Advanced Solved Problems on the Convolution Theorem and Initial Value Problems

📚 Understanding the Convolution Theorem

The Convolution Theorem is a powerful tool that simplifies solving differential equations, especially when dealing with initial value problems. It allows us to transform complex operations in the time domain into simpler algebraic operations in the frequency domain (using Laplace Transforms). Let's dive in!

• 🔍Definition: The convolution of two functions $f(t)$ and $g(t)$, denoted $f * g$, is defined as $$(f * g)(t) = \int_0^t f(\tau)g(t - \tau) d\tau$$.
• 📜History: The convolution operation has roots in probability theory and signal processing, finding its way into differential equations through the Laplace transform's properties.
• 🔑Key Principle: The Laplace Transform of a convolution is the product of the Laplace Transforms: $$\mathcal{L}\{f * g\} = F(s)G(s)$$, where $F(s) = \mathcal{L}\{f(t)\}$ and $G(s) = \mathcal{L}\{g(t)\}$.

🧪 Advanced Solved Problems

Let's tackle some challenging problems that demonstrate the Convolution Theorem in action with initial value problems.

Problem 1:

Solve the initial value problem: $$y''(t) + 4y(t) = sin(t), \quad y(0) = 1, \quad y'(0) = 0$$

1. 🍎 Taking Laplace Transforms: Applying the Laplace transform to both sides, we get $$(s^2Y(s) - sy(0) - y'(0)) + 4Y(s) = \frac{1}{s^2 + 1}$$
2. 🔢 Plugging in Initial Conditions: Substituting $y(0) = 1$ and $y'(0) = 0$, we have $$(s^2Y(s) - s) + 4Y(s) = \frac{1}{s^2 + 1}$$
3. 🧮 Solving for Y(s): Rearranging, we get $$Y(s)(s^2 + 4) = s + \frac{1}{s^2 + 1}$$ which implies $$Y(s) = \frac{s}{s^2 + 4} + \frac{1}{(s^2 + 4)(s^2 + 1)}$$
4. 💡 Using Convolution: Let $F(s) = \frac{1}{s^2 + 4}$ and $G(s) = \frac{1}{s^2 + 1}$. Then $f(t) = \frac{1}{2}sin(2t)$ and $g(t) = sin(t)$. Thus, $$\mathcal{L}^{-1}\left{\frac{1}{(s^2 + 4)(s^2 + 1)}\right\} = (f * g)(t) = \int_0^t \frac{1}{2}sin(2\tau)sin(t - \tau) d\tau$$. Evaluating this integral gives $$\frac{1}{6}sin(t) - \frac{1}{6}sin(2t)$$
5. ✅ Final Solution: Taking the inverse Laplace transform, $$y(t) = cos(2t) + \frac{1}{6}sin(t) - \frac{1}{6}sin(2t)$$

Problem 2:

Solve the integral equation: $$y(t) = t + \int_0^t y(\tau)sin(t-\tau) d\tau$$

1. 🚀 Recognizing Convolution: We can rewrite the integral as a convolution: $y(t) = t + (y * sin(t))$
2. 💫 Laplace Transform: Applying the Laplace transform: $$Y(s) = \frac{1}{s^2} + Y(s)\frac{1}{s^2 + 1}$$
3. 🔑 Solving for Y(s): Rearranging, we get $$Y(s)\left(1 - \frac{1}{s^2 + 1}\right) = \frac{1}{s^2}$$ implying $$Y(s) = \frac{s^2 + 1}{s^4}$$ or $$Y(s) = \frac{1}{s^2} + \frac{1}{s^4}$$
4. 📈 Inverse Laplace Transform: Taking the inverse Laplace transform: $$y(t) = t + \frac{t^3}{6}$$

Problem 3:

Solve: $y''(t) + y(t) = g(t)$, $y(0) = 0$, $y'(0) = 1$ where $g(t) = \begin{cases} 0, & 0 < t < 3 \\ 2, & t > 3 \end{cases}$

1. ⏱️ Representing g(t) with Heaviside: $g(t) = 2u(t-3)$ where $u(t-3)$ is the Heaviside step function.
2. 💡 Taking the Laplace Transform: $s^2Y(s) - sy(0) - y'(0) + Y(s) = \frac{2e^{-3s}}{s}$.
3. 🔑 Substituting Initial Conditions: $s^2Y(s) - 1 + Y(s) = \frac{2e^{-3s}}{s}$.
4. 🧮 Solving for Y(s): $Y(s)(s^2 + 1) = 1 + \frac{2e^{-3s}}{s}$, so $Y(s) = \frac{1}{s^2+1} + \frac{2e^{-3s}}{s(s^2+1)}$.
5. 💫 Partial Fraction Decomposition: $\frac{1}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1} \implies A=1, B=-1, C=0$. Thus, $\frac{1}{s(s^2+1)} = \frac{1}{s} - \frac{s}{s^2+1}$.
6. ✅ Inverse Laplace Transform: $y(t) = sin(t) + 2u(t-3)[1 - cos(t-3)]$.

🧠 Conclusion

The Convolution Theorem provides an elegant way to solve initial value problems and integral equations by transforming them into simpler algebraic problems. By mastering the Convolution Theorem, you'll unlock a powerful tool for solving a wide range of differential equations. Keep practicing, and you'll become a convolution expert in no time!