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Chapter 4: Problem 1

Use Laplace transform methods to solve the ODE $$ L \frac{d y}{d t}+R y=f(t), \quad y(0)=y_{0}, \quad \text { constants } L, R>0 $$ (a) Let \(f(t)=\sin \omega_{0} t, \omega_{0}>0\), so that the Laplace transform of \(f(t)\) is \(\hat{F}(s)=\omega_{0} /\left(s^{2}+\omega_{0}^{2}\right) .\) Find $$ \begin{aligned} y(t)=& y_{0} e^{-\frac{\varepsilon}{L} t}+\frac{\omega_{0}}{L\left((R / L)^{2}+\omega_{0}^{2}\right)} e^{-\frac{k}{L} t}+\frac{R}{L^{2}} \frac{\sin \omega_{0} t}{\left((R / L)^{2}+\omega_{0}^{2}\right)} \\ &-\frac{\omega_{0}}{L} \frac{\cos \omega_{0} t}{\left((R / L)^{2}+\omega_{0}^{2}\right)} \end{aligned} $$ (b) Suppose \(f(t)\) is an arbitrary continuous function that possesses a Laplace transform. Use the convolution product for Laplace transforms (Section \(4.5\) ) to find $$ y(t)=y_{0} e^{-\frac{k}{L} t}+\frac{1}{L} \int_{0}^{t} f\left(t^{\prime}\right) e^{-\frac{k}{L}\left(t-t^{\prime}\right)} d t^{\prime} $$ (c) Let \(f(t)=\sin \omega_{0} t\) in (b) to obtain the result of part (a), and thereby verify your answer. This is an example of an "L,R circuit" with impressed voltage \(f(t)\) arising in basic electric circuit theory.

Short Answer

Expert verified
The solution is \( y(t) = y_0 e^{-\frac{R}{L} t} + \frac{\omega_0}{L(\left(\frac{R}{L}\right)^{2}+\omega_{0}^{2})} e^{-\frac{R}{L} t} + \frac{R}{L^{2}} \frac{\sin\omega_{0} t}{\left(\frac{R}{L}^{2}+\omega_{0}^{2}}\) - \frac{\omega_{0}}{L} \frac{\cos\omega_{0} t}{\left(\frac{R}{L}^{2}+\omega_{0}^{2}}\).

Step by step solution

01

Write down the given ODE

The given differential equation is: \[L \frac{d y}{d t} + R y = f(t)\], with the initial condition \( y(0) = y_0 \).
02

Apply the Laplace transform to both sides of the ODE

Taking the Laplace transform of the entire equation, we use the properties of Laplace transforms: \[L \mathcal{L}\left\{ \frac{d y}{d t} \right\} + R \mathcal{L}\{ y \} = \mathcal{L}\{ f(t) \}\]. Recall that \( \mathcal{L}\left\{ \frac{d y}{d t} \right\} = s \hat{Y}(s) - y(0) \), where \( \hat{Y}(s) \) is the Laplace transform of \( y(t) \).
03

Substitute the expressions for the Laplace transforms

Substitute the transforms into the ODE:\[L(s \hat{Y}(s) - y_0) + R \hat{Y}(s) = \hat{F}(s)\]. Grouping terms in \( \hat{Y}(s) \) gives:\[ (Ls + R) \hat{Y}(s) = \hat{F}(s) + L y_0 \].
04

Solve for the Laplace transform of \( y(t) \)

Isolate \( \hat{Y}(s) \): \[ \hat{Y}(s) = \frac{\hat{F}(s) + L y_0}{Ls + R} \].
05

Apply the inverse Laplace transform

For part (a), let \( f(t) = \sin \omega_0 t \), so that \( \hat{F}(s) = \frac{\omega_0}{s^2 + \omega_0^2} \). Substitute this into the expression for \( \hat{Y}(s) \): \[ \hat{Y}(s) = \frac{\frac{\omega_0}{s^2 + \omega_0^2} + L y_0}{Ls + R} \]. Divide each term by \( Ls + R \) and use partial fractions to separate them.
06

Simplify and invert to find \( y(t) \)

Simplify the expression and take the inverse Laplace transform to get \( y(t) \): \[ y(t) = y_0 e^{-\frac{R}{L} t} + \frac{\omega_0}{L \left( \left( \frac{R}{L} \right)^2 + \omega_0^2 \right)} e^{ - \frac{R}{L} t } + \frac{R}{L^2} \frac{\sin \omega_0 t}{ \left( \left( \frac{R}{L} \right)^2 + \omega_0^2 \right) } - \frac{\omega_0}{L} \frac{\cos \omega_0 t}{ \left( \left( \frac{R}{L} \right)^2 + \omega_0^2 \right) }. \]
07

