Problem:
Which of the following satisfy the diffusion equation in one dimension (for $ t > 0 $) (including, if applicable, unbounded solutions which however satisfy the equation at any finite $ x $ and $ t $)? (Check all that apply)
Solution:
From the lecture, we see the diffusion equation is $ \frac{\partial \rho}{\partial t} = D \nabla^2 \rho $, reducing to one dimension, we have $ \frac{\partial \rho}{\partial t} = D \frac{\partial^2}{\partial x^2} \rho $, so we need to differentiate the equation once by $ t $ and twice by $ x $, then we can know the answer by simply checking if they are equals (of course, with the multiplicative constant).
Differentiating the options, we have
$ \begin{eqnarray*} \frac{\partial}{\partial t} e^{\frac{−x^2}{4 D t}} & = & \frac{x^2}{4 D t^2} e^{\frac{−x^2}{4 D t}} \\ \frac{\partial^2}{\partial x^2} e^{\frac{−x^2}{4 D t}} & = & \frac{\partial}{\partial x} \frac{-x}{2 D t} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{-1}{2 D t} e^{\frac{−x^2}{4 D t}} + \frac{-x}{2 D t} \frac{-x}{2 D t} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{-1}{2 D t} e^{\frac{−x^2}{4 D t}} + \frac{x^2}{4 D^2 t^2} e^{\frac{−x^2}{4 D t}} \\ & = & e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D^2 t^2} - \frac{1}{2 D t}) \end{eqnarray*} $
So the first options is incorrect, but it does help the computation of the below, see
$ \begin{eqnarray*} \frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} & = & \frac{1}{\sqrt{2\pi D t}} \frac{x^2}{4 D t^2} e^{\frac{−x^2}{4 D t}} + \frac{-1}{2t\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D t^2} - \frac{1}{2t}) \\ \frac{\partial^2}{\partial x^2} \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} & = & \frac{1}{\sqrt{2\pi D t}} \frac{\partial^2}{\partial x^2} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D^2 t^2} - \frac{1}{2 D t}) \end{eqnarray*} $
So the second option is right - with the complex expressions behind us, the next two is less complex.
$ \begin{eqnarray*} \frac{\partial}{\partial t} A (t+\frac{1}{2D}x^2) & = & A \\ \frac{\partial^2}{\partial x^2} A (t+\frac{1}{2D}x^2) & = & \frac{\partial}{\partial x} \frac{A}{D}x \\ & = & \frac{A}{D} \end{eqnarray*} $
So the third option is also right! Last but not least for the fourth option, we have
$ \begin{eqnarray*} \frac{\partial}{\partial t} e^{i D ( p x − E t )} & = & -i D E e^{i D ( p x − E t )} \\ \frac{\partial^2}{\partial x^2} e^{i D ( p x − E t )} & = & \frac{\partial}{\partial x} i D p e^{i D ( p x − E t )} \\ & = & (i D p )(i D p )e^{i D ( p x − E t )} \\ & = & - D^2 p^2 e^{i D ( p x − E t )} \end{eqnarray*} $
So the fourth option is wrong, as if there are equal, as $ D $, $ E $ and $ p $ cannot be real number at the same time.
Phew, that's a lot of calculation!
Which of the following satisfy the diffusion equation in one dimension (for $ t > 0 $) (including, if applicable, unbounded solutions which however satisfy the equation at any finite $ x $ and $ t $)? (Check all that apply)
- $ e^{\frac{−x^2}{4 D t}} $
- $ \frac{1}{\sqrt{2\pi D t}}e^{\frac{−x^2}{4 D t}} $
- $ A⋅(t+\frac{1}{2D}x^2) $
- $ e^{i D ( p x − E t )} $ where $ p $ and $ E $ are non-zero real numbers.
Solution:
From the lecture, we see the diffusion equation is $ \frac{\partial \rho}{\partial t} = D \nabla^2 \rho $, reducing to one dimension, we have $ \frac{\partial \rho}{\partial t} = D \frac{\partial^2}{\partial x^2} \rho $, so we need to differentiate the equation once by $ t $ and twice by $ x $, then we can know the answer by simply checking if they are equals (of course, with the multiplicative constant).
Differentiating the options, we have
$ \begin{eqnarray*} \frac{\partial}{\partial t} e^{\frac{−x^2}{4 D t}} & = & \frac{x^2}{4 D t^2} e^{\frac{−x^2}{4 D t}} \\ \frac{\partial^2}{\partial x^2} e^{\frac{−x^2}{4 D t}} & = & \frac{\partial}{\partial x} \frac{-x}{2 D t} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{-1}{2 D t} e^{\frac{−x^2}{4 D t}} + \frac{-x}{2 D t} \frac{-x}{2 D t} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{-1}{2 D t} e^{\frac{−x^2}{4 D t}} + \frac{x^2}{4 D^2 t^2} e^{\frac{−x^2}{4 D t}} \\ & = & e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D^2 t^2} - \frac{1}{2 D t}) \end{eqnarray*} $
So the first options is incorrect, but it does help the computation of the below, see
$ \begin{eqnarray*} \frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} & = & \frac{1}{\sqrt{2\pi D t}} \frac{x^2}{4 D t^2} e^{\frac{−x^2}{4 D t}} + \frac{-1}{2t\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D t^2} - \frac{1}{2t}) \\ \frac{\partial^2}{\partial x^2} \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} & = & \frac{1}{\sqrt{2\pi D t}} \frac{\partial^2}{\partial x^2} e^{\frac{−x^2}{4 D t}} \\ & = & \frac{1}{\sqrt{2\pi D t}} e^{\frac{−x^2}{4 D t}} (\frac{x^2}{4 D^2 t^2} - \frac{1}{2 D t}) \end{eqnarray*} $
So the second option is right - with the complex expressions behind us, the next two is less complex.
$ \begin{eqnarray*} \frac{\partial}{\partial t} A (t+\frac{1}{2D}x^2) & = & A \\ \frac{\partial^2}{\partial x^2} A (t+\frac{1}{2D}x^2) & = & \frac{\partial}{\partial x} \frac{A}{D}x \\ & = & \frac{A}{D} \end{eqnarray*} $
So the third option is also right! Last but not least for the fourth option, we have
$ \begin{eqnarray*} \frac{\partial}{\partial t} e^{i D ( p x − E t )} & = & -i D E e^{i D ( p x − E t )} \\ \frac{\partial^2}{\partial x^2} e^{i D ( p x − E t )} & = & \frac{\partial}{\partial x} i D p e^{i D ( p x − E t )} \\ & = & (i D p )(i D p )e^{i D ( p x − E t )} \\ & = & - D^2 p^2 e^{i D ( p x − E t )} \end{eqnarray*} $
So the fourth option is wrong, as if there are equal, as $ D $, $ E $ and $ p $ cannot be real number at the same time.
Phew, that's a lot of calculation!
No comments:
Post a Comment