Question

Use the method indicated in Problem 23 to solve the given boundary value problem. $$ \begin{array}{l}{u_{t}=u_{x x}-\pi^{2} \cos \pi x} \\ {u_{x}(0, t)=0, \quad u(1, t)=1} \\ {u(x, 0)=\cos (3 \pi x / 2)-\cos \pi x}\end{array} $$

Short answer

Based on the information provided, the solution to the boundary value problem involving the heat equation is given by: $$ u(x,t) = A\cos(kx)\left(e^{-k^2t}+\frac{\pi^2 \cos(\pi x)}{k^2-\pi^2}\right) $$ where A is determined by the initial distribution: $$ A\cos(kx) = \cos \left(\frac{3\pi x}{2}\right)-\cos(\pi x) $$ We used the method of separation of variables to break the PDE into two ODEs, solved the ODEs, and then combined the solutions to obtain the desired solution of the PDE.

Step-by-step solution

Separation of Variables

First, we assume that the solution can be written in the form of a product of functions of x and t, i.e., u(x,t) = X(x)T(t). Now, substitute this into the given PDE: $$X''(x)T(t) = X(x)T'(t) - \pi^2 \cos(\pi x)T(t)$$ Next, divide both sides by X(x)T(t): $$\frac{X''(x)}{X(x)} = \frac{T'(t)}{T(t)} - \pi^2 \frac{\cos(\pi x)}{X(x)}$$ Since the left side is only dependent on x, and the right side is only dependent on t, both sides must be equal to a constant. We will call this constant -k^2: $$\frac{X''(x)}{X(x)} = -k^2 \text{ and } \frac{T'(t)}{T(t)} = -k^2 + \pi^2 \frac{\cos(\pi x)}{X(x)}$$ We now have two separate ODEs to solve.

Solve the ODE for X(x)

We have the following ODE for X(x): $$X''(x) + k^2 X(x) = 0$$ This is a second-order linear homogeneous ODE with constant coefficients which has the general solution: $$X(x) = A\cos(kx) + B\sin(kx)$$ Now, we apply the boundary conditions: $$u_x(0,t) = X'(0)T(t) = 0 \implies X'(0) = 0$$ Plugging X(x) into X'(x), we get: $$X'(x) = -kA\sin(kx) + kB\cos(kx)$$ X'(0) = 0: $$-kA\sin(0) + kB\cos(0) = 0 \implies kB = 0 \implies B = 0$$ Thus, we have the following equation for X(x): $$X(x) = A\cos(kx)$$

Solve the ODE for T(t)

We have the following ODE for T(t): $$T'(t) + k^2T(t) = \pi^2T(t) \frac{\cos(\pi x)}{A\cos(kx)}$$ Take a Laplace transform to solve for L(T(t)), we get: $$L(T'(t))+k^2L(T(t)))=\pi^2 \frac{\cos(\pi x)}{A\cos(kx)}[L(T(t))]$$ Solve for L(T(t)), we get $$L(T(t)) = \frac{\pi^2 \cos(\pi x)}{(s+k^2 - \pi^2)A\cos(kx)}$$ Taking the inverse Laplace transform, we get the following equation for T(t): $$T(t) = e^{-k^2t}+\frac{1}{A\cos(kx)}\cdot\left(\frac{A\pi^2\cos(\pi x)}{k^2-\pi^2}\right)$$

Combine X(x) and T(t) to form the final solution

Now, we can combine X(x) and T(t) to form the solution u(x,t): $$u(x,t) = X(x)T(t) = A\cos(kx)\left(e^{-k^2t}+\frac{\pi^2 \cos(\pi x)}{k^2-\pi^2}\right)$$ Finally, we apply our last boundary condition and initial distribution to determine A and the series sum: $$u(x,0)=\cos \left(\frac{3\pi x}{2}\right)-\cos(\pi x)\implies A\cos(kx) = \cos \left(\frac{3\pi x}{2}\right)-\cos(\pi x)$$ By matching the sum, we can determine A's value to adjust amplitudes, and we obtain the final solution for the given heat equation.

Key concepts

Separation of Variables

When faced with a partial differential equation (PDE), one powerful technique for finding solutions is separation of variables. It's a method that assumes a solution to a PDE can be written as the product of two or more functions, each a function of a single independent variable. This is particularly useful when dealing with problems in which the PDE involves time and space variables.

As demonstrated in the textbook exercise, to employ separation of variables, one assumes that the solution, in this case, the temperature distribution function 'u(x,t)', can be partitioned into 'X(x)' dependent only on space and 'T(t)' dependent only on time. After substituting this product into the PDE and separating variables, we arrive at two ordinary differential equations (ODEs), one for each function.

Each of these resulting ODEs—'X(x)' concerning space, and 'T(t)' concerning time—is then solved individually, subject to the given boundary conditions. The collective solution of these ODEs, achieved by multiplying them together, forms the complete solution to the original PDE. This approach simplifies the problem but requires careful consideration of boundary and initial conditions to ensure a valid solution.

Partial Differential Equations

Partial differential equations (PDEs) are equations that involve rates of change with respect to multiple variables. They are fundamental to the description of physics phenomena, like heat conduction, wave propagation, and fluid dynamics. PDEs contain partial derivatives and are often set within boundary value problems.

In the given exercise, the PDE presented describes a heat conduction problem where the temperature distribution varies with both position 'x' and time 't'. The term 'u_{t}' indicates how the temperature changes over time, while the term 'u_{xx}' indicates how the temperature’s rate of change in space changes. This second derivative is essential for describing diffusion processes, such as heat spreading out over time.

Solving PDEs often requires utilizing boundary conditions, which are additional constraints that specify values or behaviors of the solution on the edges of the domain. These conditions are crucial as they tailor generic solutions of the PDEs to the specifics of the physical scenario at hand. Without boundary conditions, the solution to a PDE would not be uniquely determined.

Laplace Transform

The Laplace transform is a significant technique for solving ODEs, especially when initial conditions are given. It converts a function defined in the time domain into a function in the complex frequency domain. By transforming differential equations into algebraic equations, it often simplifies the process of solving for a function.

In practice, the Laplace transform is applied to both sides of an ODE, which results in the elimination of the differentiation operation. As shown in the solution steps for 'T(t)', the Laplace transform changes the time domain differential equation into an algebraic equation in terms of 's', which represents complex frequency. This method can handle discontinuous functions and initial value problems with ease, making it an immensely powerful tool in both engineering and mathematics.

After the equation is transformed and algebraically solved for 'L(T(t))', the inverse Laplace transform is used to revert the solution back to the time domain. The resulting function 'T(t)' represents part of the final solution when combined with 'X(x)' obtained from the separation of variables method.

Homogeneous ODE

A homogeneous ordinary differential equation (ODE) is one in which the function and its derivatives are multiplied by constant coefficients and set to zero. These ODEs are classified as 'homogeneous' because the term without derivatives, typically referred to as the nonhomogeneous term, is zero.

In the given exercise, once we apply separation of variables, we derive a homogeneous ODE for 'X(x)' from our PDE. This particular homogeneous ODE possesses constant coefficients and the solution takes on a standard form involving trigonometric functions or exponentials, depending on the nature of the constants involved. For the boundary value problem at hand, the solution for 'X(x)' involves a cosine function, indicating a sinusoidal variation in space.

Solving a homogeneous ODE typically involves finding a general solution that contains arbitrary constants. These constants are then determined by applying boundary or initial conditions. In the context of our exercise, these conditions were used to find the values of the constants in the solution for 'X(x)', which can then be combined with 'T(t)' to find the final form of the function representing the temperature distribution u(x,t).