Question

Use the Laplace transform to solve the following systems of partial differential equations: (a) $$ \begin{cases}u_{x}+2 x u_{t}=2 x, & 0

Short answer

For (a), using Laplace transform: \( u(x, t) = e^{-xs^2} F^{-1}\{ \frac{s+1}{s^2} \} \). For (b), solve using integrating factors: \( U(x, s) = \frac{1}{s^3} \) with boundary conditions.

Step-by-step solution

Apply the Laplace Transform with respect to time

For both parts (a) and (b), we will apply the Laplace transform with respect to time. Denoting the Laplace transform of \( u(x, t) \) as \( U(x, s) \), apply:\[ L\{ u_t(x, t) \} = sU(x, s) - u(x, 0) \] for part (a) and part (b).

Apply the Laplace Transform for Part (a)

For equation \( u_x + 2xu_t = 2x \), taking the Laplace transform with respect to \( t \) gives:\[ L\{ u_x + 2xu_t \} = L\{ 2x \} \]Substitute the transform into the PDE:\[ U_x + 2xsU - 2xu(x, 0) = \frac{2x}{s} \]Using the initial condition \( u(x, 0) = 1 \), we substitute and simplify:\[ U_x + 2xsU = \frac{2x}{s} + 2x \]

Solve the Transformed Equation for Part (a)

This is a first-order linear differential equation in \( U \) with respect to \( x \). Rewrite the differential equation:\[ U_x + 2xsU = \frac{2x(s+1)}{s} \]Use an integrating factor, \( I(x) = e^{\int 2xs \, dx} = e^{xs^2} \). Multiply through by the integrating factor and integrate both sides:\[ \frac{d}{dx}(e^{xs^2}U) = \frac{2x(s+1)}{s} e^{xs^2} \]

Solve for U(x, s) in Part (a)

Integrate the right side:\[ e^{xs^2}U(x, s) = \int \frac{2x(s+1)}{s} e^{xs^2} \, dx \]Solve using integration by parts or look-up tables to find:\[ U(x, s) = \frac{s+1}{s^2} e^{-xs^2} + C_1 e^{-xs^2} \]Apply boundary condition \( u(0, t) = 1 \) to determine \( C_1 = 0 \).

Apply the Laplace Transform for Part (b)

For equation \( xu_x + u_t = xt \), taking the Laplace transform:\[ xU_x + sU - u(x, 0) = \frac{xt}{s^2} \]Using initial condition \( u(x, 0) = 0 \), substituting gives:\[ xU_x + sU = \frac{x}{s^2} \]

Solve the Transformed Equation for Part (b)

This is a first-order linear differential equation in \( U \).Rewrite the differential equation:\[ xU_x + sU = \frac{x}{s^2} \]Divide throughout by \( x \):\[ U_x + \frac{s}{x}U = \frac{1}{s^2} \]Solve this using an integrating factor, \( I(x) = x^s \), and integrate both sides:\[ \frac{d}{dx}(x^s U) = \frac{x^{s-1}}{s^2} \]

Solve for U(x, s) in Part (b)

Integrate:\[ x^sU = \int \frac{x^{s-1}}{s^2} \, dx \]After solving the integral and simplifying:\[ U(x, s) = \frac{1}{s^3} + C_2 x^{-s} \]Apply boundary condition \( u(0, t) = 0 \) to solve for \( C_2 = 0 \).

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are mathematical equations that involve multiple independent variables and their partial derivatives. These are crucial in modeling various phenomena in physics and engineering, such as heat flow, wave propagation, and fluid dynamics. In the exercise we have seen, the systems given are \( u_{x} + 2xu_{t} = 2x \) and \( xu_{x} + u_{t} = xt \), which involve derivatives with respect to both \( x \) and \( t \). PDEs are often more complex than ordinary differential equations (ODEs) due to the involvement of several variables.

To tackle PDEs, mathematicians use different techniques, with the Laplace transform being one of the most efficient when initial and boundary conditions are known. The Laplace transform essentially transforms a PDE into an algebraic equation in terms of complex variables by integrating over time, simplifying the complexity of solving the differential form.

• Multiple variables: Variables \( x \) and \( t \).
• Partial derivatives: Derivatives like \( u_{x} \) and \( u_{t} \).
• Applications vary from physics to engineering.

The key to effectively solving PDEs is understanding the relationship between the variables and employing the right transformation techniques, like the Laplace transform used in our example.

Initial Conditions

Initial conditions are a set of conditions specified for a problem to determine a unique solution to a differential equation. They provide specific values for the function or its derivatives at certain points, which help to narrow down the possible solution to the equation.

In the problem given, initial conditions are provided as \( u(x, 0) = 1 \) and \( u(0, t) = 1 \) for equation (a), and \( u(x, 0) = 0 \) and \( u(0, t) = 0 \) for equation (b). These initial conditions are essential for applying the Laplace transform, as they allow the removal of certain terms or simplify the equation during transformation.

• They define the state of the system at \( t=0 \).
• Helps in finding particular solutions after the transformation.
• Directly impact the form of the transformed function \( U(x, s) \).

With initial conditions, we can integrate transformed equations and use them to solve for integration constants, which are crucial for determining the specific solution that fits the problem.

First-Order Linear Differential Equation

A first-order linear differential equation is an equation of the form \( rac{dy}{dx} + P(x)y = Q(x) \). It is characterized by the derivative of the unknown function of first order only. These types of equations are simpler to solve compared to higher order because they do not involve higher derivatives.

In our exercise, after applying the Laplace transform, both part (a) and part (b) reduce to first-order linear differential equations: \( U_x + 2xsU = \frac{2x(s+1)}{s} \) for part (a), and \( xU_x + sU = \frac{x}{s^2} \) for part (b). Identifying these equations as first-order linear allows us to apply standard solving methods, such as finding an integrating factor.

• Equation involves first-order derivative.
• Standard form helps in identifying suitable solving techniques.
• Beneficial for simple solutions of transformed expressions.

The simplification to first-order equations is one of the main benefits of using Laplace transforms for solving PDEs, making complex equations more manageable.

Integrating Factor

An integrating factor is a function used to facilitate the finding of the solution to a differential equation. This method is extensively used for first-order linear differential equations. The integrating factor essentially transforms the differential equation into an exact differential, which can be solved straightforwardly by integration.

In the exercise, for part (a), the integrating factor is calculated as \( I(x) = e^{ extstyle xs^2} \). For part (b), it is \( I(x) = x^s \). Multiply both sides of the differential equation with this factor to transform the left-hand side into a derivative of a product of the integrating factor and the function itself, which simplifies integration greatly:

• Transforms the differential equation into an exact form.
• Makes integration straightforward.
• Essential for solving first-order linear equations efficiently.

Understanding how to determine and use an integrating factor effectively makes solving differential equations, particularly those resulting from Laplace transformations, much simpler.