Chapter 13

Using \(u=X T\) and \(-\lambda\) as a separation constant we obtain $$\begin{array}{c} X^{\prime \prime}+\lambda X=0, \\ X(0)=0 \\\X(L)=0,\end{array},$$ and $$T^{\prime}+k \lambda T=0.$$ This leads to $$X=c_{1} \sin \frac{n \pi}{L} x \quad \text { and } \quad T=c_{2} e^{-k n^{2} \pi^{2} t / L^{2}}$$ for \(n=1,2,3, \dots\) so that $$u=\sum_{n=1}^{\infty} A_{n} e^{-k n^{2} \pi^{2} t / L^{2}} \sin \frac{n \pi}{L} x.$$ Imposing $$u(x, 0)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi}{L} x$$ gives $$A_{n}=\frac{2}{L} \int_{0}^{L / 2} \sin \frac{n \pi}{L} x d x=\frac{2}{n \pi}\left(1-\cos \frac{n \pi}{2}\right)$$ for \(n=1,2,3, \ldots\) so that $$u(x, t)=\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1-\cos \frac{n \pi}{2}}{n} e^{-k n^{2} \pi^{2} t / L^{2}} \sin \frac{n \pi}{L} x.$$

$$k \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial u}{\partial t}, \quad 00$$ $$\begin{array}{l}u(0, t)=0,\left.\quad \frac{\partial u}{\partial x}\right|_{x=L}=0, \quad t>0 \\\u(x, 0)=f(x), \quad 0

Using \(u=X T\) and \(-\lambda\) as a separation constant we obtain $$\begin{array}{c} X^{\prime \prime}+\lambda X=0, \\\X(0)=0, \\\X(L)=0,\end{array}$$ and $$T^{\prime}+k \lambda T=0.$$ This leads to $$X=c_{1} \sin \frac{n \pi}{L} x \quad \text { and } \quad T=c_{2} e^{-k n^{2} \pi^{2} t / L^{2}}$$ for \(n=1,2,3, \dots\) so that $$u=\sum_{n=1}^{\infty} A_{n} e^{-k n^{2} \pi^{2} t / L^{2}} \sin \frac{n \pi}{L} x.$$ Imposing $$u(x, 0)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi}{L} x$$ gives $$A_{n}=\frac{2}{L} \int_{0}^{L} x(L-x) \sin \frac{n \pi}{L} x d x=\frac{4 L^{2}}{n^{3} \pi^{3}}\left[1-(-1)^{n}\right]$$ for \(n=1,2,3, \ldots\) so that $$u(x, t)=\frac{4 L^{2}}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n^{3}} e^{-k n^{2} \pi^{2} t / L^{2}} \sin \frac{n \pi}{L} x.$$

$$\begin{array}{ll}k \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial u}{\partial t}, & 00 \\\u(0, t)=u_{0}, & u(L, t)=u_{1}, \quad t>0 \\ u(x, 0)=0, & 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) .\\]

$$\begin{array}{l}k \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial u}{\partial t}, \quad 00 \\\u(0, t)=100,\left.\quad \frac{\partial u}{\partial x}\right|_{x=L}=-h u(L, t), \quad t>0 \\\u(x, 0)=f(x), \quad 0

Using \(u=X Y\) and \(-\lambda\) as a separation constant we obtain $$\begin{aligned}&X^{\prime \prime}+\lambda X=0,\\\&X^{\prime}(0)=0,\\\&X^{\prime}(a)=0,\end{aligned}$$ and $$\begin{aligned} &Y^{\prime \prime}-\lambda Y=0,\\\&Y(b)=0.\end{aligned}$$ With \(\lambda=\alpha^{2}>0\) the solutions of the differential equations are $$X=c_{1} \cos \alpha x+c_{2} \sin \alpha x \quad \text { and } \quad Y=c_{3} \cosh \alpha y+c_{4} \sinh \alpha y$$ The boundary and initial conditions imply $$X=c_{1} \cos \frac{n \pi}{a} x \quad \text { and } \quad Y=c_{3} \cosh \frac{n \pi}{a} y-c_{3} \frac{\cosh \frac{n \pi b}{a}}{\sinh \frac{n \pi b}{a}} \sinh \frac{n \pi}{a} y$$ for \(n=1,2,3, \ldots\) since \(\lambda=0\) is an eigenvalue for both differential equations with corresponding eigenfunctions 1 and \(y-b,\) respectively we have $$u=A_{0}(y-b)+\sum_{n=1}^{\infty} A_{n} \cos \frac{n \pi}{a} x\left(\cosh \frac{n \pi}{a} y-\frac{\cosh \frac{n \pi b}{a}}{\sinh \frac{n \pi b}{a}} \sinh \frac{n \pi}{a} y\right).$$ Imposing $$u(x, 0)=x=-A_{0} b+\sum_{n=1}^{\infty} A_{n} \cos \frac{n \pi}{a} x$$ gives $$-A_{0} b=\frac{1}{a} \int_{0}^{a} x d x=\frac{1}{2} a$$ and $$A_{n}=\frac{2}{a} \int_{0}^{a} x \cos \frac{n \pi}{a} x d x=\frac{2 a}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right]$$ so that $$u(x, y)=\frac{a}{2 b}(b-y)+\frac{2 a}{\pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2}} \cos \frac{n \pi}{a} x\left(\cosh \frac{n \pi}{a} y-\frac{\cosh \frac{n \pi b}{a}}{\sinh \frac{n \pi b}{a}} \sinh \frac{n \pi}{a} y\right).$$

