Let's solve the problem step by step: **Problem Statement**: Determine the Laplace transform of \(\frac{\partial^2 u(x, t)}{\partial t^2}\). --- ### **Step 1: Definition of Laplace Transform** The Laplace transform of a function \(f(t)\) is defined as: \[ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt \] Let \(U(x, s)\) denote the Laplace transform of \(u(x, t)\): \[ U(x, s) = \mathcal{L}\{ u(x, t) \} = \int_0^\infty e^{-st} u(x, t) \, dt \] --- ### **Step 2: Laplace Transform of the Second Derivative** The Laplace transform of the second derivative with respect to \(t\) is: \[ \mathcal{L}\left\{ \frac{\partial^2 u(x, t)}{\partial t^2} \right\} \] By the properties of Laplace transforms: \[ \mathcal{L}\left\{ \frac{\partial^2 u(x, t)}{\partial t^2} \right\} = s^2 U(x, s) - s u(x, 0) - \left.\frac{\partial u}{\partial t}\right|_{t = 0} \] --- ### **Step 3: Final Answer** \[ \boxed{ \mathcal{L}\left\{ \frac{\partial^2 u(x, t)}{\partial t^2} \right\} = s^2 U(x, s) - s u(x, 0) - \left.\frac{\partial u}{\partial t}\right|_{t = 0} } \] Where: - \(U(x, s)\) is the Laplace transform of \(u(x, t)\), - \(u(x, 0)\) is the initial value of \(u\) at \(t=0\), - \(\left.\frac{\partial u}{\partial t}\right|_{t = 0}\) is the initial value of the first derivative at \(t=0\). --- **This is the required Laplace transform.**

Certainly! Let's proceed step-by-step with detailed explanations and calculations to find the Laplace transform of \(\frac{\partial^2 u(x,t)}{\partial t^2}\). --- ### **Step 1: Recall the definition of the Laplace transform** The Laplace transform of a function \(f(t)\), denoted as \(\mathcal{L}\{f(t)\}\), is defined as: \[ \boxed{ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt } \] In our case, \(u(x,t)\) is a function of both \(x\) and \(t\). When taking the Laplace transform with respect to \(t\), \(x\) is treated as a parameter (constant during the transform). We denote: \[ U(x, s) = \mathcal{L}\{u(x,t)\} = \int_0^\infty e^{-st} u(x,t) \, dt \] --- ### **Step 2: Use the Laplace transform property for derivatives** The key property we need is the Laplace transform of derivatives with respect to \(t\): \[ \mathcal{L}\left\{ \frac{\partial^n u(x,t)}{\partial t^n} \right\} = s^n U(x,s) - s^{n-1} u(x,0) - s^{n-2} \left.\frac{\partial u}{\partial t}\right|_{t=0} - \cdots - \left.\frac{\partial^{n-1} u}{\partial t^{n-1}}\right|_{t=0} \] For the **second derivative** (\(n=2\)), this simplifies to: \[ \boxed{ \mathcal{L}\left\{ \frac{\partial^2 u(x,t)}{\partial t^2} \right\} = s^2 U(x,s) - s u(x,0) - \left.\frac{\partial u}{\partial t}\right|_{t=0} } \] --- ### **Step 3: Derive the expression explicitly** Let's verify this step-by-step. - **Start with the Laplace transform of the second derivative:** \[ \mathcal{L}\left\{ \frac{\partial^2 u}{\partial t^2} \right\} = \int_0^{\infty} e^{-st} \frac{\partial^2 u(x,t)}{\partial t^2} dt \] - **Use integration by parts twice to evaluate the integral:** **First integration by parts:** Let: \[ f(t) = e^{-st} \quad \Rightarrow \quad f'(t) = -s e^{-st} \] \[ g(t) = \frac{\partial u}{\partial t} \quad \Rightarrow \quad g'(t) = \frac{\partial^2 u}{\partial t^2} \] Rearranged, the integral becomes: \[ \int_0^{\infty} e^{-st} \frac{\partial^2 u}{\partial t^2} dt = \left[ e^{-st} \frac{\partial u}{\partial t} \right]_0^{\infty} + s \int_0^{\infty} e^{-st} \frac{\partial u}{\partial t} dt \] **Evaluating the boundary term:** As \(t \to \infty\), \[ e^{-st} \to 0 \quad \text{(for } s > 0 \text{)}, \] and assuming \(u(x,t)\) doesn't grow faster than an exponential, the boundary term at infinity is zero. At \(t=0\): \[ e^{0} \frac{\partial u}{\partial t}\bigg|_{t=0} = \left.\frac{\partial u}{\partial t}\right|_{t=0} \] So, the boundary term is: \[ - \left.\frac{\partial u}{\partial t}\right|_{t=0} \] **Now, evaluate the remaining integral:** \[ s \int_0^{\infty} e^{-st} \frac{\partial u}{\partial t} dt \] Again, integrate by parts: Let: \[ f(t) = e^{-st} \quad \Rightarrow \quad f'(t) = -s e^{-st} \] \[ g(t) = u(x,t) \quad \Rightarrow \quad g'(t) = \frac{\partial u}{\partial t} \] Then: \[ \int_0^{\infty} e^{-st} \frac{\partial u}{\partial t} dt = \left[ e^{-st} u(x,t) \right]_0^{\infty} + s \int_0^{\infty} e^{-st} u(x,t) dt \] As \(t \to \infty\), assuming \(u(x,t)\) does not grow exponentially, the boundary term is zero. At \(t=0\): \[ e^{0} u(x,0) = u(x,0) \] Thus: \[ \int_0^{\infty} e^{-st} \frac{\partial u}{\partial t} dt = - u(x,0) + s U(x,s) \] --- ### **Step 4: Assemble the results** Putting it all together: \[ \mathcal{L}\left\{ \frac{\partial^2 u}{\partial t^2} \right\} = - \left.\frac{\partial u}{\partial t}\right|_{t=0} + s \left( - u(x,0) + s U(x,s) \right) \] Simplify: \[ = - \left.\frac{\partial u}{\partial t}\right|_{t=0} - s u(x,0) + s^2 U(x,s) \] Rearranged: \[ \boxed{ \mathcal{L}\left\{ \frac{\partial^2 u}{\partial t^2} \right\} = s^2 U(x,s) - s u(x,0) - \left.\frac{\partial u}{\partial t}\right|_{t=0} } \] --- ### **### Final Answer:** \[ \boxed{ \mathcal{L}\left\{ \frac{\partial^2 u(x,t)}{\partial t^2} \right\} = s^2 U(x,s) - s u(x,0) - \left.\frac{\partial u}{\partial t}\right|_{t=0} } \] This expression relates the Laplace transform of the second time derivative of \(u(x,t)\) to the initial conditions and the Laplace transform of \(u(x,t)\) itself. --- **Note:** To fully determine the transform, you need the initial conditions \(u(x,0)\) and \(\left.\frac{\partial u}{\partial t}\right|_{t=0}\).