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I came across the following PDE for the first time today:

$$ \frac{\partial}{\partial t} (u^2(x,t))=\frac{\partial^2 u}{\partial x^2} $$

And with the help of Wolfram Alpha, I found its solution to be $u(x,t) = -\frac{c_1^2 \tanh(c_1 x +c_2 t +c_3)}{c_2}$. I checked that indeed this furnishes a solution but I was wondering a) is the solution unique given boundary / initial conditions and b) how do we find it in the first place?

Cyclone
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    One way you might be able to do this is to search for travelling wave solutions, assume $u(x,t)=f(x-ct)=f(\xi)$, then find explicitly a form for $f$. – DaveNine Nov 22 '17 at 05:44
  • Yes, you are right, if I search for solutions $u = f(c_1x+c_2t)$ then I find Wolfram Alphas answer. Thank you. – Cyclone Nov 22 '17 at 11:08
  • Generally speaking, when should I try a travelling wave solution? Are there certain properties of my PDE that should make me think of travelling wave solutions? – Cyclone Nov 22 '17 at 18:19
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    Just something to try, especially when your equation is quasi-linear. – DaveNine Nov 22 '17 at 18:42

1 Answers1

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It's solvable with separation of variables. Let $u(x,y) = X(x)T(t)$ then

$$ 2X^2 T\ T' = X''T $$

or

$$ 2T' = \frac{X''}{X^2} = \lambda $$

The $X$ equation is non-linear but can be transformed into a seperable ODE $$ X'' X' = \lambda X^2 X' $$ $$ (X')^2 = \frac{2\lambda}{3} X^3 + \mu $$

Dylan
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  • You should be able to write the solution to the nonlinear one implicitly with an integral, no? – DaveNine Nov 22 '17 at 04:56
  • As an integral form yes. Although the solution is non-elementary – Dylan Nov 22 '17 at 08:08
  • Thank you for your reply. I thought about a product ansatz, too. But this will not give us the tanh solution from Wolfram Alpha, will it? Also, the nonlinear ode $(X')^2 = 2\lambda/3 X^3 +\mu$ seems not to be solvable analytically, do you agree? – Cyclone Nov 22 '17 at 10:53
  • @Cyclone it is impossible that it would because $T(t)=\frac{\lambda}{2}t+C$ has no way, by multiplication, to me incorporated into the solution. – DaveNine Nov 22 '17 at 17:23
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    Further, it is worth noting that the hyperbolic tangent solution need not be unique. – DaveNine Nov 22 '17 at 17:25
  • Yes, I agree. Basically we have constructed to distinct solutions (although we have not specified boundary/initial conditions). So we have proved the non-uniqueness. – Cyclone Nov 22 '17 at 18:17