Question

Consider the problem $$ \begin{aligned} \alpha^{2} u_{x x}=u_{t}, & 00 \\ u(0, t)=0, \quad u_{x}(L, t)+\gamma u(L, t)=0, & t>0 \\ u(x, 0)=f(x), & 0 \leq x \leq L \end{aligned} $$ (a) Let \(u(x, t)=X(x) T(t)\) and show that $$ X^{\prime \prime}+\lambda X=0, \quad X(0)=0, \quad X^{\prime}(L)+\gamma X(L)=0 $$ and $$ T^{\prime}+\lambda \alpha^{2} T=0 $$ where \(\lambda\) is the separation constant. (b) Assume that \(\lambda\) is real, and show that problem (ii) has no nontrivial solutions if \(\lambda \leq 0\). (c) If \(\lambda>0\), let \(\lambda=\mu^{2}\) with \(\mu>0 .\) Show that problem (ii) has nontrivial solutions only if \(\mu\) is a solution of the equation $$ \mu \cos \mu L+\gamma \sin \mu L=0 $$ (d) Rewrite Eq. (iii) as \(\tan \mu L=-\mu / \gamma .\) Then, by drawing the graphs of \(y=\tan \mu L\) and \(y=-\mu L / \gamma L\) for \(\mu>0\) on the same set of axes, show that Eq. (iii) is satisfied by infinitely many positive values of \(\mu ;\) denote these by \(\mu_{1}, \mu_{2}, \ldots, \mu_{n}, \ldots,\) ordered in increasing size. (e) Determine the set of fundamental solutions \(u_{n}(x, t)\) corresponding to the values \(\mu_{n}\) found in part (d).

Short answer

#tag_title# Step 8: Combine the Fundamental Set of Solutions#tag_content# The final solution to the PDE is a linear combination of the fundamental set of solutions. We express the desired solution as \(u(x,t) = \sum_{n=1}^{\infty} u_n(x,t) = \sum_{n=1}^{\infty}B_n \sin(\mu_n x) e^{-\mu_n^2\alpha^2t}\) where \(B_n\) are the constant coefficients to be determined. #tag_title# Step 9: Determine the Coefficients B_n#tag_content# The final step is to determine the constant coefficients \(B_n\) that satisfy the initial condition \(u(x, 0) = f(x)\). Thus, \(\sum_{n=1}^{\infty} B_n \sin(\mu_n x) = f(x)\). As the series converges, a Fourier series expression for f(x) is obtained. Using the Fourier coefficients, we can determine the values of \(B_n\). #tag_title# Step 10: The Final Solution#tag_content# After determining the \(B_n\) coefficients, we can write the final solution to the PDE as \(u(x,t) = \sum_{n=1}^{\infty} B_n \sin(\mu_n x) e^{-\mu_n^2\alpha^2t}\). This solution expresses the temperature distribution u(x,t) in a one-dimensional heat conduction problem at any position x and time t, given the initial condition f(x) and the boundary conditions.

Step-by-step solution

Apply Separation of Variables

Substitute \(u(x, t)=X(x) T(t)\) into the PDE to get \(T(t) \alpha^{2} X''(x) = X(x) T'(t)\). Divide through by \(\alpha^2 X(x) T(t)\) to get the result \(\frac{T'(t)}{\alpha^2 T(t)} = \frac{X''(x)}{X(x)}\). The only way this equation holds for all x and t is if both sides are equal to a constant, say \(\lambda\). Thus, we have \(\frac{T'(t)}{\alpha^2 T(t)} = \lambda\) and \(\frac{X''(x)}{X(x)} = \lambda\).

Obtain the Spatial Differential Equation

Rewrite the spatial part of the separation of variables to get the differential equation for X, which is \(X''(x) + \lambda X(x) = 0\). Now, apply the boundary conditions \(u(0,t)=0\) and \(u_x(L,t) + \gamma u(L,t)=0\). This results in \(X(0)=0\) and \(X'(L) + \gamma X(L)=0\) respectively.

Obtain the Temporal Equation

Rewrite the temporal part of the separation of constants to get the differential equation for T, which is \(T'(t) + \lambda \alpha^2 T(t) = 0\).

Investigate lambda less than or equal to 0

Assume \(\lambda \leq 0\) and try to find non-trivial solutions for X and T. The corresponding equation for X is \(X'' + \lambda X = 0\) where \(\lambda \leq 0\). The general solution of this equation is \(X(x) = A\cos(\sqrt{-\lambda}x) + B\sin(\sqrt{-\lambda}x)\). Using the boundary condition \(X(0)=0\), we get A = 0. Therefore, \(X(x) = B \sin(\sqrt{-\lambda}x)\). Apply the boundary condition \(X'(L) + \gamma X(L) = 0\) to determine B. If \(\lambda = 0\), then \(X(x) = Bx\) and the boundary condition gives \(B = 0\) which is a trivial solution. If \(\lambda < 0\), then the solution doesn't satisfy the boundary condition.

