Math Review with Matlab: Application: Solving Differential Equations 4/16/2017 Math Review with Matlab: Laplace Transform Application: Solving Differential Equations S. Awad, Ph.D. M. Corless, M.S.E.E. E.C.E. Department University of Michigan-Dearborn

Solving Differential Equations with Initial Conditions Theorem relating time domain differentiation and the Laplace Transform Approach to solving initial value problems for linear differential equations 2nd Order Example Exponential Example Homogeneous and Unit Step Example with Complex Poles

Theorem ROC s > a Given x(t) and it’s derivative x’(t) x(t) is a continuous function x’(t) is piecewise continuous on any interval 0 £ t £ A Where there exist constants K, a, M such that: Ensures that X(s) will converge Then LT{x(t)} exists for s > a and can be determined using the following formula ROC s > a

Proof Consider the integral: Where: Let t1, t2, t3, ... tn be the points on the interval 0 £ t £ A where x’(t) is discontinuous such that: Integrating each term on the right by parts yields:

Proof Continued ROC s > a Since x(t) is continuous As A ® ¥, e-stx(A) ® 0 whenever s > a, Hence for s > a ROC s > a

Corollary Given: Function x(t) and its derivatives such that: Where there exist constants K, a, M such that: Then the Laplace transform of the nth derivative of x(t) exists for s > a and is given by:

Approach to Solving Initial Value Problems Given an nth order time domain differential equation of x(t) with initial initial conditions for x(t) and its n-1 derivatives 1. Take the Laplace Transform of the differential equation to get a function of X(s) 2. Substitute Initial Conditions 3. Write equation for desired function 4. Convert X(s) to be strictly rational and use Partial Fraction Expansion 5. Use the table method to get the Inverse Laplace Transform thus getting x(t)

2nd Order DE Example Find the solution, y(t), for the differential equation: Subject to the Conditions:

Left Side Laplace Transform Using the following transformations: Take the Laplace Transform of the Left Side of the original equation

Right Side Laplace Transform Take the General Laplace Transform of the Right Side of the original equation Given x(t) is the discontinuous Unit Step function Initial Condition of x(t) is considered x(0-) = 0 Minus subscript indicates the value of x(t) when approaching from the left (negative time)

Laplace Transform with Initial Conditions The transformed equation is then: Substituting in the initial conditions it becomes:

Substitute and Simplify x(t) was given as the unit step function u(t) whose transform is: Simplifying terms yields:

Equation for Y(s) Reorder terms to create an equation for Y(s)

Partial Fraction Expansion Perform partial fraction expansion to break into terms with known Inverse Laplace Transforms

Inverse Laplace y(t) is the inverse Laplace Transform of Y(s): LT-1

Exponential DE Example Find the solution to the following second order differential equation with a causal exponential term: Subject to the initial conditions:

Take Laplace Transform Using the transform pairs: Take the Laplace Transform of both sides

Equation for Y(s) Replace initial conditions Write an equation for Y(s)

Partial Fraction Expansion Perform partial fraction expansion to break into terms with known Inverse Laplace Transforms

Inverse Laplace Transform Take Inverse Laplace Transform to get solution for y(t) LT-1

Matlab Verification Use Matlab to verify that Is a valid solution for: » y=(1/3)*exp(-t)+2*exp(t)-(1/3)*exp(2*t); » left_side = diff(y,2)-3*diff(y)+2*y left_side = 2*exp(-t)

General 2nd Order Example Consider a system described by the differential equation with the following initial conditions: 1. Use Matlab and the Laplace Transform method to determine x(t) when f(t) = 0 (homogenous solution) 2. Build upon the previous results to determine x(t) when f(t) = u(t) (unit step) 3. Directly use Matlab to determine x(t) when f(t) = u(t)

Part 1: 2nd Order Homogeneous Consider a system described by the linear homogeneous differential equation subject to the initial conditions: Use the Laplace Transform method to determine x(t) Use Matlab to help perform each step of the Partial Fraction Expansion process There will be imaginary poles Use Matlab to verify the Inverse Laplace Transform

Take Laplace Transform Remembering the relationships: Take the Laplace Transform of both sides

Equation for X(s) Rewrite to get an equation for X(s) Substitute in the initial conditions:

Partial Fraction Expansion Use Matlab to find imaginary poles as symbolic variables » poles = roots([1 1.4 1]) poles = -0.7000 + 0.7141i -0.7000 - 0.7141i » p1=poly2sym(poles(1),'s'); » pretty(p1) 1/2 -7/10 + 1/10 i 51 » p2=poly2sym(poles(2),'s'); » pretty(p2) -7/10 - 1/10 i 51

Relationship of Complex Conjugate Poles Poles are complex conjugates Partial fraction expansion will result in coefficients also being complex conjugates where:

Calculate Coefficients » c1= (2*s-0.2)/(s-p2); » c1= subs(c1,'s',p1); » c1= expand(c1); » pretty(c1) 1/2 8/51 i 51 + 1 » c2= conj(c1); » c2= expand(c2); » pretty(c2) 1/2 - 8/51 i 51 + 1

Inverse Laplace Transform A solution in terms of complex variables can be found directly by taking the Inverse Laplace Transform

Write as a Real Function Use trigonometric identities to simplify the answer into a purely real function

Matlab Verification Use Matlab to verify the entire Inverse Laplace process » syms x s t » X=(2*s-0.2)/(s^2+1.4*s+1); » x=ilaplace(X); pretty(x) 16 1/2 1/2 - -- exp(- 7/10 t) 51 sin(1/10 51 t) 51 1/2 + 2 exp(- 7/10 t) cos(1/10 51 t)

Part 2: 2nd Order Unit Step Repeat the example for f(t) = unit step Use results of Part 1 to help determine x(t) Take the Laplace Transform of both sides

Solve for X(s)? Substitute initial conditions Write as an equation for X(s)

Homogeneous solution, f(t)=0,solved in Part 1 X(s) Solution Solution can be broken into two parts Homogeneous solution, f(t)=0,solved in Part 1 New part to solve for

Homogeneous Solution Call homogeneous solution X1(t) From previous work in Part 1:

Inverse Laplace of X2(s) Use Matlab to determine Inverse Laplace of X2(s) » syms s » X2=1/(s*(s^2+1.4*s+1)); » x2=ilaplace(X2) x2 = 1-exp(-7/10*t)*cos(1/10*51^(1/2)*t-7/51* exp(-7/10*t)*51^(1/2)*sin(1/10*51^(1/2)*t)

Unit Step Solution Combine previous results

Part 3: Direct Solution x(t) for f(t)=u(t) can also be solved directly in Matlab to yield the same results » syms s » X=(2*s-0.2+1/s)/(s^2+1.4*s+1); » x=ilaplace(X) x = 1+exp(-7/10*t)*cos(1/10*51^(1/2)*t) -23/51*exp(-7/10*t)*51^(1/2)*sin(1/10*51^(1/2)*t)

Summary Laplace Transform is a useful technique for solving linear constant coefficient differential equations with known initial conditions Laplace Transform converts time-domain differentiation to multiplication by s terms in the s-domain Equations with real poles can be solved efficiently by hand or using Matlab to verify portions Matlab is useful for solving complex pole problems, but the solution may not be intuitively interpreted