# Solve analytically coupled nonlinear PDE?

I want to solve this coupled nonlinear set of equations:

$$\frac{\partial f}{\partial t}=\frac{\partial^2 f}{\partial x^2}+ag^n$$ $$\frac{\partial g}{\partial t}=\frac{\partial^2 g}{\partial x^2}+b\frac{\partial f}{\partial x}$$ When $$n = 1$$, this is a linear system of equations and the wave ansatz $$e^{\lambda t - ikx}$$ is a solution. I thus expect waves to remain a solution, but a pretty different one. I am particularly interested in the cases $$n=2$$ and $$n=3$$. When it concerns PDEs I am very rusty, let alone nonlinear ones. Is there a recipe to solve this analytically? Note that if the solution involves an infinite sum of wave modes, I am only interested in the first mode. Small values of $$a$$ are also ok.

EDIT: I am also potentially interested in the solution of this coupled nonlinear set of equations: $$\frac{\partial f}{\partial t}=ag^n$$ $$\frac{\partial g}{\partial t}=b\frac{\partial f}{\partial x}$$ The second-order derivatives likely have an important impact on the solution, but it is not impossible for the behavior I am interested in to still occur in these simplified equations. In this case, we can differentiate the equation for $$g$$ with respect to $$t$$ and substitute the equation for $$f$$ and get an equation solely for the variable $$g$$: $$\frac{\partial^2 g}{\partial t^2}=ab\frac{\partial g^n(x,t^\prime)}{\partial x}=nabg^{n-1}\frac{\partial g(x,t^\prime)}{\partial x}.$$ How do I solve this?

• For problems of this type, you usually start with a phase plane analysis. Let $f = F(x - ct), g = G(x - ct)$ to get \begin{align} -c F' &= F'' + aG^{n} \\ -c G' &= G'' + bF' \end{align} then reduce to a system of first order ODEs by setting $F' = A, G' = B$. Commented Jun 9, 2023 at 13:25
• @MatthewCassell ah ok thanks! But how to you deal with the $G^n$ term? You integrate $G^\prime=B$? I should probably mention that this is a local analysis and there are thus no boundary conditions. Commented Jun 9, 2023 at 14:49
• When you say you are only interested in the first mode, does that mean that you are more interested in the behavior of this mode than the actual form of an analytic solution (unlikely to exist)? One can likely get some estimates on low frequency behavior much much easier than an analytic solution Commented Jun 10, 2023 at 12:55
• Yes exactly. If it can help, I'm still interested at the behavior for a small nonlinearity, ie. $a$ is small, as well as the behavior close to where this mode has zero growth rate. In fact, I already solved a more complicated version of this set of equations numerically. I get that the solution saturates as $a$ is increased from its value where the solution has zero growth rate, ie. the growth rate stays at zero. This behavior is unexpected (for $n=2$ and 3), so I wanted to see if solving this set of equations analytically also gives me this behavior, which I expect to be the case. Commented Jun 10, 2023 at 13:19
• You might see how far you get looking at the stationary solutions first. Commented Jun 10, 2023 at 14:48

HINT: This is not a complete solution as requested by OP.
We can find closed solutions for some special cases $$n=2, b=0$$ (Maple CAS):

$$f\! \left(x,t\right) = -\frac{a c_{2}^{2} c_{3}^{2} {\mathrm e}^{2 k_{1}^{2} t-2 k_{1} x}}{2 k_{1}^{2}}-\frac{a c_{1}^{2} c_{3}^{2} {\mathrm e}^{2 k_{1}^{2} t+2 k_{1} x}}{2 k_{1}^{2}}+\frac{a c_{1} c_{2} c_{3}^{2} {\mathrm e}^{2 k_{1}^{2} t}}{k_{1}^{2}}-\frac{a c_{5} c_{3}^{2} {\mathrm e}^{-\sqrt{2}\, k_{1} x+2 k_{1}^{2} t}}{2 c_{6} k_{1}^{2}}+c_{6} c_{4} {\mathrm e}^{k_{2}^{2} t+k_{2} x}+c_{6} c_{5} {\mathrm e}^{k_{2}^{2} t-k_{2} x}-\frac{a c_{4} c_{3}^{2} {\mathrm e}^{2 k_{1}^{2} t+\sqrt{2}\, k_{1} x}}{2 c_{6} k_{1}^{2}}$$

$$g\! \left(x,t\right)=c_{1} c_{3} {\mathrm e}^{k_{1}^{2} t+k_{1} x}+c_{2} c_{3} {\mathrm e}^{k_{1}^{2} t-k_{1} x}$$

