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Deane Yang
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The first thing to check isIf you switch the symbolsecond and the characteristic variety (where the symbol is degenerate). Here, the symbol is $\sigma = A_1\xi_1 + A_2\xi_2 + A_3\xi_3$, where $A_1$, $A_2$, $A_3$ are the matrix coefficientsthird rows of $\partial_x$, $\partial_y$, and $\partial_z$. For fixed $(x,y,z)$your system, the characteristic varietydifferential operator is given by the equation $$ 0 = \det \sigma = f_1f_2f_3\xi_1\xi_2\xi_3. $$ Therefore, assuming $f_1, f_2, f_3$ are all nonzero, the system is nondegenerate, andsame as the characteristic variety consists of 3 transversal planes. This is preserved under changelinearization of variablesequation (4. In particular, if3) in Existence of elastic deformations with prescribed principal strains and triply orthogonal systems at a point where \begin{align*} u &= x+y+z\\ v &= y\\ w &= z, \end{align*}$$ g_{\alpha\beta} = \delta_{\alpha\beta} \text{ and }\frac{\partial x^\alpha}{\partial y^i} = \delta_{\alpha i}. $$ thenAs mentioned immediately after the equation becomes $$ (A_1+A_2+A_3)\partial_u N + A_2\partial_vN + A_3\partial_wN = \omega. $$ Since $A_1+A_2+A_3$, this system is invertiblenot (if $f_1,f_2,f_3$ are nonzerosymmetric), you hyperbolic. It also can multiply both sides by its inverse and get an equation of the form $$ \partial_u N + B\partial_vN + C\partial_wN = \hat\omega. $$ The symbol ofnever be elliptic. It follows that this system looks like $$ \hat\sigma = I\hat\xi_1 + B\hat\xi_2 + C\hat\xi_3. $$ Since the characteristic variety still consists of 3 distinct planes, this implies that $\hat\sigma$ is smoothly diagonalizabledifficult to solve. Therefore, the system is a strictly hyperbolic system, and the initial value problem, where $N$ is prescribed on the plane $u = 0$, is well-posed and therefore has a solution on allI don't know of $\mathbb{R}^3$, if $f_1, f_2, f_3$ never vanishany work in this direction.

In factthe paper cited, by using other linear changeswe observe that the obstruction is due to a type of variables,gauge invariance and reformulate the initial value problem with $N$ prescribed on any $2$-plane, where $\xi_1\xi_2\xi_3\ne 0$in a way that eliminates this issue, is well-posedresulting in a symmetric hyperbolic system that can be solved.

If $f_1f_2f_3 = 0$ somewhere, then it is possible, with some work piecing together solutions of different initial value problems, to identify the largest domain on which there is a solution.Perhaps your problem can be treated similarly?

The first thing to check is the symbol and the characteristic variety (where the symbol is degenerate). Here, the symbol is $\sigma = A_1\xi_1 + A_2\xi_2 + A_3\xi_3$, where $A_1$, $A_2$, $A_3$ are the matrix coefficients of $\partial_x$, $\partial_y$, and $\partial_z$. For fixed $(x,y,z)$, the characteristic variety is given by the equation $$ 0 = \det \sigma = f_1f_2f_3\xi_1\xi_2\xi_3. $$ Therefore, assuming $f_1, f_2, f_3$ are all nonzero, the system is nondegenerate, and the characteristic variety consists of 3 transversal planes. This is preserved under change of variables. In particular, if \begin{align*} u &= x+y+z\\ v &= y\\ w &= z, \end{align*} then the equation becomes $$ (A_1+A_2+A_3)\partial_u N + A_2\partial_vN + A_3\partial_wN = \omega. $$ Since $A_1+A_2+A_3$ is invertible (if $f_1,f_2,f_3$ are nonzero), you can multiply both sides by its inverse and get an equation of the form $$ \partial_u N + B\partial_vN + C\partial_wN = \hat\omega. $$ The symbol of this system looks like $$ \hat\sigma = I\hat\xi_1 + B\hat\xi_2 + C\hat\xi_3. $$ Since the characteristic variety still consists of 3 distinct planes, this implies that $\hat\sigma$ is smoothly diagonalizable. Therefore, the system is a strictly hyperbolic system, and the initial value problem, where $N$ is prescribed on the plane $u = 0$, is well-posed and therefore has a solution on all of $\mathbb{R}^3$, if $f_1, f_2, f_3$ never vanish.

In fact, by using other linear changes of variables, the initial value problem with $N$ prescribed on any $2$-plane, where $\xi_1\xi_2\xi_3\ne 0$, is well-posed.

If $f_1f_2f_3 = 0$ somewhere, then it is possible, with some work piecing together solutions of different initial value problems, to identify the largest domain on which there is a solution.

If you switch the second and third rows of your system, the differential operator is the same as the linearization of equation (4.3) in Existence of elastic deformations with prescribed principal strains and triply orthogonal systems at a point where $$ g_{\alpha\beta} = \delta_{\alpha\beta} \text{ and }\frac{\partial x^\alpha}{\partial y^i} = \delta_{\alpha i}. $$ As mentioned immediately after the equation, this system is not (symmetric) hyperbolic. It also can never be elliptic. It follows that this system is difficult to solve. I don't know of any work in this direction.

In the paper cited, we observe that the obstruction is due to a type of gauge invariance and reformulate the problem in a way that eliminates this issue, resulting in a symmetric hyperbolic system that can be solved.

Perhaps your problem can be treated similarly?

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Deane Yang
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The first thing to check is the symbol and the characteristic variety (where the symbol is degenerate). Here, the symbol is $\sigma = A_1\xi_1 + A_2\xi_2 + A_3\xi_3$, where $A_1$, $A_2$, $A_3$ are the matrix coefficients of $\partial_x$, $\partial_y$, and $\partial_z$. For fixed $(x,y,z)$, the characteristic variety is given by the equation $$ 0 = \det \sigma = f_1f_2f_3\xi_1\xi_2\xi_3. $$ Therefore, assuming $f_1, f_2, f_3$ are all nonzero, the system is nondegenerate, and the characteristic variety consists of 3 transversal planes. This is preserved under change of variables. In particular, if \begin{align*} u &= x+y+z\\ v &= y\\ w &= z, \end{align*} then the equation becomes $$ (A_1+A_2+A_3)\partial_u N + A_2\partial_vN + A_3\partial_wN = \omega. $$ Since $A_1+A_2+A_3$ is invertible (if $f_1,f_2,f_3$ are nonzero), you can multiply both sides by its inverse and get an equation of the form $$ \partial_u N + B\partial_vN + C\partial_wN = \hat\omega. $$ The symbol of this system looks like $$ \hat\sigma = I\hat\xi_1 + B\hat\xi_2 + C\hat\xi_3. $$ Since the characteristic variety still consists of 3 distinct planes, this implies that $\hat\sigma$ is smoothly diagonalizable. Therefore, the system is a strictly hyperbolic system, and the initial value problem, where $N$ is prescribed on the plane $u = 0$, is well-posed and therefore has a solution on all of $\mathbb{R}^3$, if $f_1, f_2, f_3$ never vanish.

In fact, by using other linear changes of variables, the initial value problem with $N$ prescribed on any $2$-plane, where $\xi_1\xi_2\xi_3\ne 0$, is well-posed.

If $f_1f_2f_3 = 0$ somewhere, then it is possible, with some work piecing together solutions of different initial value problems, to identify the largest domain on which there is a solution.