- #1
Old Guy
- 103
- 1
Homework Statement
For ordinary 1D diffusion show that the mean value of the square of the position is equal to 2Dt
Homework Equations
[tex]\left\langle {x^2 \left( t \right)} \right\rangle \equiv \int\limits_0^\infty {x^2 p\left( {x,t} \right)dx}
[/tex]
[tex]\frac{\partial }{{\partial t}}p\left( {x,t} \right) = D\frac{{\partial ^2 }}{{\partial x^2 }}p\left( {x,t} \right)
[/tex]
The Attempt at a Solution
[tex]\begin{array}{l}
\left\langle {x^2 \left( t \right)} \right\rangle \equiv \int\limits_0^\infty {x^2 p\left( {x,t} \right)dx} \\
\frac{\partial }{{\partial t}}\left\langle {x^2 \left( t \right)} \right\rangle = \frac{\partial }{{\partial t}}\int\limits_0^\infty {x^2 p\left( {x,t} \right)dx} = \int\limits_0^\infty {x^2 \frac{\partial }{{\partial t}}p\left( {x,t} \right)dx} + \int\limits_0^\infty {p\left( {x,t} \right)\frac{\partial }{{\partial t}}x^2 dx} \\
\frac{\partial }{{\partial t}}\left\langle {x^2 \left( t \right)} \right\rangle = \int\limits_0^\infty {x^2 \frac{\partial }{{\partial t}}p\left( {x,t} \right)dx} + \int\limits_0^\infty {p\left( {x,t} \right)\frac{\partial }{{\partial t}}x^2 dx} \\
\frac{\partial }{{\partial t}}\left\langle {x^2 \left( t \right)} \right\rangle = \int\limits_0^\infty {x^2 \frac{\partial }{{\partial t}}p\left( {x,t} \right)dx} + \int\limits_0^\infty {p\left( {x,t} \right)\frac{\partial }{{\partial t}}x^2 dx} \\
\\
\int\limits_0^\infty {x^2 \frac{\partial }{{\partial t}}p\left( {x,t} \right)dx} = D\int\limits_0^\infty {x^2 \frac{{\partial ^2 }}{{\partial x^2 }}p\left( {x,t} \right)dx} \\
u = x^2 \Rightarrow du = 2xdx \\
dv = \frac{{d^2 }}{{dx^2 }}p\left( {x,t} \right)dx \Rightarrow v = \frac{d}{{dx}}p\left( {x,t} \right) \\
\int\limits_0^\infty {x^2 \frac{\partial }{{\partial t}}p\left( {x,t} \right)dx} = Dx^2 \frac{d}{{dx}}p\left( {x,t} \right) - 2D\int\limits_0^\infty {x\frac{d}{{dx}}p\left( {x,t} \right)dx} \\
\\
\int\limits_0^\infty {x\frac{d}{{dx}}p\left( {x,t} \right)dx} \\
u = x \Rightarrow du = dx \\
dv = \frac{d}{{dx}}p\left( {x,t} \right)dx \Rightarrow v = p\left( {x,t} \right) \\
\int\limits_0^\infty {x\frac{d}{{dx}}p\left( {x,t} \right)dx} = xp\left( {x,t} \right) - \int\limits_0^\infty {p\left( {x,t} \right)dx} \\
\end{array}
[/tex]
I see that I'm home free if
[tex]Dx^2 \frac{d}{{dx}}p\left( {x,t} \right) = 0[/tex]
and
[tex]
xp\left( {x,t} \right) = 0
[/tex]
but how can I justify that?
Thanks!