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.
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:
$y(x)=\int\limits_0^xf(x)\sin(\omega(x-t))dt$.
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:
$W(y_1(x_0),y_2(x_0))=0$.
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
$n=0,1,2$.
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 $y(x)=y_1(x)+y_2(x)$.
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 $y(x)=(C_1\ln(x)+C_2)x^r$.
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: $\|P_n\|^2=2/(2n+1)$.
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:
\[
P_l^0(\xi)=\frac{1}{\pi}\int\limits_0^\pi(\xi+\sqrt{\xi^2-1}\cos(\theta))^ld\theta
\]
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
\[
H^{(1)}_n(x)=J_n(x)+iY_n(x)~~,~~H^{(2)}_n(x)=J_n(x)-iY_n(x)
\]
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:
\[
J_m(x)=\frac{1}{2\pi}\int\limits_0^{2\pi}\exp[i(x\sin(\theta)-m\theta)]d\theta=
\frac{1}{\pi}\int\limits_0^\pi\cos(x\sin(\theta)-m\theta)d\theta
\]
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 $L_n(x)$:
\[
L_n(x)=\frac{{\rm e}^x}{n!}\frac{d^n}{dx^n}\left(x^n{\rm e}^{-x}\right)=
\sum_{m=0}^\infty\frac{(-1)^m}{m!}{n\choose m}x^m
\]
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 $L_n^m(x)$:
\[
L_n^m(x)=\frac{(-1)^mn!}{(n-m)!}{\rm e}^{-x}x^{-m}\frac{d^{n-m}}{dx^{n-m}}\left({\rm e}^{-x}x^n\right)
\]
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:
\[
{\rm H}_n(x)=(-1)^n\exp\left(\frac{1}{2}x^2\right)\frac{d^n(\exp(-\frac{1}{2}x^2))}{dx^n}=2^{n/2}{\rm He}_n(x\sqrt{2})
\]
\[
{\rm He}_n(x)=(-1)^n(\exp\left(x^2\right)\frac{d^n(\exp(-x^2))}{dx^n}=2^{-n/2}{\rm He}_n(x/\sqrt{2})
\]
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 $T_n(x)=\cos(n\arccos(x))$.
The LDE $W''_n(x)+(n+\frac{1}{2}-\frac{1}{4} x^2)W_n(x)=0$ has solutions:
$W_n(x)={\rm He}_n(x)\exp(-\frac{1}{4} x^2)$.
Some non-linear differential equations and a solution are:
\[
\begin{array}{lll}
y'=a\sqrt{y^2+b^2}&~~~~&y=b\sinh(a(x-x_0))\\
y'=a\sqrt{y^2-b^2}&~~~~&y=b\cosh(a(x-x_0))\\
y'=a\sqrt{b^2-y^2}&~~~~&y=b\cos(a(x-x_0))\\
y'=a(y^2+b^2) &~~~~&y=b\tan(a(x-x_0))\\
y'=a(y^2-b^2) &~~~~&y=b\coth(a(x-x_0))\\
y'=a(b^2-y^2) &~~~~&y=b\tanh(a(x-x_0))\\
\displaystyle y'=ay\left(\frac{b-y}{b}\right)&~~~~&\displaystyle y=\frac{b}{1+Cb\exp(-ax)}
\end{array}
\]
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.
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.
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:
\[
u(x,t)=\frac{1}{2}\left[\varphi(x+ct)+\varphi(x-ct)\right]+\frac{1}{2c}
\int\limits_{x-ct}^{x+ct}\Psi(\xi)d\xi
\]
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
\[
u(x,t)=\int dt' \int dx'g(x',t')P(x,x',t-t')
\]
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:
\[
v(r,\theta,\varphi)=\sum_{l=0}^\infty\sum_{m=-l}^l[A_{lm}J_{l+\frac{1}{2}}(kr)+B_{lm}J_{-l-\frac{1}{2}}(kr)]\frac{Y(\theta,\varphi)}{\sqrt{r}}
\]
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:
\[
u(\vec{\xi})=\iiint\limits_RG(\vec{x},\vec{\xi})f(\vec{x})d^3V-
\mathop{\int\hspace{-2ex}\int\hspace{-2.8ex}\bigcirc}\limits_Sg(\vec{x})\frac{\partial G(\vec{x},\vec{\xi})}{\partial n}d^2A
\]