$$\begin{array}{l}k \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial u}{\partial t}, \quad 00 \\\\\left.\frac{\partial u}{\partial x}\right|_{x=0}=h[u(0, t)-20],\left.\quad \frac{\partial u}{\partial x}\right|_{x=L}=0, \quad t>0 \\\u(x, 0)=f(x), \quad 0

Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(X^{\prime} Y=X Y^{\prime}+X Y .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime}}{X}=\frac{Y+Y^{\prime}}{Y}=-\lambda.$$ Then $$X^{\prime}+\lambda X=0 \quad \text { and } \quad y^{\prime}+(1+\lambda) Y=0$$ so that $$\begin{aligned} &X=c_{1} e^{-\lambda x} \quad \text { and }\\\ &Y=c_{2} e^{-(1+\lambda) y}=0. \end{aligned}$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{-y-\lambda(x+y)}.$$

Using \(u=X T\) and \(-\lambda\) as a separation constant leads to $$\begin{array}{c} X^{\prime \prime}+\lambda X=0, \\ X^{\prime}(0)=0, \\ X^{\prime}(L)=0, \end{array}$$ and $$T^{\prime}+(h+k \lambda) T=0.$$ Then $$X=c_{1} \cos \frac{n \pi}{L} x \quad \text { and } \quad T=c_{2} e^{-h t-k n^{2} \pi^{2} t / L^{2}}$$ for \(n=0,1,2, \ldots(\lambda=0\) is an eigenvalue in this case) so that $$u=A_{0} e^{-h t}+e^{-h t} \sum_{n=1}^{\infty} A_{n} e^{-k n^{2} \pi^{2} t / L^{2}} \cos \frac{n \pi}{L} x.$$ Imposing $$u(x, 0)=f(x)=\sum_{n=0}^{\infty} A_{n} \cos \frac{n \pi}{L} x$$ gives $$u(x, t)=\frac{e^{-h t}}{L} \int_{0}^{L} f(x) d x+\frac{2 e^{-h t}}{L} \sum_{n=1}^{\infty}\left(\int_{0}^{L} f(x) \cos \frac{n \pi}{L} x d x\right) e^{-k n^{2} \pi^{2} t / L^{2}} \cos \frac{n \pi}{L} x.$$

In Problems 5 and 6 we try \(u(x, y, z)=X(x) Y(y) Z(z)\) to separate Laplace's equation in three dimensions: \\[ \begin{array}{c} X^{\prime \prime} Y Z+X Y^{\prime \prime} Z+X Y Z^{\prime \prime}=0 \\ \frac{X^{\prime \prime}}{X}=-\frac{Y^{\prime \prime}}{Y}-\frac{Z^{\prime \prime}}{Z}=-\alpha^{2} \end{array} \\] Then \\[ X^{\prime \prime}+\alpha^{2} X=0 \\] \\[ \begin{array}{c} \frac{Y^{\prime \prime}}{Y}=-\frac{Z^{\prime \prime}}{Z}+\alpha^{2}=-\beta^{2} \\\ Y^{\prime \prime}+\beta^{2} Y=0 \\ Z^{\prime \prime}-\left(\alpha^{2}+\beta^{2}\right) Z=0 \end{array} \\] The general solutions of equations \((\mathbf{4}),(\mathbf{5}),\) and \((\mathbf{6})\) are, respectively \\[ X(x)=c_{1} \cos \alpha x+c_{2} \sin \alpha x \\] \\[ \begin{array}{l} Y(y)=c_{3} \cos \beta y+c_{4} \sin \beta y \\ Z(z)=c_{5} \cosh \sqrt{\alpha^{2}+\beta^{2}} z+c_{6} \sinh \sqrt{\alpha^{2}+\beta^{2}} z .\end{array} \\] The boundary and initial conditions are $$\begin{array}{ll} u(0, y, z)=0, & u(a, y, z)=0 \\ u(x, 0, z)=0, & u(x, b, z)=0 \\ u(x, y, 0)=0, & u(x, y, c)=f(x, y) \end{array}$$ The conditions \(X(0)=Y(0)=Z(0)=0\) give \(c_{1}=c_{3}=c_{5}=0 .\) The conditions \(X(a)=0\) and \(Y(b)=0\) yield two sets of eigenvalues: \\[ \alpha=\frac{m \pi}{a}, m=1,2,3, \ldots \quad \text { and } \quad \beta=\frac{n \pi}{b}, n=1,2,3, \ldots \\] By the superposition principle \\[ u(x, y, t)=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{m n} \sinh \omega_{m n} z \sin \frac{m \pi}{a} x \sin \frac{n \pi}{b} y \\] where \\[ \omega_{m n}^{2}=\frac{m^{2} \pi^{2}}{a^{2}}+\frac{n^{2} \pi^{2}}{b^{2}} \\] and \\[ A_{m n}=\frac{4}{a b \sinh \omega_{m n} c} \int_{0}^{b} \int_{0}^{a} f(x, y) \sin \frac{m \pi}{a} x \sin \frac{n \pi}{b} y d x d y .\\]

