User:Peter483

The main idea of series expansion method is to do series expansion at time t=0, then express each order of time partial derivative with initial condition.

Take an example. A partial derivative equation is:

$$ \dfrac{\partial f}{\partial t} =\dfrac{\partial f}{\partial x} $$

To find the solution $$f(x,t) \,$$, we express $$f \,$$ as Tylor series at time $$t=0 \,$$:

$$f(x,t)=f\mid_{t=0} + \dfrac{\partial f}{\partial t} \mid_{t=0} \times t + \dfrac{\partial^2 f}{\partial t^2} \mid_{t=0}  \times \dfrac{t^2}{2} + \dfrac{\partial^3 f}{\partial t^3} \mid_{t=0}  \times \dfrac{t^3}{3!} + \dots$$

$$f(x,t)=\sum_{j=0}^{+\infty} \dfrac{\partial^{^j} f}{\partial t^{^j}} \mid_{t=0} \times \dfrac{t^{^j}}{j!}$$

The initial condition is $$ f\mid_{t=0} \, $$. If we can use $$f\mid_{t=0} \,$$ to express $$ \dfrac{\partial^{^j} f}{\partial t^{^j}} \mid_{t=0}$$ for every $$j\ge 0$$, then we can get solution $$f(x,t) \,$$.

Using the equation $$ \dfrac{\partial f}{\partial t} =\dfrac{\partial f}{\partial x} $$, we can get:

$$ \dfrac{\partial f}{\partial t} =\dfrac{\partial f}{\partial x} $$

$$ \dfrac{\partial^2 f}{\partial t^2} = \dfrac{\partial }{\partial t} (\dfrac{\partial f}{\partial x} ) = \dfrac{\partial }{\partial x} (\dfrac{\partial f}{\partial t} ) = \dfrac{\partial }{\partial x} (\dfrac{\partial f}{\partial x} )=\dfrac{\partial^2 f}{\partial x^2} $$

In general, we can get: $$ \dfrac{\partial^{^j} f}{\partial t^{^j}} =\dfrac{\partial^{^j} f}{\partial x^{^j}} $$

Set both sides of this equation at time $$t=0 \,$$:

$$ \dfrac{\partial^{^j} f}{\partial t^{^j}} \mid _{t=0} =\dfrac{\partial^{^j} f}{\partial x^{^j}} \mid _{t=0} $$

For simplicity, we use $$g(x) \,$$ to substitute the initial condition $$ f \mid _{t=0} $$:

$$ \dfrac{\partial^{^j} f}{\partial t^{^j}} \mid _{t=0} = \dfrac{\partial^{^j} g(x) }{\partial x^{^j}} $$

Finally we can express the solution $$f(x,t) \,$$ using initial condition $$g(x)=f\mid_{t=0}$$:

$$f(x,t)=\sum_{j=0}^{+\infty} \dfrac{\partial^{^j} g}{\partial x^{^j}} \times \dfrac{t^{^j}}{j!}$$

Using this series expansion method, we can solve any kind of partial derivative equation with initial condition.

There is a way to simplify this method. We can use Fourier transfromation.

$$f_{p} (p,t)=(2 \pi)^{-\dfrac{_1}{^2}} \int f(x,t) e^{-ipx}dx$$

$$g_{p} (p)=(2 \pi)^{-\dfrac{_1}{^2}} \int g(x) e^{-ipx}dx$$

Do Fourier transfromation to both side of equation $$f(x,t)=\sum_{j=0}^{+\infty} \dfrac{\partial^{^j} g}{\partial x^{^j}} \times \dfrac{t^{^j}}{j!}$$, we will get:

$$f_p(p,t)=\sum_{j=0}^{+\infty} g_p(p) \times (ip)^j \times \dfrac{t^{^j}}{j!}$$

$$f_p(p,t)=g_p(p) \sum_{j=0}^{+\infty}    \dfrac{(ipt)^{^j}}{j!}$$

$$f_p(p,t)=g_p(p) e^{ipt} \,$$

$$e^{ipt} \,$$ is also the Fourier transfromation of the green function of $$ \dfrac{\partial f}{\partial t} =\dfrac{\partial f}{\partial x} $$. In the language of quantum field theory $$e^{ipt} \,$$ is the propagator.

The main idea of http://en.wikipedia.org/wiki/User:Peter484 is using the method above to calculate QED.