P(\xi)=0$ Regular solutions exists only if

$C=l(l+1)$ . They are of the form:

$P_l^{|m|}(\xi)=(1-\xi^2)^{m/2}\frac{d^{|m|}P^0(\xi)}{d\xi^{|m|}}= \frac{(1-\xi^2)^{|m|/2}}{2^ll!}\frac{d^{|m|+l}}{d\xi^{|m|+l}}(\xi^2-1)^l$ For

$|m|>l$ is

$P_l^{|m|}(\xi)=0$ . Some properties of

$P_l^0(\xi)$ zijn:

$\int\limits_{-1}^1P_l^0(\xi)P_{l'}^0(\xi)d\xi=\frac{2}{2l+1}\delta_{ll'}~~~,~~~ \sum_{l=0}^\infty P_l^0(\xi)t^l=\frac{1}{\sqrt{1-2\xi t+t^2}}$ This polynomial can be written as:

$P_l^0(\xi)=\frac{1}{\pi}\int\limits_0^\pi(\xi+\sqrt{\xi^2-1}\cos(\theta))^ld\theta$

4.2.5 Solutions for Bessel's equation

Given the LDE

$x^2\frac{d^2y(x)}{dx^2}+x\frac{dy(x)}{dx}+(x^2-\nu^2)y(x)=0$ also called Bessel's equation, and the Bessel functions of the first kind

$J_\nu(x)=x^\nu\sum_{m=0}^\infty\frac{(-1)^mx^{2m}}{2^{2m+\nu}m!\Gamma(\nu+m+1)}$ for

$\nu:=n\in I\hspace{-1mm}N$ this becomes:

$J_n(x)=x^n\sum_{m=0}^\infty\frac{(-1)^mx^{2m}}{2^{2m+n}m!(n+m)!}$ When

$\nu\neq Z\hspace{-1ex}Z$ the solution is given by

$y(x)=aJ_\nu(x)+bJ_{-\nu}(x)$ . But because for

$n\in Z\hspace{-1ex}Z$ holds:

$J_{-n}(x)=(-1)^nJ_n(x)$ , this does not apply to integers. The general solution of Bessel's equation is given by

$y(x)=aJ_\nu(x)+bY_\nu(x)$ , where

$Y_\nu$ are the Bessel functions of the second kind:

$Y_\nu(x)=\frac{J_\nu(x)\cos(\nu\pi)-J_{-\nu}(x)}{\sin(\nu\pi)}~~~\mbox{and}~~~ Y_n(x)=\lim_{\nu\rightarrow n}Y_\nu(x)$ The equation

$x^2y''(x)+xy'(x)-(x^2+\nu^2)y(x)=0$ has the modified Bessel functions of the first kind

$I_\nu(x)=i^{-\nu}J_\nu(ix)$ as solution, and also solutions

$K_\nu=\pi[I_{-\nu}(x)-I_\nu(x)]/[2\sin(\nu\pi)]$ .

Sometimes it can be convenient to write the solutions of Bessel's equation in terms of the Hankel functions $H^{(1)}_n(x)=J_n(x)+iY_n(x)~~,~~H^{(2)}_n(x)=J_n(x)-iY_n(x)$

4.2.6 Properties of Bessel functions

Bessel functions are orthogonal with respect to the weight function

$p(x)=x$ .

$J_{-n}(x)=(-1)^nJ_n(x)$ . The Neumann functions $N_m(x)$ are definied as: $N_m(x)=\frac{1}{2\pi}J_m(x)\ln(x)+\frac{1}{x^m}\sum_{n=0}^\infty \alpha_nx^{2n}$ The following holds: $\lim\limits_{x\rightarrow0}J_m(x)=x^m$ , $\lim\limits_{x\rightarrow0}N_m(x)=x^{-m}$ for $m\neq0$ , $\lim\limits_{x\rightarrow0}N_0(x)=\ln(x)$ . $\lim_{r\rightarrow\infty}H(r)=\frac{{\rm e}^{\pm ikr}{\rm e}^{i\omega t}}{\sqrt{r}}~~,~~ \lim_{x\rightarrow\infty}J_n(x)=\sqrt{\frac{2}{\pi x}}\cos(x-x_n)~~,~~ \lim_{x\rightarrow\infty}J_{-n}(x)=\sqrt{\frac{2}{\pi x}}\sin(x-x_n)$ with $x_n=\frac{1}{2}\pi(n+\frac{1}{2})$ . $J_{n+1}(x)+J_{n-1}(x)=\frac{2n}{x}J_n(x)~~,~~J_{n+1}(x)-J_{n-1}(x)=-2\frac{dJ_n(x)}{dx}$ The following integral relations hold: $J_m(x)=\frac{1}{2\pi}\int\limits_0^{2\pi}\exp[i(x\sin(\theta)-m\theta)]d\theta= \frac{1}{\pi}\int\limits_0^\pi\cos(x\sin(\theta)-m\theta)d\theta$

