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PAGE 1

Laplace Transform

Initial-value problems

Solve the following initial-value problem:

\[ y'' + 4y' + 13y = 5e^{-2t} \quad y(0) = 0, \, y'(0) = 6 \]

Transform both sides:

\[ s^2 Y - s y(0) - y'(0) + 4(s Y - y(0)) + 13Y = \frac{5}{s+2} \]

Note: \( y(0) = 0 \) is substituted into the equation above.

\[ (s^2 + 4s + 13)Y = 6 + \frac{5}{s+2} \]
\[ Y = \frac{6}{s^2 + 4s + 13} + \frac{5}{(s+2)(s^2 + 4s + 13)} \]

Partial Fraction Decomposition

Decomposing the second term:

\[ \frac{5}{(s+2)(s^2 + 4s + 13)} = \frac{A}{s+2} + \frac{Bs + C}{s^2 + 4s + 13} \]
\[ 5 = A(s^2 + 4s + 13) + (Bs + C)(s+2) \]
\[ 5 = (A+B)s^2 + (4A + 2B + C)s + (13A + 2C) \]

Solving the system of equations:

  • \( A + B = 0 \rightarrow B = -A \)
  • \( 4A + 2B + C = 0 \rightarrow 2A + C = 0 \rightarrow C = -2A \)
  • \( 13A + 2C = 5 \rightarrow \dots \rightarrow A = \frac{5}{9}, \, B = -\frac{5}{9}, \, C = -\frac{10}{9} \)
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\[ Y = \frac{6}{s^2 + 4s + 13} + \frac{5}{9} \cdot \frac{1}{s+2} - \frac{5}{9} \cdot \frac{s+2}{s^2 + 4s + 13} \]

Complete the square

\[ s^2 + 4s + 4 + 9 = (s+2)^2 + 3^2 \]
\[ Y = \frac{6}{(s+2)^2 + 3^2} + \frac{5}{9} \cdot \frac{1}{s+2} - \frac{5}{9} \cdot \frac{s+2}{(s+2)^2 + 3^2} \]

Taking the inverse Laplace transform:

\[ y(t) = 2e^{-2t} \sin(3t) + \frac{5}{9} e^{-2t} - \frac{5}{9} e^{-2t} \cos(3t) \]

Step function

\[ u_c(t) = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases} \]
\[ f(t) = \begin{cases} 2t & 0 \leq t < \pi \\ 0 & t \geq \pi \end{cases} = 2t + u_{\pi}(t)(-2t) = 2t - 2u_{\pi}(t)(t) \]
\[ F(s) = \mathcal{L}\{2t\} - 2\mathcal{L}\{u_{\pi}(t) \cdot t\} \]

Shift LEFT by \( \pi \) \( \rightarrow \) change \( t \) to \( t + \pi \)

\[ = \frac{2}{s^2} - 2e^{-\pi s} \mathcal{L}\{t + \pi\} = \frac{2}{s^2} - 2e^{-\pi s} \left( \frac{1}{s^2} + \frac{\pi}{s} \right) \]

Shift RIGHT: come back to \( t \)

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Laplace Transforms: Convolution and Impulse Functions

Convolution Theorem

The convolution property of the Laplace transform is defined as:

\[ \mathcal{L} \left\{ \int_{0}^{t} f(\tau) g(t-\tau) d\tau \right\} = F(s)G(s) \]

Example Application

Consider the expression:

\[ \frac{1}{s(s^2+4)} = \frac{1}{s} \cdot \frac{1}{s^2+4} \]

Where we identify:

  • \( F(s) = \frac{1}{s} \implies f(t) = 1 \)
  • \( G(s) = \frac{1}{s^2+4} \implies g(t) = \frac{1}{2} \sin(2t) \)

Applying the inverse Laplace transform using convolution:

\[ \mathcal{L}^{-1} \left\{ \frac{1}{s(s^2+4)} \right\} = \int_{0}^{t} 1 \cdot \frac{1}{2} \sin(2(t-\tau)) d\tau \]
\[ = \int_{0}^{t} 1 \cdot \frac{1}{2} \sin(2\tau) d\tau \]

Impulse Function

The unit impulse function (Dirac delta function) is denoted as \( \delta(t-c) \).

Graph of the Dirac delta function delta(t-c) showing a vertical arrow at t=c on the horizontal axis.

Properties of the impulse function:

\[ \int_{-\infty}^{\infty} \delta(t-c) dt = 1 \]
\[ \mathcal{L} \{ \delta(t-c) \} = e^{-cs} \]
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Solving Differential Equations with Impulse Functions

Problem Statement

Solve the following second-order differential equation with an impulse forcing term:

\[ y'' + y = \delta(t-1) \quad \text{with initial conditions } y(0) = y'(0) = 0 \]

Laplace Transform Solution

Taking the Laplace transform of both sides:

\[ s^2 Y - s y(0) - y'(0) + Y = e^{-s} \]

Substituting initial conditions:

\[ (s^2 + 1) Y = e^{-s} \]

Solving for \( Y \):

\[ Y = e^{-s} \cdot \frac{1}{s^2+1} \]

Note that \( \mathcal{L}^{-1} \{ \frac{1}{s^2+1} \} = \sin(t) \). Applying the time-shifting property:

Shift RIGHT by 1: \( t \to t-1 \)
\[ y = u_1(t) \sin(t-1) \]

NOT \( \delta \)

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Linear Systems of Differential Equations

\[ \vec{x}' = A\vec{x} \]

Solution: \( \vec{x} = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2 \) where \( \lambda, \vec{v} \to \) eigenvalues/eigenvector pairs.

