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# ch5 - Chapter 5 Some boundary value problems In this...

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Unformatted text preview: Chapter 5: Some boundary value problems In this chapter we apply our results and techniques to some boundary value problems. 5.1. Sturm–Liouville problem with periodic boundary conditions Assume l > 0, and consider the Sturm-Liouville problem with periodic bound- ary conditions ( P 1 ) f 00 + λf = 0 on (- l, l ) , f (- l ) = f ( l ) , f (- l ) = f ( l ) . We wish to find all eigenvalues and eigenfunctions for this problem. Below, let a, b ∈ IR be arbitrary real numbers. If λ = 0, then the general solution of the equation is f = ax + b . From the first boundary condition we have that- al + b = al + b , which implies a = 0. Hence f ( x ) = b , ( b 6 = 0), is an eigenfunction with corresponding eigenvalue λ = 0. Next assume that λ < 0. Setting μ := √- λ , the general solution of the equation is f ( x ) = a cosh μx + b sinh μx . Then f ( x ) = aμ sinh μx + bμ cosh μx . The boundary conditions give a cosh μl- b sinh μl = a cosh μl + b sinh μl , and- a sinh μl + b cosh μl = a sinh μl + b cosh μl . The only solution of this linear system is a = b = 0, so that the problem does not have negative eigenvalues. Finally assume that λ > 0. Setting ν := √ λ , the general solution of the equation is f ( x ) = a cos νx + b sin νx . Then f ( x ) =- aν sin νx + bν cos νx , and the boundary conditions give a cos νl- b sin νl = a cos νl + b sin νl , and a sin νl + b cos νl =- a sin νl + b cos νl . Hence a sin νl = b sin νl = 0, which implies that νl = nπ , ( n = 1 , 2 , . . . , ). In other words, λ = ( n 2 π 2 /l 2 ) is an eigenvalue, with corresponding eigenfunction f ( x ) = a cos( nπx/l ) + b sin( nπx/l ), ( n = 1 , 2 , . . . , ). Notice that, in accordance with Theorem 4.3, the eigenspace for each positive eigenvalue is two . 34 5.2. Dirichlet problem in a rectangle Let l > 0, L > 0, D = [0 , l ] × [0 , L ] = { ( x, y ) : 0 ≤ x ≤ l, ≤ y ≤ L } , and let f 1 , f 2 ∈ C ([0 , l ]), g 1 , g 2 ∈ C ([0 , L ]) be given functions. We are looking for the solution of the following Dirichlet problem for Laplace’s equation in the rectangle D , ( P 2 ) Δ u = u xx + u yy = 0 , (1) u ( x, 0) = f 1 ( x ) , u ( x, L ) = f 2 ( x ) , u (0 , y ) = g 1 ( y ) , u ( l, y ) = g 2 ( y ) . Below we only consider the special case g 1 = g 2 = 0, and we leave the general case to the reader. In the first step we look for a nontrivial separated solution , u ( x, y ) = X ( x ) Y ( y ), which satisfies the equation, together with the boundary conditions u (0 , y ) = u ( l, y ) = 0. From (1) we find X 00 y + XY 00 = 0, and X (0) Y ( y ) = X ( l ) Y ( y ) = 0. Since we only consider solutions which do not vanish ev- erywhere, this implies X 00 + λX = 0, Y 00- λY = 0, for some λ ∈ IR, and X (0) = X ( l ) = 0. The last boundary condition implies that the only solu- tions are given by X = sin νx , Y = a cosh νy + b sinh νy , where λ = ν 2 , and ν = ( nπ/l ), ( n = 1 , 2 , . . . , ). By the superposition principle this implies...
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ch5 - Chapter 5 Some boundary value problems In this...

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