4.2.7 Laguerre's equation

Given the LDE

$x\frac{d^2y(x)}{dx^2}+(1-x)\frac{dy(x)}{dx}+ny(x)=0$ Solutions of this equation are the Laguerre polynomials

$L_n(x)$ :

$L_n(x)=\frac{{\rm e}^x}{n!}\frac{d^n}{dx^n}\left(x^n{\rm e}^{-x}\right)= \sum_{m=0}^\infty\frac{(-1)^m}{m!}{n\choose m}x^m$

4.2.8 The associated Laguerre equation

Given the LDE

$\frac{d^2y(x)}{dx^2}+\left(\frac{m+1}{x}-1\right)\frac{dy(x)}{dx}+\left(\frac{n+\frac{1}{2}(m+1)}{x}\right)y(x)=0$ Solutions of this equation are the associated Laguerre polynomials

$L_n^m(x)$ :

$L_n^m(x)=\frac{(-1)^mn!}{(n-m)!}{\rm e}^{-x}x^{-m}\frac{d^{n-m}}{dx^{n-m}}\left({\rm e}^{-x}x^n\right)$

4.2.9 Hermite

The differential equations of Hermite are:

$\frac{d^2 {\rm H}_n(x)}{dx^2}-2x\frac{d{\rm H}_n(x)}{dx}+2n{\rm H}_n(x)=0~~\mbox{and}~~ \frac{d^2{\rm He}_n(x)}{dx^2}-x\frac{d{\rm He}_n(x)}{dx}+n{\rm He}_n(x)=0$ Solutions of these equations are the Hermite polynomials, given by:

${\rm H}_n(x)=(-1)^n\exp\left(\frac{1}{2}x^2\right)\frac{d^n(\exp(-\frac{1}{2}x^2))}{dx^n}=2^{n/2}{\rm He}_n(x\sqrt{2})$

${\rm He}_n(x)=(-1)^n(\exp\left(x^2\right)\frac{d^n(\exp(-x^2))}{dx^n}=2^{-n/2}{\rm He}_n(x/\sqrt{2})$

4.2.10 Chebyshev

The LDE

$(1-x^2)\frac{d^2U_n(x)}{dx^2}-3x\frac{dU_n(x)}{dx}+n(n+2)U_n(x)=0$ has solutions of the form

$U_n(x)=\frac{\sin[(n+1)\arccos(x)]}{\sqrt{1-x^2}}$ The LDE

$(1-x^2)\frac{d^2T_n(x)}{dx^2}-x\frac{dT_n(x)}{dx}+n^2T_n(x)=0$ has solutions

$T_n(x)=\cos(n\arccos(x))$ .

4.2.11 Weber

The LDE

$W''_n(x)+(n+\frac{1}{2}-\frac{1}{4} x^2)W_n(x)=0$ has solutions:

$W_n(x)={\rm He}_n(x)\exp(-\frac{1}{4} x^2)$ .

4.3 Non-linear differential equations

Some non-linear differential equations and a solution are:

$\begin{array}{lll} y'=a\sqrt{y^2+b^2}&~~~~&y=b\sinh(a(x-x_0))\\ y'=a\sqrt{y^2-b^2}&~~~~&y=b\cosh(a(x-x_0))\\ y'=a\sqrt{b^2-y^2}&~~~~&y=b\cos(a(x-x_0))\\ y'=a(y^2+b^2) &~~~~&y=b\tan(a(x-x_0))\\ y'=a(y^2-b^2) &~~~~&y=b\coth(a(x-x_0))\\ y'=a(b^2-y^2) &~~~~&y=b\tanh(a(x-x_0))\\ \displaystyle y'=ay\left(\frac{b-y}{b}\right)&~~~~&\displaystyle y=\frac{b}{1+Cb\exp(-ax)} \end{array}$

4.4 Sturm-Liouville equations

Sturm-Liouville equations are second order LDE's of the form:

$-\frac{d}{dx}\left(p(x)\frac{dy(x)}{dx}\right)+q(x)y(x)=\lambda m(x)y(x)$ The boundary conditions are chosen so that the operator

