How to prove solution to Fick's second law by substitution

[neromax]

Homework Statement

I am to prove that a solution to the differential equation Fick's second law is valid by substitution.

Homework Equations

Fick's second law:
\frac{\partial C}{\partial t} = \frac{\partial}{\partial x} \left( D \frac{C}{\partial x} \right)
Solution to Fick's second law:
C(x,t) = \left( \frac{C_1 + C_2}{2} \right) - \left( \frac{C_1 - C_2}{2} \right) \text{erf} \left( \frac{x}{2 \sqrt{Dt}} \right)
The (Gauss) error function (erf) which I found online:
\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_o^z e^{-y^2} dy

The Attempt at a Solution

I presume I am supposed to differentiate the proposed solution with respect to t once and compare it to the proposed solution differentiated twice with respect to x? But I am not sure how i shall handle the integral.

Can someone help me/point me to literature or give me some pointers on how to proceed?

 

[HallsofIvy]

If you have taken Calculus, then you should know "the fundamental theorem of Calculus"-
\frac{d}{dx} \int_a^x f(t) dt= f(x)

 

[neromax]

@HallsofIvy - Thanks for a reply. The problem bugging me though, is that i do not have x, but \frac{x}{2\sqrt{Dt}}, such that:

\frac{2}{\sqrt{\pi}} \frac{\partial}{\partial x} \int_0^{\frac{x}{2\sqrt{Dt}}} e^{-y^2}dy

 

[neromax]

Is this the correct change of limits and function of the integral?
<br /> <br /> \frac{\partial}{\partial x} C(x,t) = -B \frac{\partial}{\partial x}\int_0^{\frac{x}{2\sqrt{Dt}}} e^{-y^2}dy = -B \frac{\partial}{\partial x} \int_0^x e^{-y^2\left(4Dt \right)^{-1}}dy = -B e^{-x^2\left(4Dt \right)^{-1}}\\<br /> \frac{\partial^2}{\partial^2 x}C(x,t) = -B \frac{\partial}{\partial x} e^{-x^2\left(4Dt \right)^{-1}} = -B e^{-x^2\left(4Dt \right)^{-1}} \frac{\partial}{\partial x}\left( -x^2\left(4Dt \right)^{-1} \right) = B \frac{2}{Dt}\cdot x e^{-x^2\left(4Dt \right)^{-1}}<br />

 

[neromax]

Solved it!

I finally understood how it worked

<br /> <br /> D\frac{\partial}{\partial x} C(x,t) = -BD \frac{\partial}{\partial x}\int_0^{\frac{x}{2\sqrt{Dt}}} e^{-y^2}dy = -BD e^{-\left(\frac{x}{2\sqrt{Dt}}\right)^2} \frac{\partial}{\partial x} \left(\frac{x}{2\sqrt{Dt}}\right) = -BD e^{-\frac{x^2}{4Dt}} \left(\frac{1}{2\sqrt{Dt}}\right) \\= \frac{-BD}{2\sqrt{Dt}} e^{-\frac{x^2}{4Dt}} \\<br /> \frac{\partial}{\partial x} \left( D \frac{C(x,t)}{\partial x} \right) = \frac{\partial}{\partial x} \frac{-BD}{2\sqrt{Dt}} e^{-\frac{x^2}{4Dt}} = \frac{-BD}{2\sqrt{Dt}} e^{-\frac{x^2}{4Dt}} \frac{\partial}{\partial x} \left( - \frac{x^2}{4Dt} \right) \\= \frac{BDx}{4}\left( Dt \right)^{-\frac{3}{2}}e^{-\frac{x^2}{4Dt}}\\<br /> <br /> \frac{\partial}{\partial t} = -B \frac{\partial}{\partial t}\int_0^{\frac{x}{2\sqrt{Dt}}} e^{-y^2}dy = -B e^{-\left(\frac{x}{2\sqrt{Dt}}\right)^2} \frac{\partial}{\partial t} \left(\frac{x}{2\sqrt{Dt}}\right) = -B e^{-\left(\frac{x}{2\sqrt{Dt}}\right)^2} \left( - \frac{Dx}{4(Dt)^{3/2}} \right) \\= \frac{BDx}{4}\left( Dt \right)^{-\frac{3}{2}}e^{-\frac{x^2}{4Dt}} \qquad \square<br /> <br />