Lecture 2 Differential equations Physics for informatics Lecture 2 Differential equations Ing. Jaroslav Jíra, CSc.

Basic division of differential equations According to type of derivation Ordinary Differential Equations – ODE They contain a function of one independent variable and its derivatives. Example: Partial Differential Equations - PDE They contain unknown multivariable functions and their partial derivatives. Example:

Types of Ordinary Differential Equations General definition of an ODE First order differential equations The highest derivative of the function y of independent variable they contain is one. Example: Higher order differential equations They contain at least second derivative of the function y Example:

Types of Ordinary Differential Equations Linear differential equations The function y appears just linearly here. There are no powers of the function y and its derivatives, there are no products of the funciton y and its derivatives. There are also no functions of the function y like sin(y), exp(y) etc. Example: Nonlinear differential equations If any of conditions previously defined for the linear DE is not met, then we talk about nonlinear DE. Example:

Types of Ordinary Differential Equations Homogeneous differential equations There is no constant or function of x on the right side of the equation. Example: Inhomogenous differential equations Exact opposite to the homogenous DE. Example:

Types of Ordinary Differential Equations Differential equations with constant coefficients The function y and all its derivatives are mutiplied just by constants. Example: Differential equations with variable coefficients The function y and its derivatives are mutiplied either by constants or by functions of x. Example:

How to solve differential equations The following methods will be mentioned subsequently Separation of variables – applicable to homogeneous first order ODEs and to specific types of inhomogeneous first order ODEs. Characteristic equation – applicable to homogeneous first order and higher order linear ODEs with constant coefficients Integrating factor – applicable to inhomogeneous first order and higher order linear ODEs. This method usually completes the solution of previous two methods for inhomogeneous equations.

First order inhomogenous linear differential equation with constant coefficients – separation of variables The most simple differential equation: where Integrating the last equation Example: Where x2+C is general solution of the differential equation Sometimes an additional condition is given like that means the function y(x) must pass through a point We have obtained a particular solution y(x)=x2 -1

First order homogenous linear differential equation with constant coefficients – separation of variables The general formula for such equation is Now we integrate both sides of the equation and then we apply an exponential function to it substituting we obtain general solution Where C is a constant resulting from the initial condition

First order homogenous linear differential equation with constant coefficients – characteristic equation The general formula for such equation is To solve this equation we assume the solution in the form of exponential function. If then and the equation changes into after dividing by the eλx we obtain This is the characteristic equation the solution is Where C is a constant resulting from the initial condition

Example of the first order linear ODE – RC circuit Find the time dependence of the electric current i(t) in the given circuit. Now we take the first derivative of the last equation with respect to time characteristic equation is Constant K can be calculated from initial conditions. We know that general solution is particular solution is

Solution of the RC circuit in the Mathematica Given values are R=1 kΩ; C=100 μF; u=10 V

First order inhomogenous linear differential equation with constant coefficients – integrating factor The general formula for such equation is Where P(x) and Q(x) are continuous functions of variable x. In the first step we omit the right side taking Q(x)=0 and than we can solve the homogeneous equation by separation of variables. Separation of variables gives us: Having solution of homogeneous part of the equation we can continue by looking for the integrating factor. Instead of constant C we are looking for a function C(x), which satisfies both homogeneous solution and the original differential equation.

If C(x) should be a function, then Substituting for y and y’ into the original equation we obtain and after small adjustment: Our integrating factor is: The general solution of this inhomogeneous equation is:

Example of the integrating factor solution – LR circuit Find the time dependence of the electric current i(t) in the given circuit. Firstly we have to solve homogeneous equation Solution of charact. equation In the second step we are looking for the integrating factor First derivative of the current

Substituting for i and di/dt into the original equation gives us General solution Knowing that initial current is zero i(0)=0, we can determine K2. Particular solution of the equation

Solution of the RC circuit in the Mathematica Given values are R=1 kΩ; L=100 mH; U=10 V

The general formula for such equation is Second order homogenous linear differential equation with constant coefficients The general formula for such equation is To solve this equation we assume the solution in the form of exponential function: If then and and the equation will change into after dividing by the eλx we obtain We obtained a quadratic characteristic equation. The roots are

There exist three types of solutions according to the discriminant D 1) If D>0, the roots λ1, λ2 are real and distinct 2) If D=0, the roots are real and identical λ12 =λ 3) If D<0, the roots are complex conjugate λ1, λ2 where α and ω are real and imaginary parts of the root

If we substitute we obtain This is general solution in some cases, but … Further substitution is sometimes used and then considering formula we finally obtain where amplitude A and phase φ are constants which can be obtained from initial conditions and ω is angular frequency. This example leads to an oscillatory motion.

Example of the second order LDE – a simple harmonic oscillator Evaluate the displacement x(t) of a body of mass m on a horizontal spring with spring constant k. There are no passive resistances. If the body is displaced from its equilibrium position (x=0), it experiences a restoring force F, proportional to the displacement x: From the second Newtons law of motion we know Characteristic equation is We have two complex conjugate roots with no real part

The general solution for our symbols is No real part of λ means α=0, and omega in our case The final general solution of this example is Answer: the body performs simple harmonic motion with amplitude A and phase φ. We need two initial conditions for determination of these constants. These conditions can be for example From the first condition From the second condition The particular solution is

Example 2 of the second order LDE – a damped harmonic oscillator The basic theory is the same like in case of the simple harmonic oscillator, but this time we take into account also damping. The damping is represented by the frictional force Ff, which is proportional to the velocity v. The total force acting on the body is The following substitutions are commonly used Characteristic equation is

Solution of the characteristic equation where δ is damping constant and ω is angular frequency There are three basic solutions according to the δ and ω. 1) δ>ω. Overdamped oscillator. The roots are real and distinct 2) δ=ω. Critical damping. The roots are real and identical. 3) δ<ω. Underdamped oscillator. The roots are complex conjugate.

Damped harmonic oscillator in the Mathematica Damping constant δ=1 [s-1], angular frequency ω=10 [s-1]

Damped harmonic oscillator in the Mathematica All three basic solutions together for ω=10 s-1 Overdamped oscillator, δ=20 s-1 Critically damped oscillator, δ=10 s-1 Underdamped oscillator, δ=1 s-1