파동 방정식

\({ \partial^2 u \over \partial t^2 } = v^2 \nabla^2 u\)

 

맥스웰방정식 으로부터 전기장이 파동방정식을 만족시킴을 알 수 있다

\( \nabla^2 \mathbf{E}= \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}} {\partial t^2}\)

 

 

일반해

\(Y=f(x+at)+g(x-at)\). Let \(a\) be a constant.

Show \(\frac{\partial^2 Y}{\partial t^2}=a^2\frac{\partial^2 Y}{\partial x^2}\).

 

Let \(u=x+at\), \(v=x-at\).

Then \(Y=f(u)+g(v)\).

\(\frac{\partial Y}{\partial t}=\frac{\partial Y}{\partial u}\frac{\partial u}{\partial t} +\frac{\partial Y}{\partial v}\frac{\partial v}{\partial t}=f'(u)a+g'(v)(-a)=af'(u)-ag'(v)\)

Let \(W(u,v)=\frac{\partial Y}{\partial t}=af'(u)-ag'(v)\).

\(\frac{\partial^2 Y}{\partial t^2}=\frac{\partial W}{\partial t}=\frac{\partial W}{\partial u}\frac{\partial u}{\partial t} +\frac{\partial W}{\partial v}\frac{\partial v}{\partial t}=af''(u)a-ag''(v)(-a)=a^2(f''(u)+g''(v))\)

 

Now turn to the right hand side.

\(\frac{\partial Y}{\partial x}=\frac{\partial Y}{\partial u}\frac{\partial u}{\partial x} +\frac{\partial Y}{\partial v}\frac{\partial v}{\partial x}=f'(u)+g'(v)\)

Let \(Z(u,v)=\frac{\partial Y}{\partial x}=f'(u)+g'(v)\)

\(\frac{\partial^2 Y}{\partial x^2}=\frac{\partial Z}{\partial x}=\frac{\partial Z}{\partial u}\frac{\partial u}{\partial x} +\frac{\partial Z}{\partial v}\frac{\partial v}{\partial x}=f''(u)+g''(v)\)

 

Therefore 

\(\frac{\partial^2 Y}{\partial t^2}=a^2\frac{\partial^2 Y}{\partial x^2}=a^2(f''(u)+g''(v))\)