Electric field

Vector electric field

Electrostatics study the force of two Q, Q’ point electrical charges between them; if we look only at one of the charges, say the Q, we realize that if we place any Q’ load in any position of space surrounding Q, a force will be exerted on Q’, it is as if the space surrounding Q has a new property, that of exerting a force on any Q’ load in any position that we place it.

Let’s place a small known value q test load at any position in the space given by the position vector v; if we see that an F-force (a vector) is exerted on the charge simply because it is situated at that point, we infer that space is affected by some other unknown Q load that creates that effect, and then the law of Coulomb will be fulfilled,

F-R-qQ/r3[1]

where r is the position vector of q relative to Q. Figure 1 shows the loads q, Q, their position vectors v, w, the relative position vector of q to Q, r s v – w, and the resulting force F, in the direction of r, which is assumed repulsive (loads are of the same sign and repel each other).

Fig.1: electrostatic force between two charges

Assuming the view that the space around Q is affected by that load, since a force is exerted at a distance on any load q at any position v, we can define a vector E such that the force exerted on any load q is simply F s qE from the expression [1]:

E s k(r/r3) Q[2]

This vector is called the electric field vector, and does not depend on the test load q, only on the Q-causing load. If we take two test loads q, q’, and measure the vectors force F, F’, looking for the intersection of the support lines of the forces we will find the origin of the field: the position of the Q load that generates it (Figure 2). In addition, the magnitude of the force F, known the q load, also determines the Q load through the equation,[1] and therefore determines the E field.

Fig.2: two test loads q, q’ determine the origin of the electric field, the Q load, and the vector field E

The property that all the support lines of the electrostatic forces are cut at one point can be expressed by saying that the electric field is a central field because all forces start from a central point of space.

In Physics the concept field is used to describe how a measurable physical magnitude is “distributed” by space; so, we can talk about electric fields, magnetic fields, and even velocity fields in a fluid. If the magnitude is scalar, the field will be, if vector, the field is vector, and if the magnitude is a tensor, the field will be tensory. Specific mathematics for describing fields has been studied in depth, giving rise to the branch of Mathematical Physics known as Field Theory.

Electric field produced by several point loads

Fig.3: Single-load single-charge electric field lines are straight cut into the load

In the case of a single point charge we have seen that all the straight supports of the forces, straight that we will call lines of force of the field or simply lines of the field start from a central point where the single load is located (fig. 3).

What happens to lines if the field is generated by two loads? Since the electrostatic force is cumulative (the contributions of all loads are added) the electric field E will be, and at each point in space the field vectors corresponding to each load will be added. In addition, the force F and field E decrease with the square of the distance, so the lines of the field curve; a field line passes through each point in space, so that the electric field at that point is tangent to the line. Figure 4 represents two equal positive loads, and a small test load in which the two vectors E, E’ generated by the field sources are added. The resulting vectors do not point to any center, the resulting field is no longer central, and the field lines (in black in the figure) are curved.

Fig. 4: field and field lines generated by two equal point loads

It should be borne in mind that any load will generate its own electric field that will overlap existing ones; that is why we are talking about placing “test burdens” which are supposed to be much smaller than the field-generating loads, so that their contribution can be disregarded. In addition, source loads, even though they are much larger than test loads, are supposed to be of point dimensions to avoid mathematical complications.

Electric dipole field

An important particular case is that of the so-called electric dipole, which are two loads of equal magnitude q and counter sign separated by a small d distance (fig. 5).

Fig. 5: field lines produced by an electric dipole, source: By Geek3 , from [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0)]Wikimedia CommonsIt def

ines the electrical moment of the dipole, p, by the vector

p s qd,[3]

being d the vector that starts from the negative charge and ends up in the positive charge. The distance d is supposed to be much smaller than the distances at which we will place the test loads where we will measure field E, this allows to simplify its mathematical expression, which is obtained from adding the contributions of each load to the total field, turning out to be:

\boldsymbol E=qk\left(3\frac{d\cos\left(\theta\right)}{r^4}\boldsymbol r-\frac{\boldsymbol d}{r^3}\right) [4]

where r is the vector that starts from the center point of the dipole (between its two charges); alternatively, using[3] :

\boldsymbol E=k\left(3p\frac{d\cos\left(\theta\right)}{r^4}\boldsymbol r-\frac{\boldsymbol p}{r^3}\right) [5]


Example 1: We place a load of +10⁻3C at the coordinate origin, and another load of -10⁻3C at the point (1, 0, 0). Calculate the electric field at point P(0.5, 1, 1). If we place at that point a small load of +10⁻⁶C, what force will the field exert on it?

The E’ field due to the first load will be, applying [2]:

\begin{array}{l}E'=k\cdot\lbrack{(0.5,1,1)-(0,0,0)}/\sqrt{(0.5²+1²+1²)\rbrack}^3\cdot(10⁻³)=\\\;(0.5,\;1,\;1)\cdot\frac{10⁻³k}{\left(3/2\right)^3}\end{array}

For the second load:

\begin{array}{l}E''=k\cdot\lbrack{(0.5,1,1)-(1,0,0)}/\sqrt{(0.5²+1²+1²)\rbrack}^3\cdot(-10⁻³)=\\-\;(-0.5,\;1,\;1)\cdot\frac{10⁻³k}{\left(3/2\right)^3}\end{array}

The total field will be the sum of the above:

E=10^{-3}k\left(\frac23\right)^3\left(1,0,0\right).

The force exerted on the test load q is given by F s qE:

\boldsymbol F=q\boldsymbol E=10^{-6}\cdot10^{-3}\cdot k\left(\frac23\right)^3\left(1,0,0\right)=9\cdot\cancel{10^9}\cdot\bcancel{10^{-9}}\frac8{27}\left(1,0,0\right)=\left(1,0,0\right)

a force of 1 Newton in the direction of the X axis.


