Given the boundary-value problem

2ux2=2ut2,0<x<1,0<t<1u(0,t)=0,u(1,t)=0,0t1\begin{array}{ll}{\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}},} & {0<x<1, \quad 0<t<1} \\ {u(0, t)=0,} & {u(1, t)=0, \quad 0 \leq t \leq 1}\end{array}

u(x,0)=x(1x),utt=0=0,0x1u(x, 0)=x(1-x),\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=0, \quad 0 \leq x \leq 1 use h=k=15h=k=\frac{1}{5} in equation (6) to compute the values of ui,1u_{i, 1} by hand.


Answered 2 years ago
Answered 2 years ago
Step 1
1 of 2

The given problem is


u(0,t)=0;u(1,t)=0u(0,t)=0; u(1,t)=0

u(x,0)=x(1x),ut(x,0)=0u(x,0)=x(1-x), u_t(x,0)=0

Now from equation (6), we have

ui,1=0.5(ui+1,0+ui1,0)u_{i,1}=0.5\cdot (u_{i+1,0}+u_{i-1,0})

Here g(xi)=0g(x_i)=0 Now,

u1,1=0.5(u2,0+u0,0)=0.08u_{1,1}=0.5\cdot (u_{2,0}+u_{0,0})=0.08

u2,1=0.5(u3,0+u1,0)=0.12u_{2,1}=0.5\cdot (u_{3,0}+u_{1,0})=0.12

u3,1=0.5(u4,0+u2,0)=0.20u_{3,1}=0.5\cdot (u_{4,0}+u_{2,0})=0.20

u4,1=0.5(u5,0+u3,0)=0.20u_{4,1}=0.5\cdot (u_{5,0}+u_{3,0})=0.20

\cdots \cdots so on.. here we use the initial condition u(x,0)=x(1x).u(x,0)=x(1-x).

Create an account to view solutions

Create an account to view solutions

Recommended textbook solutions

Advanced Engineering Mathematics 10th Edition by Erwin Kreyszig

Advanced Engineering Mathematics

10th EditionISBN: 9780470458365 (3 more)Erwin Kreyszig
4,134 solutions
Advanced Engineering Mathematics 9th Edition by Erwin Kreyszig

Advanced Engineering Mathematics

9th EditionISBN: 9780471488859 (1 more)Erwin Kreyszig
4,201 solutions
Advanced Engineering Mathematics 7th Edition by Peter V. O'Neil

Advanced Engineering Mathematics

7th EditionISBN: 9781111427412 (4 more)Peter V. O'Neil
1,608 solutions
Advanced Engineering Mathematics 6th Edition by Dennis G. Zill

Advanced Engineering Mathematics

6th EditionISBN: 9781284105902 (7 more)Dennis G. Zill
5,294 solutions

More related questions