Investigate lambda greater than 0

Now assume \(\lambda > 0\). For this case, let \(\lambda = \mu^2\) to help manage the notation. The corresponding equation for X is now \(X'' + \mu^2X = 0\). The general solution of this equation is \(X(x) = C\cos(\mu x) + D\sin(\mu x)\). Using the boundary condition \(X(0)=0\), we get C = 0. Therefore, \(X(x) = D \sin(\mu x)\). Apply the boundary condition \(X'(L) + \gamma X(L) = 0\) to get \(\mu D\cos(\mu L) + \gamma D\sin(\mu L) = 0\). Dividing through by D and rearranging gives \(\mu\cos(\mu L) + \gamma\sin(\mu L) = 0\).

Identify the Transcendental Equation

A different way of expressing the equation from step 5 is \(\tan(\mu L) = -\mu / \gamma\), which is a transcendental equation that may have infinitely many solutions. These solutions depend on how the graphs of the two functions on the left and right sides cross, and are all positive roots.

Determine the Fundamental Set of Solutions

The transcendental equation is schematically solved to yield the values \(\mu_1, \mu_2, ..., \mu_n, ...\) . Corresponding to each \(\mu_n\), the function \(X(x) = D \sin(\mu x)\) satisfies the spatial differential equation and the boundary conditions. The differential equation for T has the solution \(T(t) = E\exp(-\mu^2\alpha^2 t)\). Hence the fundamental set of solutions is \(u_n(x,t) = X_n(x)T_n(t)\) where \(X_n(x) = D_n \sin(\mu_n x)\) and \(T_n(t) = E_n \exp(-\mu_n^2\alpha^2 t)\).

Key concepts

Separation of Variables

Separation of variables is a method often used to solve partial differential equations (PDEs). This technique allows us to simplify a complex PDE into simpler, ordinary differential equations. The fundamental assumption here is that a solution can be represented as a product of functions, each dependent on a single coordinate.

In our case, we assume a solution of the form \(u(x, t) = X(x) T(t)\). By substituting this into the given PDE, \(\alpha^2 u_{xx} = u_t\), and performing algebraic manipulations, we separate the equation into two independent functions: one involving only \(x\) and the other only \(t\).

• The spatial part becomes \(X''(x) + \lambda X(x) = 0\).
• The temporal part becomes \(T'(t) + \lambda \alpha^2 T(t) = 0\).

This separation allows us to focus on solving simpler ordinary differential equations, one for \(X\) and one for \(T\), under the assumption that \(\lambda\), the separation constant, is constant.

Boundary Conditions

Boundary conditions are essential components of a differential equation problem as they help determine the specific solution that fits the physical situation being modeled.

In the problem under consideration, the boundary conditions are:

• \(u(0, t) = 0\)
• \(u_x(L, t) + \gamma u(L, t) = 0\)

These conditions apply constraints on the function \(X(x)\) since they depend only on the spatial variable and not on \(t\).

Applying \(u(0, t) = 0\) gives us \(X(0) = 0\). The condition \(u_x(L, t) + \gamma u(L, t) = 0\) leads to the relation \(X'(L) + \gamma X(L) = 0\). Together, these boundary conditions ensure that we find the correct form of the spatial solution \(X(x)\) that satisfies the physical constraints of the problem.

Transcendental Equations

Transcendental equations appear naturally when solving differential equations with specific boundary conditions. They are non-algebraic equations, often involving trigonometric functions, exponential functions, or logarithms.

In this problem, after applying boundary conditions to the spatial equation \( X''(x) + \lambda X(x) = 0\), we derive the transcendental equation:

\[\mu \cos(\mu L) + \gamma \sin(\mu L) = 0\]
This can be rewritten as:

\[\tan(\mu L) = -\frac{\mu}{\gamma}\]
The transcendental equation helps determine the specific values of \(\mu\) that satisfy the physical constraints. These \(\mu\) values are crucial as they define the characteristics of the solution. Since the function \(\tan(\mu L)\) is periodic, the transcendental equation typically yields infinitely many solutions, each associated with a distinct mode of the wave equation.

Initial Value Problem

An initial value problem involves finding a solution to a differential equation that satisfies specific conditions at the initial point in time, often at \(t = 0\).

For our exercise, the initial condition is given by \(u(x, 0) = f(x)\). This condition prescribes the state of the system at time \(t = 0\).

Once the spatial solutions \(X_n(x)\) corresponding to the calculated \(\mu_n\) are determined, the temporal part of the solution at \(t = 0\) is used to match the initial condition. This implication helps us find constants in the complete solution, \(u_n(x,t) = X_n(x) T_n(t)\). The function \(X_n(x) = D_n \sin(\mu_n x)\), along with the determined \(T_n(t) = E_n \exp(- \mu_n^2 \alpha^2 t)\), forms the full solution that solves the initial value problem accordingly.
This ensures that our mathematical solution accurately describes the initial conditions laid out by the problem.