Electric field created by a load distribution

When instead of point loads we have loaded material bodies we model them as if they contained point loads distributed throughout the body, so that the electric field produced by the body is obtained by adding the contributions of the point loads; Depending on the mathematical form we give to the distribution, the sum can be more or less direct, simple, or complicated. In the limit case, which is in fact the usual, in which we consider that there are a myriad of point loads it will be necessary to use differential calculation and integral.

Electric field created by a flat and homogeneous load distribution

The simplest case of distribution of countless point loads is that of a thin bar of 2L length that has loads only on one side and are also evenly distributed; let’s call the electrical charge density per unit of length, which will obviously be Q/2L being Q the total load of the bar. and let’s just calculate the electric field at a point P above the bisector of the bar, at a height h:

Fig. 6: Geometry for the calculation of field E produced by a homogeneous load line at a point above the bisector of the line

Taking a differential element of length dx and located at a distance x from the center of the bar, by the geometry of the problem we see that the distance r2 will be equal to x2 + h2, and the differential load of that element will be dQ s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

\operatorname dE_1=k\frac{dQ}{r^2}=k\frac{\rho\cdot dx}{x^2+h^2}

Given the symmetrical situation point P with respect to the bar, there will be another dx element located in -x that will produce a differential field dE_2 such that when you sum the vectors dE_1 with dE_2 the horizontal components will be overridden and the resulting will be vertical, dE with a module:

\operatorname dE=\operatorname dE_1\sin\left(\theta\right)+\operatorname dE_2\sin\left(\theta\right)=2\cdot k\frac{\rho\cdot dx}{x^2+h^2}\cdot\sin\left(\theta\right)

This will happen along the entire bar, so we conclude that the resulting field must be vertical. We now have to add all the differential contributions along the bar:

E=\int_0^L\operatorname dE=2k\int_0^L\frac{\rho\cdot dx}{x^2+h^2}\cdot\sin\left(\theta\right)

Note that the integration limits are and a[0, L]re not[-L, L] for each dx element to the right of the bar (in x) we have already added the contribution of the symmetrical element to the left (in -x). Considering that -sin-left(-theta-right)-frac hr-frac h-sqrt-x-2+h-2- we reach the integral

E=2k\int_0^L\frac{\rho\cdot dx}{x^2+h^2}\cdot\frac h{\sqrt{x^2+h^2}}=2kh\rho\int_0^L\frac{dx}{\left(x^2+h^2\right)^{3/2}}

Without going into details of the integral calculation (we can calculate it for example using WolframAlpha), we will have:

E=2kh\rho\frac L{h^2\sqrt{h^2+L^2}}=\frac{2k\rho L}{h\sqrt{h^2+L^2}}

In the event that the bar is very long compared to the distance h, i.e. L >> h, we can simplify the value of the field:

\lim_{L\rightarrow\infty}E=\frac{2k\rho}h\lim_{L\rightarrow\infty}\frac L{\sqrt{h^2+L^2}}=\frac{2k\rho}h [6]

The value of the field will vary depending on 1/h.


Vector field flow. Solid Angle

In the previous section we have seen that even in a simple case (one-dimensional load line, homogeneous, field at a point of the bisector…) when calculating fields due to continuous distributions of loads immediately appear complicated integrals, or very complicated. In this section and the next we see a view of the geomeath-based issue that often greatly simplifies calculations,

Consider a central vector field, which is the one that makes each P point of the space correspond to a vector V that follows the OP direction, being O the center of the field, a fixed point (Figure 7).

Fig.7: Central field, the agent mass located in O

The special case of the form V = r·c/r² where r is the OP module (the distance to the center of the field), r the unit vector indicating the direction of OP, and c is a constant that we call the field’s agent mass; it’s called the Newtonian field. They are Newtonian fields, the electric field, the magnetic field and the gravitational field. Let’s define the flow of the vector field through an infinitesimal dS surface, which we calld\phi, as the scalar product V·dS, where dS is the vector perpendicular to the surface; being infinitesimal dS (very small), we will consider its curvature to be negligible, and therefore flat (Figure 8):

Fig. 8: surface differential element, vector dS, and vector field V that we assume passes through the center of the surface

\operatorname d\phi=V\cdot dS=V\cdot dS\cdot\cos\left(\theta\right) [7]

Next, let’s define the solid angle of the dS surface relative to the origin of the center field O. Let’s join the ends of the dS surface with the O point using OP lines, and define a C sphere of radius 1 with center in O; the OP lines will cut to the sphere by defining on it a small surface of dS over C (Figure 9).

Fig. 9: solid angle subtended on sphere C by surface element dS

Now let’s look at a geometric property of Newtonian fields: the flow of the field through dS is:

\operatorname d\phi=V\cdot dS=V\cdot dS\cdot\cos\left(\theta\right)=\frac c{r^2}dS\cdot\cos\left(\theta\right)=\frac c{r^2}dS' [8]

where we’ve matched dS\cdot\cos\left(\theta\right)=dS', which is the projection of the dS surface over the perpendicular to the field vector V. The surfaces dS’ and dΩ (Fig. 10) are parallel, and are joined by the same lines to the center point O, then it is true that the reason for their areas is equal to the reason for their distances to center O squared:

Fig. 10: dS’ and d’ are parallel, and are attached by the same lines to the central point O

\frac{\operatorname d\phi}{dS'}=\frac1{r^2}\Leftrightarrow dS'=r^2\operatorname d\phi

If the reader don’t see why, think that the area of the sphere is 4πr², the area of the unit sphere is 4π, and the area of a sphere that passes by dS’ is 4πr², so the ratio of areas is 4πr² : 4π = r². Using this ratio in the equatio[8]n we get:

\operatorname d\phi=\frac c{r^2}r^2\cdot d\Omega=c\cdot d\Omega [9]

which tells us that the flow of the Newtonian field vector across any differential surface does not depend on the distance r, and is directly equal to the product of the agent mass of the field by the solid angle subtended by the surface over the unit sphere centered on the origin of the field.

