1. Reduce to two ordinary differential equations, one an eigenvalue problem, the other having one initial condition, and find the particular solutions: (a) u= 0 for 00, 1 ar ax (x, 0)% 0, u (0, г) - и(1, 1)-0. at
1. Reduce to two ordinary differential equations, one an eigenvalue problem, the other having one initial condition, and find the particular solutions: (a) u= 0 for 00, 1 ar ax (x, 0)% 0, u (0, г) - и(1, 1)-0. at
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![1. Reduce to two ordinary differential equations, one an eigenvalue problem, the other having one initial condition, and find the particular solutions:
(a)
\[
\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - u = 0 \quad \text{for} \quad 0 < x < 1, \, t > 0,
\]
\[
\frac{\partial u}{\partial t}(x, 0) = 0,
\]
\[
u(0, t) = u(1, t) = 0.
\]
(b)
\[
\frac{\partial^2 u}{\partial t^2} + 2 \frac{\partial u}{\partial t} - 4 \frac{\partial^2 u}{\partial x^2} + u = 0 \quad \text{for} \quad 0 < x < 1, \, t > 0,
\]
\[
u(x, 0) = 0,
\]
\[
\frac{\partial u}{\partial x}(0, t) = u(1, t) = 0.
\]
(c)
\[
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial u}{\partial y} - u = 0 \quad \text{for} \quad a < x < b, \, 0 < y < 1,
\]
\[
u(a, y) = 0,
\]
\[
u(b, y) = 0,
\]
\[
u(x, 0) = u(x, 1) = 0.
\]
(d)
\[
\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = u = 0 \quad \text{for} \quad 0 < x < 1, \, t > 0,
\]
\[
u(0, t) = 0,
\]
\[
u(1, t) = 0.
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a00eb18-131f-41a2-8922-58a31d7db03b%2F432ee305-c4cf-49eb-9e5c-cc30d46c9245%2Foos7pa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. Reduce to two ordinary differential equations, one an eigenvalue problem, the other having one initial condition, and find the particular solutions:
(a)
\[
\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} - u = 0 \quad \text{for} \quad 0 < x < 1, \, t > 0,
\]
\[
\frac{\partial u}{\partial t}(x, 0) = 0,
\]
\[
u(0, t) = u(1, t) = 0.
\]
(b)
\[
\frac{\partial^2 u}{\partial t^2} + 2 \frac{\partial u}{\partial t} - 4 \frac{\partial^2 u}{\partial x^2} + u = 0 \quad \text{for} \quad 0 < x < 1, \, t > 0,
\]
\[
u(x, 0) = 0,
\]
\[
\frac{\partial u}{\partial x}(0, t) = u(1, t) = 0.
\]
(c)
\[
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial u}{\partial y} - u = 0 \quad \text{for} \quad a < x < b, \, 0 < y < 1,
\]
\[
u(a, y) = 0,
\]
\[
u(b, y) = 0,
\]
\[
u(x, 0) = u(x, 1) = 0.
\]
(d)
\[
\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = u = 0 \quad \text{for} \quad 0 < x < 1, \, t > 0,
\]
\[
u(0, t) = 0,
\]
\[
u(1, t) = 0.
\]
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