I've come across the following simple family of PDE's and am wondering if they fit into a better known class or if they are attack-able by any standard techniques. The equation for $u(x, t)$ with $t \geq 0,x \geq 1$ is $$ u_{tx} = - (1 - u)^n $$ with boundary conditions $$ \begin{cases} u_t(x, 0) = 1 - x \\ u(x, 0) = 0 \\ u(1, t) = 0 \end{cases} $$ with $n \geq 1$ an integer.
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$\begingroup$ what about power series in $t,(x-1)$? but you're missing anyway a boundary condition at $x\to\infty$... $\endgroup$– leo monsaingeonCommented Jun 2, 2014 at 18:53
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$\begingroup$ How can you have $u(x,0)=0$ and $u_x(x,0)=1-x$? $\endgroup$– Michael RenardyCommented Jun 2, 2014 at 18:55
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$\begingroup$ Quite unlikely, indeed... Thanks Michael Renardy for pointing that out. Is there a typo, or should we close? $\endgroup$– leo monsaingeonCommented Jun 2, 2014 at 19:05
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$\begingroup$ It was a typo - thanks for catching it. $\endgroup$– guestCommented Jun 2, 2014 at 19:13
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1$\begingroup$ Ummm.. This is just the wave equation in 2d in null coordinates with nonlinearity $-(1-u)^n$. Is that where you got it from? $\endgroup$– k3thompsCommented Jun 2, 2014 at 19:14
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1 Answer
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I would set $u_x=v$ and reformulate the problem as $$v_t(x,t)=-(1-\int_1^x v(y,t)\,dy)^n, v(x,0)=0.$$ In this formulation, existence, uniqueness and numerical solution all become fairly straightforward. I doubt that there is much hope for a more explicit solution.
answered Jun 2, 2014 at 21:44
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$\begingroup$ The constraint $u(1,t)=0=u(x,0)$ are used, but what about $u_t(x,0)=1-x$? I don't see it right away. $\endgroup$– usernameCommented Jun 2, 2014 at 23:25
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1$\begingroup$ It is automatic. You have $v_t(x,0)=u_{xt}(x,0)=-1$, and then you integrate using the boundary condition at $x=1$. $\endgroup$ Commented Jun 2, 2014 at 23:31