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Chapter 13: Problem 19
Identifying \(A=1, B=6,\) and \(C=9,\) we compute \(B^{2}-4 A C=0 .\) The equation is parabolic.
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$$\begin{array}{l}k \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial
u}{\partial t}, \quad 0
From (8) in the text we have \\[ u(x, t)=\sum_{n=1}^{\infty}\left(A_{n} \cos \frac{n \pi a}{L} t+B_{n} \sin \frac{n \pi a}{L} t\right) \sin \frac{n \pi}{L} x \\] since \(u_{t}(x, 0)=g(x)=0\) we have \(B_{n}=0\) and \\[ \begin{aligned} u(x, t) &=\sum_{n=1}^{\infty} A_{n} \cos \frac{n \pi a}{L} t \sin \frac{n \pi}{L} x \\ &=\sum_{n=1}^{\infty} A_{n} \frac{1}{2}\left[\sin \left(\frac{n \pi}{L} x+\frac{n \pi a}{L} t\right)+\sin \left(\frac{n \pi}{L} x-\frac{n \pi a}{L} t\right)\right] \\ &=\frac{1}{2} \sum_{n=1}^{\infty} A_{n}\left[\sin \frac{n \pi}{L}(x+a t)+\sin \frac{n \pi}{L}(x-a t)\right] \end{aligned} \\] From \\[ u(x, 0)=f(x)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi}{L} x \\] we identify \\[ f(x+a t)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi}{L}(x+a t) \\] and \\[ f(x-a t)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi}{L}(x-a t) \\] so that \\[ u(x, t)=\frac{1}{2}[f(x+a t)+f(x-a t)] \\]
Substituting \(u(x, t)=X(x) T(t)\) into the partial differential equation yields \(a^{2} X^{\prime \prime} T=X T^{\prime \prime} .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime \prime}}{X}=\frac{T^{\prime \prime}}{a^{2} T}=-\lambda.$$ Then $$X^{\prime \prime}+\lambda X=0 \quad \text { and } \quad T^{\prime \prime}+a^{2} \lambda T=0.$$ We consider three cases: I. If \(\lambda=0\) then \(X^{\prime \prime}=0\) and \(X(x)=c_{1} x+c_{2} .\) Also, \(T^{\prime \prime}=0\) and \(T(t)=c_{3} t+c_{4},\) so $$u=X T=\left(c_{1} x+c_{2}\right)\left(c_{3} t+c_{4}\right).$$ II. If \(\lambda=-\alpha^{2}<0,\) then \(X^{\prime \prime}-\alpha^{2} X=0,\) and \(X(x)=c_{5} \cosh \alpha x+c_{6} \sinh \alpha x .\) Also, \(T^{\prime \prime}-\alpha^{2} a^{2} T=0\) and \(T(t)=c_{7} \cosh \alpha a t+c_{8} \sinh \alpha a t,\) so $$u=X T=\left(c_{5} \cosh \alpha x+c_{6} \sinh \alpha x\right)\left(c_{7} \cosh \alpha a t+c_{8} \sinh \alpha a t\right).$$ III. If \(\lambda=\alpha^{2}>0\), then \(X^{\prime \prime}+\alpha^{2} X=0\), and \(X(x)=c_{9} \cos \alpha x+c_{10} \sin \alpha x\). Also, \(T^{\prime \prime}+\alpha^{2} a^{2} T=0\) and \(T(t)=c_{11} \cos \alpha a t+c_{12} \sin \alpha a t,\) so $$u=X T=\left(c_{9} \cos \alpha x+c_{10} \sin \alpha x\right)\left(c_{11} \cos \alpha a t+c_{12} \sin \alpha a t\right).$$
$$\begin{array}{l}k \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial
u}{\partial t}, \quad 0
As shown in Example 1 in the text, separation of variables leads to \\[ \begin{array}{l} X(x)=c_{1} \cos \alpha x+c_{2} \sin \alpha x \\ Y(y)=c_{3} \cos \beta y+c_{4} \sin \beta y \end{array} \\] and \\[ T(t)+c_{5} e^{-k\left(\alpha^{2}+\beta^{2}\right) t} \\] The boundary conditions \\[ \left.\begin{array}{l} u_{x}(0, y, t)=0, \quad u_{x}(1, y, t)=0 \\ u_{y}(x, 0, t)=0, \quad u_{y}(x, 1, t)=0 \end{array}\right\\} \quad \text { imply } \quad\left\\{\begin{array}{ll} X^{\prime}(0)=0, & X^{\prime}(1)=0 \\ Y^{\prime}(0)=0, & Y^{\prime}(1)=0 \end{array}\right. \\] Applying these conditions to \\[ X^{\prime}(x)=-\alpha c_{1} \sin \alpha x+\alpha c_{2} \cos \alpha x \\] and \\[ Y^{\prime}(y)=-\beta c_{3} \sin \beta y+\beta c_{4} \cos \beta y \\] gives \(c_{2}=c_{4}=0\) and \(\sin \alpha=\sin \beta=0 .\) Then \\[ \alpha=m \pi, m=0,1,2, \ldots \quad \text { and } \quad \beta=n \pi, n=0,1,2, \ldots \\] By the superposition principle \\[ u(x, y, t)=A_{00}+\sum_{m=1}^{\infty} A_{m 0} e^{-k m^{2} \pi^{2} t} \cos m \pi x+\sum_{n=1}^{\infty} A_{0 n} e^{-k n^{2} \pi^{2} t} \cos n \pi y \\] \\[ +\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{m n} e^{-k\left(m^{2}+n^{2}\right) \pi^{2} t} \cos m \pi x \cos n \pi y \\] We now compute the coefficients of the double cosine series: Identifying \(b=c=1\) and \(f(x, y)=x y\) we have \\[ \begin{aligned} A_{00} &=\int_{0}^{1} \int_{0}^{1} x y d x d y=\left.\int_{0}^{1} \frac{1}{2} x^{2} y\right|_{0} ^{1} d y=\frac{1}{2} \int_{0}^{1} y d y=\frac{1}{4} \\ A_{m 0} &=2 \int_{0}^{1} \int_{0}^{1} x y \cos m \pi x d x d y=\left.2 \int_{0}^{1} \frac{1}{m^{2} \pi^{2}}(\cos m \pi x+m \pi x \sin m \pi x)\right|_{0} ^{1} y d y \\ &=2 \int_{0}^{1} \frac{\cos m \pi-1}{m^{2} \pi^{2}} y d y=\frac{\cos m \pi-1}{m^{2} \pi^{2}}=\frac{(-1)^{m}-1}{m^{2} \pi^{2}} \\ A_{0 n} &=2 \int_{0}^{1} \int_{0}^{1} x y \cos n \pi y d x d y=\frac{(-1)^{n}-1}{n^{2} \pi^{2}} \end{aligned} \\] and \\[ A_{m n}=4 \int_{0}^{1} \int_{0}^{1} x y \cos m \pi x \cos n \pi y d x d y=4 \int_{0}^{1} x \cos m \pi x d x \int_{0}^{1} y \cos n \pi y d y \\] \\[ =4\left(\frac{(-1)^{m}-1}{m^{2} \pi^{2}}\right)\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right) .\\]
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