$$\begin{aligned} &k \frac{\partial^{2} u}{\partial x^{2}}-h u=\frac{\partial u}{\partial t}, \quad 00, \quad h \text { a constant }\\\ &\begin{array}{l} u(0, t)=\sin \frac{\pi t}{L}, \quad u(L, t)=0, \quad t>0 \\ u(x, 0)=f(x), \quad 0

In Problems 5 and 6 we try \(u(x, y, z)=X(x) Y(y) Z(z)\) to separate Laplace's equation in three dimensions: \\[ \begin{array}{c} X^{\prime \prime} Y Z+X Y^{\prime \prime} Z+X Y Z^{\prime \prime}=0 \\ \frac{X^{\prime \prime}}{X}=-\frac{Y^{\prime \prime}}{Y}-\frac{Z^{\prime \prime}}{Z}=-\alpha^{2} \end{array} \\] Then \\[ X^{\prime \prime}+\alpha^{2} X=0 \\] \\[ \begin{array}{c} \frac{Y^{\prime \prime}}{Y}=-\frac{Z^{\prime \prime}}{Z}+\alpha^{2}=-\beta^{2} \\\ Y^{\prime \prime}+\beta^{2} Y=0 \\ Z^{\prime \prime}-\left(\alpha^{2}+\beta^{2}\right) Z=0 \end{array} \\] The general solutions of equations \((\mathbf{4}),(\mathbf{5}),\) and \((\mathbf{6})\) are, respectively \\[ X(x)=c_{1} \cos \alpha x+c_{2} \sin \alpha x \\] \\[ \begin{array}{l} Y(y)=c_{3} \cos \beta y+c_{4} \sin \beta y \\ Z(z)=c_{5} \cosh \sqrt{\alpha^{2}+\beta^{2}} z+c_{6} \sinh \sqrt{\alpha^{2}+\beta^{2}} z .\end{array} \\] The boundary and initial conditions are \\[ \begin{array}{ll} u(0, y, z)=0, & u(a, y, z)=0 \\ u(x, 0, z)=0, & u(x, b, z)=0 \\ u(x, y, 0)=f(x, y), & u(x, y, c)=0 \end{array} \\] The conditions \(X(0)=Y(0)=0\) give \(c_{1}=c_{3}=0 .\) The conditions \(X(a)=Y(b)=0\) yield two sets of eigenvalues: \\[ \alpha=\frac{m \pi}{a}, m=1,2,3, \ldots \quad \text { and } \quad \beta=\frac{n \pi}{b}, n=1,2,3, \ldots \\] Let \\[ \omega_{m n}^{2}=\frac{m^{2} \pi^{2}}{a^{2}}+\frac{n^{2} \pi^{2}}{b^{2}} \\] Then the boundary condition \(Z(c)=0\) gives \\[ c_{5} \cosh c \omega_{m n}+c_{6} \sinh c \omega_{m n}=0 \\] from which we obtain \\[ \begin{aligned} Z(z) &=c_{5}\left(\cosh w_{m n} z-\frac{\cosh c \omega_{m n}}{\sinh c \omega_{m n}} \sinh \omega z\right) \\ &=\frac{c_{5}}{\sinh c \omega_{m n}}\left(\sinh c \omega_{m n} \cosh \omega_{m n} z-\cosh c \omega_{m n} \sinh \omega_{m n} z\right)=c_{m n} \sinh \omega_{m n}(c-z) \end{aligned} \\] By the superposition principle \\[ u(x, y, t)=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{m n} \sinh \omega_{m n}(c-z) \sin \frac{m \pi}{a} x \sin \frac{n \pi}{b} y \\] where \\[ A_{m n}=\frac{4}{a b \sinh c \omega_{m n}} \int_{0}^{b} \int_{0}^{a} f(x, y) \sin \frac{m \pi}{a} x \sin \frac{n \pi}{b} y d x d y .\\]

