Unizor - Partial Differential Equations - Heat Equation

Notes to a video lecture on http://www.unizor.com

Partial Differential Equations
Overview - Heat Equation

Since we talked about ordinary differential equations, where a function of one argument together with its derivatives participates in the equation, we might as well touch on differential equations, where function of two (or more) arguments together with its partial derivatives participates in the equation. These equations are called partial differential equations.

In order to present this material in a more practical light and to emphasize the importance of differential equations (particularly, partial) in practical applications, let's discuss one particular physical process that naturally leads to partial differential equations.

Let's imagine an insulated thin metal rod heated on one end and examine how its temperature T at different distance from the edge changes with time.
We assume that the rod is stretched along the X-axis with its heated end at origin. The distance from this end to any point on a rod is X-coordinate of this point. Time t will be measured from the moment the heat is applied to the edge at x=0.

Considering our rod is insulated and thin, we can safely assume that the heat dissipates only along its length from point x=0 towards the other edge, and the rod's temperature is a function of two arguments:
distance x from the edge
and time t:
T = T(t,x)
The illustration below will be handy in our analysis.


Before embarking on calculations let us remind a few physical properties that affect the process of heat dissipation.

1. Heat Q - a form of energy manifested in vibration of molecules of an object.
It is measured in the same units as energy.

2. Temperature T - a measure of intensity of the molecular vibration.
It is measured in degrees of different scales.

3. Specific heat capacity C - an amount of energy (heat) needed to increase temperature of a unit of mass by a unit of temperature.
It is assumed to be a constant for any specific material within reasonable range of temperatures and precision.
If an object of mass m increased in temperature by ΔT, it consumed ΔQ=C·m·ΔT units of heat (energy).

4. Thermal conductivity k - a measure of how fast molecular vibration is transferred within an object.
It was experimentally established that the amount of heat transferred from one zone of an object to another per unit of time through a unit of area on the boundary between these zones is proportional to a rate of change of temperature at this boundary and depends on the qualities of the material. For every material it is a constant within a reasonable range of temperatures.
In one-dimensional case on an above picture of a thin rod that has area S of crosscut, when the heat is transformed through a point x and the rate of change of temperature at that point and at that time is ∂T(t,x)/∂x, the amount of heat that goes through during time Δt can be calculated as
ΔQ = −k·S·Δt·∂T(t,x)/∂x
Minus in this equality reflects that heat is transferred from hot to cold area and, therefore, the partial derivative of temperature T by x is negative, while energy must be positive.

Now we are ready to connect all the parameters mentioned above into one heat equation.

Consider a part of a thin rod from point x to point x+Δx.
As the heat dissipates from left to right, during the time Δt certain amount of heat enters this area through crossing at point x and certain amount of heat exits this area through crossing at point x+Δx.

The heat entering through crossing at point x measures
ΔQ_in = −k·S·Δt·∂T(t,x)/∂x
The heat exiting through crossing at point x+Δx measures
ΔQ_out = −k·S·Δt·∂T(t,x+Δx)/∂x

The difference between them is an amount of heat that contributed to a rise in temperature of the part of a rod from x to x+Δx. This difference equals to
(A) ΔQ₊ = k·S·Δt·[∂T(t,x+Δx)/∂x − ∂T(t,x)/∂x]

On the other hand, the heat ΔQ₊, consumed by a part of a rod of length Δx, area of crosscut S and specific heat capacity C should increase the temperature by ΔT related to this heat as
ΔQ₊ = C·m·ΔT
where m is a mass of this part of a rod, that can be calculated as
m = ρ·S·Δx
where ρ is a mass of a unit of volume of the material this rod is made of.
Therefore,
(B) ΔQ₊ = C·ρ·S·Δx·ΔT

From equations (A) and (B) we derive the following equality:
C·ρ·S·Δx·ΔT = k·S·Δt·[∂T(t,x+Δx)/∂x − ∂T(t,x)/∂x]
This can be transformed into
ΔT/Δt = [k/(C·ρ)]·[∂T(t,x+Δx)/∂x−∂T(t,x)/∂x]/Δx

When Δt→0 and Δx→0 our equation can be represented as
∂T(t,x)/∂t = [k/(C·ρ)]·∂²T(t,x)/∂x²

Traditionally, this heat equation is written as
∂T(t,x)/∂t = a²·∂²T(t,x)/∂x²
where a² = k/(C·ρ)

Unizor - Ordinary Differential Equations - Hooke's Law

Notes to a video lecture on http://www.unizor.com

Higher Order Ordinary
Differential Equations -
Hooke'sLaw

Our next subject is Hooke's Law.
This law describes the force of a stretched or compressed spring.

Let's assume that we have a weightless spring horizontally lying on the frictionless table along an imaginary X-axis and fixed at the left end. Its free right end is at coordinate x=0 and there is a point mass m attached to this free end of a spring.
Then we stretch this spring by pulling the right end from a neutral position by certain length x.

Obviously, the spring exerts a force to compress back to a neutral position. The Hooke's Law states that within certain reasonable boundaries (no over-stretching) this force is proportional to a difference in length between a stretched string and a string in a neutral position.
This is expressed by the formula
F = −k·x
where F is the force exerted by a spring, x is a displacement of the free end of a spring from a neutral position, k is a positive constant that characterizes a spring (called a spring constant) and the minus sign signifies that the direction of force is opposite to the direction of displacement because, if displacement is positive (stretching), the force is directed towards negative direction of the X-axis and, if displacement is negative (compression), the force is directed towards positive direction of the X-axis.

Now recall Newton's Second Law that related the force and acceleration
F = m·a
where F is the force, m is the mass of an object and a is its acceleration.

From these two laws we conclude that
m·a = −k·x

Since x is a distance along the X-axis and a is an acceleration along this axis, that is a second derivative from a distance by time, we came up with the following differential equation
m·x''(t) = −k·x or
m·x''(t) + k·x(t) = 0 or
x''(t) + (k/m)·x(t) = 0
This is a second order ordinary differential equation. It is a little more complex than we considered in a lecture about acceleration and Newton's Second Law.
Let's try to solve it.

First of all, let us mention that even a simple guessing in this and many other cases is a good choice. Recall that first derivative of sin() is cos() and the first derivative of cos(), that is the same as the second derivative of sin(), is −sin(). So, the equation x''(t)+x(t)=0 has a solution x=sin(t). This is very close to what we have. Adding a factor α to an argument might help to satisfy multipliers in our equation:
if x(t)=sin(α·t) then
x'(t)=α·cos(α·t) and
x''(t) = −α²·sin(α·t)
and, therefore,
x''(t)+α²·x(t) = 0
Now we can choose α to satisfy α²=k/m, and the solution to our equation is found.

Guessing is good, when we can guess (as in this case), but guessing might not be successful and, even if you managed to guess one solution, it's not a guarantee that all solutions are found. By the way, if we start with cos(), we will also find a solution.
So, let's have some theory.

Our differential equation belongs to a class of linear ordinary differential equations of second order with constant coefficients and can be generalized as
x''(t) + p·x'(t) + q·x(t) = 0

As we saw above, functions sin() and cos() might be involved in a solution. Analogous quality of derivative being similar to a function itself is possessed by exponential functions. Recall also that exponential functions with complex exponent is related to trigonometric function through famous Euler's formula
e^it = cos(t) + i·sin(t)
So, exponential functions, in some way, are more general than trigonometric, they encompass them.
Therefore, it's only natural to look for a solution in terms of exponential functions.
Let's try.