Flow of the field through closed surfaces

The scalar product defined in[7] can be positive or negative depending on the relative orientation of the field vectors V and dS; consider a closed surface S, let’s calculate the total flow of the field through S integrating for each dS element:

\phi=\int_S\operatorname d\phi=\int_SV\cdot dS  [10]

If the total flow is positive, we will say that it is an incoming flow in S, and if it is negative, it will be an outgoing flow of S. If the center of field O is on the outside of S, the OP lines from the center passing through S will cut to S by an even number of points; however, if O is inside S, the lines will cut to S by an odd number of points (Figure 11).

Fig.11: flows from the O center through closed surfaces

Therefore in the outer case each line will create a succession of inbound-outbound-inbound-outgoing-etc flows in even number, that is, of alternating positive and negative flows; we have seen that the flow does not depend on the distance to the center (equation [9]) but only on the solid angle subtended, which will be the same for each surface element over the lines starting from O. Therefore, for an outer O point, inbound and outbound flows have the same value and override each other, resulting in a total null flow:

In a Newtonian field, the total flow through a closed surface that does not contain the center of the field is null.

However, if the surface contains the center of the field we will have an odd number of incoming-outgoing flows, and their sum will not be overridden; in fact its value is given by the theorem of Gauss, which is derived from the equations [9]and[10] :

\phi=\int_S\operatorname d\phi=\int_Sc\cdot\operatorname d\Omega=c\cdot\int_S\operatorname d\Omega=4\pi c  [11]

Expressed in words:

In a Newtonian field, the total flow through a closed surface containing the center of the field is equal to 4\pi multiplied by the value of the agent mass.

In the case of the electric field, the agent mass is worth k-Q, Q being the electrical charge.

If we have a set of loads, each load will create its field flow, and the total flow will be the sum of all of them.

Field created by an infinite distribution of loads, flat and homogeneous, using the Gaussian theorem

As an example of the usefulness of the concept of field flow and theoriema of Gauss will calculate the electric field created by a uniformly loaded plate, which we will assume to be very large, over a P point located at a height h of the plane.  Imagine another P’ point located on the other side of the plane, symmetrical to P, and think of a cylinder with PP’ axis and upper area dS (in Figure 12 we see a side view of the loaded plane and the situation).

Fig. 12: Using gauss’s theorem to calculate electric field E

By symmetry the E field must be vertical and in the directions indicated in the figure; the normal dS vectors to the cylinder bases will have the same direction as E, then the resulting scalar product, taking into account that there will be no flow of E on the sidewalls of the cylinder because it is parallel to the field (then the vector normal to the walls is perpendicular to E and its scalar product, null) is \operatorname d\phi=E\cdot dS+E\cdot dS=2E\cdot dS.  If we call sigma the load density per surface unit of the board, the load enclosed inside the cylinder will be \operatorname dQ=\sigma\cdot\operatorname dS. Because of the Gaussian theorem, the flow must then be \operatorname d\phi=4\mathrm\pi\cdot\mathrm k\cdot\mathrm\sigma\cdot\operatorname d\mathrm S; matching the two expressions for the flow we get:

d\phi=4\pi k\sigma\cdot\cancel{dS}=2E\cdot\cancel{dS}\Leftrightarrow\boxed{E=2\pi k\sigma}

We see that the ingensity of field E created by a homogeneously charged infinite plate does not depend on the distance h to the plate, a remarkable result.

Problems

  1. Calculate the electric field produced by a sphere loaded with homogeneous load density, both inside the sphere and outside.
  2. A sphere loaded with homogen load density has a spherical cavity inside it, the centers of the loaded sphere and the cavity are at a distance d. Calculate the electric field in the cavity.


Solutions

Problem 1 – In the figure we see the geometry of the problem: we represent a spherical surface S inside and concentric to the loaded sphere (in blue) and on S any point P, by which we draw a line that passes through the center O and divide the spheres into two symmetrical halves; by symmetry, the vector field E(P) at point P cannot be directed towards either of the two halves in particular, so it must be radial. In addition, point P could be any point located in S because we have spherical symmetry, then the module of field E will be the same throughout S, that is, the value of E depends only on the radius of S.

By symmetry of the problem with any line that passes through O, the E field must have the same module on every inner S sphere, and it must be radial

Let’s apply gauss’s theorem to surface S: first of all we give mathematical form to the above considerations about field E:

\overrightarrow E(P)=\frac1rE(r)\cdot\overrightarrow{OP},

that is the vector field E is equal to the E(r) module by the radial vector OP divided by the OP module, which is r. We can define the radial unit vector , and the expression of the field at all point P of S is more compact: \overrightarrow E(P)=E(r)\cdot\widehat r.

Now we calculate the flow of E through the S surface, applying [10]:

\phi=\int_S\operatorname d\phi=\int_S\overrightarrow E\cdot d\overrightarrow S=\int_SE\left(r\right)\widehat r\cdot d\overrightarrow S=E\left(r\right)\int_SdS=E\left(r\right)\cdot4\pi r^2

where we have applied that the vector differential element of surface dS is a radial vector of dS module, and therefore its scalar product with the unit radial vector is simply dS, in addition E(r) is constant over S, then it can exit outside the integral, and this integral over S of the dS element is simply the surface of the sphere S. This flow that we have calculated, according to the Gaussian theorem [11], must be equal to \phi=4\pi kQ, where Q is the load contained in the inner volume to S; calling \rho to the load density per unit volume, we have:

Q=\int_V\rho\cdot\operatorname dV=\frac43\rho\pi r^3\Rightarrow\phi=4\pi k\cdot\frac43\rho\pi r^3=\frac{16}3k\rho\pi^2r^3 [12]

By matching this flow given by the Gaussian theorem with which we calculated earlier, we find the module of field E:

E(r)\cdot4\pi r^2=\frac{16}3k\rho\pi^2r^3\Rightarrow\boxed{E(r)=\frac43k\rho\pi r} [13]

