Solve the following PDE : $$\frac{\partial x}{\partial t}+\frac{3}{2}\sqrt{\frac{g}{h_0}}\bigg(x\frac{\partial x}{\partial y}+\frac{2h_0}{3}\frac{\partial x}{\partial y}+\frac{h_0^3}{9}\frac{\partial^3x}{\partial y^3}\bigg)=0$$ where $x$ is function of $t,y$ and $g,h_0$ are constants.

I have tried all the methods that are available to me, i.e. separation of variables, integral transforms, certain transformations but to no avail. Any help on this one is appreciated.

  • 1
    $\begingroup$ It is not strange. It is nonlinear. $\endgroup$
    – Jon
    Mar 4, 2020 at 12:38
  • $\begingroup$ I appreciate the joke. Do you have a solution, by the way? $\endgroup$ Mar 4, 2020 at 13:26
  • $\begingroup$ It depends on your needs. I can provide you a small and a large perturbation series on the nonlinear term. $\endgroup$
    – Jon
    Mar 4, 2020 at 13:31
  • $\begingroup$ No problem. I just want an approach to begin with. $\endgroup$ Mar 4, 2020 at 13:35
  • $\begingroup$ I think separating out time is a very easy first step ... write $x(y,t)=A e^{\lambda t} \psi (y)$ and then you're left with $$\lambda \psi (y)+\frac{3}{2}\sqrt{\frac{g}{h_0}}\bigg(\psi \frac{\partial \psi}{\partial y}+\frac{2h_0}{3}\frac{\partial \psi}{\partial y}+\frac{h_0^3}{9}\frac{\partial^3\psi}{\partial y^3}\bigg)=0$$ $\endgroup$ Mar 4, 2020 at 14:29

1 Answer 1


Firstly, rescale the variables as $$ x\rightarrow ax \quad t\rightarrow bt \quad y\rightarrow cy $$ and choose $$ a=\frac{2h_0}{3}\sqrt{\frac{h_0}{g}}\quad b=h_0c\quad \frac{h_0^3c^2}{9}=1, $$ the equation takes the form $$ \frac{\partial x}{\partial t}+\lambda x\frac{\partial x}{\partial y}+\frac{\partial x}{\partial y}+\frac{\partial^3 x}{\partial y^3}=0 $$ being now $\lambda$ an ordering parameter, introduced by convenience, taken to be 1 at the end of the computation.

Then, you can consider two different situations: $\lambda\rightarrow 0$ and $\lambda\rightarrow\infty$. For the former case, take the series $$ x=x_0+\lambda x_1+O(\lambda^2). $$ I limit the computation at the first order just to explain the technique. Put this into the equation and you will obtain the set of equations $$ \frac{\partial x_0}{\partial t}+\frac{\partial x_0}{\partial y}+\frac{\partial^3 x_0}{\partial y^3}=0, $$ $$ \frac{\partial x_1}{\partial t}+\frac{\partial x_1}{\partial y}+\frac{\partial^3 x_1}{\partial y^3}=-x_0\frac{\partial x_1}{\partial y}-x_1\frac{\partial x_0}{\partial y}, $$ $$ \vdots. $$ Now, you can realize that the equation for $x_0$ is linear and can be solved exactly, e.g. with a Fourier series, given the proper boundary and initial conditions. The procedure can be iterated at whatever order you like, just consider that going at higher orders implies more involved computations.

Finally, the opposite limit can be evaluated by rescaling the time variable as $t\rightarrow \lambda t$. Then, take the series $$ x=x_0+\frac{1}{\lambda}x_1+O\left(\frac{1}{\lambda^2}\right). $$ This kind of approach is known in fluid dynamics as a boundary layer problem. Then you get the set of equations $$ \frac{\partial x_0}{\partial t}+x_0\frac{\partial x_0}{\partial y}=0, $$ $$ \frac{\partial x_1}{\partial t}+x_0\frac{\partial x_1}{\partial y}+x_1\frac{\partial x_0}{\partial y}=-\frac{\partial x_0}{\partial y}-\frac{\partial^3x_0}{\partial y^3}, $$ $$ \vdots. $$ I do not enter too much into these equations as in boundary layer problems there is a further complication arising by the boundary conditions. I just notice that the leading order equation can be solved by the characteristic method.

This should give you some starting point to work with.

  • $\begingroup$ see my comment under original post for easy way to separate out time $\endgroup$ Mar 4, 2020 at 14:40
  • $\begingroup$ That does not help. You are left with a nonlinear equation to solve. Perturbation theory is more general than what you proposed. $\endgroup$
    – Jon
    Mar 4, 2020 at 14:43
  • $\begingroup$ it does help actually ... still left with a difficult differential equation sans time but the time portion absolutely simplifies out $\endgroup$ Mar 4, 2020 at 14:53
  • $\begingroup$ Please, explain how. $\endgroup$
    – Jon
    Mar 4, 2020 at 14:54
  • $\begingroup$ reduces the differential equation to dependence on single variable $y$ ... again see my comment in original post $\endgroup$ Mar 4, 2020 at 15:03

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