{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Week_02

# Week_02 - ACM 100c Methods of Applied Mathematics Week 2...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ACM 100c - Methods of Applied Mathematics Week 2 April 4, 2009 1 Solving the initial boundary value problem We can use the solutions generated by separation of variables to solve boundary value problems of the form ∂T ∂t = k ∂ 2 T ∂x 2 < x < L with boundary conditions T (0 ,t ) = 0 , T ( L,t ) = 0 , and with initial conditions T ( x, 0) = f ( x ) . Note we previously got solutions of the form u ( x,t ) = exp(- kλ 2 t ) sin( λx ) u ( x,t ) = exp(- kλ 2 t ) cos( λx ) . We would like to satisfy the boundary conditions at x = 0 and x = L . The cosine form of the solution cannot do this, but the sine form will at least satisfy the condition at x = 0 . To satisfy the condition at x = L we can pick the value of λ . In fact this is just the Sturm- Liouville eigenvalue problem we encountered last term. By setting λ = nπ L , n = 1 , 2 , 3 , ··· , we see that the solutions u n ( x,t ) = A n exp- k ( n 2 π 2 /L 2 ) t sin( nπx/L ) satisfy both boundary conditions. However these solutions u n , while satisfying the bound- ary conditions, satisfy the initial condition u n ( x, 0) = A n sin( nπ/L ) . 1 In order to solve the initial value problem with u ( x, 0) = f ( x ) we note that we can actually use any superposition of the solutions u n ( x,t ) : u ( x,t ) = ∞ X n =1 A n exp- n 2 π 2 kt L 2 sin( nπx/L ) , which corresponds to an initial condition given by u ( x, 0) = ∞ X n =1 A n sin( nπx/L ) . We recognize here the Fourier sine series over the interval < x < L and we know from ACM 95b that such a Fourier series is complete and can be used to represent any square integrable function over the interval < x < L . So setting f ( x ) = ∞ X n =0 A n sin( nπx/L ) , we can use the orthogonality of the Fourier sine series to solve for A n : A n = 2 L Z L f ( x ) sin( nπx/L ) dx. This series now satisfies the boundary conditions and the initial conditions. We have therefore derived a series solution to the heat equation of the form u ( x,t ) = ∞ X n =1 A n exp- k ( n 2 π 2 /L 2 ) t sin( nπx/L ) < x < L, t > . One question that immediately arises is whether this series makes any sense. In other words, does it converge? We recall that while the Fourier series coefficients for a generic function f ( x ) exist as long as f ( x ) is integrable, meaning that Z L | f ( x ) | dx exists, the series coefficients A n can decay with n very slowly leading to nonuniform convergence of the series. However, we see that even if this is the case for a given f ( x ) , as soon as t > the argument of the exponential terms in the series terms in the series u ( x,t ) = ∞ X n =1 A n exp- k ( n 2 π 2 /L 2 ) t sin( nπx/L ) becomes nonzero and the series coefficients decay as A n exp- k ( n 2 π 2 /L 2 ) t 2 as n → ∞ . This implies the resulting series converges uniformly and rapidly and all the derivatives of u also converge uniformly and rapidly. This is actually a generic feature of the heat equation for these boundary conditions. The smoothness of the solution isof the heat equation for these boundary conditions....
View Full Document

{[ snackBarMessage ]}

### Page1 / 16

Week_02 - ACM 100c Methods of Applied Mathematics Week 2...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online