We see that the E(r) dependency is linear: it increases linearly with r. In the international system of units the constant k is expressed in function of the so-called electrical permittivity of the vacuum -v</em>arepsilon_0, and the field is reduced to

E(r)=\frac43\frac1{4\pi\varepsilon_0}\rho\pi r=\frac1{3\varepsilon_0}\rho r. [14]

This expression is valid for r\leq a, being a t the radius of the loaded sphere. For distances r to the center of the sphere that are greater than the radius to the calculation is very similar, only that the value of the Q load is constant, being the total load of the sphere, Q-frac43-pi a-3-rho, and when replacing it in the Gaussian theorem, the total flow across a radio surface r > to -phi-frac43-pi a-3-rho-cdot4-pi k , matching this flow with the vectorly calculated:

'phi'frac43's'4'p'r'2'Leftrightarrow's 'left's's'left')right's'right's''frac's'left')'s'3'''2''2'''leftrightarrow''[14b] which, based on-varepsilon_0will be worth E-left(r-right)-frac43-pi-frac-a-3-r-2-rho-frac1-4-pi-varepsilon_0-frac-rho-a-3-3-varepsilon_0r-.. [15]

So inside the sphere the E field grows linearly with the distance to center, and on the outside it decreases squarely:

Variation of field E with distance r to center in the case of a radio sphere to evenly loaded

Problem 2 – The complication of this problem is to see how to treat the cavity inside the sphere; the easiest way is to realize that the situation is equivalent, in terms of the calculation of the E field, to assume that the sphere, from radius R and center to O, is evenly loaded in its entirety with a load density of -rho, and that the cavity is another spherical surface, radio r< R y centro O’, a la que añadimos, superponiéndola, otra densidad de carga de igual valor pero signo contrario, -rho, con ello, la carga neta en la cavidad será cero. So, because the electric field is cumulative, we can calculate the field E by overlapping the field created by the entire radius sphere R, which we call E_O, plus the field created by the negative load in the radius sphere r, which we call E'_O. The geometry of this scheme is seen in the image, where we have drawn a P point either within the cavity, the two fields generated at that point, and the total field.

Vectors E and geometry of the problem

To calculate each field we apply t[13]he P point as it is interior to the two spheres considered:

'overrightarrow E'left(P'right)'overrightarrow E'_O'left(P'right)+'overrightarrow E'_O'left(P'right)'frac43k's "cdot"widehat r-'frac43k''ah'rho r''cdot'widehat r'''frac43k''p'2'left(r'cdot'widehat r-r''cdot'widehat r''right)

We need to r-cdot-widehat r-overrightarrow r,-;r'&r'&gt'>cdot'widehat r''overrightarrow r' and looking at the figure we deduce that r'cdot'widehat r' -r'-cdot-widehat r-overrightarrow r--overrightarrow r'-overrightarrow-OP--overrightarrow-O'P-overrightarrow-OO-'.  Therefore we have the following expression for the field inside the cavity:

'overrightarrow E'left(P'right)'frac43k''th'cdot'overrightarrow'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''frac43k''-th-rho-cdot-overrightarrow d

Note that E is a constant vector directed according to the OO’ line (the module vector d equal to the distance d between centers).

The field inside a cavity within a uniformly charged sphere is constant

Posted in Uncategorised | Leave a comment

1st-order ordinary differential equations

separator2

Separable equations

The separable differential equations are those of the form dy / dx = H (x, y) where H (x, y) can be expressed either as the quotient H (x, y) = f (x) / g (y) or as the product H (x, y) = f (x) text {·} g (y).

In the first case, dy / dx = f (x) / g (y) Rightarrow g (y) dy = f (x) dx Rightarrow int g (y) dy = int f (x) dx + C, while in the second case dy/dx=f(x) text{·} g(y) Rightarrow g^{-1}(y)dy=f(x)dx Rightarrow int g^{-1}(y)dy = int f(x)dx+C.

Example 1: Solve the differential equation x^{2}frac{dy}{dx}=frac{x^{2}+1}{3y^{2}+1}.

Let's take H(x,y)=frac{left(x^{2}+1right)/x^{2}}{3y^{2}+1}=frac{f(x)}{g(y)} so that int f(x)dx=intfrac{x^{2}+1}{x^{2}}dx=x-frac{1}{x}, int g(y)dy=intleft(3y^{2}+1right)dy=y^{3}+y and therefore the solution will be given implicitly by y^{3}+y=x-frac{1}{x}+C.

Example 2: Find all the solutions of the equation dy/dx=2xsqrt{y-1}.

Let's H(x,y) = 2x sqrt{y-1} = f(x) text{·} g(y), and let's integrate each function: int f(x)dx=int2xdx=x^{2}; int g^{-1}(y)dy=intfrac{1}{sqrt{y-1}}dy=2sqrt{y-1}. Solutions are 2sqrt{y-1}=x^{2}+CRightarrow y=left(frac{x^{2}+C}{2}right)^{2}+1 . Note that the function y (x) = 1 is also a solution of the equation, but it is not included in the general solution y = left ( frac {x ^ {2} + C} {2} right) ^ {2} +1, that is to say that for all C we will have y = left (frac {x ^ {2} + C} {2} right) ^ {2} +1 neq1. We call solutions not included in the general solution singular solutions of the differential equation.

separator2

Unique solutions

In example 2 we have seen that we can have solutions that are not included in the general solution, called singular solutions. When will this happen and how can we find those other solutions? Let's remember what we have seen in the post Ordinary differential equations: introduction, section "Bundle of curves and differential equations of the first order": the general solution of the ordinary differential equation y'=F(x,y) represents a bundle of curves y=f(C,x) such that each curve of that beam satisfies the equation, that is, for each point (x, y) the slope of any of the curves y=f(C,x) at that point fulfills y'=F(x,y).

