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How to solve this partial differential equation $$\frac{\partial p(k,t)}{\partial t}+k\frac{\partial p(k,t)}{\partial k}+k^2p(k,t)=0$$

I'm a beginner to PDE, I think I need to construct the characteristic equation but I have no clue.

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    $\begingroup$ Yes, the method of characteristics is the way to go. Why don't you try it? $\endgroup$ – Robert Israel Jan 27 at 2:24
  • $\begingroup$ I have no idea how to construct the equation by the method of characteristics. Can you give a hint? $\endgroup$ – kinder chen Jan 27 at 2:54
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If you have never used this method before you need some help setting you on the correct path.

The problem is well posed if we are given initial data of the form $p(k,0) = f(k)$.

With the method of characteristics we find a family of curves $k(t;\alpha)$ that intersects the $k$-axis in the $kt$-plane at a unique point $(\alpha,0)$ and such that the PDE reduces to an ODE along the curves.

The characteristics here solve the IVP,

$$\displaystyle \frac{dk}{dt} = k, \,\, k(0) = \alpha,$$

with solutions $k(t) = \alpha e^t$.

Taking $v(t) = u(\alpha e^t,t)$ you must determine the ODE satisfied by $v$ and solve by applying the intial condition $v(0) = u(\alpha,0) = f(\alpha)$. Finally transform back to $(k,t)$ coordinates using $\alpha = ke^{-t}$.

This leaves you with a fair amount of work to do.

The ultimate solution will be

$$ u(k,t) = f(ke^{-t})\exp\left(- \frac{k^2}{2} \right)$$

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  • $\begingroup$ Thx, but I don't understand how to get $\frac{dk}{dt}=k$? $\endgroup$ – kinder chen Jan 28 at 19:56
  • $\begingroup$ I suggest reading about the method of characteristics for the full story. I tried to convey some of the intuition in my answer. So the essence of what we are doing here is to find a curve $\hat{k}(t)$ such that along this curve, the PDE reduces to an ODE which is easier to solve. Notice that $\frac{d}{dt}u(\hat{k}(t),t) = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial k} \frac{d\hat{k}}{dt}$ and the RHS matches the derivative terms in the PDE when $\frac{d\hat{k}}{dt }= \hat{k}$. Then we get the ODE $\frac{d}{dt}u(\hat{k}(t),t) + \hat{k}^2(t)u(\hat{k},t) = 0$ which is easy $\endgroup$ – RRL Jan 28 at 20:08
  • $\begingroup$ Again I have tried to explain why this work and not just give you the mechanical recipe you can find in Wikipedia. If that doesn't help you then ... $\endgroup$ – RRL Jan 28 at 20:11
  • $\begingroup$ Also I gave you the solution so that should help you understand the process by working backwards. Try taking the partial derivatives and see why this satisfies the PDE. $\endgroup$ – RRL Jan 28 at 20:16
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This method poses an auxiliary system of ODE's for the curves (the characteristics) forming the solution, in this case and put in form of proportions:

$$\dfrac{\mathbb dt}{1}=\dfrac{\mathbb dk}{k}=-\dfrac{\mathbb dp}{k^2p}$$

You may prefer to solve $\dfrac{\mathbb dt}{1}=\dfrac{\mathbb dk}{k}$ and $\dfrac{\mathbb dk}{k}=-\dfrac{\mathbb dp}{k^2p}$. Giving two equations with an arbitrary constant for each and defining the characteristics (you can get the general solution from here).

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  • $\begingroup$ Thx, how to get the proportions listed in your answer? $\endgroup$ – kinder chen Jan 28 at 19:26
  • $\begingroup$ It's a standard procedure for quasilinear PDEs:The differentials are divided by their factors following this: $a(x;y,u)u_x+b(x,y,u)u_y=c(x,y,u)$ corresponds with $\dfrac{\mathbb dx}{a}=\dfrac{\mathbb dy}{b}=\dfrac{\mathbb du}{c}$ The theory for this is very easy to find. $\endgroup$ – Rafa Budría Jan 28 at 19:58
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Correction: I thank RRL for catching my mistake.

Hint: Consider \begin{align} u(t, k) = e^{k^2/2}p(t, k) \end{align} then check that \begin{align} \partial_t u+k\partial_k u = 0 \end{align} which is just a transport equation.

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  • $\begingroup$ @RRL Thanks for your comment. I have made the correction. $\endgroup$ – Jacky Chong Jan 27 at 15:42

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