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I've reduced the PDE $x^2 u_{xx}-y^2 u_{yy} -xu_x - yu_y $ to canonical form, the solution in general is $u = f(\phi) + g(\psi)$, where $\phi = xy$, $\psi = \frac{y}{x}$

Now I am given boundary conditions $u = x^6 + x^{-1}$ and $u_x = 2x^5 + x^{-2}$ when $y=x^2$ and $x\ \ge 1$.

I can impose these boundary conditions fine, and find a solution that satisfies them. I'm now trying to find the region of the plane where the solution is uniquely determined and I am having some difficulty.

I know intuitively that the information on the boundary has to "travel" along the characteristic curves and so the solution should be defined wherever you can sort of trace your way back to the initial data (is this the right intuition?) However I am having trouble applying that to this problem and finding the correct region of the plane.

Thank you for any help

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  • $\begingroup$ What are $\phi$ and $\psi$? $\endgroup$ – user147263 Jun 10 '14 at 23:19
  • $\begingroup$ Have added this to the post, thanks $\endgroup$ – Wooster Jun 11 '14 at 7:38
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Let's first deal with this formally. On the curve $y=x^2$, $x\ge 1$, we have $\phi\ge 1$ and $\psi\ge 1$. By imposing the given conditions, you determine the values of $f(\phi)$ and $g(\psi)$ for $\phi,\psi\ge 1$. These function can do whatever they want on $(-\infty,-1)$. In terms of $xy$-plane, the region $\phi,\psi\ge 1$ is the disjoint union
$$\{(x,y): y\ge x, \ y\ge 1/x,\ x>0 \} \cup \{(x,y): y\le x,\ y\le 1/x, \ x<0 \} $$ The solution is determined here and not elsewhere.

But... isn't this weird? How does the boundary condition propagate to the 3rd quadrant from the first, across a sizable gap in which it has no influence?

The issue is the interpretation of the PDE at $(0,0)$, where it degenerates due to coefficients being zero. At this point we don't really have a PDE. Observe that the term $g(y/x)$ is not even continuous there, unless it's constant. This leads to an observation: $g(y/x)$ is not a most general form of that term. As written, it must have the same value at antipodal points $(x,y)$ and $(-x,-y)$; but this constraint isn't present in the problem. The term $g(y/x)$ should be replaced with $g(\theta)$, where $\theta$ is the polar angle ans $g$ is a smooth $2\pi$-periodic function. Then we see that $g$ is determined for $\pi/4 < \theta < \pi/2$, and not elsewhere. The ultimate conclusion is that the region of uniqueness is $$\{(x,y): y\ge x, \ y\ge 1/x,\ x>0 \} $$

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  • $\begingroup$ I'm having trouble understanding the first part of this answer. How does the condition on the boundary affect where the solution is determined in other parts of the plane? $\endgroup$ – Wooster Jun 11 '14 at 15:33
  • $\begingroup$ @Wooster When you "impose these boundary conditions fine, and find a solution that satisfies them", think of what those computations actually mean. This is not just "finding a solution": you make a conclusion about the behavior of $f$ and $g$ in a certain interval. $\endgroup$ – user147263 Jun 11 '14 at 15:35
  • $\begingroup$ Ah yes, so we have only defined $f$ and $g$ on a certain region. The boundary conditions tell us nothing about what $f$ and $g$ might be doing elsewhere. That makes sense, thank you! $\endgroup$ – Wooster Jun 11 '14 at 15:43

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