Enveloping curve of a bundle of curves and singular solution

Two one-beam lines (blue and red) and beam envelope curve (green)

Two one-beam lines (blue and red) and beam envelope curve (green)

Given a bundle of curves F(x,y,C)=0 the curve f(x,y) is called the beam envelope curve such that it is tangent to all the curves of the beam.

Example 3: The image shows two lines of the line bundle y=(1-10/C)*x+(10-C), specifically the two lines corresponding to the values C=3 in red and C=6 in blue, and the enveloping curve of the beam, which is tangent to the first line at a point close to x=1,  y=5 and the second line about x=3.5, y=1.5.

Remember that we can associate to any bundle of curves F(x,y,C)=0 a differential equation y'=f(x,y) such that the beam is the general solution of the equation: for each value of C, we have a curve of the beam such that its derivative verifiesy'=f(x,y). But we have seen that the enveloping curve is tangent to all the curves of the beam, that is, its derivative at each point coincides with the derivative of one of the curves of the beam; this is why the enveloping curve is also a solution of the differential equation of the beam.

Example 4.Consider the parabolal bundle 4y=(x + C)^2, all curves are tangent to the X axis at some point; the image depicts three of the beam curves, tangent to y=0 at points x=-3, -2, -1.

Parabola beam 4y=(x + C)²

Parabola beam 4y=(x + C)²

What is the differential equation of the curve beam? We derive with respect to x the equation of the beam: 4y'=2(x + C), and eliminate the constant C using this equality and the equation of the beam:

left.begin{array}{r}4y'=2(x+C)Rightarrow C=2y'-x\4y=(x;+; C)^2end{array}right}Rightarrow4y=(2y'{)^2=4y'^2}Rightarrow y=y'^2.

Note that the line y=0, being tangent to all the curves of the beam, has the same derivative as the curves of the beam at each point on the X axis; therefore, it also verifies the differential equation of the beam. In general: if there exists an enveloping curve that is tangent to all the curves of a beam, then that curve will be a singular solution of the differential equation of the beam.

Calculating the envelope of a beam of curves

Given any value x=x_0, the envelope will have the same derivative at that point as some curve of the beam, that is, there will be a constant C, dependent on x=x_0, such that the derivative of the beam curve F(x_0,y,C)=0 will have the same value at that point. We dispense with x_0: in general for each x we will have a C(x). Let us derive with respect to x the expression of the beam, taking into account that y also depends on x, using the rule of the chain and the implicit derivation:

F(x,y,C)=0Rightarrowfrac{operatorname d{F(x,y,C)}}{operatorname dx}=frac{partial F(x,y,C)}{partial x}+frac{partial F(x,y,C)}{partial y}frac{operatorname dy}{operatorname dx}=frac{partial F(x,y,C)}{partial x}+frac{partial F(x,y,C)}{partial y}y'=0

We clear y' to obtain the derivative at any point of any curve of the beam:

y'=-frac{partial F(x,y,C)}{partial x}left(frac{partial F(x,y,C)}{partial y}right)^{-1}.

We now turn to the enveloping curve; at the contact points (x,y) between each curve and the envelope the derivatives match the curves of the beam; in each given value x we will have determined a C(x), as the value y(x) also coincides with that of the beam, we will have that at the point of contact F(x,y,C(x)) is fulfilled. We derive again with respect to x:

begin{array}{l}F(x,y,Cleft(xright))=0Rightarrow\frac{operatorname d{F(x,y,Cleft(xright))}}{operatorname dx}=frac{partial F(x,y,C)}{partial x}+frac{partial F(x,y,C)}{partial y}frac{operatorname dy}{operatorname dx}+frac{partial F(x,y,C)}{partial C}frac{operatorname dC}{operatorname dx}\=frac{partial F(x, y,C)}{partial x}+frac{partial F(x,y,C)}{partial y}y'+frac{partial F(x,y,C)}{partial C}C'=0end{array}.

We clear and':

y'=left(frac{partial F(x,y,C)}{partial y}right)^{-1}left[frac{partial F(x,y,C)}{partial x}+frac{partial F(x,y,C)}{partial C}C'right].

The two expressions for the derivative y' must be equal:

y'=left(frac{partial F(x,y,C)}{partial y}right)^{-1}left[frac{partial F(x,y,C)}{partial x}+frac{partial F(x,y,C)}{partial C}C'right]=left(frac{partial F(x,y,C)}{partial y}right)^{-1}left[frac{partial F(x,y,C)}{partial x}right].

We see that it must be: frac{partial F(x,y,C)}{partial C}C'=0, that has two possible solutions, the trivial C'=0Rightarrow C=cte and the condition frac{partial F(x,y,C)}{partial C}=0. The first is an unimportant solution because it tells us that the constant C will not depend
of x (degenerate curve beam: all curves are equal), while the second is the general condition to be met by the envelope.

Example 5: Find the expression of the enveloping curve of the line bundle of example 3, y=(1-10/C)*x+(10-C).

We express it in the form F(x,y,C)=y-(1-10/C)ast x-(10-C)=0 to derive with respect to C and equal to zero:

frac{partial{F(x,y,C)}}{partial C}=frac{-10}{C^2}x+1=0Rightarrow C=sqrt{10x}.

We substitute this value in the expression of the beam to remove C and get the equation of the envelope:

F(x,y)=y(1-frac{10}{sqrt{10x}})x+(10-sqrt{10x})=y-10-x+2sqrt{10x}=0.

The envelope is y=10+x-2sqrt{10x}.

Example 6: Calculate the envelope of the curve beam 4y=(x + C)^2 of example 4.

begin{array}{l}F(x,y,C)=4y-(x+C)^2=0;\frac{partial F}{partial C}=2(x+C)=0Rightarrow C=-x;\F(x,y)=4y-(x-x)^2=4y=0Rightarrowboxed{y=0}.end{array}

separator2

Homogeneous functions. Application to differential equations.

