Как да използваме якобийци

2023 Автор: Virginia Parson | [email protected]. Последно модифициран: 2023-08-03 17:57

Якобианската промяна на променливите е техника, която може да се използва за решаване на интеграционни проблеми, които иначе биха били трудни с помощта на нормални техники. Якобианът е матрица от частични производни от първи ред на векторнозначна функция.

Целта на якобианската промяна на променливите е да се преобразува от физическо пространство, определено по отношение на x { displaystyle x}

and y{displaystyle y}

variables to a parameter space defined in terms of u(x, y){displaystyle u(x, y)}

and v(x, y){displaystyle v(x, y)}

when applied to integration, finding the determinant of the jacobian will be essential to ensure that the magnitude is correct.

steps

part 1 of 3: derivation of the jacobian matrix

step 1. consider a position vector r=xi+yj{displaystyle \mathbf {r} =x\mathbf {i} +y\mathbf {j} }

here, i{displaystyle \mathbf {i} }

and j{displaystyle \mathbf {j} }

are the unit vectors in a two-dimensional cartesian coordinate system.

step 2. take partial derivatives of r{displaystyle \mathbf {r} }

with respect to each of the parameters.

this is the first step in converting to parameter space.

dru=(∂x∂ui+∂y∂uj)du{displaystyle \mathrm {d} mathbf {r} _{u}=\left({frac {partial x}{partial u}}\mathbf {i} +{frac {partial y}{partial u}}\mathbf {j} right)\mathrm {d} u}
drv=(∂x∂vi+∂y∂vj)dv{displaystyle \mathrm {d} mathbf {r} _{v}=\left({frac {partial x}{partial v}}\mathbf {i} +{frac {partial y}{partial v}}\mathbf {j} right)\mathrm {d} v}

step 3. find the area defined by the above infinitesimal vectors

recall that the area can be written in terms of the magnitude of the cross product of the two vectors.

da=|dru×drv|=|ijk∂x∂u∂y∂u0∂x∂v∂y∂v0|dudv=|∂x∂u∂y∂u∂x∂v∂y∂v|dudv{displaystyle {begin{aligned}\mathrm {d} a&=|\mathrm {d} mathbf {r} _{u}\times \mathrm {d} mathbf {r} _{v}|\\&={begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \{dfrac {partial x}{partial u}}&{dfrac {partial y}{partial u}}&0\\{dfrac {partial x}{partial v}}&{dfrac {partial y}{partial v}}&0\end{vmatrix}}\mathrm {d} u\mathrm {d} v\\&={begin{vmatrix}{dfrac {partial x}{partial u}}&{dfrac {partial y}{partial u}}\\{dfrac {partial x}{partial v}}&{dfrac {partial y}{partial v}}\end{vmatrix}}\mathrm {d} u\mathrm {d} v\end{aligned}}}

step 4. arrive at the jacobian

the determinant above is the jacobian determinant. a shorthand notation can be written as below, where we remember that we convert to parameter space as defined by the variables on the bottom. should you end up with a negative determinant, neglect the negative sign - only the magnitude matters.

|∂(x, y)∂(u, v)|{displaystyle {begin{vmatrix}{dfrac {partial (x, y)}{partial (u, v)}}\end{vmatrix}}}

step 5. write the area da{displaystyle \mathrm {d} a}

in terms of the inverse jacobian.

the reason why this is more applicable is because normally, we would define our parameters in terms of the physical variables, but then have to solve for the physical variables in order to take partial derivatives. recognizing that the determinant of an inverse is the multiplicative inverse of the determinant detj−1=1detj, {displaystyle \det j^{-1}={frac {1}{det j}}, }

we can skip a step by taking the inverse jacobian determinant first, and then finding its reciprocal to recover the actual determinant that we want.

|∂(u, v)∂(x, y)|{displaystyle {begin{vmatrix}{dfrac {partial (u, v)}{partial (x, y)}}\end{vmatrix}}}

part 2 of 3: example 1

step 1. find ∫d(2x+3y)da{displaystyle \int _{d}(2x+3y)\mathrm {d} a}

over d{displaystyle d}

bounded by the following.

{2x+3y=02x+3y=5{displaystyle {begin{cases}2x+3y=0\\2x+3y=5\end{cases}}}
{3x−y=03x−y=3{displaystyle {begin{cases}3x-y=0\\3x-y=3\end{cases}}}
plotting this on a graph, we see that the domain is a rotated rectangle. integrating over this domain by normal means would be rather tedious, but using jacobian change of variables, this problem is trivial.

step 2. define parameters u{displaystyle u}

and v{displaystyle v}

notice that using our definition, we have changed the integrand to simply u.{displaystyle u.}

u=2x+3y{displaystyle u=2x+3y}
v=3x−y{displaystyle v=3x-y}

step 3. find the inverse jacobian determinant

take partial derivatives with respect to each of the physical variables x{displaystyle x}

and y, {displaystyle y, }

plug them into the inverse jacobian matrix, and take its determinant.