Separation of variables leads to \\[ \begin{array}{l} Y^{\prime \prime}+\alpha^{2} Y=0 \\ X^{\prime \prime}-\alpha^{2} X=0 \end{array} \\] and \\[ Y(y)=c_{1} \cos \alpha y+c_{2} \sin \alpha y \\] \\[ X(x)=c_{3} \cosh \alpha x+c_{4} \sinh \alpha x \\] From \(Y(0)=0\) we find \(c_{1}=0 .\) From \(Y^{\prime}(1)=0\) we obtain \(\cos \alpha=0\) and \\[ \alpha=\frac{\pi(2 n-1)}{2}, n=1,2,3, \dots \\] Thus \\[ Y(y)=c_{2} \sin \left(\frac{2 n-1}{2}\right) \pi y \\] From \(X^{\prime}(0)=0\) we find \(c_{4}=0 .\) Then \\[ u(x, y)=\sum_{n=1}^{\infty} A_{n} \cosh \left(\frac{2 n-1}{2}\right) \pi x \sin \left(\frac{2 n-1}{2}\right) \pi y \\] where \\[ u_{0}=u(1, y)=\sum_{n=1}^{\infty} A_{n} \cosh \left(\frac{2 n-1}{2}\right) \pi \sin \left(\frac{2 n-1}{2}\right) \pi y \\] and \\[ A_{n} \cosh \left(\frac{2 n-1}{2}\right) \pi=\frac{\int_{0}^{1} u_{0} \sin \left(\frac{2 n-1}{2}\right) \pi y d y}{\int_{0}^{1} \sin ^{2}\left(\frac{2 n-1}{2}\right) \pi y d y}=\frac{4 u_{0}}{(2 n-1) \pi} \\] Thus \\[ u(x, y)=\frac{4 u_{0}}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2 n-1) \cosh \left(\frac{2 n-1}{2}\right) \pi} \cosh \left(\frac{2 n-1}{2}\right) \pi x \sin \left(\frac{2 n-1}{2}\right) \pi y \\]

$$\begin{array}{l} a^{2} \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}}, \quad 00 \\ u(0, t)=0, \quad u(L, t)=0, \quad t>0 \\ u(x, 0)=x(L-x),\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=0, \quad 0

Using \(u=X T\) and \(-\lambda\) as a separation constant we obtain \\[ \begin{array}{c} X^{\prime \prime}+\lambda X=0 \\ X^{\prime}(0)=0 \\ X^{\prime}(L)=0 \end{array} \\] and \\[ \begin{array}{c} T^{\prime \prime}+\lambda a^{2} T=0 \\ T^{\prime}(0)=0 \end{array} \\] Solving the differential equations we get $$X=c_{1} \sin \frac{n \pi}{L} x+c_{2} \cos \frac{n \pi}{L} x \quad \text { and } \quad T=c_{3} \cos \frac{n \pi a}{L} t+c_{4} \sin \frac{n \pi a}{L} t$$ for \(n=1,2,3, \ldots .\) The boundary and initial conditions, together with the fact that \(\lambda=0\) is an eigenvalue with eigenfunction \(X(x)=1,\) give $$u=A_{0}+\sum_{n=1}^{\infty} A_{n} \cos \frac{n \pi a}{L} t \sin \frac{n \pi}{L} x$$ Imposing \\[ u(x, 0)=x=A_{0}+\sum_{n=1}^{\infty} A_{n} \cos \frac{n \pi}{L} x \\] gives \\[ A_{0}=\frac{1}{L} \int_{0}^{L} x d x=\frac{L}{2} \\] and \\[ A_{n}=\frac{2}{L} \int_{0}^{L} x \cos \frac{n \pi}{L} x d x=\frac{2 L}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \\] for \(n=1,2,3, \dots,\) so that \\[ u(x, t)=\frac{L}{2}+\frac{2 L}{\pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n}-1}{n^{2}} \cos \frac{n \pi a}{L} t \cos \frac{n \pi}{L} x \\]

$$\begin{array}{l} a^{2} \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}}, \quad 00 \\ u(0, t)=0, \quad u(L, t)=0, \quad t>0 \\ u(x, 0)=0,\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=\sin \frac{\pi x}{L}, \quad 0

Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(y X^{\prime} Y^{\prime}+X Y=0 .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime}}{X}=-\frac{Y}{y Y^{\prime}}=-\lambda.$$ When \(\lambda \neq 0\) $$X^{\prime}+\lambda X=0 \quad \text { and } \quad \lambda y Y^{\prime}-Y=0$$ so that $$X=c_{1} e^{-\lambda x} \quad \text { and } \quad Y=c_{2} y^{1 / \lambda}.$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{-\lambda x} y^{1 / \lambda}.$$ In this case \(\lambda=0\) yields no solution.

$$\begin{array}{l} a^{2} \frac{\partial^{2} u}{\partial x^{2}}-2 \beta \frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial t^{2}}, \quad 00 \\ u(0, t)=0, \quad u(L, t)=\sin \pi t, \quad t>0 \\ u(x, 0)=f(x),\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=0, \quad 0

Substituting \(u(x, t)=X(x) T(t)\) into the partial differential equation yields \(k X^{\prime \prime} T-X T=X T^{\prime} .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{k X^{\prime \prime}-X}{X}=\frac{T^{\prime}}{T}=-\lambda.$$ Then $$X^{\prime \prime}+\frac{\lambda-1}{k} X=0 \quad \text { and } \quad T^{\prime}+\lambda T=0.$$ The second differential equation implies \(T(t)=c_{1} e^{-\lambda t}\). For the first differential equation we consider three cases: $$\begin{array}{l} \text { I. If }(\lambda-1) / k=0 \text { then } \lambda=1, X^{\prime \prime}=0 \text { , and } X(x)=c_{2} x+c_{3} \text { , so } \\ \qquad u=X T=e^{-t}\left(A_{1} x+A_{2}\right) \end{array}$$

$$\begin{aligned} &a^{2} \frac{\partial^{2} u}{\partial x^{2}}+A x=\frac{\partial^{2} u}{\partial t^{2}}, \quad 00, A \text { a constant }\\\ &\begin{array}{l} u(0, t)=0, \quad u(L, t)=0, \quad t>0 \\ u(x, 0)=0,\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=0, \quad 0

Substituting \(u(x, t)=X(x) T(t)\) into the partial differential equation yields \(k X^{\prime \prime} T=X T^{\prime} .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime \prime}}{X}=\frac{T^{\prime}}{k T}=-\lambda.$$ Then $$X^{\prime \prime}+\lambda X=0 \quad \text { and } \quad T^{\prime}+\lambda k T=0.$$ The second differential equation implies \(T(t)=c_{1} e^{-\lambda k t}\). For the first differential equation we consider three cases: I. If \(\lambda=0\) then \(X^{\prime \prime}=0\) and \(X(x)=c_{2} x+c_{3},\) so $$u=X T=A_{1} x+A_{2}.$$ II. If \(\lambda=-\alpha^{2}<0,\) then \(X^{\prime \prime}-\alpha^{2} X=0,\) and \(X(x)=c_{4} \cosh \alpha x+c_{5} \sinh \alpha x,\) so $$u=X T=\left(A_{3} \cosh \alpha x+A_{4} \sinh \alpha x\right) e^{k \alpha^{2} t}.$$ III. If \(\lambda=\alpha^{2}>0,\) then \(X^{\prime \prime}+\alpha^{2} X=0,\) and \(X(x)=c_{6} \cos \alpha x+c_{7} \sin \alpha x,\) so $$u=X T=\left(A_{3} \cos \alpha x+A_{6} \sin \alpha x\right) e^{-k \alpha^{2} t}.$$

Using \(u=X Y\) and \(-\lambda\) as a separation constant we obtain $$\begin{array}{c}X^{\prime \prime}+\lambda X=0, \\\X(0)=0, \\\X(\pi)=0,\end{array}$$ and $$Y^{\prime \prime}-\lambda Y=0.$$ With \(\lambda=\alpha^{2}>0\) the solutions of the differential equations are $$X=c_{1} \cos \alpha x+c_{2} \sin \alpha x \quad \text { and } \quad Y=c_{3} e^{\alpha y}+c_{4} e^{-\alpha y}$$ Then the boundedness of \(u\) as \(y \rightarrow \infty\) implies \(c_{3}=0,\) so \(Y=c_{4} e^{-n y} .\) The boundary conditions at \(x=0\) and \(x=\pi\) imply \(c_{1}=0\) so \(X=c_{2} \sin n x\) for \(n=1,2,3, \ldots\) and $$u=\sum_{n=1}^{\infty} A_{n} e^{-n y} \sin n x.$$ Imposing $$u(x, 0)=f(x)=\sum_{n=1}^{\infty} A_{n} \sin n x$$ gives $$A_{n}=\frac{2}{\pi} \int_{0}^{\pi} f(x) \sin n x d x$$ so that $$u(x, y)=\sum_{n=1}^{\infty}\left(\frac{2}{\pi} \int_{0}^{\pi} f(x) \sin n x d x\right) e^{-n y} \sin n x.$$