A function F(x,y) is homogeneous of degree n if F(lambda x,lambda y)=lambda^nF(x,y) for every parameter lambdaneq0. Per example, the function F(x,y)=xy-x^2 is homogeneous of degree 2, since F(lambda x,lambda y)=lambda xlambda y- ( lambda x)^2=lambda^2F(x,y).

In the special case that F(x,y) is homogeneous of degree 0 we have F(lambda x,lambda y)=F(x,y), and gives us a method of solving differential equations y'=F(x,y) in which the F is homogeneous of degree 0: with the change of variable u=y/x they become equations of separate variables. Indeed:

begin{array}{l}u=y/xLeftrightarrow y=uxRightarrow y'=u'x+u;\y'=F(x,y)Leftrightarrow u'x+u=F(x,ux)=F(1,u)end{array}

where we have used the x instead of the lambda as a parameter of the homogeneous. Then

begin{array}{l}frac{du}{dx}x+u=F(u)Leftrightarrowfrac{du}{dx}x=F(u)-uLeftrightarrowfrac{du}{F(u)-u}=frac{dx}xLeftrightarrow\intfrac{du}{F(u)-u}=intfrac{dx}x=lnleft(xright)+Cend{array}.

Example 7: Solve y'=frac y{sqrt{x^2+y^2}-x}.

F(lambda x,lambda y)=frac{lambda y}{sqrt{lambda^2x^2+lambda^2y^2}-lambda x}=frac{lambda y}{lambdaleft(sqrt{x^2+y^2}-xright)}=frac y{sqrt{x^2+y^2}-x} then it is homogeneous of degree 0. We make the change u=y/x to get it to be of separate variables:

begin{array}{l}y'=u'x+u=frac{ux}{sqrt{x^2+u^2x^2}-x}=frac u{sqrt{1+u^2}-1};\frac{du}{dx}x=frac u{sqrt{1+u^2}-1}-u=frac{uleft(sqrt{1+u^2}+1right)}{1+u^2-1}-u=frac{sqrt{1+u^2}+1}u-u=\frac{sqrt{1+u^2}+1-u^2}u;\frac{udu}{sqrt{1+u^2}+1-u^2}= frac{dx}x\end{array}.

Then intfrac{udu}{sqrt{1+u^2}+1-u^2}=intfrac{dx}x=lnleft(xright)+C. In the first integral we make the variable change 1+u^2=t^2 that makes it:

intfrac{udu}{sqrt{1+u^2}+1-u^2}=intfrac{toperatorname dt}{t+1-left(t^2-1right)}=intfrac{toperatorname dt}{-t^2+t+2},

that it is of a rational type; we decompose into simple fractions:

begin{array}{l}intfrac{toperatorname dt}{-t^2+t+2}=intfrac{-1/3}{t+1}+intfrac{-2/3}{t-2}=-frac13lnleft(t+1right)-frac23lnleft(t-2right)\=-frac13lnleft(left(t+1right)left(t-2right)^2right)=lnleft(left(t+1right)left(t-2right)^2right)^{-1/3}end{array}.

We equate the two integrals, and undo the changes to obtain the general solution implicitly:

begin{array}{l}lnleft(left(t+1right)left(t-2right)^2right)^{-1/3}=lnleft(xright)+CLeftrightarrow\Cx=left(left(t+1right)left(t-2right)^2right)^{-1/3}=left(t^3-3t^2+4right)^{-1/3}Leftrightarrow\frac C{x^3}=left(u^2+1right)^{3/2}-3u^2+1Leftrightarrow\C=x^3left[left(left(frac yxright)^2+1right)^frac32-3left(frac yxright)^2+1right].end{array}.

Reduction to homogeneous

The rational functions of form

F(x,y)=frac{ax+by+c}{a'x+b'y+c'}

they are only homogeneous of degree 0 when c=c'=0, however if we think of the numerator and denominator of the function as if they were two lines ax+by+c=0,;a'x+b'y+c'=0, we can try to find its intersection point (x_0,y_0), if it exists, then the variable change X = x - x_0,  Y = y - y_0 (which is a translation of the xy axes to the XY axes, taking as the coordinate origin the point (x_0,y_0)), converts the function into homogeneous of degree 0.

Example 8: Solve y'=frac{y-x-1}{x+y-1}.

We find the intersection of the lines:

left.begin{array}{r}y-x-1=0\x+y-1=0end{array}right}y=1,;x=0

we make the change X=x, Y=y-1 which transforms the equation into:

y'=frac{y-x-1}{x+y-1}Leftrightarrow Y'=frac{Y-X}{X+Y}

containing a homogeneous f(X,Y) function of degree 0; we make the change u=Y/X to separate the variables:

begin{array}{l}u'X+u=frac{u-1}{1+u}Leftrightarrowfrac{operatorname du}{operatorname dX}X=frac{u-1}{1+u}-u=frac{-1-u^2}{1+u}=-frac{1+u^2}{1+u};\intfrac{1+u}{1+u^2}operatorname du=intfrac{operatorname dX}X=lnleft(Xright)+Cend{array}.

We decompose the first integral:

begin{array}{l}intfrac{1+u}{1+u^2}=intfrac1{1+u^2}+intfrac u{1+u^2}=tan^{-1}left(uright)+frac12intfrac{2u}{1+u^2}=\tan^{-1}left(uright)+frac12lnleft(1+u^2right)end{array}.