Assume, we are looking for a solution to our equation in the form x(t)=e^λ·t, where λ might be any (including complex to accommodate trigonometric functions) number.
Then derivatives of this function are:
x'(t) = λ·e^λ·t
x''(t) = λ²·e^λ·t
Putting this into our equations, we get
λ²·e^λ·t + λ·p·e^λ·t + q·e^λ·t = 0
Canceling e^λ·t, we get a simple quadratic equation for λ called a characteristic polynomial of a given differential equation:
λ² + p·λ + q = 0
Since this equation always has two solutions λ₁ and λ₂ among complex numbers, we will have two particular solutions to our differential equation:
e^λ₁·t and e^λ₂·t
Finally, any linear combination of these two particular solutions will also be a solution (since our differential equation is linear and a derivative of linear combination of functions is a linear combination of derivatives).
Therefore, we can state the general solution to our differential equation:
x(t) = C₁·e^λ₁·t+C₂·e^λ₂·t
which depends on two unknown complex constants C₁and C₂, their values can be defined only if some initial conditions of the movement are given.

Let's get back to a movement of a spring.
Our initial equation
x''(t) + (k/m)·x(t) = 0
has a characteristic polynomial
λ² + k/m = 0
(where both k and m are positive) with two solutions:
λ₁ = √(k/m)·i and
λ₂ = −√(k/m)·i
where i²=−1 is an imaginary unit in the field of complex numbers.

Let ω = √(k/m).
Now we can represent the general solution to a movement of a spring based on the Hooke's Law as follows:
x(t) = C₁·e^iωt + C₂·e^−iωt
This expression can be easily transformed using Euler's formula into
x(t) = C₁·cos(ωt)+i·C₁·sin(ωt)+
+C₂·cos(−ωt)+i·C₂·sin(−ωt)

Since C₁ can be represented as A₁+i·B₁ and C₂ can be represented as A₂+i·B₂, where A₁, B₁, A₂ and B₂ are undefined unknown real numbers, the whole expression can be represented as
D₁·cos(ωt) + D₂·sin(ωt) + i·Z
where coefficients D₁ and D₂are any real numbers and i·Z represents purely imaginary part.

Since we deal with physics, we should exclude all imaginary solutions and leave only those, where D₁ and D₂ are real numbers.
So, the general physical solution looks like
x(t) = D₁·cos(ωt) + D₂·sin(ωt)
where D₁ and D₂ are undefined unknown real numbers.

In our experiment we have stretched a string by some known distance from a neutral position and let it spring back. That means, we know the initial position x(0)=d and initial speed x'(0)=0.
These initial conditions are sufficient to determine two unknown constants in our equation of a motion:
x(0)=d ⇒
⇒ D₁cos(0)+D₂sin(0)=d
⇒ D₁ = d
x'(0)=0 ⇒
⇒ −ω·D₁sin(0)+ω·D₂cos(0)=0
⇒ D₂ = 0

The final form of an equation of motion is
x(t) = d·cos(ωt)
where ω = √(k/m), k is a spring constant, m is a point mass at its free end and d is the initial distance we have stretched a spring from its neutral position.
As seen from this equation of motion, a free end of a spring with a mass attached to it will indefinitely oscillate around the neutral point.
The end.

Unizor - Ordinary Differential Equations - Acceleration

Notes to a video lecture on http://www.unizor.com

Higher Order Ordinary
Differential Equations -
Acceleration

Differential equations can include derivatives of higher order - second derivative, third, etc.
Probably, most common equations of this type are those with the second order derivative.
These equations are very often occur in science, especially in Physics. Let's address these equations and approaches to solve them.

Our first subject is a concept of acceleration and Newton's Second Law.

Recall that speed measures how fast a distance from some starting point changes, that is, if this distance is represented as a function x(t) of time t, speedv(t) at any moment t is the first derivative of distance by time:
v(t) = x'(t) = dx/dt

But speed does not have to be constant, we can move faster, increasing our speed (accelerating), or slower, decreasing it (decelerating).
To measure how fast our speed changes with time, as usually, when we want to measure how fast anything changes with time, we use a derivative.
Differentiating speed (a function of time) by time we obtain this measure of change of speed at any moment. This derivative of speed by time is called acceleration a(t):
a(t) = v'(t) = dv/dt =
= x''(t) = d²x/dx²
That is, acceleration is the second derivative of distance by time.

Newton's Second Law states that the force F applied to an object and the acceleration a this object obtains as a result of this application of force are related as follows:
F = m·a
where m is the object's mass (presumed constant).

Assuming that our motion occurs along a straight line with coordinates and, therefore, the position of an object is defined by its X-coordinate x(t), Newton's Second Law is an ordinary differential equation of second order because acceleration is the second derivative of the X-coordinate of an object:
F(t) = m·x''(t)
Usually our task is to find where exactly our object is located (that is, its X-coordinate), if the force, as a function of time, is given.

Consider a case when there is no force applied to an object, that is F(t)=0.
Then, according to the Newton's Second Law,
0 = m·a(t)
from which we derive a(t)=0
Since a(t)=v'(t), we can find the speed:
v'(t)=0
⇒ v(t) = C
(where C is an unknown constant)
⇒ x'(t) = C
⇒ x(t) = C·t + D
(where D is another unknown constant)

That concludes the solution of our differential equation of the second order, and the solution includes two unknown constants that cannot be determined from the equation alone. It's understandable since we don't know initial position of an object on the coordinate axis x(0) and the initial speed it moved v(0). These two additional pieces of information (initial conditions) are needed to determine unknown constants participating in the solution.
If x(0)=x₀ and v(0)=v₀, we can easily determine
x₀ = D and
v₀ = C
which results in the final equation of motion of an object, to which no forces are applied (or, more generally, all forces applied to it are balancing each other).
x(t) = x₀ + v₀·t

By solving the above differential equation of the second order, we have mathematically derived the Newton's First Law as a consequence of the Second Law.
Newton's First Law (law of inertia) states that if the sum of all forces applied to an object is zero, then the object at rest will continue to stay at rest (its speed is and will be 0) and objects moving at some speed will continue to move with the same speed and direction (its speed is constant).

Now consider a case when the force applied to an object is not zero, but constant, that is F(t)=P (const). Let's attempt to solve our differential equation in this case to determine the coordinate of an object as a function of time x(t).
F(t)=P=m·a(t)=m·v'(t)
where P is a known constant.
This implies that acceleration a(t) must be a known constant and equals to P/m. Let's use symbol a instead of a(t) to signify this.
Since a=v'(t), we derive
v(t) = a·t + C
where C is an unknown constant.
Then
x'(t) = a·t + C
⇒ x(t) = a·t²/2 + C·t + D
where D is another unknown constant.
To determine two unknown constants we need additional information - initial conditions.
Assume that the original position of an object is x(0)=x₀. This allows to determine D=x₀.
If initial speed v(0)=v₀ is known, we can determine C=v₀.
So, the final equation of the motion, when a constant force is applied is
x(t) = a·t²/2 + v₀·t + x₀

In general, if the force is variable and/or the mass is variable, from Newton's Second Law we can construct a differential equation of the second order, where the second derivative is explicitly represented by a known function:
F(t) = m(t)·x''(t)
⇒ x''(t) = F(t)/m(t)
⇒ d/dt[x'(t)] = F(t)/m(t)
⇒ x'(t) = ∫[F(t)/m(t)]dt
⇒ x(t) = ∫{∫[F(t)/m(t)]dt}dt

As we see, Newton's Second Law presents the simplest kind of ordinary differential equation of the second order. It can be solved by double integration.
It should not be forgotten that in the process of each integration there will appear an unknown constant, to get its value an initial condition should be known and applied.