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)] \\]

$$\begin{array}{l}\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0

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 $$u=X T=\left(c_{9} \cos \alpha x+c_{10} \sin \alpha x\right)\left(c_{11} \cos \alpha a t+c_{12} \sin \alpha a t\right).$$

Using \(u=X Y\) and \(-\lambda\) as a separation constant we obtain $$\begin{array}{c}X^{\prime \prime}+\lambda X=0, \\\X^{\prime}(0)=0, \\\X^{\prime}(\pi)=0,\end{array}$$ and $$Y^{\prime \prime}-\lambda Y=0.$$ With \(\lambda=\alpha^{2}>0\) the solutions of the differential equations are $$X=c_{1} \cos \alpha x+c_{2} \sin \alpha x \quad \text { and } \quad Y=c_{3} e^{\alpha y}+c_{4} e^{-\alpha y}$$ The boundary conditions at \(x=0\) and \(x=\pi\) imply \(c_{2}=0\) so \(X=c_{1} \cos n x\) for \(n=1,2,3, \ldots .\) Now the boundedness of \(u\) as \(y \rightarrow \infty\) implies \(c_{3}=0,\) so \(Y=c_{4} e^{-n y} .\) In this problem \(\lambda=0\) is also an eigenvalue with corresponding eigenfunction 1 so that $$u=A_{0}+\sum_{n=1}^{\infty} A_{n} e^{-n y} \cos n x.$$ Imposing $$u(x, 0)=f(x)=A_{0}+\sum_{n=1}^{\infty} A_{n} \cos n x$$ gives $$A_{0}=\frac{1}{\pi} \int_{0}^{\pi} f(x) d x \quad \text { and } \quad A_{n}=\frac{2}{\pi} \int_{0}^{\pi} f(x) \cos n x d x$$ so that $$u(x, y)=\frac{1}{\pi} \int_{0}^{\pi} f(x) d x+\sum_{n=1}^{\infty}\left(\frac{2}{\pi} \int_{0}^{\pi} f(x) \cos n x d x\right) e^{-n y} \cos n x.$$

Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(x^{2} X^{\prime \prime} Y+X Y^{\prime \prime}=0 .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$-\frac{x^{2} X^{\prime \prime}}{X}=\frac{Y^{\prime \prime}}{Y}=-\lambda.$$ Then $$x^{2} X^{\prime \prime}-\lambda X=0 \quad \text { and } \quad Y^{\prime \prime}+\lambda Y=0.$$ We consider three cases: I. If \(\lambda=0\) then \(x^{2} X^{\prime \prime}=0\) and \(X(x)=c_{1} x+c_{2} .\) Also, \(Y^{\prime \prime}=0\) and \(Y(y)=c_{3} y+c_{4}\) so $$u=X Y=\left(c_{1} x+c_{2}\right)\left(c_{3} y+c_{4}\right).$$ II. If \(\lambda=-\alpha^{2}<0\) then \(x^{2} X^{\prime \prime}+\alpha^{2} X=0\) and \(Y^{\prime \prime}-\alpha^{2} Y=0 .\) The solution of the second differential equation is \(Y(y)=c_{5} \cosh \alpha y+c_{6} \sinh \alpha y .\) The first equation is Cauchy-Euler with auxiliary equation \(m^{2}-m+\alpha^{2}=0 .\) Solving for \(m\) we obtain \(m=\frac{1}{2} \pm \frac{1}{2} \sqrt{1-4 \alpha^{2}} . \quad\) We consider three possibilities for the discriminant \(1-4 \alpha^{2}\) (i) If \(1-4 \alpha^{2}=0\) then \(X(x)=c_{7} x^{1 / 2}+c_{8} x^{1 / 2} \ln x\) and \\[ u=X Y=x^{1 / 2}\left(c_{7}+c_{8} \ln x\right)\left(c_{5} \cosh \alpha y+c_{6} \sinh \alpha y\right) \\] (ii) If \(1-4 \alpha^{2}<0\) then \(X(x)=x^{1 / 2}\left[c_{9} \cos (\sqrt{4 \alpha^{2}-1} \ln x)+c_{10} \sin (\sqrt{4 \alpha^{2}-1} \ln x)\right]\) and \\[ u=X Y=x^{1 / 2}\left[c_{9} \cos (\sqrt{4 \alpha^{2}-1} \ln x)+c_{10} \sin (\sqrt{4 \alpha^{2}-1} \ln x)\right]\left(c_{5} \cosh \alpha y+c_{6} \sinh \alpha y\right) \\] (iii) If \(1-4 \alpha^{2}>0\) then \(X(x)=x^{1 / 2}\left(c_{11} x^{\sqrt{1-4 \alpha^{2}} / 2}+c_{12} x^{-\sqrt{1-4 \alpha^{2}} / 2}\right)\) and \\[ u=X Y=x^{1 / 2}\left(c_{11} x^{\sqrt{1-4 \alpha^{2} / 2}}+c_{12} x^{-\sqrt{1-4 \alpha^{2} / 2}}\right)\left(c_{5} \cosh \alpha y+c_{6} \sinh \alpha y\right) \\] III. If \(\lambda=\alpha^{2}>0\) then \(x^{2} X^{\prime \prime}-\alpha^{2} X=0\) and \(Y^{\prime \prime}+\alpha^{2} Y=0 .\) The solution of the second differential equation is \(Y(y)=c_{13} \cos \alpha y+c_{14} \sin \alpha y .\) The first equation is Cauchy-Euler with auxiliary equation \(m^{2}-m-\alpha^{2}=0\) Solving for \(m\) we obtain \(m=\frac{1}{2} \pm \frac{1}{2} \sqrt{1+4 \alpha^{2}}\). In this case the discriminant is always positive so the solution of the differential equation is \(X(x)=x^{1 / 2}\left(c_{15} x^{\sqrt{1+4 \alpha^{2}} / 2}+c_{16} x^{-\sqrt{1+4 \alpha^{2} / 2}}\right)\) and \\[ u=X Y=x^{1 / 2}\left(c_{15} x^{\sqrt{1+4 \alpha^{2}} / 2}+c_{16} x^{-\sqrt{1+4 \alpha^{2}} / 2}\right)\left(c_{13} \cos \alpha y+c_{14} \sin \alpha y\right) \\]

Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(X^{\prime \prime} Y+X Y^{\prime \prime}=X Y .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime \prime}}{X}=\frac{Y-Y^{\prime \prime}}{Y}=-\lambda.$$ Then $$X^{\prime \prime}+\lambda X=0 \quad \text { and } \quad Y^{\prime \prime}-(1+\lambda) Y=0.$$ We consider three cases: I. If \(\lambda=0\) then \(X^{\prime \prime}=0\) and \(X(x)=c_{1} x+c_{2}\). Also \(Y^{\prime \prime}-Y=0\) and \(Y(y)=c_{3} \cosh y+c_{4} \sinh y\) so \\[ u=X Y=\left(c_{1} x+c_{2}\right)\left(c_{3} \cosh y+c_{4} \sinh y\right) \\] II. If \(\lambda=-\alpha^{2}<0\) then \(X^{\prime \prime}-\alpha^{2} X=0\) and \(Y^{\prime \prime}+\left(\alpha^{2}-1\right) Y=0 .\) The solution of the first differential equation is \(X(x)=c_{5} \cosh \alpha x+c_{6} \sinh \alpha x .\) The solution of the second differential equation depends on the nature of \(\alpha^{2}-1 .\) We consider three cases: (i) If \(\alpha^{2}-1=0,\) or \(\alpha^{2}=1,\) then \(Y(y)=c_{7} y+c_{8}\) nad \\[ u=X Y=\left(c_{5} \cosh \alpha x+c_{6} \sinh \alpha x\right)\left(c_{7} y+c_{8}\right) \\] (ii) If \(\alpha^{2}-1<0\), or \(0<\alpha^{2}<1\), then \(Y(y)=c_{9} \cosh \sqrt{1-\alpha^{2}} y+c_{10} \sinh \sqrt{1-\alpha^{2}} y\) and \\[ u=X Y=\left(c_{5} \cosh \alpha x+c_{6} \sinh \alpha x\right)\left(c_{9} \cosh \sqrt{1-\alpha^{2}} y+c_{10} \sinh \sqrt{1-\alpha^{2}} y\right) \\] (iii) If \(\alpha^{2}-1>0,\) or \(\alpha^{2}>1,\) then \(Y(y)=c_{11} \cos \sqrt{\alpha^{2}-1} y+c_{12} \sin \sqrt{\alpha^{2}-1} y\) and \\[ u=X Y=\left(c_{5} \cosh \alpha x+c_{6} \sinh \alpha x\right)\left(c_{11} \cos \sqrt{\alpha^{2}-1} y+c_{12} \sin \sqrt{\alpha^{2}-1} y\right) \\] III. If \(\lambda=\alpha^{2}>0,\) then \(X^{\prime \prime}+\alpha^{2} X=0\) and \(X(x)=c_{13} \cos \alpha x+c_{14} \sin \alpha x .\) Also, $$\begin{array}{c} Y^{\prime \prime}-\left(1+\alpha^{2}\right) Y=0 \text { and } Y(y)=c_{15} \cosh \sqrt{1+\alpha^{2}} y+c_{16} \sinh \sqrt{1+\alpha^{2}} y \mathrm{so} \\ u=X Y=\left(c_{13} \cos \alpha x+c_{14} \sin \alpha x\right)\left(c_{15} \cosh \sqrt{1+\alpha^{2}} y+c_{16} \sinh \sqrt{1+\alpha^{2}} y\right) \end{array}$$

