Analytic Solution to: $ n_t = C_1(n_x^2 + nn_{xx}) \text{ where }n(-L,t)=n(L,t)=0 \text{ and }n(x,0)=e^{-x^2} $

Question

I am working with a nonlinear PDE and am looking for an analytical solution. I'm unsure how to figure out if a PDE has a known solution, so I figured that someone here may know.

This is the PDE: $$ \frac{\partial n}{\partial t} = C_1 \frac{\partial}{\partial x}\left[n \frac{\partial n}{\partial x}\right], \quad \text{where $C_1$ is some constant}. $$ Put another way, $$ n_t = C_1(n_x^2 + nn_{xx}), \quad \text{where $C_1$ is some constant}. $$ For this particular problem: $$ n(-L, t) = n(L, t) = 0 \quad\text{ and }\quad n(x,0) = e^{-x^2}.$$

I've done some rudimentary numerical analysis in Mathematica and have obtained the following information about the PDE

3D Plot of n(x,t) vs. x vs. t

2D Plot of n(x,t) vs. x with traces in time

The distance between the roots of n(x,t) as a function of time

From this analysis, the solution appears to be gaussian for small values of $t$ and take on characteristics of a downward opening quadratic centered at the origin for larger times. Additionally, the roots of this "quadratic" have a distance that varies as: $$ |x^*_2 - x_1^*|(t) \propto \alpha(C_1)t^{1/3}, $$ where $x^*$ is the location of the roots and $\alpha(C_1)$ describes some dependence on the coefficient of this relation. For $C_1=1$, we found that $\alpha(1)\approx 4$.

What was also analyzed, but not included because I can't access Mathematica when at home, was the way that the peak of the "quadratic" changes in time. The maximum value of our function's solution occurs at $x=0$, and the way that it varies in time goes as: $$ h(0, t) \propto \beta(C_1)t^{-1/3}, $$ where, similar to the distance between the roots function, $\beta(C_1)$ describes some dependence of the coefficient of the height function on $C_1$. We also found that $\beta(1)\approx\frac{2}{3}$.

If you have made it this far, you can see that I have no idea what I'm doing, so any help or guidance would be greatly appreciated. Thank you!

Answer 1

By a product ansatz you get separation of variables

$$ \frac{1}{f(t)^2 g(x)} \left( \ \partial_t \ (f(t) \ g(x)), \quad \partial_x\ \left(\ f(t)\ g(x) \ \partial_x \ (f(t) \ g(x)) \ \right) \ \right) = \left(\ \frac{f'(t)}{f(t)^2},\quad g''(x)+\frac{g'(x)^2}{g(x)}\ \right) $$

The two terms can be equal for all $t,x$ only if they are both the same constant.

The time dependent equation is easy

    ft[t_, \[Sigma]_, t0_] = 
    DSolveValue[Derivative[1][f][t]/f[t]^2 == -1/\[Sigma]^3, f[t], t] /
     {C[1] -> Subscript[t, 0]/\[Sigma]^3}

$$ft(t,\sigma ,t_0)= \frac{\sigma ^3}{t-t_0}$$

The solution of the x dependent part applies the technique to exchange the variables $t\to g(t) , g\to t(g)$

   DSolveValue[g'[x]^2/g[x] + g''[x] == -k^2, g[x], x]

$$\text{InverseFunction}\left[-\frac{\text{$\#$1}^2 \sqrt{1-\frac{2 \text{$\#$1}^3 k^2}{3 c_1}} \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{5}{3};\frac{2 k^2 \text{$\#$1}^3}{3 c_1}\right)}{2 \sqrt{-\frac{2 \text{$\#$1}^3 k^2}{3}+c_1}}\&\right][x+c_2]$$

This can be simplied to

$$\text{InverseFunction}\left[ g \ \mapsto \ \frac{ g^2 \ \sqrt{3 \sigma }} {2 \ k \ \sigma ^2} \quad _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{5}{3};\frac{2 g^3}{\sigma ^3}\right) \right][x-\text{x0}]$$

In numerical graphics with parametric plots, such solutions can be used easily by exchanging the axes.