Unizor - Ordinary Differential Equations - Linear Equations

Notes to a video lecture on http://www.unizor.com

Linear Ordinary Differential Equations

Standard form of linear ordinary differential equations is
f(x)·y' + g(x)·y + h(x) = 0
As the first step, we can divide all members of this equation by f(x) (assuming it's not identically equal to 0), getting a simpler equation
y'+u(x)·y+v(x) = 0
The suggested solution lies in the substitution y(x)=p(x)·q(x), where p(x) and q(x) are unknown (for now) functions.
Express y'(x) in terms of p(x) and q(x):
y' = p·q'+q·p'
Substitute this into our equation:
p·q'+q·p'+u·p·q+v = 0
Let's simplify this
p(q'+u·q)+q·p'+v = 0
If there are such functions p(x) and q(x) that satisfy conditions
(1) q'+u·q = 0 and
(2) q·p'+v = 0
our job would be finished.
Let's try to find such functions.
From the equation (1) in our pair of equations we derive
q'/q = −u,
which can be converted into
dq(x)/q(x) = −u(x)·dx
that can be solved by integrating:
ln(q(x)) = −∫u(x)·dx
q(x) = e^{−∫u(x)·dx}
Once q(x) is found, we solve the equation (2) for p(x):
p'(x) = −v(x)/q(x),
which can be integrated to find
p(x) = −∫v(x)/q(x) dx
and, consequently, y(x)=p(x)·q(x) can be fully determined.
Let's consider a few examples.

Example 1

Solve the following linear differential equation
y' + y + x = 0

Let's look for a solution in a form
y(x)=p(x)·q(x)
Then
y'(x) = p'(x)·q(x)+p(x)·q'(x)
Our equation looks like this now
p'·q+p·q' + p·q + x = 0
Factor out p, getting
p·(q'+q) + (p'·q+x) = 0
We will try to find p(x) and q(x) to separately bring to zero q'+q and p'·q+x.
Let's look for a function q(x) that brings expression q'+q to zero:
q'+q = 0
dq/q = −dx
∫dq/q = −∫dx
ln(q) = −x + A
(where A is any constant)
q(x) = B·e^−x
(where B=e^A, so it represents any positive number)
Next, let's find a solution to
p'·q+x = 0
p'·B·e^−x+x = 0
p(x) = −(1/B)·∫x·e^xdx
This integral can be found using the "by parts" technique:
p(x) = −(1/B)·(x·e^x−∫e^xdx) =
= −(1/B)·(x·e^x−e^x−C) =
= −(1/B)·(x−1)·e^x+C/B
(where B is any positive constant and C is any constants)
Now let's find y(x)=p(x)·q(x):
y(x) = [−(1/B)·(x−1)·e^x+C/B]·[B·e^−x] =
= 1−x+C·e^-x

Checking:
y'(x) = −1 − C·e^-x
y'+y+x = −1 − C·e^-x+1−x+C·e^-x+x = 0
Solution was correct.

Example 2

Solve the following linear differential equation
y'·cos(x) + y·sin(x) − 1 = 0

First of all, let's normalize it by dividing by cos(x), noticing that sin(x)/cos(x)=tan(x) and 1/cos(x)=sec(x):
y' + y·tan(x) − sec(x) = 0
Let's look for a solution to this equation in a form
y(x)=p(x)·q(x)
Then
y'(x) = p'(x)·q(x)+p(x)·q'(x)
Our equation looks like this now:
p'·q+p·q'+p·q·tan(x)−sec(x) = 0
Factor out p, getting
p·(q'+q·tan(x))+p'·q−sec(x) = 0
First, let's find function q(x) such that
q' + q·tan(x) = 0
It can be solved using the technique of separation:
dq/q = −tan(x)dx
Since [ln(x)]' = 1/x and [cos(x)]' = −sin(x), the last equation can be transformed into
d(ln(q)) = d(cos(x))/cos(x)
d(ln(q)) = d(ln(cos(x)))
Now it's easy to integrate, the result is
ln(q) = ln(cos(x))+C
where C - any real number, from which, raising number e to both left and right sides, follows that
q = D·cos(x)
(new constant D=e^C represents any positive number)
Now let's find function p(x) such that
p'·q − sec(x) = 0
Substitute already found q(x) getting
p'·D·cos(x) − sec(x) = 0
p'(x) = (1/D)·(1/cos²(x))
p(x) = (1/D)·∫dx/cos²(x) =
= (1/D)·[tan(x) + E]
where D and E are constants (D - any positive, E - any real)
Let's determine y=p·q now.
y(x) = p(x)·q(x) =
= (1/D)·[tan(x) + E]·D·cos(x) =
= sin(x) + E·cos(x)

Let's check this result.
y'(x) = cos(x)−E·sin(x)
y'·cos(x) + y·sin(x) − 1 =
= cos²(x)−E·sin(x)·cos(x) + sin²(x)+E·cos(x)·sin(x)−1 =
= sin²(x) + cos²(x) −1 = 0
which proves the correctness of our answer.


Example 3

Solve the following differential equation
ln(x·y'+y) = ln(2x)+x²

It's not linear, but can be made linear if we raise e to a power defined by its left and right sides, getting
x·y'+y = 2x·e^x²
Let's normalize it by dividing by x:
y' + y/x = 2e^x²
Now it's a linear equation that we know how to solve.
Let's look for a solution in a form
y(x)=p(x)·q(x)
Then
y'(x) = p'(x)·q(x)+p(x)·q'(x)
Our equation looks like this now
p'·q+p·q' + p·q/x = 2e^x²
Factor out p, getting
p·(q'+q/x) + p'·q = 2e^x²
We will try to find p(x) and q(x) to separately
(a) bring to zero q'+q/x and
(b) equalize p'·q with 2e^x².
Let's solve equation (a) and look for a function q(x) that brings expression q'+q/x to zero:
q'+q/x = 0
dq/q = −dx/x
∫dq/q = −∫dx/x
ln(|q|) = −ln(|x|) + A
where A - any constant.
Raising e to both sides of this equation, we get
|q| = B/|x|
where B=e^A - any positive number.
Let's get rid of absolute values in the above equation by allowing B to be any non-zero real number, so
q = B/x
Substitute it to equation (b):
p'·B/x = 2e^x²
p(x) = (1/B)∫2x·e^x²·dx
Since derivative of x² is 2x,
p(x) = (1/B)∫e^x²·d(x²)
Now we can integrate directly:
p(x) = (1/B)e^x² + C
where C is any real number.
This allows to express the solution to our differential equation in the form
y(x) = p(x)·q(x)
where
p(x) = (1/B)e^x² + C and
q(x) = B/x
That produces
y = e^x²/x + C/x = (e^x²+C)/x
where C - any real number.

Checking:
y' = −(e^x²+C)/x² + e^x²·2x/x =
= e^x²·(2−1/x²) −C/x²
y/x = e^x²/x² + C/x²
y' + y/x = 2e^x²,
which corresponds to the original equation after multiplying both sides by x and taking logarithm.
The end.

Unizor - Ordinary Differential Equations - Homogeneous Equations

Notes to a video lecture on http://www.unizor.com

Homogeneous Ordinary Differential Equations

We have defined homogeneous ordinary differential equations of the first order as an equation
F(x, y, y')=0
which does not change if we replace x with λ·x and y with λ·y, where λ - any real number not equal to zero.
In other words,
F(x, y, y') = F(λ·x, λ·y, y')
Examples:
F(x, y, y') = y'+y/x
F(x, y, y') = 3y'+x·y/(x²+y²)
etc.