We equalize integrals and undo the change:

begin{array}{l}-tan^{-1}left(uright)-frac12lnleft(1+u^2right)=lnleft(Xright)+C=lnleft(CXright)Leftrightarrow\-tan^{-1}left(frac YXright)=lnleft(CXleft(1+left(frac YXright)^2right)right);\lnleft(Cxleft(1+left(frac{y-1}xright)^2right)right)+tan^{-1}left(frac{y-1}xright)=0.\end{array}

Special case: the lines are parallel

In the special case that the two lines have no intersection, then they are parallel, with proportional director vectors: frac ba=frac{b'}{a'}=r; but then we can do

F(x,y)=frac{ax+by+c}{a'x+b'y+c}=frac{a'rx+b'ry+c}{a'x+b'y+c}=frac{rleft(a'x+b'yright)+c}{a'x+b'y+c},

with variable change Y=a'x+b'y becomes

begin{array}{l}F(x,Y)=frac{rY+c}{Y+c'};;y=frac{Y-a'x}{b'}Rightarrowfrac{operatorname dy}{operatorname dx}=frac1{b'}left(frac{operatorname dY}{operatorname dx}-a'right);\frac{operatorname dy}{operatorname dx}=F(x,y)Leftrightarrowfrac1{b'}left(frac{operatorname dY}{operatorname dx}-a'right)=frac{rY+c}{Y+c'}Leftrightarrow\frac{operatorname dY}{operatorname dx}=b'frac{rY+c}{ Y+c'}+a'end{array}

which is of separate variables.

Example 9: y'=frac{x+y+1}{x+y-1}.

The lines x+y+1=0 and x+y-1 have no intersection, they are parallel; we make the change Y=x+y, Y'=1+y' to obtain:

begin{array}{l}Y'-1=frac{Y+1}{Y-1};;frac{operatorname dY}{operatorname dx}=frac{Y+1}{Y-1}+1=frac{Y+1+Y-1}{Y-1}=frac{2Y}{Y-1};\intfrac{Y-1}{2Y}operatorname dY=intoperatorname dx;\frac12intleft(1-frac1Yright)operatorname dY=x+C;\Y-lnleft(Yright)=2x+C;\x+y-lnleft(x+yright)=2x+C;\y=x+lnleft(x+yright)+C.end{array}

separator2

Exact equations

The equations of the form M (x, y) + N (x, y) frac {dy} {dx} = 0, or equivalently, M (x, y) dx + N (x, y) dy = 0, we will say that they are exact as long as M and N are continuous functions and
check the condition frac {partial M} {partial y} = frac {partial N} {partial x}.

The solution will be given implicitly by F (x, y) = C where frac {partial F} {partial x} = M, : frac {partial F} {partial y} = N.

Example 10: The equation y^{3} dx + 3xy^{2} dy = 0 is exact, because N = 3xy^{2}, frac{partial N} {partial x} = 3y^{2}, : M = y^{3}, : frac {partial M} {partial y} = 3y^{2}. As frac {partial F} {partial x} = M, will be F (x, y) = int MDX = xy^{3} + C (y) and also F (x, y) = int Ndy = xy^{3} + C (x); equalizing the two expressions we have F (x,  y) = xy^{3}, C(x) = C(y) = 0, so the general solution is xy^{3} = C.

Example 11: Solve the differential equation left (6xy-y^{3} right) dx + left (4y + 3x^{2} -3xy^{2} right) dy = 0.

As M = 6xy-y^{3}, N = 4y + 3x^{2} -3xy^{2}, frac {partial N} {partial x} = 6x-3y^{2} = frac {partial M} {partial y}, the equation is exact. Then

F (x, y) = int Ndy = 2y^{2} + 3x^{2} y-xy^{3} + C (x), F (x, y) = int MDX = 3x^{2} y-xy^{3} + C(y),

and when we equalize the two expressions we get C (y) = 2y^{2}.

Posted in Mathematics | Leave a comment

Ordinary Differential Equations: Introduction

Motivation

The equations studied in Algebra, such as x^{3}+7x^{2}+41=0,
they express static numerical relationships, and their solution is numbers that fulfill the equation. Now, the most interesting natural phenomena involve dynamic relationships and are best expressed with relationships between variable quantities, that is, with equations that are not static, but express variations and relationships between changing magnitudes: they are the differential equations.

Recall that the derivative of a function dy/dt=f'(t) can be interpreted as the proportion of change of the dependent variable y with respect to the changes in the dependent variable t. That is why equations describing changes use derivatives of functions.

Definition 1: An equation containing an unknown function and one or more of its derivatives is called a differential equation. By solving it, we get functions y = f(x) that verify the equation. If the function has only one independent variable, the equation is called the ordinary differential equation. If the function depends on two or more variables, the derivatives will be partial and the equation is called the partial differential equation. The order of a differential equation is that of the highest derivative that appears in the equation.

In this post we will only introduce ordinary differential equations, giving some definitions, and solving the most immediate ones.

Example 1: The variation of the temperature T of a body with respect to time is proportional to the difference (T-A) where A is the temperature of the environment (Newton's cooling law). The physical law is represented by a first-order ordinary differential equation (EDO of order 1):

frac{dT}{dt}=kleft(T-Aright)

Example 2: The variation with respect to the time of a population P(t) with constant rates of birth and mortality is proportional to the size of the population, and is also an EDO of order 1:

frac{dP}{dt}=kP

Example 3: Torricelli's law states that the variation with respect to the time of volume V of water in a tank being emptied is proportional to the square root of the depth and water of the reservoir:

frac{dV}{dt}=-ky^{1/2}

Example 4: The distance x(t) traveled in the accelerated motion of a mass body m subjected to a variable force F(t) is given by an EDO of order 2:

frac{d^{2}x}{dt}=frac{F(t)}{m}

Definition 2: A function y = f(x) is called a solution of a differential equation if the equation is fulfilled when y and its derivatives are replaced by f(x) and their derivatives, respectively. A particular solution of a differential equation is any solution that is obtained by assigning concrete values to the constants in the general solution. In practice, particular solutions are obtained from initial conditions that provide the value of the dependent variable or any of its derivatives for a particular value of the independent variable.

Example 5: of the differential equation y''-y=0 are solutions: a) y=sin(x), b) y=4e^{-x}  c) y=Ce^{-x} for any real C value. Solution (b) is a particular solution obtained from general solution (c). Although less obvious, also solution a) is a particular solution obtained from the general solution (it can be seen using Taylor's developments of the exponential function and the sine function).

Bundle of curves and first-order differential equations

Another way to introduce differential equations is from the geometric point of view. Consider the graph of the function y=Cx^2 for all possible real values of C. The image represents the values C=1, 2, 4, 8.