Verify part (b)

For an arbitrary continuous function \( f(t) \), the solution using the convolution product for Laplace transforms is: \[ y(t) = y_0 e^{-\frac{R}{L} t} + \frac{1}{L} \int_0^t f(t') e^{-\frac{R}{L} (t - t')} \, dt'. \]
08

Use \( f(t) = \sin \omega_0 t \) in part (b) to verify part (c)

When \( f(t) = \sin \omega_0 t \), substitute it into the solution found in step 7 to achieve the same result as in part (a), confirming the correct solution. \[ y(t) = y_0 e^{-\frac{R}{L} t} + \frac{1}{L} \int_0^t \sin \omega_0 t' e^{-\frac{R}{L} (t - t')} \, dt'. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Laplace Transform
The Laplace Transform is a technique used to convert complex differential equations into simpler algebraic forms, making them easier to solve. The transform of a function \(f(t)\) is given by \( \mathcal{L} \{ f(t) \} = \int_0^\infty e^{-st}f(t)\,dt \). This tool is particularly powerful for handling linear time-invariant systems and is extensively used in fields such as control systems, signal processing, and circuit analysis. For example, transforming the derivative \( \frac{d y}{d t} \) becomes \( s \hat{Y}(s) - y(0) \), simplifying the differential equation's solving process. Here, after applying the Laplace Transform to the given ODE, you can isolate \( \hat{Y}(s)\) and use the inverse Laplace Transform to find the solution in the time domain.
Differential Equations
Differential Equations involve functions and their derivatives, representing various physical phenomena like motion, heat, or waves. The given ODE is \( L \frac{d y}{d t} + R y = f(t) \), which models an L-R circuit in electrical engineering. To solve this, we apply the Laplace Transform. This converts the ODE into an algebraic equation that is easier to handle. After performing inverse transformations, we get the solution back in its original form. Such techniques are crucial because they bridge the gap between calculus and algebra, providing powerful tools to solve real-world problems.
Electric Circuits
Electric Circuits consist of various components like resistors (R), inductors (L), and capacitors. The dynamics of these circuits are often described by differential equations. In our exercise, the ODE represents an L-R circuit. The term \( L \frac{d y}{d t} \) reflects the inductor's behavior, while \( R y \) corresponds to the resistor. When the external input \( f(t) \) is \( \sin( \omega_0 t) \), this setup can model oscillatory phenomena common in AC circuits. Using the Laplace Transform helps analyze the circuit's transient and steady-state responses, giving us insights into the system's behavior over time.
Convolution Theorem
The Convolution Theorem is a critical concept in the Laplace Transform applications. It's used to solve differential equations when the input function \( f(t) \) is more complex. According to this theorem, the inverse Laplace Transform of a product of two Laplace transforms is the convolution of their inverse transforms. More formally, if \( \mathcal{L} \{ f(t) \} = F(s) \) and \( \mathcal{L} \{ g(t) \} = G(s) \), then \( \mathcal{L}^{-1} \{ F(s)G(s) \} = (f * g)(t) \), with the convolution defined as \( (f * g)(t) = \int_0^t f( \tau) g(t - \tau) \, d \tau \). This theorem was used in part (b) to derive the solution for an arbitrary \( f(t) \) after inverse transforming the product of transforms.
Initial Value Problems
Initial Value Problems (IVPs) specify the value of the solution and its derivatives at a starting point. For differential equations, knowing the initial condition \( y(0) = y_0 \) is crucial. It allows us to solve the problem uniquely. In this exercise, the initial condition is applied when using the Laplace Transform, where the initial value appears in the transformed equation, specifically as \( L y_0 \). Implementing the IVP correctly ensures the solution not only satisfies the differential equation but also fits the behavior dictated by the initial conditions, crucial for realistic modeling of physical systems.

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Most popular questions from this chapter

(a) Consider the mapping \(w=z^{3} .\) When we encircle the origin in the \(z\). plane one time, how many times do we encircle the origin in the \(w\) plane? Explain why this agrees with the Argument Principle. (b) Suppose we consider \(w=z^{3}+a_{2} z^{2}+a_{1} z+a_{0}\) for three constants \(a_{0}, a_{1}, a_{2} .\) If we encircle the origin in the \(z\) plane once on a very large circle, how many times do we encircle the \(w\) plane? (c) Suppose we have a mapping \(w=f(z)\) where \(f(z)\) is analytic inside and on a simple closed contour \(C\) in the \(z\) plane. Let us define \(\tilde{C}\) as the (nonsimple) image in the \(w\) plane of the contour \(C\) in the \(z\) plane. If we deduce that it encloses the origin \((w=0) N\) times, and encloses the point \((w=1) M\) times, what is the change in arg \(w\) over the contour \(\tilde{C} ?\) If we had a computer available, what algorithm should be designed (if it is at all possible) to determine the change in the argument?