The recommended technique to solve these equations is to substitute function y(x) with x·z(x) and solve the equation for z(x), after which determine y(x)=x·z(x).

Let's solve a few equations of this kind.

Example 1

Check for homogeneousness and solve the following equation:
x·y' = x·sin(y/x) + y

Checking for homogeneousness.
Substitute x with λ·x and y with λ·y:
λ·x·y' = λ·x·sin(λ·y/(λ·x)) + λ·y
Obviously, λ cancels out completely, which proves homogeneous character of the equation.
Now let's solve this equation using the substitution z(x)=y(x)/x, which results in y(x)=z(x)·x, and express the initial equation in terms of x, z and z'.
x·(z'·x + z) = x·sin(z) + z·x
Simplifying:
z'·x + z = sin(z) + z
z'·x = sin(z)
dz/sin(z) = dx/x
∫dz/sin(z) = ∫dx/x
The right side is easy, the integral equals to ln(|x|)+C.
The left side is more involved.
∫dz/sin(z) =
= ∫dz/(2sin(z/2)·cos(z/2)) =
= ∫d(z/2)/(sin(z/2)·cos(z/2))
Substitute u=z/2, getting
∫du/(sin(u)·cos(u)) =
= ∫cos(u)d(u)/(sin(u)·cos²(u)) =
= ∫d(sin(u))/(sin(u)·cos²(u))
Substitute t=sin(u), getting
∫dt/[t·(1−t²)]
The polynomial in the denominator is
t·(1−t²) = t·(1−t)·(1+t)
Its inverse can be represented as
1/t − 1/[2(1+t)] + 1/[2(1−t)]
which makes our integral equal to
∫{[2/t−1/(1+t)+1/(1−t)]/2}dt
The last expression can be represented as a sum of three integrals, the result of integration is:
ln(|t|) − ln(|1+t|)/2 − ln(|1−t|)/2
where t=sin(z/2)
This leads us to a final solution of our differential equation.
ln(|sin(z/2)|) − ln(1+sin(z/2))/2 − ln(1−sin(z/2))/2 = ln(|x|)+C
and then we should substitute z=y/x to get the final expression
ln(|sin(y/2x)|) − ln(1+sin(y/2x))/2 − ln(1−sin(y/2x))/2 = ln(|x|)+C
Using this as an exponent, we come up with an expression without logarithms
|sin(y/2x)| /[(1+sin(y/2x))·(1−sin(y/2x))] = C·|x|
A simplification in the denominator results in
|sin(y/2x)| / cos²(y/2x) = C·|x|
We leave it "as is" without resolving for y(x).

Example 2

Check for homogeneousness and solve the following equation:
[(y − x·y')/x]^x = e^y

Checking for homogeneousness.
Substitute x with λ·x and y with λ·y:
[(λy − λx·y')/λx]^λx = e^λy
Cancel λ in the ratio, getting:
[(y − x·y')/x]^λx = e^λy
This can be written as
{[(y − x·y')/x]^x}^λ = [e^y]^λ
Raising both sides to power 1/λ (or, which is the same, extracting a root of power λ) we come to the original equation, which proves homogeneous character of the equation.
Now we will solve it using the recommended technique.
Substitute z(x)=y(x)/x, which results in y(x)=z(x)·x and express the initial equation in terms of x, z and z'.
The expression for a derivative y' is:
y' = (z·x)' = z'·x+z
New equation is, therefore,
[(z·x − x·(z'·x+z))/x]^x = e^z·x
Simplifying it by raising to power 1/x both sides (or, equivalently, extracting a root of power x):
[(z·x − x·(z'·x+z))/x] = e^z
Cancel x:
z − (z'·x+z) = e^z
Cancel z:
−z'·x = e^z
This equation is separable, let's separate x from x, getting
−e^−z·dz = dx/x
Ready to integrate:
∫−e^−z·dz = ∫dx/x
e^-z = ln(x)+C
(assuming for simplicity positive only sign for x, so integral on the right is ln(x) instead of ln(|x|))
From the last equation we derive:
−z = ln(ln(x)) + C
z = −ln(ln(x)) + C
Now we can use it to find an expression for y:
y = −x·ln(ln(x)) + C

Solution must be checked.
It's easier, instead of checking the original equation
[(y − x·y')/x]^x = e^y
to check the equality of logarithms from both sides:
x·ln[(y − x·y')/x] = y
or, simpler,
ln(y/x − y') = y/x
where we should substitute
y = −x·ln(ln(x)) + C
and
y' = −ln(ln(x))−x·(1/ln(x))·(1/x)
or, simpler,
y' = −ln(ln(x)) − 1/ln(x)
Let's disregard constant C in this checking to make manipulations simpler.
Then, since
y/x = −ln(ln(x))
we will have to check that ln(−ln(ln(x)) + ln(ln(x)) + 1/ln(x)) = −ln(ln(x))
Canceling opposite positive and negative members under logarithm on the left, we come to an obvious equality
ln(1/ln(x)) = −ln(ln(x))
which proves the correctness of our solution.

Example 3

Check for homogeneousness and solve the following equation:
x·y·y' = (x+y)²

Checking for homogeneousness.
Substitute x with λ·x and y with λ·y:
λx·λy·y' = (λx+λy)²
λ²x·y·y' = λ²(x+y)²
Obviously, λ cancels out, and we get the same original equation.
Now let's solve it by substituting z(x)=y(x)/x, which results in y(x)=z(x)·x and express the initial equation in terms of x, z and z'.
The expression for a derivative y' is:
y' = (z·x)' = z'·x+z
So, our equation looks like
x·(z·x)·(z'·x+z) = (x+z·x)²
Simplifying by opening all parenthesis, we get
x²·(x·z·z'+z²) = x²·(1+z)²
x·z·z'+z² = (1+z)²
x·z·z' = 1+2z
This equation can be solved using the method of separation.
z·dz/(1+2z) = dx/x
Integrating the left side of this equation:
∫z·dz/(1+2z) =
= (1/2)∫(1+2z−1)·dz/(1+2z) =
= (1/2)[∫dz − ∫dz/(1+2z)] =
= (1/2)[z−(1/2)ln(1+2z)] + C
Integrating the right side of the equation:
∫dx/x = ln(x) + C
Since integral of both sides are equal,
(1/2)[z−(1/2)ln(1+2z)] =
= ln(x)+ C
which can be simplified
2z − ln(1+2z) = 4ln(x) + C
Though this equation for z(x) cannot be easily solve for z, it allows to replace the original differential equation for y with purely algebraic one, replacing z with y/x:
(A) 2y/x − ln(1+2y/x) =
= 4ln(x) + C
This is the final algebraic answer to our differential equation. Though it's not resolved for y(x), it's still the best solution we can come up with.

Solution must be checked.
If this equality that includes function y(x) is correct, derivatives of both parts are also equal. Let's differentiate them both.
−2y/x² + 2y'/x − (1/(1+2y/x))·(−2y/x²+2y'/x) = 4/x
Simplifying by multiplying by x²:
−2y + 2xy' − x·(−2y+2xy')/(x+2y) = 4x
Multiplying by x+2y:
−2xy−4y²+2x²y'+4xyy'+2xy−2x²y' = 4x²+8xy
After cancellation of mutually opposing by sign members and dividing by 4 we get:
−y²+xyy' = x²+2xy
which easily transforms into
xyy' = (x+y)²
that corresponds to original differential equation.
This proves the correctness of the answer (A) as an equation that includes x and y(x) without derivatives that we obtained above.