$L=-\frac{d}{dx}\left(p(x)\frac{d}{dx}\right)+q(x)$ is Hermitean. The normalization function

$m(x)$ must satisfy

$\int\limits_a^bm(x)y_i(x)y_j(x)dx=\delta_{ij}$ When

$y_1(x)$ and

$y_2(x)$ are two linear independent solutions one can write the Wronskian in this form:

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\ y_1'&y_2' \end{array}\right|= \frac{C}{p(x)}$ where

$C$ is constant. By changing to another dependent variable

$u(x)$ , given by:

$u(x)=y(x)\sqrt{p(x)}$ , the LDE transforms into the normal form:

$\frac{d^2u(x)}{dx^2}+I(x)u(x)=0~~~\mbox{with}~~~ I(x)=\frac{1}{4}\left(\frac{p'(x)}{p(x)}\right)^2-\frac{1}{2}\frac{p''(x)}{p(x)}-\frac{q(x)-\lambda m(x)}{p(x)}$ If

$I(x)>0$ , than

$y''/y<0$ and the solution has an oscillatory behaviour, if

$I(x)<0$ , than

$y''/y>0$ and the solution has an exponential behaviour.

4.5 Linear partial differential equations

4.5.1 General

The normal derivative is defined by:

$\frac{\partial u}{\partial n}=(\vec{\nabla}u,\vec{n})$ A frequently used solution method for PDE's is separation of variables: one assumes that the solution can be written as

$u(x,t)=X(x)T(t)$ . When this is substituted two ordinary DE's for

$X(x)$ and

$T(t)$ are obtained.

4.5.2 Special cases

4.5.2.1 The wave equation

The wave equation in 1 dimension is given by

$\frac{\partial^2u}{\partial t^2}=c^2\frac{\partial^2u}{\partial x^2}$ When the initial conditions

$u(x,0)=\varphi(x)$ and

$\partial u(x,0)/\partial t=\Psi(x)$ apply, the general solution is given by:

$u(x,t)=\frac{1}{2}\left[\varphi(x+ct)+\varphi(x-ct)\right]+\frac{1}{2c} \int\limits_{x-ct}^{x+ct}\Psi(\xi)d\xi$

4.5.2.2 The diffusion equation

The diffusion equation is:

$\frac{\partial u}{\partial t}=D\nabla^2u$ Its solutions can be written in terms of the propagators

$P(x,x',t)$ . These have the property that

$P(x,x',0)=\delta(x-x')$ . In 1 dimension it reads:

$P(x,x',t)=\frac{1}{2\sqrt{\pi Dt}}\exp\left(\frac{-(x-x')^2}{4Dt}\right)$ In 3 dimensions it reads:

$P(x,x',t)=\frac{1}{8(\pi Dt)^{3/2}}\exp\left(\frac{-(\vec{x}-\vec{x}')^2}{4Dt}\right)$ With initial condition

$u(x,0)=f(x)$ the solution is:

$u(x,t)=\int\limits_{\cal G}f(x')P(x,x',t)dx'$ The solution of the equation

$\frac{\partial u}{\partial t}-D\frac{\partial^2u}{\partial x^2}=g(x,t)$ is given by

$u(x,t)=\int dt' \int dx'g(x',t')P(x,x',t-t')$

4.5.2.3 The equation of Helmholtz

The equation of Helmholtz is obtained by substitution of

$u(\vec{x},t)=v(\vec{x})\exp(i\omega t)$ in the wave equation. This gives for

$v$ :

$\nabla^2v(\vec{x},\omega)+k^2v(\vec{x},\omega)=0$ This gives as solutions for

$v$ :

In cartesian coordinates: substitution of $v=A\exp(i\vec{k}\cdot\vec{x})$ gives: $v(\vec{x})=\int\cdots\int A(k){\rm e}^{i\vec{k}\cdot\vec{x}}dk$ with the integrals over $\vec{k}^2=k^2$ . \item In polar coordinates: $v(r,\varphi)=\sum_{m=0}^\infty(A_mJ_m(kr)+B_mN_m(kr)){\rm e}^{im\varphi}$
In spherical coordinates: $v(r,\theta,\varphi)=\sum_{l=0}^\infty\sum_{m=-l}^l[A_{lm}J_{l+\frac{1}{2}}(kr)+B_{lm}J_{-l-\frac{1}{2}}(kr)]\frac{Y(\theta,\varphi)}{\sqrt{r}}$

4.5.3 Potential theory and Green's theorem

Subject of the potential theory are the Poisson equation

$\nabla^2u=-f(\vec{x})$ where

$f$ is a given function, and the Laplace equation

$\nabla^2u=0$ . The solutions of these can often be interpreted as a potential. The solutions of Laplace's equation are called harmonic functions.