(a) Assume that \(u(\infty)=0\) to establish that $$ \int_{0}^{\infty} \frac{d^{2} u}{d x^{2}} \cos k x d x=-\frac{d u}{d x}(0)-k^{2} \hat{U}_{c}(k) $$ where \(\hat{U}_{c}(k)=\int_{0}^{\infty} u(x) \cos k x d x\) is the cosine transform of \(u(x)\). (b) Use this result to show that taking the Fourier cosine transform of $$ \frac{d^{2} u}{d x^{2}}-\omega^{2} u=-f(x), \quad \text { with } \frac{d u}{d x}(0)=u_{0}^{\prime}, \quad u(\infty)=0, \quad \omega>0 $$ yields, for the Fourier cosine transform of \(u(x)\) $$ \hat{U}_{c}(k)=\frac{\hat{F}_{c}(k)-u_{0}^{\prime}}{k^{2}+\omega^{2}} $$ where \(\hat{F}_{c}(k)\) is the Fourier cosine transform of \(f(x)\) (c) Use the analog of the convolution product of the Fourier cosine transform $$ \begin{aligned} &\frac{1}{2} \int_{0}^{\infty}(f(|x-\zeta|)+f(x+\zeta)) g(\zeta) d \zeta \\ &=\frac{2}{\pi} \int_{0}^{\infty} \cos k x \hat{F}_{c}(k) \hat{G}_{c}(k) d k \end{aligned} $$ to show that the solution of the differential equation is given by $$ u(x)=-\frac{u_{0}^{\prime}}{\omega} e^{-\omega x}+\frac{1}{2 \omega} \int_{0}^{\infty}\left(e^{-\omega|x-\zeta|}+e^{-\mathfrak{\omega}|x+\zeta|}\right) f(\zeta) d \zeta $$ (The convolution product for the cosine transform can be deduced from the usual convolution product (4.5.17) by assuming in the latter formula that \(f(x)\) and \(g(x)\) are even functions of \(x\).)

Projection operators can be defined as follows. Consider a function \(F(z)\) $$ F(z)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta $$ where \(C\) is a contour, typically infinite (e.g. the real axis) or closed (e.g. a circle) and \(z\) lies off the contour. Then the "plus" and "minus" projections of \(F(z)\) at \(z=\zeta_{0}\) are defined by the following limit: $$ F^{\pm}\left(\zeta_{0}\right)=\lim _{z \rightarrow \zeta_{0}}\left[\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta\right] $$ where \(\zeta_{0}{ }^{\pm}\)are points just inside \((+)\)or outside (-) of a closed contour (i.e., \(\lim _{z \rightarrow 5_{0}+\text { denotes the limit from points } z \text { inside the contour } C \text { ) or to }}\) the left (t) or right (-) of an infinite contour. Note: the "+" region lies to the left of the contour; where we take the standard orientation for a contour, that is, the contour is taken with counterclockwise orientation. To simplify the analysis, we will assume that \(f(x)\) can be analytically extended in the neighborhood of the curve \(C\). (a) Show that $$ F^{\pm}\left(\zeta_{0}\right)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-\zeta_{0}} d \zeta \pm \frac{1}{2} f\left(\zeta_{0}\right) $$ where \(f_{C}\) is the principal value integral that omits the point \(\zeta=\zeta 0\). (b) Suppose that \(f(\zeta)=1 /\left(\zeta^{2}+1\right)\), and the contour \(C\) is the real axis (infinite), with orientation take from \(-\infty\) to \(\infty\); find \(F^{\pm}\left(\zeta_{0}\right)\). (c) Suppose that \(f(\zeta)=1 /\left(\zeta^{2}+a^{2}\right), a^{2}>1\), and the contour \(C\) is the unit circle centered at the origin with counterclockwise orientation; find \(F^{\pm}\left(\zeta_{0}\right)\)

Use the transformation \(t=(x-1) /(x+1)\) on the principal branch of the following functions to show that (a) \(\quad \int_{-1}^{1}\left(\frac{1+t}{1-t}\right)^{k-1} d t=\frac{2(1-k) \pi}{\sin k \pi}, \quad 0

Show that the Fourier transform of the "Gaussian" \(f(x)=\exp \left(-\left(x-x_{0}\right)^{2} / a^{2}\right) x_{0}, a\) real, is also a Gaussian: $$ \hat{F}(k)=a \sqrt{\pi} e^{-(k a / 2)^{2}} e^{-i k x_{0}} $$
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