To solve the given problem step by step, let us proceed:

Step 1: Verify the Laplace transform identity

The problem provides the Laplace transform of the fourth derivative:

$\mathcal{L}\left\{\frac{d^4y}{dx^4}\right\} = s^4 \tilde{y}(s) - s^3 y(0) - s^2 y'(0) - s y''(0) - y'''(0)$

This matches the standard Laplace transform property for derivatives: $\mathcal{L}\left\{\frac{d^n f(x)}{dx^n}\right\} = s^n \mathcal{L}\{f(x)\} - s^{n-1} f(0) - s^{n-2} f'(0) - \dots - f^{(n-1)}(0)$ So the identity holds.

Step 2: Take Laplace transforms of the given equation

The differential equation provided is: $\frac{d^4y}{dx^4} = \frac{P_0}{EI} \delta\left(x - \frac{l}{3}\right)$

Taking the Laplace transform of both sides: $\mathcal{L}\left\{\frac{d^4y}{dx^4}\right\} = \mathcal{L}\left\{\frac{P_0}{EI} \delta\left(x - \frac{l}{3}\right)\right\}$

Using the Laplace transform for a delta function $\delta(x - a)$, we have: $\mathcal{L}\{\delta(x - a)\} = e^{-as}$

Thus: $s^4 \tilde{y}(s) - s^3 y(0) - s^2 y'(0) - s y''(0) - y'''(0) = \frac{P_0}{EI} e^{-\frac{l}{3}s}$

Step 3: Substitute the boundary conditions

The boundary conditions are: $y(0) = 0, \quad y'(0) = 0, \quad y(l) = 0, \quad y'(l) = 0$

Using $y(0) = 0$ and $y'(0) = 0$, the Laplace equation simplifies to: $s^4 \tilde{y}(s) - s y''(0) - y'''(0) = \frac{P_0}{EI} e^{-\frac{l}{3}s}$

Let: $y''(0) = A, \quad y'''(0) = B$

Substitute these unknown constants: $s^4 \tilde{y}(s) - s A - B = \frac{P_0}{EI} e^{-\frac{l}{3}s}$

Step 4: Solve for $\tilde{y}(s)$

Rearranging for $\tilde{y}(s)$: $\tilde{y}(s) = \frac{\frac{P_0}{EI} e^{-\frac{l}{3}s} + s A + B}{s^4}$

Step 5: Take the inverse Laplace transform

The inverse Laplace transform of $\tilde{y}(s)$ involves breaking the terms into simpler components:

1. The first term: $\mathcal{L}^{-1}\left\{\frac{e^{-\frac{l}{3}s}}{s^4}\right\}$ Using the time-shifting property and the Laplace transform of $\frac{1}{s^4}$, this term becomes: $\frac{P_0}{EI} \cdot \frac{\left(x - \frac{l}{3}\right)^3}{6} \quad \text{for } x \geq \frac{l}{3}, \text{ and } 0 \text{ otherwise.}$

2. The second term: $\mathcal{L}^{-1}\left\{\frac{s}{s^4}\right\} = \mathcal{L}^{-1}\left\{\frac{1}{s^3}\right\} = \frac{A x^2}{2}.$

3. The third term: $\mathcal{L}^{-1}\left\{\frac{1}{s^4}\right\} = \frac{B x^3}{6}.$

Step 6: Apply boundary conditions at $x = l$

From $y(l) = 0$ and $y'(l) = 0$, solve for the constants $A$ and $B$.

Let me know if you'd like me to compute the exact values of these constants! Would you like further elaboration?

Questions for Deeper Understanding

1. How is the Laplace transform helpful in solving differential equations involving boundary value problems?
2. Can the method be extended to beams with different loading conditions, like distributed loads?
3. What assumptions are made about the flexural rigidity $EI$ in this problem, and how could they be relaxed?
4. Why is the delta function $\delta(x - a)$ used to model point loads in engineering?
5. How do higher-order derivatives affect the smoothness and behavior of the beam deflection curve?

Tip

When solving boundary value problems, always cross-check boundary conditions after obtaining the solution