When a vector field $\vec{v}$ is given by $\vec{v}={\rm grad}\varphi$ holds: $\int\limits_a^b(\vec{v},\vec{t})ds=\varphi(\vec{b})-\varphi(\vec{a})$ In this case there exist functions $\varphi$ and $\vec{w}$ so that $\vec{v}={\rm grad}\varphi+{\rm curl}\vec{w}$ .

The field lines of the field $\vec{v}(\vec{x})$ follow from: $\dot{\vec{x}}(t)=\lambda\vec{v}(\vec{x})$ The first theorem of Green is: $\iiint\limits_{\cal\!\!\! G}[u\nabla^2v+(\nabla u,\nabla v)]d^3V=\mathop{\int\hspace{-2ex}\int\hspace{-2.8ex}\bigcirc}\limits_{\cal\!\!\! S}u\frac{\partial v}{\partial n}d^2A$ The second theorem of Green is: $\iiint\limits_{\cal\!\!\! G}[u\nabla^2v-v\nabla^2u]d^3V=\mathop{\int\hspace{-2ex}\int\hspace{-2.8ex}\bigcirc}\limits_{\cal\!\!\! S}\left(u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\right)d^2A$ A harmonic function which is 0 on the boundary of an area is also 0 within that area. A harmonic function with a normal derivative of 0 on the boundary of an area is constant within that area.

The Dirichlet problem is: $\nabla^2u(\vec{x})=-f(\vec{x})~~,~~\vec{x}\in R~~,~~u(\vec{x})=g(\vec{x})~~\mbox{for all}~~\vec{x}\in S.$ It has a unique solution.

The Neumann problem is: $\nabla^2u(\vec{x})=-f(\vec{x})~~,~~\vec{x}\in R~~,~~\frac{\partial u(\vec{x})}{\partial n}=h(\vec{x})~~\mbox{for all}~~\vec{x}\in S.$ The solution is unique except for a constant. The solution exists if: $-\iiint\limits_{\!\!\!R}f(\vec{x})d^3V=\mathop{\int\hspace{-2ex}\int\hspace{-2.8ex}\bigcirc}_{\!\!\!S}h(\vec{x})d^2A$ A fundamental solution of the Laplace equation satisfies: $\nabla^2u(\vec{x})=-\delta(\vec{x})$ This has in 2 dimensions in polar coordinates the following solution: $u(r)=\frac{\ln(r)}{2\pi}$ This has in 3 dimensions in spherical coordinates the following solution: $u(r)=\frac{1}{4\pi r}$ The equation $\nabla^2v=-\delta(\vec{x}-\vec{\xi})$ has the solution $v(\vec{x})=\frac{1}{4\pi|\vec{x}-\vec{\xi}|}$ After substituting this in Green's 2nd theorem and applying the sieve property of the $\delta$ function one can derive Green's 3rd theorem: $u(\vec{\xi})=-\frac{1}{4\pi}\iiint\limits_R\frac{\nabla^2u}{r}d^3V+ \frac{1}{4\pi}\mathop{\int\hspace{-2ex}\int\hspace{-2.8ex}\bigcirc}\limits_S\left[\frac{1}{r}\frac{\partial u}{\partial n}-u\frac{\partial}{\partial n}\left(\frac{1}{r}\right)\right]d^2A$ The Green function $G(\vec{x},\vec{\xi})$ is defined by: $\nabla^2G=-\delta(\vec{x}-\vec{\xi})$ , and on boundary $S$ holds $G(\vec{x},\vec{\xi})=0$ . Than $G$ can be written as: $G(\vec{x},\vec{\xi})=\frac{1}{4\pi|\vec{x}-\vec{\xi}|}+g(\vec{x},\vec{\xi})$ Than $g(\vec{x},\vec{\xi})$ is a solution of Dirichlet's problem. The solution of Poisson's equation $\nabla^2u=-f(\vec{x})$ when on the boundary $S$ holds: $u(\vec{x})=g(\vec{x})$ , is: $u(\vec{\xi})=\iiint\limits_RG(\vec{x},\vec{\xi})f(\vec{x})d^3V- \mathop{\int\hspace{-2ex}\int\hspace{-2.8ex}\bigcirc}\limits_Sg(\vec{x})\frac{\partial G(\vec{x},\vec{\xi})}{\partial n}d^2A$