When solving a problem, I got stuck in solving this equation. Is there any Mathematician who could solve it?

$ \partial_{t} p(k,t) = -|k|^{\alpha} p(k,t) + i k \sum_{n=1}^{4} a_{n} e^{-t |b_{n}(k,k_0)|}$

$|b_{1}(k,k_0)|= {|k+k_0|^{\alpha}}$

$|b_{2}(k,k_0)|= {|k-k_0|^{\alpha}}$

$|b_{3}(k,k_0)|= {|k-2k_0|^{\alpha}}$

$|b_{4}(k,k_0)|= {|k+2k_0|^{\alpha}}$

Where $a_{n}$ is a constant and $i $ is the imaginary unit. initial condition: p(k,0)=1

  • 1
    $\begingroup$ There is only one derivative, your PDE is actually an inhomogeneous ODE in time, so use an integrating factor. $\endgroup$ – mattos Nov 19 '16 at 8:43
  • $\begingroup$ Sorry, What do you mean by integrating factor? @Mattos $\endgroup$ – Mona phys Nov 19 '16 at 9:20

Since there is only one derivative (with respect to only one variable $t$), the equation is a simple ODE. The general solution is shown below.

The arbitrary function $f(k)$ involved in the solution cannot be determined because the initial conditions are not specified in the wording of the question. (The initial condition have been added latter, then the calculus of $f(k)$ is obvious).

enter image description here

Note that $t,k, k_0, \alpha, a_n$ are involved in the solution. So a complete symbolism would not be $p(k,t)$ but $p(k, k_0, \alpha, a_n, t)$.

Nevertheless, in the differential equations, it is of use to name "variables" those that are involved as derivatives and name "parameters" those that are not involved as derivatives. In the present case there is only one variable $t$ and several parameters $k,k_0,\alpha,a_n$. Also, it is of use to use the symbol $\partial$ if they are several variables (in sens of differential equation) and the symbol $d$ if there is only one variable (which is the case in the present problem). Of course, this is valid even if some parameters are considered as variables in further calculus (such as $k$ in the present case).

  • $\begingroup$ But where are a_{n} in your solution? I also added initial condition $\endgroup$ – Mona phys Nov 19 '16 at 10:06
  • $\begingroup$ Since they are forgotten in first place, they are forgotten everywhere (Now corrected). $\endgroup$ – JJacquelin Nov 19 '16 at 10:32
  • $\begingroup$ Just put $t=0$ and $p=1$ into the general solution and $f(k)$ is obtained straightforward. $\endgroup$ – JJacquelin Nov 19 '16 at 10:35

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