Question

Show that the Laplace transform of the derivative of a function is given by \(L\left[F^{\prime}\right](s)=s L[F](s)-F(0) .\) Similarly, show that for the second derivative the transform is $$L\left[F^{\prime \prime}\right](s)=s^{2} L[F](s)-s F(0)-F^{\prime}(0)$$ Use these results to solve the differential equation \(u^{\prime \prime}(t)+\omega^{2} u(t)=0\) subject to the boundary conditions \(u(0)=a, u^{\prime}(0)=0\).

Short answer

The solution to the differential equation \( u''(t)+ω^2u(t)=0 \) with boundary conditions \( u(0)=a, u'(0)=0 \) is \( u(t) = a*sin(ωt) \).

Step-by-step solution

Laplace Transform of First Derivative

By definition, the Laplace Transform of a function \( F' \) is: \( L[F'](s)=\int_{0}^{∞}e^{-st}F'(t)dt. \) With integration by parts, take \( u=e^{-st} \) and \( dv=F'(t)dt \). Then \( du=-se^{-st}dt \) and \( v=F(t) \). Hence, \( L[F'](s)= [e^{-st}F(t)]_{0}^{∞} - \int_{0}^{∞}(-se^{-st})F(t)dt. \) As t approaches infinity, \( F(t) \) is finite, hence the infinity term becomes zero, and at \( t=0, F(t)=F(0) \). This results in \( sL[F](s)-F(0) \).

Laplace Transform of Second Derivative

Repeat the previous process on \( F'(t) \) to obtain the Laplace Transform of the second derivative. The Laplace Transform of \( F''(t) \) by definition is: \( L[F''](s)=\int_{0}^{∞}e^{-st}F''(t)dt. \) Apply integration by parts like in Step 1 to obtain \( L[F''](s)=sL[F'](s)-F'(0) \). Substituting \( L[F'](s)=sL[F](s)-F(0) \) into this gives \( L[F''(t)](s)=s^2L[F](s)-sF(0)-F'(0) \).

Applying to Differential Equation

To solve the given differential equation \( u''(t)+ω^2u(t)=0 \), take the Laplace transform of both sides. The Laplace transform of the left hand side using results from Steps 1 and 2 becomes: \( L[u''+ω^2u](s)= s^2L[u](s)-su(0)-u'(0)+ω^2L[u](s). \) Given \( u(0)=a, u'(0)=0 \) it simplifies to: \( L[u''+ω^2u](s)=(s^2 + ω^2)L[u](s)-sa. \) Hence, the Laplace Transform \( L[u](s) \) of the function can be obtained as: \( L[u](s) = a/(s^2+ω^2). \) To finally get \( u(t) \), take the inverse Laplace transform \( L^{-1}[L[u](s)]=u(t) \). According to standard Laplace transforms, \( L^{-1}[1/(s^2+ω^2)] = sin(ωt) \). So, the final solutionis \( u(t)=a*sin(ωt) \).

Verification

Check the solution by substituting back \( u(t)=a*sin(ωt) \) into the differential equation. This should satisfy both the equation and the boundary conditions.

Key concepts

Integration by Parts

Integration by parts is a mathematical technique used when faced with the task of integrating the product of two functions. The strategy is based on the product rule for differentiation and is often the method of choice for integrations that involve polynomial and exponential functions, or in our case, finding the Laplace transform of a derivative.

When applying integration by parts to find the Laplace transform of a function's derivative, we choose one function to differentiate (usually the derivative of the original function) and another to integrate (usually the exponential decay function from the Laplace transform). It simplifies the integral into smaller, more manageable parts. This process can be represented by the formula:
\( \text{Let } u = e^{-st} \text{ and } dv = F'(t)dt, \text{then} \)
\( u dv = u v - \text{\[\int v du\]} \)

This method is used in our exercise to derive the formulas for the Laplace transform of the first and second derivatives of a function.

Laplace Transform of Derivatives

The Laplace transform of derivatives is particularly valuable when we are dealing with differential equations. In the context of our exercise, we use the Laplace transform to convert a differential equation into an algebraic equation which is often easier to solve.

The rule for the Laplace transform of the first derivative \(L[F'](s)\) is the outcome of applying the technique of integration by parts. Similarly, to find the Laplace transform of the second derivative \(L[F''](s)\), we apply the process in succession. This powerful technique not only makes solving differential equations more manageable but also helps us incorporate initial value and boundary conditions directly into the algebraic equation.

Boundary Conditions

Boundary conditions are critical to solving differential equations, notably when the solution involves the Laplace transform. They are the values of the function or its derivatives at specific points, often the start or end of the interval we are examining. In the context of the given exercise, we are given initial boundary conditions for \(u(t)\) and \(u'(t)\) at \(t=0\).

Incorporating boundary conditions into our Laplace transformed equations allows us to solve for constants and functions in terms of known values, providing specific solutions tailored to the initial state of the system. These conditions can directly replace the corresponding parts in the equation resulting from the Laplace transforms of the derivatives, greatly streamlining the process towards finding the final solution.

Inverse Laplace Transform

Once we have applied the Laplace transform to turn a differential equation into an algebraic equation and have solved that equation for \(L[u](s)\), we need to revert back to the time domain to find the function \(u(t)\) itself. This is where the inverse Laplace transform comes into play.

The inverse Laplace transform is basically a method of matching the transformed function \(L[u](s)\) to a known corresponding time-domain function whose Laplace transform we can identify. By consulting a table of standard Laplace transforms, we find the matching function that gives us our solution in terms of time, \(t\). In the case of our exercise, the inverse transform of \(1/(s^2 + \text{\(\omega\)}^2)\) is known to be \(sin(\text{\(\omega\)}t)\), which gives us the final solution of \(u(t)=a\text{\(\ast\)}sin(\text{\(\omega\)}t)\).