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liliy
 one year 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
liliy
 one year 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

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SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0I'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.

liliy
 one year ago
Best ResponseYou've already chosen the best response.0how do i find the partial derivative?

liliy
 one year ago
Best ResponseYou've already chosen the best response.0bec i think i got to dp/dt part but d^2/ dx^2 i dont know how to do it

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0It'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.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0For this particular function, you'll have your work cut out for you. Plenty of chain rule and product rule.

liliy
 one year ago
Best ResponseYou've already chosen the best response.0I actually just need to code this in mathematica, not figure it all out. lol

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0In 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%5Bx%5E2%2F+4+D*t%5D%2Ct%5D And here's the 2ndorder 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%5Bx%5E2%2F+4+D*t%5D%2Cx%5D%2Cx%5D
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