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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
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

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SithsAndGiggles
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0how do i find the partial derivative?

liliy
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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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