Question

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

Short answer

Question: Determine the general form of the solution to the given boundary value problem involving the following partial differential equation (PDE) and boundary and initial conditions: $$ u_t = u_{xx} - 2 \\ u(0,t) = 1 \\ u(1,t) = 0 \\ u(x,0) = x^2 - 2x + 2 $$ Answer: The general form of the solution to the given boundary value problem is: $$ u(x,t) = (x - 1) + \sum_{n=1}^{\infty} B_n \sin(n\pi x) e^{-n^2\pi^2 t} $$ Where \(B_n\) are the coefficients determined as: $$ B_n = 2\int_0^1 (x^2 - x) \sin (n\pi x) dx $$

Step-by-step solution

1. Find the separated solutions

Assume u(x,t) can be separated into two functions, one depending on x (X(x)), and one depending on t (T(t)), i.e, \(u(x,t) = X(x) T(t)\). Divide both sides by \(X(x)T(t)\): $$ \frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} - \frac{2}{X(x)T(t)} $$ Since the left side depends only on t and the right side depends only on x, both must be equal to a constant, say \(\lambda\). Thus, we obtain two ordinary differential equations: $$ \frac{T'(t)}{T(t)} = \lambda \\ \frac{X''(x)}{X(x)} - \frac{2}{X(x)T(t)} = \lambda $$

2. Solve the eigenvalue problem

The eigenvalue problem involves the second equation: $$ X''(x) - (2 + \lambda)X(x) = 0 $$ Depending on the value of \(\lambda\), we can have three cases based on the characteristic equation \(r^2 - (2 + \lambda) = 0\). Let's analyze the three cases: i. If \(\lambda = -k^2\) with \(k > 0\), the equation becomes: $$ X''(x) + k^2X(x) = 2 $$ The solution to this equation is: $$ X_k(x) = A \cos(kx) + B \sin(kx) $$ ii. If \(\lambda = 0\), the equation becomes: $$ X''(x) = 2 $$ The solution to this equation is: $$ X_0(x) = Cx + D $$ iii. If \(\lambda = k^2\) with \(k > 0\), the equation becomes: $$ X''(x) - k^2X(x) = 2 $$ The solution to this equation is: $$ X_{-k}(x) = E e^{kx} + F e^{-kx} $$

3. Apply the boundary conditions

Boundary conditions are \(u(0,t) = 1\) and \(u(1,t) = 0\). This gives us: $$ T(t) X(0) = 1 \\ T(t) X(1) = 0 $$ For the first equation, we have that \(X(0) = C + D\). For the second equation, since T(t) is nonzero, \(X(1) = 0\). Apply these to each case: i. $$ A + B\sin(k) = 0 \\ A\cos(k) + B\sin(k) = 1 $$ Nontrivial solutions exist if and only if the determinant is zero: \(-\sin^2(k) - \cos^2(k) = 0\). Since this equation is always satisfied, case i will give infinitely many solutions. ii. $$ C = 1 \\ C + D = 0 $$ We get \(C = 1\) and \(D = -1\). iii. No nontrivial solutions exist in this case.

4. Combine the separated solutions

To fully construct a solution, we need to add together the solutions we found in each case. The general solution is given by: $$ u(x,t) = X_0(x)T_0(t) + \sum_{n=1}^{\infty} X_k(x) T_k(t) $$ We have \(X_0(x) = x - 1\), and using the initial condition \(u(x,0) = x^2 - 2x + 2\), we can determine the corresponding coefficients.

5. Determine coefficients using the initial condition

We have to match the initial condition: $$ x^2 - 2x + 2 = (x - 1)T_0(0) + \sum_{n=1}^{\infty} X_k(x) T_k(0) $$ Notice that \(T_0(0)\) should not be zero, so we can set \(T_0(0) = 1\). To find the other coefficients, we can use the method of Fourier series for the periodic terms. Let \(Y(x) = x^2 - x\). We can write \(Y(x)\) as the following Fourier sine series: $$ Y(x) = \sum_{n=1}^{\infty} B_n \sin(n\pi x) $$ The coefficients \(B_n\) are given by: $$ B_n = 2\int_0^1 (x^2 - x) \sin (n\pi x) dx $$ Compute this integral to find the coefficients.

6. Final solution

The final solution u(x,t) can now be written as: $$ u(x,t) = (x - 1) + \sum_{n=1}^{\infty} B_n \sin(n\pi x) e^{-n^2\pi^2 t} $$ Where \(B_n\) are the coefficients found in step 5.

Key concepts

Separated Solutions

In solving complex problems such as boundary value problems, sometimes a method called separation of variables is employed. Separation of variables is a strategy where the solution is expressed as the product of functions that each depend on only one of the variables. In your exercise, the function u(x, t) is assumed to be a product of two functions: X(x), which only depends on the spatial variable x, and T(t), which only depends on the time variable t.

This assumption allows us to break down the original partial differential equation into simpler ordinary differential equations (ODEs), which are easier to solve. The process of separation simplifies the problem into algebraic equations that are dependent on a single variable, and these equations are often more approachable than the original partial differential equation.

Eigenvalue Problems

When solving the separated equations from the boundary value problem, we encounter a special type of ODE called an eigenvalue problem. An eigenvalue problem determines the permissible values of a parameter—called eigenvalues—for which the differential equation has non-trivial solutions. Non-trivial solutions are those that are not identically zero and satisfy the given constraints or boundary conditions.

In the context of your exercise, the value of \( \lambda \) determines the form of the solution for X(x). Eigenvalue problems are foundational in understanding physical phenomena, such as vibrations in mechanical systems or quantum mechanics, where eigenvalues can represent frequencies of vibration or energy levels respectively. Identifying eigenvalues allows us to find the specific solutions that are 'allowed' by the boundaries of the problem, linking the mathematical theory to the physical reality of the problem at hand.

Fourier Series

Finally, the Fourier series is an indispensable tool for analyzing and solving boundary value problems. It allows us to express periodic functions as an infinite sum of sines and cosines, each with specific coefficients that depend on the function being represented. This is particularly useful when dealing with problems defined in a spatial domain, where the initial conditions can be periodic in nature.

In the given problem, after applying the boundary conditions, you are led to a series solution involving a Fourier sine series to match the initial conditions at t=0. The Fourier coefficients \( B_n \) are determined using an integration formula that projects the initial condition function onto the basis functions—sine and cosine. These coefficients capture the amplitude of each sine component in the series, and with these, you can accurately construct the solution space for your boundary value problem. The power of the Fourier series lies in its ability to approximate complex periodic phenomena using simple, periodic basis functions.