Referring to the discussion in this section of the text we identify \(a=b=2, f(x)=0,\) $$g(x)=\left\\{\begin{array}{ll} x, & 0

Substituting \(u(x, t)=X(x) T(t)\) into the partial differential equation yields \(a^{2} X^{\prime \prime} T-g=X T^{\prime \prime},\) which is not separable.

Identifying \(A=B=C=1,\) we compute \(B^{2}-4 A C=-3 < 0 .\) The equation is elliptic.

Identifying \(A=3, B=5,\) and \(C=1,\) we compute \(B^{2}-4 A C=13 > 0 .\) The equation is hyperbolic.

Identifying \(A=1, B=6,\) and \(C=9,\) we compute \(B^{2}-4 A C=0 .\) The equation is parabolic.

Identifying \(A=1, B=2,\) and \(C=1,\) we compute \(B^{2}-4 A C=0 .\) The equation is parabolic.

Identifying \(A=k > 0, B=0,\) and \(C=0,\) we compute \(B^{2}-4 A C=-4 k < 0 .\) The equation is elliptic.

Substituting \(u(r, t)=R(r) T(t)\) into the partial differential equation yields $$k\left(R^{\prime \prime} T+\frac{1}{r} R^{\prime} T\right)=R T^{\prime}.$$ Separating variables and using the separation constant \(-\lambda\) we obtain. $$\frac{r R^{\prime \prime}+R^{\prime}}{r R}=\frac{T^{\prime}}{k T}=-\lambda.$$ Then $$r R^{\prime \prime}+R^{\prime}+\lambda r R=0 \quad \text { and } \quad T^{\prime}+\lambda k T=0.$$ Letting \(\lambda=\alpha^{2}\) and writing the first equation as \(r^{2} R^{\prime \prime}+r R^{\prime}=\alpha^{2} r^{2} R=0\) we see that it is a parametric Bessel equation of order \(0 .\) As discussed in Chapter 5 of the text, it has solution \(R(r)=c_{1} J_{0}(\alpha r)+c_{2} Y_{0}(\alpha r) .\) since a solution of \(T^{\prime}+\alpha^{2} k T\) is \(T(t)=e^{-k \alpha^{2} t},\) we see that a solution of the partial differential equation is $$u=R T=e^{-k \alpha^{2} t}\left[c_{1} J_{0}(\alpha r)+c_{2} Y_{0}(\alpha r)\right].$$

We identify \(A=x y+1, B=x+2 y,\) and \(C=1 .\) Then \(B^{2}-4 A C=x^{2}+4 y^{2}-4 .\) The equation \(x^{2}+4 y^{2}=4\) defines an ellipse. The partial differential equation is hyperbolic outside the ellipse, parabolic on the ellipse, and elliptic inside the ellipse.

Assuming \(u(x, y)=X(x) Y(y)\) and substituting into \(\partial^{2} u / \partial x^{2}-u=0\) we get \(X^{\prime \prime} Y-X Y=0\) or \(Y\left(X^{\prime \prime}-X\right)=0\) This implies \(X(x)=c_{1} e^{x}\) or \(X(x)=c_{2} e^{-x}\). For these choices of \(X, Y\) can be any function of \(y .\) Two solutions of the partial differential cquation are then $$u_{1}(x, y)=A(y) e^{x} \quad \text { and } \quad u_{2}(x, y)=H(y) e^{x}.$$ since the partial differential equation is linear and homogencous the superposition principle indicates that another solution is $$u(x, y)=u_{1}(x, y)+u_{2}(x, y)=A(y) e^{x}+B(y) e^{-x}.$$