Case 1: Real Eigenvalues with Opposite Signs

If \( \lambda \)'s are real and opposite signs:

Phase portrait of a saddle point with trajectories moving toward and away from the origin along eigenvectors.

The origin is a saddle point.

  • \( \vec{v}_1, \lambda > 0 \)
  • \( \vec{v}_2, \lambda < 0 \)

Case 2: Real Eigenvalues with the Same Sign

If \( \lambda \)'s have the same sign (\( \lambda > 0 \)):

Phase portrait showing a nodal source where all trajectories originate from the origin.

If \( \lambda < 0 \), reverse the direction.

The origin is a nodal source (\( \lambda > 0 \)) or nodal sink (\( \lambda < 0 \)).

Note: Trajectories follow the eigenvector with the smaller eigenvalue more closely near the origin.

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Repeated Eigenvalues

If \( \lambda \) is repeated but there are enough eigenvectors, nothing special happens \( \to \) see two straight line solutions.

If there are not enough eigenvectors, then:

\[ \vec{x} = c_1 e^{\lambda t} \vec{v} + c_2 e^{\lambda t} (t\vec{v} + \vec{u}) \]

Only one straight line solution is visible on the graph.

Where \( \vec{u} \) is the generalized eigenvector: \( (A - \lambda I)\vec{u} = \vec{v} \)

And \( \vec{v} \) is the true eigenvector.

Complex Eigenvalues

If \( \lambda = a + bi \):

\[ \vec{x} = c_1 e^{at} \vec{u} + c_2 e^{at} \vec{v} \]

Where \( \vec{u}, \vec{v} \) are the real and imaginary parts of one solution.

Spiral source phase portrait for complex eigenvalues with a positive real part.

\( a > 0 \) (Spiral Source)

Spiral sink phase portrait for complex eigenvalues with a negative real part.

\( a < 0 \) (Spiral Sink)

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Solving Systems with Complex Eigenvalues

\[ \vec{x}' = \begin{bmatrix} -1 & 2 \\ -1 & -3 \end{bmatrix} \vec{x} \]

Eigenvalues and eigenvectors:

\[ \lambda = -2 + i, \quad -2 - i \]\[ \vec{v} = \begin{bmatrix} 2 \\ -1 + i \end{bmatrix}, \quad \begin{bmatrix} 2 \\ -1 - i \end{bmatrix} \]

Euler's Formula Reference

\[ e^{ibt} = \cos(bt) + i \sin(bt) \]\[ e^{it} = \cos(t) + i \sin(t) \]

Forming the Solution

Pick a pair, form one solution:

\[ e^{(-2+i)t} \begin{bmatrix} 2 \\ -1 + i \end{bmatrix} = e^{-2t} e^{it} \begin{bmatrix} 2 \\ -1 + i \end{bmatrix} \]\[ e^{-2t} (\cos t + i \sin t) \begin{bmatrix} 2 \\ -1 + i \end{bmatrix} \]\[ = e^{-2t} \begin{bmatrix} 2 \cos t + i 2 \sin t \\ -\cos t - \sin t - i \sin t + i \cos t \end{bmatrix} \]\[ = e^{-2t} \begin{bmatrix} 2 \cos t \\ -\cos t - \sin t \end{bmatrix} + i e^{-2t} \begin{bmatrix} 2 \sin t \\ \cos t - \sin t \end{bmatrix} \]

Form solutions with these real and imaginary parts.

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Nonhomogeneous Systems

  • Undetermined coefficients
  • Variation of parameters

Undetermined Coefficients

Follow the form of the right side:

\[ \vec{x}' = \begin{bmatrix} 2 & 1 \\ 4 & -1 \end{bmatrix} \vec{x} + \begin{bmatrix} 5 \\ 1 \end{bmatrix} \]\[ \vec{x}_c = c_1 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} 1 \\ -4 \end{bmatrix} \]

Right side: \( \begin{bmatrix} 5 \\ 1 \end{bmatrix} \) so, \( \vec{y}_p = \vec{a} \)

Variation of Parameters

\[ \vec{x} = \Psi \vec{u} \]\[ = \begin{bmatrix} e^{3t} & e^{-2t} \\ e^{3t} & -4e^{-2t} \end{bmatrix} \vec{u} \]

Where \( \Psi \) contains fundamental solutions as columns.

\[ \Psi \vec{u}' = \vec{g} \]

Where \( \vec{g} \) is the right side of the original equation.