Curve beam y=Cx²

Curve beam y=Cx²

A bundle of plane curves is the set of all curves that are graphs of a general function y=F(x,C). When we give C all possible values, the generated curves fill an R region of the plane; in the case of the image above that region is the entire upper half-plane xinleft(-infty,inftyright),;yinleft[0,inftyright). In general the R region will depend on the beam.

We ask ourselves now: for each point (x,y) of the plane there will be a single value C such that it defines us unequivocally the function y=F(x,C) that passes through that point? We can put it in these terms: fixing x=x_0 the function F becomes dependent only on C: y=F(x_0,C)=f(C); For each value of and will there be a unique value of C? This will be the case as long as the function f is strictly increasing or decreasing in R, and that will happen when its derivative is not overridden: frac{operatorname df}{operatorname dC}neq0. In this case it is as if we had another function C=psi(x,y) of two variables that determines C for each point in the plane.

In the example of the previous image, fixing x=x_0 any we obtain y=f(C)=Cx_0^2, f'(C)=x_0^2, this value is always non-zero except in the origin of coordinates, therefore for the beam y=Cx^2 we have uniqueness in the sense that given a left(x, yright)neqleft(0,0right) there is a single value C=psi(x,y)=y/x^2 that determines the curve that passes through that point; by the origin (0,0) instead all the curves of the beam pass.

If we derive the equation from the beam we obtain y'=2Cx, then, substituting the previous value C=psi(x,y) we eliminate the C to obtain: y'=2Cx=2left(frac y{x^2}right)x=frac{2y}x, which is the differential equation of the bundle of curves, which is a first-order equation.

Example 6: Find the differential equation of the plane curved beam y=Csin(x).

The R region is the union of the +X+Y and -X-Y quadrants, some curves are shown in the figure.

Beam of curves y=C·Sin(x)

Beam of curves y=C·Sin(x)

Setting x=x_0 any we get y=Csinleft(x_0right)=fleft(Cright);f'left(Cright)=sinleft(x_0right). This value will be zero whenever x_0=kmathrmpi,;mathrm k=0,1,dots. In this set of points all the curves of the beam coincide, and in the rest of the points we have uniqueness: a single curve for each point, given by: C=frac y{sinleft(xright)}=psi(x,y).

Deriving the equation of the beam: y'=Ccdotcosleft(xright). We substitute the value of C:

left.begin{array}{r}y'=Ccdotcosleft(xright)\C=frac y{sinleft(xright)}end{array}right}Rightarrow y'=frac{ycosleft(xright)}{sinleft(xright)}=frac y{tanleft(xright)}

Definition 2: The beam of curves y=f(C,x) is the general solution of the ordinary differential equation; if we fix a value of C=C_0, we obtain a single curve of the beam, which we call the particular solution of the ordinary differential equation.

Example 7: The curve beam y=Csin(x) is the general solution of the differential equation y'=frac y{tanleft(xright)}. By the point left(frac{mathrmpi}2,3right) passes a single curve, y=3Sin(x), which is a particular solution of the differential equation.

Existence and uniqueness of the solution of a first-order differential equation

We have seen that to obtain the differential equation of a beam of curves you have to derive and eliminate the constant. We now pose the inverse problem: given any differential equation, is there "its" bundle of curves as we have defined it? The following theorem answers us for the case of first-order equations.

Theorem 1: existence and uniqueness. If we have a differential equation given in the form y'=f(x,y) such that the function f is derivable from all orders (there are all derivatives of any order) in an environment of (x_0,y_0), then there exists a single curve y(x) such that it passes through the point (x_0,y_0) and satisfies the equation y'=f(x, and).

Immediate first-order ordinary differential equations

We see in this section only the 1st order OEDs simplest to solve, in other posts we will see the more general cases.

Equations of type dy/dx=fleft(xright)

They are integrable directly, writing them as dy=f(x)Rightarrow y=int f(x)dx+C.

Example 8: Solving the differential equation y'+x^{2}-1=e^{x}.

The equation is equivalent to dy/dx=e^{x}-x^{2}+1Rightarrow y=intleft(e^{x}-x^{2}+1right)dx+C=e^{x}-frac{1}{3}x^{3}+x+C.

Equations of type dy/dx=ky

If y = f(x) is a function, the equations of type dy / dx = ky have as a solution the set of functions y (x) = Ce ^ {kx} where C is any real number. In general, a differential equation has infinite solutions.

Example 9: The solution of the equation frac {dP} {dt} = kp which establishes the evolution of a population P(t) with constant birth and mortality rates is any function of the form y(x) = Ce^{kx}.

Example 10: Suppose P(t) is the population of a colony of bacteria at time t, that the population in t = 0 is 1000, and that the population doubles in one hour. Then we can say that

1000 = P(0)=C
2000 = P(1)=Ce^{k}

therefore C=1000 and k=ln2, then P(t)=1000e^{xtext{·} ln2}.

The condition P(0) =1000 is called the initial condition because normally the value t =0 is taken as the initial state. When we give an initial condition, the solution of the differential equation will no longer have infinite solutions in general, but will have only one, or perhaps none if the conditions are incompatible. Equivalently, by giving an initial condition we pass, if it exists, from the general solution to the particular solution that satisfies that condition. Thus, the initial condition P (0) = -1000 has no solution of the type P(t) = Ce^{kt}.

Example 11: To solve the equation dy / dx = frac{x} {left (x^{2} +9 right)^{1/2}} with the initial condition y(4)=2 we do y(x)=intfrac{x}{left(x^{2}+9right)^{1/2}}dx=left(x^{2}+9right)^{1/2}+C, as y(4)=left(16+9right)^{1/2}+C=5+C must be C=-3.

separator2

Bibliography

DIFFERENTIAL EQUATIONS – Theoretical summary and collection of solved and proposed exercises.

 

Posted in Mathematics | Leave a comment