Special case $$n=3, b=0$$:

$$f\! \left(x,t\right) = -\frac{a c_{1}^{3} c_{3}^{3} {\mathrm e}^{3 k_{1}^{2} t+3 k_{1} x}}{6 k_{1}^{2}}-\frac{a c_{2}^{3} c_{3}^{3} {\mathrm e}^{3 k_{1}^{2} t-3 k_{1} x}}{6 k_{1}^{2}}+\frac{3 a c_{2} c_{1}^{2} c_{3}^{3} {\mathrm e}^{3 k_{1}^{2} t+k_{1} x}}{2 k_{1}^{2}}+\frac{3 a c_{1} c_{2}^{2} c_{3}^{3} {\mathrm e}^{3 k_{1}^{2} t-k_{1} x}}{2 k_{1}^{2}}+\frac{a c_{4} c_{3}^{3} {\mathrm e}^{3 k_{1}^{2} t+\sqrt{3}\, k_{1} x}}{6 c_{6} k_{1}^{2}}+\frac{a c_{5} c_{3}^{3} {\mathrm e}^{-\sqrt{3}\, k_{1} x+3 k_{1}^{2} t}}{6 c_{6} k_{1}^{2}}+c_{6} c_{4} {\mathrm e}^{k_{2}^{2} t+k_{2} x}+c_{6} c_{5} {\mathrm e}^{k_{2}^{2} t-k_{2} x}$$

$$g\! \left(x,t\right)=c_{1} c_{3} {\mathrm e}^{k_{1}^{2} t+k_{1} x}+c_{2} c_{3} {\mathrm e}^{k_{1}^{2} t-k_{1} x}$$

But don't know if there are closed solutions for $$b\neq 0$$ or general $$n\in \mathbb{Z}$$.

Visualization $$[f(x,t),g(x,t)]$$ with example values $$n=2, a=1, b=0, k_1=i, k_2=2 i, c_1=c_2=c_3=c_4=c_5=c_6=1$$:

For case $$n=1, b\neq 0$$ we'll get expressions like this in the solution:
$${\mathrm e}^{\frac{\left(\mathit{RootOf}\left(\textit{_}Z^{3} c_{1}^{6}+4 c_{2} \textit{_}Z^{2} c_{1}^{4}+5 c_{2}^{2} \textit{_}Z c_{1}^{2}-c_{1}^{3} a b+2 c_{2}^{3}, \mathit{index}=1\right) c_{1}^{2}+2 c_{2}\right) \left(c_{1} x+c_{2} t+c_{3}\right)}{c_{1}^{2}}}$$
It means that we have to determine the roots of a third order polynomial in $$\textit{_}Z$$.
• Well, with $b=0$, the equation for $g$ is no longer coupled to the one for $f$ and is a simple heat equation. So you solve it by separation of variables and then you plug that solution into the equation for $f$, which in turn becomes an inhomogeneous heat equation, whose solution can also be easily found on the internet. But with $b\neq 0$, which is absolutely necessary in my case, things change completely. Like I said, waves are supposed to be solutions of this equation, and in your plot we see that the the solutions with $b=0$ decay; they are as expected heat equation solutions. Commented Jun 10, 2023 at 11:25
• Hmm, for your addendum, I am a bit confused. This is the linear problem so I used the ansätze $f(x,t)=f_0 exp(\lambda t-ikx)$ and $g(x,t)=g_0 exp(\lambda t-ikx+i\phi)$ and could get something much simpler. However, to be more exact maybe they should be sums for exponents with opposite signs. But I followed what was done in a textbook and some papers. Or maybe this is just another way to write the solution. I think you can make your life harder and have to solve for a cubic equation actually... In any case thanks for the help! Commented Jun 10, 2023 at 13:28