Unizor - Ordinary Differential Equations - Separable Equations

Notes to a video lecture on http://www.unizor.com

Separable Ordinary Differential Equations

The process of "separation" as a method of solving differential equation of the first order F(x,y,y')=0 should result in the following equality:
f(y)·dy = g(x)·dx
which allows for separate integration of left and right sides.

This can be assured if our initial equation F(x,y,y')=0 can be transformed into y'=P(x)·Q(y).
Indeed, from the last equation follows
dy/dx = P(x)·Q(x)
and
dy/Q(y) = P(x)·dx
which can be integrated separately, left side - by y and right side - by x.

Examples below use exactly this approach.

Example 1

y' + x·y + y − x = 1

Perform the transformation:
y' = −x·y − y + x + 1
y' = −y·(x+1) + (x + 1)
y' = (1−y)·(x+1)
dy/(1−y) = (x+1)dx
∫dy/(1−y) = ∫(x+1)dx
Both integrals are trivial.
−∫d(y−1)/(y−1)=∫(x+1)d(x+1)
−ln(y−1) = (x+1)²/2 + C
y = 1 + C·e^−(x+1)²/2

Example 2

y' − (x+1)·e^(x+y) = 0

Perform the transformation:
y' = (x+1)·e^x·e^y
y'·e^−y = (x+1)·e^x
e^−y·dy = (x+1)·e^x·dx
∫e^−y·dy = ∫(x+1)·e^x·dx
Integral on the left is straight forward.
Integral on the right can be calculated using the integration "by-part":
−e^−y = (x+1)·e^x −∫e^x·d(x+1)
−e^−y = (x+1)·e^x − e^x + C
−e^−y = x·e^x + C
e^−y = C−x·e^x
y = −ln(C−x·e^x)

Example 3

ln(y') = x + y

Perform the transformation:
y' = e^x · e^y
e^−y·dy = e^x·dx
∫e^−y·dy = ∫e^x·dx
−e^−y = e^x+C
e^−y = −e^x−C
y = −ln(C−e^x)

Example 4

sin(y)·y' = sin(x+y) + sin(x−y)

First of all, recall the trigonometric identities
sin(x+y) =
= sin(x)·cos(y)+cos(x)·sin(y)
sin(x−y) =
= sin(x)·cos(y)−cos(x)·sin(y)
from which follows
sin(x+y)+sin(x−y) =
= 2·sin(x)·cos(y)
Perform the transformation of our equation using the last expression:
sin(y)·y' = 2·sin(x)·cos(y)
Now we can separate:
sin(y)·dy/cos(y) = 2·sin(x)·dx
Continue transformation:
−dcos(y)/cos(y) = −2·dcos(x)
Easy to integrate now:
ln(|cos(y)|) = 2·cos(x) + C
Ignoring difficulties with absolute value and periodicity to shorten the presentation of an idea, it can be solved for y
|cos(y)| = e^2·cos(x)+C
y = arccos(e^2·cos(x)+C)

Unizor - Ordinary Differential Equations - Major Types

Notes to a video lecture on http://www.unizor.com

Ordinary Differential Equations
Major Types of Equations

In this lecture we will only consider first order ordinary differential equations for a function of one argument y(x) (no higher order derivatives). The general form of these equations is
F(x, y, dy/dx) = 0

We will consider three major types of these differential equations with known approaches to integration:
- separable equations,
- homogeneous equations,
- linear non-homogeneous equations.

Separable Ordinary Differential Equations

A few examples we were working with in the introductory lecture to ordinary differential equations are separable in a sense that the original differential equation, that can be generally expressed as F(x, y, dy/dx) = 0, can be transformed into
f(y)·dy = g(x)·dx
that can be separately integrated, using the techniques of calculating indefinite integrals, and, hopefully, resolved for y.
Even if it will not be possible to resolve it for y, the result of integration will be a simpler formula G(x,y)=0 (it will also include a constant as a result of integration, which can be found if some initial condition on a function y(x) is imposed).
In any case, whether the result of integration can or cannot be resolved for y, it's still a significantly better than original equation that includes a derivative.

Example

y' + x·y = 0
Let's use the Leibniz notation for derivatives to facilitate the separation of function from its argument and resolve the equation for a derivative.
dy/dx = −x·y
Separate x and y:
dy/y = −x·dx
Now we can apply an indefinite integral to both sides to solve the equation.

Homogeneous Ordinary Differential Equations

Homogeneous equations can be defined using the following criterion.
Replace all occurrences of x with λ·x and all occurrences of y with λ·y. Do not change anything with derivative dy/dx. If, as a result, all λ's cancel each other out, the equation is homogeneous.
For example, consider the following equation:
y' + x/y + x²/y² = 0
Substitute x with λ·x and y with λ·y:
y' + (λ·x)/(λ·y) + (λ·x)²/(λ·y)² = 0
Obviously, we can reduce both ratios, getting exactly the same equation as before.
Now we will use the above example to explain the method of solving homogeneous equations.
Let's introduce a new function z(x)=y(x)/x, which results in y(x)=z(x)·x and express the initial equation in terms of x, z and z'.
The expression for a derivative y' is:
y' = (z·x)' = z'·x+z
So, our equation looks like
z'·x + z + x/(z·x) + x²/(z·x)² = 0
Simplifying by reducing the ratios by x and x², we get
z'·x + z + 1/z + 1/z²= 0
This equation can be solved using the method of separation.
z'·x = −(z + 1/z + 1/z²)
dz/(z+1/z+1/z²) = −dx/x
Now we can apply an indefinite integral to both sides to solve the equation for z(x) and then multiply it by x to get y(x).

Linear Non-Homogeneous Ordinary Differential Equations

Standard form of this type of differential equations is
f(x)·y' + g(x)·y + h(x) = 0
As the first step, we can divide all members of this equation by f(x) (assuming it's not identically equal to 0), getting a simpler equation
y'+u(x)·y+v(x) = 0
The suggested solution lies in the substitution y(x)=p(x)·q(x), where p(x) and q(x) are unknown (for now) functions.
Express y'(x) in terms of p(x) and q(x):
y' = p·q'+q·p'
Substitute this into our equation:
p·q'+q·p'+u·p·q+v = 0
Let's simplify this
p(q'+u·q)+q·p'+v = 0
If there are such functions p(x) and q(x) that satisfy conditions
(1) q'+u·q = 0 and
(2) q·p'+v = 0
our job would be finished.
Let's try to find such functions.
From the equation (1) in our pair of equations we derive
q'/q = −u,
which can be converted into
dq(x)/q(x) = −u(x)·dx
that can be solved by integrating:
ln(q(x)) = −∫u(x)·dx
q(x) = e^{−∫u(x)·dx}
Once q(x) is found, we solve the equation (2) for p(x):
p'(x) = −v(x)/q(x),
which can be integrated to find
p(x) = −∫v(x)/g(x) dx
and, consequently, y(x)=p(x)·q(x) can be fully determined.
Let's consider an example.
y' + x·y + x² = 0
If y(x)=p(x)·q(x), our equation looks like this:
p'·q+p·q'+x·p·q+x² = 0
(q'+x·q)·p+(q·p'+x²) = 0
Now we have to solve the following equation to nullify the first term:
q'+x·q = 0
(which is solvable through separation)
and substitute the resulting function q(x) into
q·p'+x² = 0
to solve it for p(x)
(which is a simple integration).

Unizor - Ordinary Differential Equations - Introduction

Notes to a video lecture on http://www.unizor.com

Ordinary Differential Equations
Introduction

Ordinary differential equations are equations, where derivatives of some function participate in the equation.
Assuming that y(x) is some unknown function, a differential equation, in its general form, looks like this:
F(x, y, y', y'',...) = 0
where F(...) is some function of many arguments.
The goal is to find the function y(x) that satisfies this equation.

