Problem 20
Question
In the general case the associated Sturm-Liouville problem is \(X^{\prime \prime}+\lambda X=0, \quad X^{\prime}(0)=0, \quad X^{\prime}(L)=0\) with eigenvalues and eigenfunctions \(\lambda_{0}=0, X_{0}=1,\) and \(\lambda_{n}=n^{2} \pi^{2} / L^{2}, X_{n}=\cos n \pi x / L, n=1,2,3, \ldots .\) The entire set of eigenfunctions can be written as \(X_{n}=\cos n \pi x / L, n=0,1,2, \ldots,\) which serves as the basis for the Fourier cosine series. Hence, we assume in this problem that \(u(x, t)=\frac{1}{2} u_{0}(t)+\sum_{n=1}^{\infty} u_{n}(t) \cos \frac{n \pi}{L} x\) and \(F(x, t)=\frac{1}{2} F_{0}(t)+\sum_{n=1}^{\infty} F_{n}(t) \cos \frac{n \pi}{L} x\). Taking \(k=1, L=1, F(x, t)=t x,\) and \(f(x)=0\) we have \(u(x, t)=\frac{1}{2} u_{0}(t)+\sum_{n=1}^{\infty} u_{n}(t) \cos n \pi x\) and \(F(x, t)=\frac{1}{2} F_{0}(t)+\sum_{n=1}^{\infty} F_{n}(t) \cos n \pi x\). By treating \(t\) as a parameter, the coefficients \(F_{n}\) can be computed: \(F_{0}(t)=2 \int_{0}^{1} t x d x=t\) \(F_{n}(t)=2 \int_{0}^{1} t x \cos n \pi x d x=2 t \int_{0}^{1} x \cos n \pi x d x=2 t \frac{(-1)^{n}-1}{n^{2} \pi^{2}}\). Hence \(t x=\frac{1}{2} t+2 t \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos n \pi x\). Now, using the series representation for \(u(x, t),\) we have \(\frac{\partial^{2} u}{\partial x^{2}}=\sum_{n=1}^{\infty} u_{n}(t)\left(-n^{2} \pi^{2}\right) \cos n \pi x \quad\) and \(\quad \frac{\partial u}{\partial t}=\frac{1}{2} u_{0}^{\prime}(t)+\sum_{n=1}^{\infty} u_{n}^{\prime}(t) \cos n \pi x\). Writing the partial differential equation as \(u_{t}-u_{x x}=t x\) and using the above results we have \(\frac{1}{2} u_{0}^{\prime}(t)+\sum_{n=1}^{\infty}\left[u_{n}^{\prime}(t)+n^{2} \pi^{2} u_{n}(t)\right] \cos n \pi x=\frac{1}{2} t+2 t \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos n \pi x\). Equating coefficients we get \(u_{0}^{\prime}(t)=t \quad\) and \(\quad u_{n}^{\prime}(t)+n^{2} \pi^{2} u_{n}(t)=2 t \frac{(-1)^{n}-1}{n^{2} \pi^{2}}\). From the first equation we obtain \(u_{0}(t)=\frac{1}{2} t^{2}+C_{0} .\) The second equation is a linear, first-order differential equation whose general solution is \(u_{n}(t)=2\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right)\left(\frac{n^{2} \pi^{2} t-1}{n^{4} \pi^{4}}\right)+C_{n} e^{-n^{2} \pi^{2} t}\). Thus \(u(x, t)=\frac{1}{4} t^{2}+\frac{1}{2} C_{0}+\sum_{n=1}^{\infty}\left[2\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right)\left(\frac{n^{2} \pi^{2} t-1}{n^{4} \pi^{4}}\right)+C_{n} e^{-n^{2} \pi^{2} t}\right] \cos n \pi x\). The initial condition \(u(x, 0)=0\) implies \(\frac{1}{2} C_{0}+\sum_{n=1}^{\infty}\left[2\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right)\left(\frac{-1}{n^{4} \pi^{4}}\right)+C_{n}\right] \cos n \pi x=\frac{1}{2} C_{0}+\sum_{n=1}^{\infty}\left[2 \frac{1-(-1)^{n}}{n^{6} \pi^{6}}+C_{n}\right] \cos n \pi x=0\) so that \(C_{0}=0 \quad\) and \(\quad C_{n}=2 \frac{(-1)^{n}-1}{n^{6} \pi^{6}}\). Therefore \(\begin{aligned} u(x, t) &=\frac{1}{4} t^{2}+\sum_{n=1}^{\infty}\left[2\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right)\left(\frac{n^{2} \pi^{2} t-1}{n^{4} \pi^{4}}\right)+2 \frac{(-1)^{n}-1}{n^{6} \pi^{6}} e^{-n^{2} \pi^{2} t}\right] \cos n \pi x \\\ &=\frac{1}{4} t^{2}+\frac{2}{\pi^{6}} \sum_{n=1}^{\infty}\left[\frac{\left[(-1)^{n}-1\right]\left[n^{2} \pi^{2} t-1+e^{-n^{2} \pi^{2} t}\right]}{n^{6}}\right] \cos n \pi x \end{aligned}\).
Step-by-Step Solution
VerifiedKey Concepts
Eigenvalues and Eigenfunctions
To calculate these, we apply the given boundary conditions \(X'(0) = 0\) and \(X'(L) = 0\). This leads us to the set of eigenvalues \(\lambda_n = n^2 \pi^2 / L^2\) and corresponding eigenfunctions \(X_n = \cos(n \pi x / L)\), where \(n = 0, 1, 2, \ldots\).
These eigenfunctions form an orthogonal basis, allowing us to represent solutions to more complex problems through their linear combinations. They play a significant role in constructing series solutions such as Fourier series.
Fourier Cosine Series
In the context of the Sturm-Liouville problem, the eigenfunctions \(X_n = \cos(n \pi x / L)\) provide the basis for forming the Fourier cosine series. These are utilized to represent the solution \(u(x, t)\) as a combination of these cosines, allowing us to solve the given partial differential equations.
For a function \(u(x, t)\) or \(F(x, t)\) defined on \(0 \leq x \leq L\), the Fourier cosine series takes the form:
- \(u(x, t) = \frac{1}{2} u_0(t) + \sum_{n=1}^{\infty} u_n(t) \cos(n \pi x / L)\)
- \(F(x, t) = \frac{1}{2} F_0(t) + \sum_{n=1}^{\infty} F_n(t) \cos(n \pi x / L)\)
Partial Differential Equations (PDEs)
In this problem, we encounter a PDE of the form \(u_t - u_{xx} = tx\), which combines both temporal and spatial derivatives of the function \(u(x, t)\). The objective is to determine \(u(x, t)\) that satisfies both the PDE and the given initial/boundary conditions.
By representing \(u(x, t)\) in terms of a Fourier cosine series, one can separate variables and solve a sequence of ordinary differential equations for the coefficients \(u_n(t)\).
- The temporal equation satisfies \(u_0'(t) = t\), leading to a solution \(u_0(t) = \frac{1}{2}t^2 + C_0\).
- For \(n \geq 1\), the differential equation \(u_n'(t) + n^2\pi^2 u_n(t) = 2t\frac{(-1)^n-1}{n^2\pi^2}\) provides \(u_n(t)\) solutions.