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Let $\phi:\mathbb{R}_+\times \mathbb{R}_+\to\mathbb{R}$ be a function with continuous partial derivatives in both arguments which we denote by $\partial_1$ and $\partial_2$. Consider the following differential equation: \begin{align} &\partial_1[\phi(t,x)+2\phi(t,y)]+\partial_2[\phi(t,x)+2\phi(t,y)] +6\phi(t,0)-2\phi(t,x)\phi(t,y)-\phi(t,y)^2=0\;,\\[1ex] &0\leq t<\infty\;,\;0\leq x<y\;. \end{align} This is not a PDE in the classical sense.

Question: Does the equation have a solution?

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  • $\begingroup$ I do not fully understand your notation: why don' t you consider a map $t\mapsto \phi_t$, with $\phi_t=\phi_t(x,y)$, instead? You could try to define your PDE involving the derivatives w.r.t. $x$ and $y$ for $any$ $t\geq 0$. Does it convince you? $\endgroup$
    – Avitus
    Aug 13, 2013 at 8:15
  • $\begingroup$ By continuity you can take the limit $x\to y$ and show that necessarily your function $\phi$ must solve also the equation $$ \partial_t \phi + \partial_x\phi - \phi^2 = -2 f$$ where $f(t) = \phi(t,0)$ is independent of $x$. So it may help to consider the solvability of the reduced equation above. $\endgroup$ Aug 13, 2013 at 8:33

1 Answer 1

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Define

$$ F(t,x,y) = \phi_1(t,x) + 2\phi_1(t,y) + \phi_2(t,x) + 2 \phi_2(t,y) + 6 \phi(t,0) - 2\phi(t,x) \phi(t,y) - \phi(t,y)^2 $$

Your equation requires $F \equiv 0$ for $t\in [0,\infty)$ and $0\leq x < y$. For any point in the domain we must have that $\mathrm{d}F = 0$ This is equivalent to

$$ \begin{align} \phi_{11}(t,x) + 2\phi_{11}(t,y) + \phi_{12}(t,x) + 2 \phi_{12}(t,y) + 6 \phi_1(t,0) - 2 \phi_1(t,x) \phi(t,y) - 2 \phi(t,x) \phi_1(t,y) - 2 \phi(t,y)\phi_1(t,y) &= 0 \\ \phi_{12}(t,x) + \phi_{22}(t,x) - 2\phi_2(t,x) \phi(t,y) &= 0\\ \phi_{12}(t,y) + \phi_{22}(t,y) - \phi_2(t,y) \phi(t,x) - \phi_2(t,y) \phi(t,y) &= 0 \end{align} $$

Consider the second of the equations. Again this holds for all points $(t,x,y)$ in the relevant domain. Taking a derivative in $y$ then requires that $$ \phi_2(t,x) \phi_2(t,y) = 0 $$ (note that for this we only need to assume that $\phi$ is twice continuously differentiable and not thrice!) Taking the limit $x \to y$ in the domain, by continuity we must have that $\phi_2 = 0$ everywhere!

This implies that $\phi$ is constant in its second variable, and you equation reduces to just a simple ODE $$ \phi'(t) + 2 \phi(t) - \phi(t)^2 = 0 $$ which we can solve to be $$ \frac12 (\log (2 - \phi) - \log\phi) ) + C = t \implies C e^{2t} = \frac{2-\phi}{\phi} \implies \phi(t) = \frac{2}{C e^{2t} + 1}$$

where the constant $C$ on the far right hand side is determined by $\phi(0)$: $$ C = \frac{2}{\phi(0)} - 1$$


In summary, if we assume $\phi$ is twice continuously differentiable, then $\phi$ must be independent of its second argument and

$$ \phi(t,x) = \frac{2}{\left(\frac{2}{\phi(0,0)} - 1\right) e^{2t} + 1} $$

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