Let's start with a simple example of an ordinary differential equation.
y'(x) = 2x
We can easily guess that, if a derivative of a function equals to 2x, the function must be y(x)=x²+C, where C - any constant.

On the other hand, we can represent this equation in the form
dy/dx = 2x
and transform it into
dy = 2x·dx
This is a relationship between two infinitesimals that signifies that these infinitesimals are equal in a sense that the difference between them is an infinitesimal of a higher order than themselves.
Now we can apply an operation of integration to both getting the following
∫ 1·dy = ∫ 2x·dx

Integration results in the following equality
y + C₁ = x² + C₂,
where C₁ and C₂ are any constants, and therefore, can be combined into one, getting
y = x² + C
This method of integration is a little more "scientific" than straight guessing that we employed above, though, by itself, might be difficult since it involves the operation of integration.

Notice the presence of any constant in the result. This is typical for differential equations and is similar to indefinite integrals.

Arguably, the method of separation of argument x and function y into different sides of an equation with subsequent integration is the most effective way to solve differential equations. Those equations that allow solution of this type are called separable differential equations

Let's consider a few more examples.

Example 1

x²·y'(x) = y(x)
Let's represent y'(x) as a ratio of differentials dy/dx, our equation will look like
x²·dy/dx = y(x)
Now we can separate argument x and function y into different sides of an equation
dy/y = dx/x²
Integrate both sides
∫dy/y = ∫dx/x²
which results in
ln(y) = −1/x + C
(where C is any constant) or, since we have to find an expression for y in terms of x, we can use this equality as exponents and raise e into it, getting
y = C·e^−1/x

Let's check this result.
y'(x) = C·e^−1/x·(1/x²)
x²·y'(x) = C·e^−1/x = y(x)
All is correct.

Example 2

tan(x)·y'(x) = y²(x)
Let's represent y'(x) as a ratio of differentials dy/dx, our equation will look like
tan(x)·dy/dx = y²(x)
Now we can separate argument x and function y into different sides of an equation
dy/y² = dx/tan(x)
Integrate both sides
∫dy/y² = ∫dx/tan(x)
which results in
−1/y + C = ∫cos(x)dx/sin(x)
or, equivalently, since
cos(x)·dx = dsin(x),
it can be transformed into
−1/y + C = ∫dsin(x)/sin(x)
The integral on the right can be calculated and the result is
−1/y + C = ln(sin(x))
(where C is any constant) or, since we have to find an expression for y in terms of x, we can transform it into
y = −1/[ln(sin(x))+C]
It would look better if we bring the constant under a logarithm, getting
y = −1/ln(C·sin(x))

Let's check this result.
y'(x) = [1/ln²(C·sin(x))] · [1/(C·sin(x))] · C·cos(x) = cot(x)/ln²(C·sin(x))
tan(x)·y'(x) = 1/ln²(C·sin(x)) = y²(x)
All is correct.

As you see, in all examples above there is a constant that can take any value, as in the case of indefinite integrals. That's because we explicitly use integration as a tool to solve our differential equation. That's why the term "solving" as related to differential equations sometimes is replaced with term "integrate". So, to integrate a differential equation means to solve it.

Without any additional information, as we see, a differential equation can have infinite number of solutions. But we need only one, that corresponds to some practical problem, from which this equation was obtained. Therefore, we need some condition imposed on our solution to determine this constant that is present in the general solution.

Consider Example 1 above
x²·y'(x) = y(x)
and its solution
y = C·e^−1/x
This solution represents a whole family of functions, each satisfying our differential equation.
To determine a particular solution we are interested in, we have to define what we are interested in using some additional information about function y(x). For example, we know that our function y(x) equals to 1 if x=1.
Let's substitute this into a general solution to our differential equation to find the value of constant C needed to satisfy our condition.
y(1) = C·e^−1/1 = 1
from which we can find constant C:
C·e^-1 = 1
C/e = 1
C=e
Therefore, particular solution we are looking for is
y = e·e^−1/x = e^1-(1/x)

In the Example 2 let's determine constant C by a condition y(π/2)=1
That results in the following
y(π/2) = −1/ln(C·sin(π/2)) = 1
ln(C) = −1
C = 1/e
So, our particular solution is
y = −1/ln(sin(x)/e)

Unizor - Partial Derivatives - Stationary Points

Notes to a video lecture on http://www.unizor.com

Partial Derivatives Properties - Stationary Points

We will mostly be concerned with partial derivatives of functions with two arguments.
The theory can be extended to functions of any number of arguments, but it's outside of the scope of this course.
Besides, functions of two arguments can be visualized as surfaces in three-dimensional space to better understand their properties.

Stationary points are those, where both partial derivatives of function f(x,y) of two arguments are equal to zero.

Let
g(x,y)=∂f(x,y)/∂x
h(x,y)=∂f(x,y)/∂y

Definition:
Point (a,b) is a stationary point for function f(x,y) if g(a,b)=0 and h(a,b)=0.

Theorem
A smooth function f(x,y) of two variables that has a local maximum at point (a,b) has both of its partial derivatives at this point equal to zero.

Proof

Let's prove that ∂f(x,y)/∂x=0 for x=a and y=b. The proof for other partial derivative ∂f(x,y)/∂y is analogous.
So, we fix variable y=b and calculate the partial derivative of f(x,y) by x at point x=a as follows:
∂f(x,y)/∂x = {at x=a,y=b} = lim[f(a+Δx,b)−f(a,b)]/Δx
(the limit is taken as Δx→0)
Since point (a,b) is a local maximum, the numerator [f(a+Δx,b)−f(a,b)] is negative, while the denominator Δx=(x+Δx)−x is non-positive for x+Δx ≤ x and non-negative for x+Δx ≥ x.
For a sufficiently smooth function (at least, we need the continuity of partial derivatives) this implies that the limit above must be equal to zero.

So, we have proven that for a smooth function of two variables the necessary condition for having a local maximum at point (a,b) is the equality of its partial derivatives to zero at this point.

The situation with local minimum is analogous and the equality of partial derivatives to zero at some point is a necessary condition for having a local minimum at this point.


IMPORTANT NOTE
The equality of partial derivatives to zero at some point is only a necessary condition for a function to have a local maximum or minimum at that point. It's not a sufficient condition.
This is similar to a situation with functions of one variable, when a derivative can be zero at some point, but a function can have an inflection point like function y=x³ at point x=0.
For a function of two variables a situation like this might occur when it has a saddle point.
Here is an example:

At the point in the middle of this "saddle" both partial derivatives are equal to zero, but this point is not a local minimum or maximum of a function.

Obviously, we would like to differentiate cases of a stationary point being a local maximum, a local minimum or a saddle point similarly to a situation with functions of one argument, where the second derivative sign (positive or negative) indicated whether a stationary point is minimum, maximum or inflection point.

Here is the rule, which we provide without rigorous proof.
Let's assume that function f(x,y) can be partially differentiated twice (that is, ∂f(x,y)/∂x, ∂f(x,y)/∂y, ∂²f(x,y)/∂x², ∂²f(x,y)/∂y², ∂²f(x,y)/∂x∂y exist) and all second partial derivatives are continuous.
Let's further assume that at point (a,b) both first partial derivatives equal to zero:
∂f(x,y)/∂x = 0 at x=a, y=b
∂f(x,y)/∂y = 0 at x=a, y=b
Consider the expression
Δ = ∂²f(x,y)/∂x² · ∂²f(x,y)/∂y² − [∂²f(x,y)/∂x∂y]²
at point x=a, y=b.
The rule is:
if Δ < 0, (a,b) is a saddle point;
if Δ > 0, (a,b) is a local minimum or local maximum point and the sign of ∂²f(x,y)/∂x² or ∂²f(x,y)/∂y² can be used to distinguish minimum from maximum (positive for minimum, negative for maximum and these two second derivatives must have the same sign since otherwise Δ would be negative).
All other cases are not sufficient to determine the behavior of the function at this point.

