I am seeking for the analytic solution of partial differential equation $$ \begin{aligned} \frac{{\partial y\left( {x,t} \right)}}{{\partial t}} &= {D}\frac{{{\partial ^2}y\left( {x,t} \right)}}{{\partial {x^2}}} +{\mu}\frac{\partial}{{\partial x}} \left[ y(x,t)(E_1x+E_0) \right] -\frac{y(x,t)}{{{\tau}}} + G \\&= {D}\frac{{{\partial ^2}y\left( {x,t} \right)}}{{\partial {x^2}}} + \mu(E_1x+E_0)\frac{\partial y(x,t)}{\partial x} + \left( \mu E_1-\frac{1}{\tau} \right)y(x,t) +G \end{aligned} $$ This eq. is defined at $-\infty<x<\infty$, $t>0$.

I resorted to fourier transform method to solve the pde. The definition of Fourier transform with respect to x is $$ F(y(x,t))=\int_{-\infty}^{\infty}e^{-ikx}u(x,t)dx=U(k,t) $$ The list of derivatives I used are $$ F(\frac{\partial y(x,t)}{\partial t})=\frac{\partial U(k,t)}{\partial t} \qquad F(\frac{\partial y(x,t)}{\partial x})=ikU(k,t) \qquad F(\frac{\partial^2 y(x,t)}{\partial x^2})=-k^2U(k,t) $$ Also, $$ F(G)=d\int_{-\infty}^{\infty}e^{-ikx}dx =2\pi G\delta(k), $$ $$ F(x^ny(x,t))=i^n\frac{d^n U(k,t))}{dk^n} $$ $$ F\left(x\frac{\partial y(x,t)}{\partial x}\right) =i\frac{\partial}{\partial k}\left(ikU(k,t)\right) =-U(k,t)-k\frac{\partial U(k,t)}{\partial x} $$ Equation looks $$ \frac{\partial U(k,t)}{\partial t} = \left[-Dk^2+ik\mu E_0-\frac{1}{\tau}\right]U(k,t)+2\pi G\delta(k) -\mu E_1k\frac{\partial U(k,t)}{\partial k} $$ Now, If $E_1$ was not present, then the solution would be very easy to derive. However, with $E_1$ present, there is derivative at RHS.

How do I proceed from this?


The second order EDP with unknown $y(x,t)$ has been reduced to a first order EDP with unknown $U(k,t)$.

The general solution can be found thanks to the method of characteristics.

To go further, one have to know the boundary conditions.

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  • $\begingroup$ I can see that method of characteristic is working on my problem, but I couldn't really understand from what you have written. So I have to resort to another literature on methods of characteristics. How did it go from general solution to U(k,t)? $\endgroup$ – user65452 Feb 25 '16 at 22:48
  • $\begingroup$ It seems little bit restrictive that in order to solve PDE via method of characteristic you must have initial condition. To me it feels it is asking too much. do you have other method? I looked up and there was method of change of variable. Does this work? $\endgroup$ – user65452 Feb 26 '16 at 2:00
  • $\begingroup$ I think that you misunderstood the meaning of my answer. The method of characteristics gives the general solution $U(k,t)$ of the first order PDE. Nothing can be more general. All the methods are not able to gives the general solution, but only give particular or less general solutions. On the contrary in the present case, the use of Fourier transform is less general because it requires to compute the inverse Fourier transform of an arbitrary function $\Psi(...)$ : this is more restrictive in cases where you don't know how to do the inverse Fourier transform. $\endgroup$ – JJacquelin Feb 27 '16 at 7:58

I think it might be easier if use $$Y(x,t)=y(x,t)+\frac{G}{\mu E_1-\frac{1}{\tau}}$$ so that $$\frac{\partial Y}{\partial t}=D\frac{\partial ^2 Y}{\partial x^2}+\mu (E_1 x+E_0)\frac{\partial Y}{\partial x}+\left(\mu E_1-\frac{1}{\tau}\right)Y$$ such a form make it easier to separate as $Y(x,t)=X(x)\times T(t)$ so $$\frac{T'(t)}{T(t)}=\frac{DX''+\mu (E_1 x+E_0)X'+\left(\mu E_1-\frac{1}{\tau}\right)X}{X}=\lambda$$ Then solve $$DX''+\mu (E_1 x+E_0)X'+\left(\mu E_1-\frac{1}{\tau}-\lambda\right)X=0$$ with $\lambda$ a constant use your method.

Use $Y$, I think you can also obtain another $U$ without $\delta (k)$, for example (I did not check this though) something like $$\frac{\partial U}{\partial t}=\left(...\right)U+\left(...\right)\frac{\partial U}{\partial k}$$ and then use $U=K(k)\times T(t)$. Hope this can help.

  • $\begingroup$ I followed your method, but I got stuck. How do you get rid of $\lambda$ in the $\hat{X}$? $\endgroup$ – user65452 Feb 25 '16 at 18:46
  • $\begingroup$ @user65452 I think $\lambda$ is preserved until you have some other condition to constrain it, because any constant $\lambda$ correspond to a possible solution of the equation. There is an example established here en.wikipedia.org/wiki/Separation_of_variables , see the "Partial differential equations" section. Hope it helps $\endgroup$ – Alexis Feb 26 '16 at 1:42
  • $\begingroup$ well.. I went further, but your method doesn't really help. I went further down to get X function with the help of mathematica, but I got super weird function and doesn't even cancel out $\lambda$ $\endgroup$ – user65452 Feb 26 '16 at 1:56

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