∂u∂x=2; ∂u∂y=3; ∂v∂x=3; ∂v∂y=−1{displaystyle {frac {partial u}{partial x}}=2;\ {frac {partial u}{partial y}}=3;\ {frac {partial v}{partial x}}=3;\ {frac {partial v}{partial y}}=-1}
detj−1=|∂(u, v)∂(x, y)|=|233−1|=−11{displaystyle {begin{aligned}\det j^{-1}&={begin{vmatrix}{dfrac {partial (u, v)}{partial (x, y)}}\end{vmatrix}}\\&={begin{vmatrix}2&3\\3&-1\end{vmatrix}}\\&=-11\end{aligned}}}

step 4. reinvert the determinant

take its magnitude (neglect any negative signs) and relate it to the infinitesimal area.

|∂(x, y)∂(u, v)|=111{displaystyle {begin{vmatrix}{dfrac {partial (x, y)}{partial (u, v)}}\end{vmatrix}}={frac {1}{11}}}
da=11dudv{displaystyle \mathrm {d} a=11\mathrm {d} u\mathrm {d} v}

step 5. evaluate the integral using any means possible

∫d(2x+3y)da=111∫05udu∫03dv=111252(3)=7522{displaystyle {begin{aligned}\int _{d}(2x+3y)\mathrm {d} a&={frac {1}{11}}\int _{0}^{5}u\mathrm {d} u\int _{0}^{3}\mathrm {d} v\\&={frac {1}{11}}{frac {25}{2}}(3)\\&={frac {75}{22}}\end{aligned}}}

part 3 of 3: example 2

step 1. find the centroid of the region d{displaystyle d}

bounded by the following.

{xy=1xy=3{displaystyle {begin{cases}xy=1\\xy=3\end{cases}}}
{x2y=1x2y=2{displaystyle {begin{cases}x^{2}y=1\\x^{2}y=2\end{cases}}}
recall that the centroid is the mean of all the points of the region. the region is defined in such a way as to involve three separate integrals just to find the area. to find the centroid would mean to take several more integrals. this is obviously not the way to go, so we use jacobians to convert this into an easier problem.
- c=(∫dxda∫dda, ∫dyda∫dda){displaystyle c=\left({frac {int _{d}x\mathrm {d} a}{int _{d}\mathrm {d} a}}, {frac {int _{d}y\mathrm {d} a}{int _{d}\mathrm {d} a}}\right)}

step 2. define parameters u{displaystyle u}

and v{displaystyle v}

u=xy{displaystyle u=xy}
v=x2y{displaystyle v=x^{2}y}

step 3. take partial derivatives

use them to find the determinant of the inverse jacobian.

∂u∂x=y; ∂u∂y=x; ∂v∂x=2xy; ∂v∂y=x2{displaystyle {frac {partial u}{partial x}}=y;\ {frac {partial u}{partial y}}=x;\ {frac {partial v}{partial x}}=2xy;\ {frac {partial v}{partial y}}=x^{2}}
|y2xyxx2|=x2y−2x2y=−v{displaystyle {begin{vmatrix}y&2xy\\x&x^{2}\end{vmatrix}}=x^{2}y-2x^{2}y=-v}

step 4. invert the determinant and neglect any negative signs

then plug it into the area integral.

∫dda=∬d1vdvdu{displaystyle \int _{d}\mathrm {d} a=\iint _{d}{frac {1}{v}}\mathrm {d} v\mathrm {d} u}

step 5. evaluate the area integral using any means possible

∫13du∫12dv1v=2ln⁡2{displaystyle \int _{1}^{3}\mathrm {d} u\int _{1}^{2}\mathrm {d} v{frac {1}{v}}=2\ln 2}

step 6. solve for x{displaystyle x}

and y{displaystyle y}

to obtain the integrands in terms of u{displaystyle u}

and v{displaystyle v}

y=ux=vx2{displaystyle y={frac {u}{x}}={frac {v}{x^{2}}}}
- x=vu{displaystyle x={frac {v}{u}}}
x=uy=vy{displaystyle x={frac {u}{y}}={sqrt {frac {v}{y}}}}
- y=u2v{displaystyle y={frac {u^{2}}{v}}}

step 7. evaluate the other integrals to find the centroid

∫dxda=∬dvu1vdvdu=∫131udu∫12dv=ln⁡3{displaystyle {begin{aligned}\int _{d}x\mathrm {d} a&=\iint _{d}{frac {v}{u}}{frac {1}{v}}\mathrm {d} v\mathrm {d} u\\&=\int _{1}^{3}{frac {1}{u}}\mathrm {d} u\int _{1}^{2}\mathrm {d} v\\&=\ln 3\end{aligned}}}
∫dyda=∬du2v2dvdu=∫13u2du∫121v2dv=(273−13)(−12+1)=133{displaystyle {begin{aligned}\int _{d}y\mathrm {d} a&=\iint _{d}{frac {u^{2}}{v^{2}}}\mathrm {d} v\mathrm {d} u\\&=\int _{1}^{3}u^{2}\mathrm {d} u\int _{1}^{2}{frac {1}{v^{2}}}\mathrm {d} v\\&=\left({frac {27}{3}}-{frac {1}{3}}\right)\left(-{frac {1}{2}}+1\right)\\&={frac {13}{3}}\end{aligned}}}

step 8. arrive at the centroid

the centroid is the center of mass of the region. if one were to balance an object whose shape was defined by that region using a tip of a pin, the only way it would work is if it were balanced at the centroid.