Example 1

f(x,y)=1/(1+x²+y²)

∂f(x,y)/∂x = −2x/(1+x²+y²)²
∂f(x,y)/∂y = −2y/(1+x²+y²)²
At point (0,0) both partial derivatives are equal to zero, therefore (0,0) is a stationary point.
Examine the second derivatives.
∂²f(x,y)/∂x² = (6x²−2y²−2)/(1+x²+y²)³
∂²f(x,y)/∂y² = (6y²−2x²−2)/(1+x²+y²)³
∂²f(x,y)/∂x∂y = 8x·y/(1+x²+y²)³
At point x=0, y=0 the three expressions above can be used to calculate
Δ = (−2)·(−2)−0² = 4
Since Δ is positive, we have a local minimum or maximum at point (0,0). To distinguish between them, look at the sign of the second partial derivative by x. It is negative. Therefore, we have a local maximum as is obvious from the graph above.

Example 2

f(x,y)=x·y

∂f(x,y)/∂x = y
∂f(x,y)/∂y = x
At point (0,0) both partial derivatives are equal to zero, therefore (0,0) is a stationary point.
Examine the second derivatives.
∂²f(x,y)/∂x² = 0
∂²f(x,y)/∂y² = 0
∂²f(x,y)/∂x∂y = 1
At point x=0, y=0 the three expressions above can be used to calculate
Δ = 0·0−1² = −1
Since Δ is negative, we have a local saddle point (0,0), as is obvious from the graph above.

Unizor - Partial Derivatives - Basic Properties

Notes to a video lecture on http://www.unizor.com

Partial Derivatives -
Basic Properties

We will mostly be concerned with partial derivatives of functions of two variable arguments.
The theory can be extended to functions of any number of arguments, but it's outside of the scope of this course.
Besides, functions of two arguments can be visualized as surfaces in three-dimensional space to better understand their properties.

Certain simple properties of partial derivatives of multivariable functions coincide with corresponding properties of derivatives of functions of one variable because the process of partial differentiation is, actually, a differentiation by one variable, keeping the others as constants.

So, we will not bother with proofs since they are based on corresponding properties of derivatives of functions of a single variable.
Also, whatever property is listed below for one argument of partial differentiation of multivariable function is valid for any other argument.

1. ∂[f(x,y)+g(x,y)]/∂x = ∂f(x,y)/∂x + ∂f(x,y)/∂x

Example 1
∂[ln(x²+y²)+x·y]/∂x = 2x/(x²+y²)+y

2. ∂[K·f(x,y)]/∂x = K·∂f(x,y)/∂x

Example 2
∂[2·ln(x²+y²)]/∂x = 2·2x/(x²+y²) = 4x/(x²+y²)

3. ∂[f(x,y)·g(x,y)]/∂x = g(x,y)·∂f(x,y)/∂x + f(x,y)·∂g(x,y)/∂x

Example 3
∂[(x−y)·(x²+xy+y²)]/∂x = (x−y)·(2x+y)+1·(x²+xy+y²) = 3x²
At the same time,
(x−y)·(x²+xy+y²) = x³−y³ and a partial derivative by x of this expression equals to 3x², which corresponds to a previous result.

4. ∂f(x(t),y)/∂t = ∂f(x,y)/∂x · dx(t)/dt

Example 4
∂[ln²(t)+y²]/∂t = 2ln(t)/t

Let's consider properties specific for multivariable functions.
Recall that for a function of a single variable f(x) the following approximation can be established:
Δf(x) = f(x+Δx) − f(x) ≅ (df(x)/dx)·Δx
As Δx→0, this approximation is transformed into a relationship between infinitesimals:
df(x) = (df(x)/dx)·dx

Similarly, we can establish a relationship in case of multivariable functions for each of its arguments:
f(x+Δx,y) − f(x,y) ≅ (∂f(x,y)/∂x)·Δx
and
f(x,y+Δy) − f(x,y) ≅ (∂f(x,y)/∂y)·Δy

Using the above properties, let's consider a general case when both arguments of function f(x,y) are incremented.
f(x+Δx,y+Δy) − f(x,y) =
= f(x+Δx,y+Δy) − f(x,y+Δy) + f(x,y+Δy) − f(x,y) ≅
≅ (∂f(x,y+Δy)/∂x)·Δx + (∂f(x,y)/∂y)·Δy

As Δx→0 and Δy→0, we can replace these increments with infinitesimal differentials and, using the smoothness of our functions (in particular, continuity of partial derivatives), we can write the following equivalence between infinitesimals:
df(x,y)=f(x+dx,y+dy)−f(x,y) =
= (∂f(x,y)/∂x)·dx + (∂f(x,y)/∂y)·dy
This expression is called total differential of a function of two variables.

Consider now that both arguments of our function f(x,y) are, in turn, functions of some argument:
x = x(t)
y = y(t)
We can use the same expression for a total differential
df(x,y) = (∂f(x,y)/∂x)·dx + (∂f(x,y)/∂y)·dy
but, considering that x and y are functions of t and, therefore,
dx = x'(t)·dt and
dy = y'(t)·dt,
we get the following:
df(x,y) = (∂f(x,y)/∂x)·x'(t)dt + (∂f(x,y)/∂y)·y'(t)dt

From the last equivalence we derive the formula for total derivative
df(x,y)/dt = (∂f(x,y)/∂x)·x'(t) + (∂f(x,y)/∂y)·y'(t)
This formula represents the chain rule for a function of two arguments when these arguments are, in turn, functions of some other same argument.

Example 5

Consider certain quantity of ideal gas in a reservoir with a piston, so we can change its volume, and a heater, so we can change its temperature.
The law of physics tells that the pressure P, volume V and absolute temperature T are connected by a formula
P·V/T = const
where the constant on the right depends on quantity of gas and in our case can be fixed and equal to C.

Let's apply some pressure to a piston to squeeze the gas.
The function that describes the change of volume with time is V(t).
Let's heat the gas.
The function that describes the change of temperature with time is T(t).
How fast the pressure would rise?

Since P(t)·V(t)/T(t)=C,
P(t) = C·T(t)/V(t)
The speed, with which the pressure is rising is a derivative dP(t)/dt
To calculate this derivative, we can use the formula of total derivative of a function of two arguments:
dP(t)/dt = (∂P(V,T)/∂V)·V'(t) + (∂P(V,T)/∂T)·T'(t) =
= −C·T(t)·V'(t)/V²(t) + C·T'(t)/V(t)

Unizor - Partial Derivatives - ∂²/∂x ∂y = ∂²/∂y ∂x

Notes to a video lecture on http://www.unizor.com

Partial Derivatives -
∂²/∂x ∂y = ∂²/∂y ∂x

In the following examples compare ∂²z/∂x ∂y and ∂²z/∂y ∂x.
They should be identical.

