Partial Differential Equations


 Clarence Riley
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1 Prtil Differentil Equtions Notes by Robert Piché, Tmpere University of Technology reen s Functions. reen s Function for OneDimensionl Eqution The reen s function provides complete solution to boundry vlue problem in much the sme wy tht n inverse mtri provides generl solution for systems of liner equtions. In this section the reen s function is introduced in the contet of simple onedimensionl problem. Some of the proofs use the identity b (uv vu ) d b uv vu. This cn be obtined by integrting (uv vu ) uv vu. A singulrity function K(, ξ) of the opertor L defined by Lu() u () c()u() is chrcterised by three properties:. K is continuous; 2. K is continuous in < ξ nd in > ξ, nd K ( +, ) K (, ) ; 3. K is continuous nd LK for ξ. Note tht the three properties do not define singulrity function uniquely: if K is singulrity function then so is K + H, where H(, ξ) is ny function with continuous H nd H nd with LH. The reen s function (, ξ) for the opertor L nd the domin (, b) with Dirichlet boundry conditions is the singulrity function tht stisfies the homogeneous Dirichlet conditions (, ξ) nd (b, ξ). The reen s function provides the solution to the boundry vlue problem with Dirichlet boundry conditions: Theorem If u stisfies the differentil eqution u + cu f on (, b) nd the boundry conditions u(), u(b), then for ll ξ (, b). u(ξ) b (, ξ)f() d () Proof. Letting v() (, ξ), we hve b vf d ξ b uv vu d + uv vu d ξ + ξ uv vu + b ξ + uv vu b uv vu u(ξ) ξ+ ξ v + ξ+ ξ vu, which completes the proof.
2 The lod (or source) function f in the differentil eqution u cu f cn be thought of s superposition of point lods f() δ( ξ)f(ξ)dξ, where δ( ξ) is concentrted t ξ nd hs unit mgnitude (i.e. δ d ). Then formul () represents the solution s weighted sum of reen s functions, where ech (ξ, ) is the solution to u () δ( ξ). The reen s function cn thus be thought of s the response to unit point lod. The reen s function lso provides the solution of the boundry vlue problem with nonhomogeneous boundry conditions: Theorem 2 The solution of u +cu with boundry conditions u() h nd u(b) h stisfies u(ξ) (, ξ)h (b, ξ)h (2) for ll ξ (, b). Proof. Denoting v() (, ξ), we hve ξ which completes the proof. b (uv vu ) d + (uv vu ) d ξ + ξ (uv vu ) + b ξ + (uv vu ) b (uv vu ) ξ+ ξ (uv vu ) b uv + u(ξ), The following result tells us tht the response t to point lod pplied t ξ is equl to the response t ξ to point lod pplied t. Theorem 3 (Reciprocity Principle) (, ξ) (ξ, ) for ll, ξ (, b). Proof. Let y nd η be distinct points in (, b) with y < η, let u() (, η) nd v() (, y). Then y y b uv vu η b uv vu d + uv vu d + uv vu d y + η + uv vu + η y + uv vu + b η + uv vu y+ y uv vu η+ η uv vu, u(y) v(η) leving u(y) v(η), tht is, (y, η) (η, y). The proof for y > η is similr. As consequence of Theorem 2, we cn rewrite formul () s nd formul (2) s u() b (, ξ)f(ξ) dξ. u() ξ (, )h ξ (, b)h. A singulrity function of d2 is given by d 2 K(, ξ) ξ, 2 s cn redily be verified. The reen s function for the intervl (, ) cn be 2
3 found by solving H with boundry conditions H(, ξ) K(, ξ) nd H(, ξ) K(, ξ), then setting K + H. This yields.25 (, ξ) ξ + ( + ξ) ξ 2 2 { ( )ξ for ξ < ( ξ) for < ξ. The solution of u f stisfying u() h nd u() h is given by ξ u() ( ) (, ξ)f(ξ) dξ + ξ (, )h ξ (, )h ξf(ξ) dξ + ( ξ)f(ξ) dξ + ( )h + h..2 reen s Function for TwoDimensionl Poisson Eqution Now we go through the sme discussion in two dimensions. Some of the proofs use reen s second identity u v v u da (u v v u) n dl, Ω This cn be derived by writing reen s first identity twice, with u nd v interchnged the second time, nd subtrcting. A singulrity function K(, ) of the opertor is chrcterised by the three properties. For ny fied, lim disk centred t ; ɛ 2. For ny fied, lim ɛ outwrd unit norml to B ɛ; 3. K is hrmonic s function of for. K(, ) dl, where B ɛ denotes the rdiusɛ K(, ) n dl, where n denotes the Note tht these three properties do not define singulrity function uniquely: if K is singulrity function then so is K + H, where H(, ) is ny function tht is hrmonic s function of. A singulrity function of is given by K(, ) 2π ln. (3) This ssertion cn be verified s follows. Without loss of generlity we cn tke. In polr coordintes, we hve K ln r with r, which is 2π hrmonic in R 2 \ {} (see section 9.4). Also, K dl 2π ln ɛ ɛ dθ ɛ ln ɛ 2π nd (becuse K n K/ r) K n dl 2π 3 2π ɛ dθ. ɛ
