## liliy 2 years ago using derivatives verify that: p(x,t)= 1/( Sqrt[ 4 Pi Dt]) * Exp[-x^2/ 4 Dt] is a solution of hte diffusion equation: dp/dt= D d^2p/dx^2

1. SithsAndGiggles

I'm sure you mean the partial derivatives where you wrote "dp/dt": $\frac{\partial p}{\partial t}=D\frac{\partial^2p}{\partial x^2}$ Given $$p(x,t)=\dfrac{1}{\sqrt{4\pi Dt}}\exp\left({\dfrac{-x^2}{4}Dt}\right)$$, all you have to do is find the partial derivatives $$\dfrac{\partial p}{\partial t}$$ and $$\dfrac{\partial^2 p}{\partial x^2}$$ and plug them into the equation.

2. liliy

how do i find the partial derivative?

3. liliy

bec i think i got to dp/dt part but d^2/ dx^2 i dont know how to do it

4. SithsAndGiggles

It's a lot like taking a regular derivative. The partial derivative of a function $$f(x,t)$$ with respect to $$t$$, or $$\dfrac{\partial f}{\partial t}$$, is obtained by differentiating the function with respect to $$t$$, but fixing $$x$$ as a constant. As an example: Suppose $$f(x,t)=xt$$. Then $$\dfrac{\partial f}{\partial x}=t$$ and $$\dfrac{\partial f}{\partial t}=x$$, since $$t$$ is considered constant in the first and $$x$$ is considered to be constant in the second.

5. SithsAndGiggles

For this particular function, you'll have your work cut out for you. Plenty of chain rule and product rule.

6. liliy

I actually just need to code this in mathematica, not figure it all out. lol

7. SithsAndGiggles

In that case, just use the differentiation operator. In Mathematica, you'd write $$\dfrac{\partial p}{\partial x}=$$ D[p(x,t),t] $$\dfrac{\partial^2 p}{\partial x^2}=$$ D[D[p(x,t), x],x] where p(x,t) is the given function, obviously. Here's the partial with respect to t: http://www.wolframalpha.com/input/?i=D%5B1%2F%28+Sqrt%5B+4+Pi*+D*t%5D%29++*+Exp%5B-x%5E2%2F+4+D*t%5D%2Ct%5D And here's the 2nd-order partial with respect to x: http://www.wolframalpha.com/input/?i=D%5BD%5B1%2F%28+Sqrt%5B+4+Pi*+D*t%5D%29++*+Exp%5B-x%5E2%2F+4+D*t%5D%2Cx%5D%2Cx%5D