Mixing
The Jacobian and Particular Solutions to an Underdetermined Equation
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I was wondering if the factor $sqrt{(x')^2 + (y')^2}$ in the line integral formula $$int_a^b f(x, y) sqrt{(x')^2 + (y')^2} dt$$ can also be thought of as a Jacobian determinant, due to the resemblance of parameterization to changing variables in multiple integrals.
For this to be the case, there must be at least one solution to the partial differential equation $$(x')^2 + (y')^2 = a(x, y)^2d(x, y)^2 - 2a(x, y)b(x, y)c(x, y)d(x, y) + b(x, y)^2c(x, y)^2,$$ which would correspond to the Jacobian $$begin{bmatrix}a(x, y) c(x, y) \ b(x, y) d(x, y)end{bmatrix}.$$
It should be noted that I don't know any methods for solving PDEs other than guessing, but I doubt analytic methods exist for anything this multivariate. My initial attempt was to let $a^2d^2 = x'$ and $b^2c^2 = y'$, but it follows that $-2abcd = 0$, so at least one of $a$, $b$, $c$, and $d$, must equal $0$, which messes up at least one of the other terms. I am aware that the Jacobian in a change of variables is interchangeable with a certain differential manipulation involving the exterior product, but since $dt Lambda dt = 0$, its not clear that this process is more easily reversible than the Jacobian method. Is there a way to solve this PDE and/or the equivalent exterior algebra problem?
pde exterior-algebra line-integrals jacobian multivariate-polynomial
edited Jan 23 at 21:05
gt6989b
34.9k22557
asked Jan 23 at 21:00
user10478user10478
472211
$endgroup$
add a comment |
$begingroup$
I was wondering if the factor $sqrt{(x')^2 + (y')^2}$ in the line integral formula $$int_a^b f(x, y) sqrt{(x')^2 + (y')^2} dt$$ can also be thought of as a Jacobian determinant, due to the resemblance of parameterization to changing variables in multiple integrals.
For this to be the case, there must be at least one solution to the partial differential equation $$(x')^2 + (y')^2 = a(x, y)^2d(x, y)^2 - 2a(x, y)b(x, y)c(x, y)d(x, y) + b(x, y)^2c(x, y)^2,$$ which would correspond to the Jacobian $$begin{bmatrix}a(x, y) c(x, y) \ b(x, y) d(x, y)end{bmatrix}.$$
It should be noted that I don't know any methods for solving PDEs other than guessing, but I doubt analytic methods exist for anything this multivariate. My initial attempt was to let $a^2d^2 = x'$ and $b^2c^2 = y'$, but it follows that $-2abcd = 0$, so at least one of $a$, $b$, $c$, and $d$, must equal $0$, which messes up at least one of the other terms. I am aware that the Jacobian in a change of variables is interchangeable with a certain differential manipulation involving the exterior product, but since $dt Lambda dt = 0$, its not clear that this process is more easily reversible than the Jacobian method. Is there a way to solve this PDE and/or the equivalent exterior algebra problem?
pde exterior-algebra line-integrals jacobian multivariate-polynomial
edited Jan 23 at 21:05
gt6989b
34.9k22557
asked Jan 23 at 21:00
user10478user10478
472211
$endgroup$
add a comment |
$begingroup$
I was wondering if the factor $sqrt{(x')^2 + (y')^2}$ in the line integral formula $$int_a^b f(x, y) sqrt{(x')^2 + (y')^2} dt$$ can also be thought of as a Jacobian determinant, due to the resemblance of parameterization to changing variables in multiple integrals.
For this to be the case, there must be at least one solution to the partial differential equation $$(x')^2 + (y')^2 = a(x, y)^2d(x, y)^2 - 2a(x, y)b(x, y)c(x, y)d(x, y) + b(x, y)^2c(x, y)^2,$$ which would correspond to the Jacobian $$begin{bmatrix}a(x, y) c(x, y) \ b(x, y) d(x, y)end{bmatrix}.$$
It should be noted that I don't know any methods for solving PDEs other than guessing, but I doubt analytic methods exist for anything this multivariate. My initial attempt was to let $a^2d^2 = x'$ and $b^2c^2 = y'$, but it follows that $-2abcd = 0$, so at least one of $a$, $b$, $c$, and $d$, must equal $0$, which messes up at least one of the other terms. I am aware that the Jacobian in a change of variables is interchangeable with a certain differential manipulation involving the exterior product, but since $dt Lambda dt = 0$, its not clear that this process is more easily reversible than the Jacobian method. Is there a way to solve this PDE and/or the equivalent exterior algebra problem?
pde exterior-algebra line-integrals jacobian multivariate-polynomial
edited Jan 23 at 21:05
gt6989b
34.9k22557
asked Jan 23 at 21:00
user10478user10478
472211
$endgroup$
I was wondering if the factor $sqrt{(x')^2 + (y')^2}$ in the line integral formula $$int_a^b f(x, y) sqrt{(x')^2 + (y')^2} dt$$ can also be thought of as a Jacobian determinant, due to the resemblance of parameterization to changing variables in multiple integrals.
For this to be the case, there must be at least one solution to the partial differential equation $$(x')^2 + (y')^2 = a(x, y)^2d(x, y)^2 - 2a(x, y)b(x, y)c(x, y)d(x, y) + b(x, y)^2c(x, y)^2,$$ which would correspond to the Jacobian $$begin{bmatrix}a(x, y) c(x, y) \ b(x, y) d(x, y)end{bmatrix}.$$
It should be noted that I don't know any methods for solving PDEs other than guessing, but I doubt analytic methods exist for anything this multivariate. My initial attempt was to let $a^2d^2 = x'$ and $b^2c^2 = y'$, but it follows that $-2abcd = 0$, so at least one of $a$, $b$, $c$, and $d$, must equal $0$, which messes up at least one of the other terms. I am aware that the Jacobian in a change of variables is interchangeable with a certain differential manipulation involving the exterior product, but since $dt Lambda dt = 0$, its not clear that this process is more easily reversible than the Jacobian method. Is there a way to solve this PDE and/or the equivalent exterior algebra problem?
pde exterior-algebra line-integrals jacobian multivariate-polynomial
pde exterior-algebra line-integrals jacobian multivariate-polynomial
edited Jan 23 at 21:05
gt6989b
34.9k22557
asked Jan 23 at 21:00
user10478user10478
472211
edited Jan 23 at 21:05
gt6989b
34.9k22557
asked Jan 23 at 21:00
user10478user10478
472211
edited Jan 23 at 21:05
gt6989b
34.9k22557
edited Jan 23 at 21:05
gt6989b
34.9k22557
edited Jan 23 at 21:05
gt6989b
34.9k22557
34.9k22557
asked Jan 23 at 21:00
user10478user10478
472211
asked Jan 23 at 21:00
user10478user10478
472211
asked Jan 23 at 21:00
user10478user10478
472211
472211
add a comment |
add a comment |
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