Question

(Partial differentiation equation involving erf)

http://f.imgtmp.com/OPQot.jpg

The erf thing throws me off. Other than that, it's basically just plugging into the differential equation to see if it works, right?

Answer 1

yes, and you can use the fundamental theorem of calculus to deal with the error function

Answer 2

How do I make erf x as given become erf [x/sqrt(4kt)]?

Answer 3

\[\large \text{erf}[\frac x{\sqrt{4kt}}]=\int_{0}^{\frac x{\sqrt{4kt}}}e^{-s^2}ds\]

Answer 4

Could you show me? The erf is still confusing me.

Is it something along the lines of having the antiderivative difference F(x/sqrt(4kt)) - F(0) and plugging that into the erf part of the psi equation and then partial differentiating to the first and second order and checking if the first order partial is a constant multiple of the second order partial?

Answer 5

using the fundamental theorem of calculus\[\large \frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]=e^{-(\frac x{\sqrt4kt})^2}[-\frac12x(4kt)^{-3/2}](4k)\]\[\large\frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]=-2kxe^{-(\frac x{\sqrt4kt})^2}(4kt)^{-3/2}\]

Answer 6

when we do the derivative of the error function we can use the fundamental theorem, watch carefully:\[\large \frac{\partial}{\partial x}\text{erf}[\alpha]=\frac{\partial}{\partial x}\frac2{\sqrt\pi}\int_{0}^{\alpha}e^{-s^2}ds=\frac2{\sqrt\pi}e^{-s^2}\frac{\partial\alpha}{\partial x}\]if you don't remember the logic here, realize that the evaluation of the integral at 0 is a constant, so it disappears from the formula after we take the derivative

Answer 7

\[\large \frac{\partial}{\partial x}\text{erf}[\alpha]=\frac{\partial}{\partial x}\frac2{\sqrt\pi}\int_{0}^{\alpha}e^{-s^2}ds=\frac2{\sqrt\pi}e^{-\alpha^2}\frac{\partial\alpha}{\partial x}\]typo in the previous post, should be e^(-alpha...

Answer 8

I also forgot the 2/sqrt(pi) in the error function a few posts up
I'll write this out but it will take a while...

Answer 9

I'm trying to read what you say but it keeps freezing!

Answer 10

Also, no need to retype, just state where the error is and if I don't get it, I'll tell you.

Answer 11

I'm not retyping much, I can pull out things I have already written, so don't worry about that
\[\large \psi(x,y)=A+B\text{erf}[\frac x{\sqrt{4kt}}]\]\[\large\frac{\partial\psi}{\partial t}=k\frac{\partial^2\psi}{\partial x^2}\]since A drops out from the derivative we can write this as\[\large B\frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]=Bk\frac{\partial^2}{\partial t^2}\text{erf}[\frac x{\sqrt{4kt}}]\]and since we can cancel B on both sides all we need to show is that\[\large \frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]=k\frac{\partial^2}{\partial t^2}\text{erf}[\frac x{\sqrt{4kt}}]\]so we need the partial wrt t of erf:\[\large \frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]=\frac2{\sqrt\pi}e^{-(\frac x{\sqrt4kt})^2}[-\frac12x(4kt)^{-3/2}](4k)\]\[\large\frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]=-\frac{4kx}{\sqrt\pi}e^{-(\frac x{\sqrt4kt})^2}(4kt)^{-3/2}\]and now the second partial wrt x of erf...

Answer 12

\[\large\frac{\partial}{\partial x}\text{erf}[\frac x{\sqrt{4kt}}]=\frac2{\sqrt\pi}e^{-(\frac x{\sqrt{4kt}})^2}(\frac1{\sqrt{4kt}})\]\[\large\frac{\partial^2}{\partial x^2}\text{erf}[\frac x{\sqrt{4kt}}]=\frac2{\sqrt\pi}[-2(\frac x{\sqrt{4kt}})e^{-(\frac x{\sqrt{4kt}})^2}(\frac1{\sqrt{4kt}})^2]\]\[\large=\frac{-4x}{\sqrt\pi}e^{-(\frac x{\sqrt{4kt}})^2}(4kt)^{-3/2}=\frac1k\frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]\]so we see that\[\large\frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]=k\frac{\partial^2}{\partial x^2}\text{erf}[\frac x{\sqrt{4kt}}]\]\[\large B\frac{\partial}{\partial t}\text{erf}[\frac x{\sqrt{4kt}}]=kB\frac{\partial^2}{\partial x^2}\text{erf}[\frac x{\sqrt{4kt}}]\]\[\large\frac{\partial\psi}{\partial t}=k\frac{\partial^2\psi}{\partial x^2}\]which is what we wanted to show

Answer 13

Okay, I get the concept/idea but I will look more thoroughly at the detailed algebra later because I am very tired at the moment but I should be okay. Thanks :).

Answer 14

sure thing
the only tricky part is using the fundamental theorem, but I think I explained that
good luck!