Example 1

Let z=√x·y
Then
∂z/∂x = y/(2√x·y)
∂²z/∂y ∂x = 1/(4√x·y)
∂z/∂y = x/(2√x·y)
∂²z/∂x ∂y = 1/(4√x·y)

Example 2

Let z=e^x·y
Then
∂z/∂x = y·e^x·y
∂²z/∂y ∂x = (x·y+1)·e^x·y
∂z/∂y = x·e^x·y
∂²z/∂x ∂y = (x·y+1)·e^x·y

Example 3

Let z=1/(x²+y²)
Then
∂z/∂x = −2x/(x²+y²)²
∂²z/∂y ∂x = 8x·y/(x²+y²)³
∂z/∂y = −2y/(x²+y²)²
∂²z/∂x ∂y = 8x·y/(x²+y²)³

Example 4

Let z=sin(x)/y²
Then
∂z/∂x = cos(x)/y²
∂²z/∂y ∂x = −2cos(x)/y³
∂z/∂y = −2sin(x)/y³
∂²z/∂x ∂y = −2cos(x)/y³

Example 5

Let z=arctan(x√y)
Then
∂z/∂x = √y/(1+x²·y)
∂²z/∂y ∂x = (1−x²·y)/2√y(1+x²·y)²
∂z/∂y = x/[2√y(1+x²·y)]
∂²z/∂x ∂y = (1−x²·y)/2√y(1+x²·y)²

Example 6

Let z=y^x
Then
∂z/∂x = y^x·ln(y)
∂²z/∂y ∂x = y^x−1·(x·ln(y)+1)
∂z/∂y = x·y^x−1
∂²z/∂x ∂y = y^x−1·(x·ln(y)+1)

Unizor - Partial Derivatives - Examples

Notes to a video lecture on http://www.unizor.com

Partial Derivatives - Examples

Example 1

Let z=√x·y
Then
∂z/∂x = assuming y constant = y/(2√x·y)
∂z/∂y = assuming x constant = x/(2√x·y)

Example 2

Let z=e^x·y
Then
∂z/∂x = assuming y constant = y·e^x·y
∂z/∂y = assuming x constant = x·e^x·y

Example 3

Let z=1/(x²+y²)
Then
∂z/∂x = assuming y constant = −2x/(x²+y²)²
∂z/∂y = assuming x constant = −2y/(x²+y²)²

Example 4

Let z=sin(x)/y²
Then
∂z/∂x = assuming y constant = cos(x)/y²
∂z/∂y = assuming x constant = −2sin(x)/y³

Example 5

Let z=arctan(x√y)
Then
∂z/∂x = assuming y constant = √y/(1+x²·y)
∂z/∂y = assuming x constant = x/[2√y(1+x²·y)]

Example 6

Let z=y^x
Then
∂z/∂x = assuming y constant = y^x·ln(y)
∂z/∂y = assuming x constant = x·y^x−1

Unizor - Definite Integrals - Length of Curve

Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Length of Curve

Our task is to calculate a length of a curve on a coordinate plane, defined parametrically as a set of points (x(t),y(t)), where x(t) and y(t) are smooth functions of parameter t∈[a,b].

This is one of the typical integration problem and can be approached similarly to our approach to calculate the area under a curve.

Let's break down a segment [a,b] into N intervals by points a=t₀, t₁,...t_N=b, which results in corresponding breaking of our curve into smaller pieces by points
(x₀,y₀)=(x(t₀),y(t₀)),
(x₁,y₁)=(x(t₁),y(t₁)),...
...
(x_N,y_N)=(x(t_N),y(t_N)),

Each small piece of a curve can be approximated with a straight line from one of its ends to another, and this approximation will be better with the density of break points increasing.
So, the n-th piece of a curve is approximated with a segment from point (x_n−1,y_n−1) to point (x_n,y_n).

The length of this n-th piece of a curve L_n can be calculated using the regular formula of a length of a segment between two points on a plane:
L_n² = (x_n−x_n−1)²+(y_n−y_n−1)²

Taking into account that both coordinates are functions of parameter t, we can express it differently:
L_n² = (x(t_n)−x(t_n−1))² +
+ (y(t_n)−y(t_n−1))²

Notice that
x(t_n)−x(t_n−1) ≅ x^I(t_n)·(t_n−t_n−1)
where x^I(t) is a derivative of function x(t) by t and approximation gets better and better as we increase the density of points t_n.

So, we can express the length of the n-th piece of a curve as
L_n² ≅ [x^I(t_n)² + y^I(t_n)²] ·
· (t_n−t_n−1)²

From this we get
L_n ≅ SQRT[x^I(t_n)² + y^I(t_n)²] ·
· (t_n−t_n−1)

As before, we use Δt_n=t_n−t_n−1.
Now the length of a curve can be approximated by this sum:
L ≅ Σ_n∈[1,N] L_n ≅
≅ Σ_n∈[1,N] f(x_n)·Δt_n
where
f(t)=SQRT[x^I(t)² + y^I(t)²]

Recall the definition of the definite integral:
∫_a^bf(x) dx =
= lim Σ_i∈[1,N] f(x_i)·Δx_i
where Δx_i=x_i−x_i−1 represents partitioning of segment [a,b] into N parts, and it is assumed that the widest interval Δx_i is shrinking to zero by length as N→∞.

Clearly, we are dealing with an integral in our case. The sum of pieces of our curve represents Riemann sum, and the limit of this sum, as the density of points t_n increases, is the following integral:
∫_a^bSQRT[x^I(t)² + y^I(t)²] dt

So, the length of the curve on a coordinate plane, defined parametrically by coordinate functions x(t) and y(t), where t∈[a,b], equals to
∫_a^bSQRT[x^I(t)² + y^I(t)²] dt

Let's apply this to a couple of practical problems.

Problem 1
Calculate the length of a circle of radius R.

Solution
To define a circle parametrically, let's choose an angle between its radius and a positive direction of the X-axis as a parameter t∈[0,2π].
Then the X-coordinate of a point on a circle, whose radius forms an angle t with positive direction of the X-axis, equals to x(t)=R·cos(t) and Y-coordinate equals to y(t)=R·sin(t).

Now we can use the above formula to calculate the length of circle.
∫₀^2πSQRT[x^I(t)² + y^I(t)²] dt =
= ∫₀^2πSQRT[R²sin²(t) +
+ R²cos²(t)] dt =
= ∫₀^2πR dt =
= R·2π − R·0 = 2πR
As we see, the length, as we counted, equals to the one in a classical formula for the length of the circle. No surprise here.
The end.

Problem 2
Calculate the length of an astroid given parametrically as
x=R·cos³(t), y=R·sin³(t)
where t∈[0,2π]

Solution
Astroid looks like this:

Its vertices are at distance R from the origin, and its four arcs correspond to parameter t changing in each quadrant from 0 to 2π.
Using the symmetry, lets' calculate the length of only one arc for t∈[0,π/2] and then multiply it by four.
First, calculate the derivatives by t:
x^I(t) = −3R·cos²(t)·sin(t)
y^I(t) = 3R·sin²(t)·cos(t)
Now we use the formula for the length of a curve.
SQRT[x^I(t)² + y^I(t)²] =
= 3R·|sin(t)·cos(t)| =
= (3/2)R·|sin(2t)|
This should be integrated within margins t∈[0,π/2].
Within these margins 2t changes from 0 to π and sin(t) is non-negative, so we can drop absolute value.
So, we need to find the following integral:
∫₀^π/2(3/2)R·sin(2t) dt
Indefinite integral of sin(2t) is −cos(2t)/2, which means that we have to calculate the following
(−cos(π)/2) − (−cos(0)/2) =
= 1/2 + 1/2 = 1.
Therefore, the length of one quarter of an astroid equals to (3/2)R·1 = (3/2)R and the length of an entire astroid is (3/2)R·4=6R.
The end.