Solution Found!
Use the results of Exercise 15 to approximate the solution to u t 2u x2 = 2, 0 < x < 1
Chapter 12, Problem 16(choose chapter or problem)
Use the results of Exercise 15 to approximate the solution to
\(\begin{aligned}
\frac{\partial u}{\partial t}-\frac{\partial^{2} u}{\partial x^{2}} & =2, \quad 0<x<1,0<t ; \\
u(0, t) & =u(1, t)=0, \quad 0<t ; \\
u(x, 0) & =\sin \pi x+x(1-x),
\end{aligned}\)
with h = 0.1 and k = 0.01. Compare your answer at t = 0.25 to the actual solution u(x, t) = \(e^{-\pi^{2} t}\) sin \(\pi x+x(1-x)\).
Questions & Answers
QUESTION:
Use the results of Exercise 15 to approximate the solution to
\(\begin{aligned}
\frac{\partial u}{\partial t}-\frac{\partial^{2} u}{\partial x^{2}} & =2, \quad 0<x<1,0<t ; \\
u(0, t) & =u(1, t)=0, \quad 0<t ; \\
u(x, 0) & =\sin \pi x+x(1-x),
\end{aligned}\)
with h = 0.1 and k = 0.01. Compare your answer at t = 0.25 to the actual solution u(x, t) = \(e^{-\pi^{2} t}\) sin \(\pi x+x(1-x)\).
ANSWER:Step 1 of 4
Referring to the exercise 15, the modification of the algorithm described in 12.2 for the parabolic partial differential equation:
\(\begin{array}{c}
\frac{\partial u}{\partial t}-\frac{\partial^{2} u}{\partial x^{2}}=F(x), 0<x<I, t>0 \\
u(0, t)=u(1, t)=0, t>0 \\
u(x, 0)=f(x), 0<x<1
\end{array}\)
is given by
\(z_{i}=\frac{\left[w_{i}+k F(i h)+\lambda z_{i-1}\right]}{l_{i}}, i=2, \ldots, m-1\)
The modification of the algorithm described in 12.3.for the above problem is expressed by
\(z_{i}=\frac{\left[(1-\lambda) w_{i}+\frac{\lambda}{2}\left(w_{i+1}+w_{i-1}+z_{i-1}\right)+k F(i h)\right]}{l_{i}}, i=1,2, \ldots, m-1\)