4 The reen s function for nd domin Ω with Dirichlet boundry conditions is singulrity function tht stisfies (, ) for. The reen s function provides the solution to the Poisson eqution with homogeneous Dirichlet boundry conditions: Theorem 4 If u f in Ω nd u on then u( ) (, )f() da ( Ω). (4) Ω Proof. Letting v() (, ), we hve vf da u v v u da Ω\B ɛ Ω\B ɛ (u v v u) n dl (u v v u) n dl u( ) v n dl + vu n dl, } {{ } nd the second term goes to zero becuse min u n v dl which completes the proof. vu n dl m u n v dl, The formul (4) represents the solution s the superposition of reen s functions, which cn be thought of s point source responses. The following result tells us tht the response t to point source locted t is equl to the response t to point source locted t. Theorem 5 (Reciprocity Principle) (, ) (, ) for. Proof. Let y nd y be distinct points in Ω, let B ɛ nd B ɛ denote ɛrdius disks centred t y nd y, nd let u() (, y ) nd v() (, y). Then u v v u da Ω\B ɛ\b ɛ (u v v u) n dl B ɛ u(y) v n dl + vu n dl B } ɛ B {{}} ɛ {{} uv n dl + v(y ) u n dl, leving us with u(y) v(y ), tht is, (y, y ) (y, y). 4
5 As consequence of Theorem 4, formul (4) cn be written s u() (, )f( ) da. Ω The singulrity function (3) is clled the freespce reen s function for Poisson s eqution. It is reen s function for Poisson s eqution with the boundry condition u() s. The freespce reen s function doesn t stisfy this boundry condition, but the boundry condition does ensure tht the term (u v v u) n dl in the proof of Theorem 3 goes to zero when Ω is tken to be disk of rdius R centred t nd R. The method of imges cn be used to find reen s functions for other domins. For emple, using the ide tht unit point source locted t (, y ) with y > nd point source of strength locted t (, y ) will cncel ech other on the is, we find the reen s function (, ) 2π ln ( ( ) 2 + (y y ) 2) 2 + 2π ln ( ( ) 2 + (y + y ) 2) 2 for the Poisson eqution in the hlfplne y >. Using similr ides, one cn derive the formul for reen s function for the origincentred disk of rdius s (, ) 2π ln π ln. Representing nd in polr coordintes s (r, θ) nd (ρ, θ ), nd using the cosine lw cos γ with γ θ θ, the disk reen s function cn be written (, ) 2π ln 4 + r 2 ρ rρ cos γ 2 (r 2 + ρ 2 2rρ cos γ). (5) The reen s function ((, y), (,.5)) for the hlfplne nd the unitrdius disk re shown below y.5 y The reen s function lso provides the solution of Lplce s eqution with nonhomogeneous Dirichlet boundry conditions: 5
6 Theorem 6 If u in Ω nd u h on then u( ) h() (, ) n dl. Proof. Similr to the proof of Theorem 4. For the hlfplne y > the unit norml points in the negtive y direction nd we hve n y /π y y ( ) 2 + (y ). 2 The solution of Lplce s eqution u in the hlfplne with u(, ) h() is therefore u(, y ) y h() π ( ) 2 + (y ) d. 2 For the origincentred disk, the unit outwrd norml points in the rdil direction nd we hve n r r 2π 2 ρ ρ 2 2ρ cos γ. The solution of Lplce s eqution on the disk with u h on the boundry r is therefore u(ρ, θ ) 2π ( 2 ρ 2 )h(θ) 2π 2 + ρ 2 2ρ cos(θ θ ) dθ, which is Poisson s integrl formul (section.)..3 reen s functions from eigenfunctions The Poisson problem cn lso be solved by the method of eigenfunctions. To introduce the technique, we strt with onedimensionl problem. Consider the differentil eqution u f with boundry conditions u(), u(). The ssocited eigenvlue problem is φ + λφ with the sme boundry conditions. The eigenvlues nd eigenfunctions re λ m m2 π 2 2, φ m () sin mπ. Substituting tril solution of the form m A m φ m () into the differentil eqution, multiplying through by φ m (), nd integrting gives A m λ m sin 2 mπ d /2 m f() sin mπ Thus the A m re λ m the Fourier sine coefficients of f, nd the solution t point ξ (, ) is [ ] 2 mπ u(ξ) f() sin m 2 π2 sin mπξ d. d, 6
7 Compring this with formul (), we deduce the reen s function to be the term in brckets, tht is, (, ξ) m 2 mπ mπξ sin sin m 2 π2. When, this is the Fourier sine epnsion of the reen s function presented in section. Net, consider the twodimensionl Poisson eqution u f on the domin (, ) (, b), with u on the boundry. The ssocited eigenvlue problem is u + λu, with the sme boundry conditions. Assuming solution of the form u(, y) X()Y (y) yields, with α 2 s seprtion constnt, the two eigenvlue problems X + α 2 X, Y + (λ α 2 )Y with homogeneous boundry conditions X() X() Y () Y (b). The eigenfunctions of the onedimensionl problems re nd the eigenvlues re X m () sin mπ, Y n(y) sin nπy b. λ mn π 2 ( m n2 b 2 Substituting tril solution of the form m the Poisson eqution, multiplying through by X m Y n, nd integrting gives A mn λ mn sin 2 mπ d sin 2 nπy dy b /2 b b/2 ). n A m,n X m ()Y n (y) into b f(, y) sin mπ nπy sin d dy b Thus the A mn re λ mn the twodimensionl Fourier sine coefficients of f, nd the solution t point is with (, y,, y ) 4b π 2 u(, y ) m n b f(, y)(, y,, y ) d dy mπ sin m 2 b 2 + n 2 2 sin nπy sin mπ b b sin nπy. b The reen s function (, y, /2, /4) for squre domin is plotted below y/.25.5 / 7
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