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Suppose I want to solve $u_{xx} + 2u_{xy} + u_{yy} = 0$ by operator factoring. This is $(D^2 + 2DE + E^2)[u] = 0$, where $D$ is a differential operator with respect to $x$ and $E$ is a differential operator with respect to $y$. The substitution $v = (D + E)[u]$ yields $v_x + v_y = 0$, whose solution is $v = f(y - x)$.

Thus $u_x + u_y = f(y - x)$. The homogeneous solution to this PDE is $u_{_h} = g(y - x)$. I am trying to guess a particular solution. I believe the answer should be $u_p = xF(w)$, where $w = y - x$ and $F$ is the antiderivative of $f$ with respect to $w$. However, when I plug this in I get $F(w) + xf(w)w_x + xf(w)w_y = f(w)$, and thus $F(w) = f(w)$. What am I doing wrong here?

user10478
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  • I literally gave you the answer on your other question. Anyway, if $u_{p} = x F(y-x)$ then $$(u_{p}){x} + (u{p})_{y} = F(y-x) + x(-F'(y-x)) + x(+F'(y-x)) = F(y-x)$$ and so yes, $F = f$. – Matthew Cassell Nov 29 '21 at 06:50
  • Sorry I didn't ask in that question first. That question was focused on a different technique and so I didn't recall that you had commented on operator factoring. The part that confuses me is that it feels odd to get $F = f$. If the solution satisfies the PDE, then I should get a truism when I plug it back in. – user10478 Nov 29 '21 at 07:11
  • I'm not sure why you think it feels odd, nor what you mean by getting a truism. I also don't understand why you 'believe the answer should be $u_{p} = x F(w)$, where $w=y-x$ and $F$ is the antiderivative of $f$ with respect to $w$. There is no reason to assume this. Note that the inhomogeneity $f(y-x)$ of the PDE lies in the nullspace of the PDE and with that in mind, consider what the solutions to, say, the ODE $u' + u = e^{-x}$ look like. What happens to the inhomogeneity here? Is the solution proportional to $x e^{-x}$ or $x \int e^{-x} dx$? – Matthew Cassell Nov 29 '21 at 10:33
  • @mattos I see your point on the comparison to the ODE. I was following this instruction (https://youtu.be/KFS_Fs1ZGRw?t=547) when I guessed the antiderivative factor. The instructor also introduced a constant and absorbed it into $F$ later, but that shouldn't account for the difference. Would you say this part of the video is incorrect? – user10478 Nov 29 '21 at 17:23
  • As for "feeling odd" and "truism," I should more clearly say that when I plug a particular solution into the LHS, I expect to get an expression identically equal to the RHS, as opposed to $F = f$. – user10478 Nov 29 '21 at 17:25
  • The person in the other video is looking at a problem that is different to ours. That person is solving $$u_{t} + cu_{x} = f(x+ct)$$ Notice that $f(x+ct)$ is not a solution to the homogeneous problem $u_{t} + cu_{x} = 0$. You are solving (letting $y = t$ and introducing $c$ for consistency with the video problem, where in our problem $c=1$) $$u_{t}+cu_{x}=f(x-ct)$$ Notice that $f(x-ct)$ is a solution to the homogeneous problem. Now if you guess the solution $F(x-ct)$ like in the video with $F'=f$, then you end up with $$0=f(x-ct)$$ which doesn't satisfy the equation for arbitrary $f$. – Matthew Cassell Nov 29 '21 at 17:53
  • @mattos Okay, this is where my understanding of the heuristics for guessing from ODE world break down. I see the nullspace issue and the need to multiply the guess by $x$, but not why the solution should be an antiderivative in one case and not the other. Undetermined Coefficients seems like a dark art quickly turning into "solve the DE in Mathematica and make your guess according to the form of the solution." – user10478 Nov 29 '21 at 18:40

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