=0, (x, y) Ω (10.1) Depending on the nature of these boundary conditions, forced, natural or mixed type, the elliptic problems are classified as

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1 Chapte 1 Elliptic Equations 1.1 Intoduction The mathematical modeling of steady state o equilibium phenomena geneally esult in to elliptic equations. The best example is the steady diffusion of heat in any twodomain Ω bounded by Ω. In the absence of any souces, the govening equation is the Laplace s equation given by x + u =, (x, y) Ω (1.1) y Due to the absence of time deivative tems in the equation (1.1), unlike the poblems given in the ealie two chaptes, these ae pue bounday value poblems. Theefoe, bounday conditions alone (no initial conditions) have to be pescibed ove the entie bounday Ω. Depending on the natue of these bounday conditions, foced, natual o mixed type, the elliptic poblems ae classified as 1. Diichelt poblem : The diffeential equation along with fixed (foced) bounday conditions on the bounday, that is, u = f(x, y) oveω.. Neumann poblem : The diffeential equation and deivative bounday conditions given by u x n = f(x, y) oveω, whee x n is the nomal to Ω. 3. Robin o Mixed poblem : The diffeential equation along with a combination of foced and natual bounday conditions given by αu + β u x n = f(x, y) ove Ω, whee α, β ae constants. 111

2 11 Sanyasiaju V S S Yedida syedida@iitm.ac.in 1. Sepaation of Vaiables Decomposing u(x, y) = X(x)Y (y) esult in to odinay diffeential equations d X kx dx = (1.) d Y + ky dy = (1.3) whee k is a constant. The solution u can be witten as Case i: k is positive, that is k = λ. u(x, y) =(c 1 e λx + c e λx ) (c 3 cos λy + c 4 sin λy) (1.4) Case ii: k is negative, that is k = λ. u(x, y) =(c 1 cos λx + c sin λx) (c 3 e λy + c 4 e λy ) (1.5) Case iii: k = u(x, y) =(c 1 x + c ) (c 3 y + c 4 ) (1.6) 1.3 Diichlet Poblem in a Rectangula Domain Solve u + u =, (x, y) Ω whee Ω = (,a) (,b) with bounday conditions x y u(x, ) = u(x, b) =,u(,y)=andu(a, y) =f(y). Solution : Due to the homogenous natue of bounday conditions, in the y diection, that is, at y =andy = b, non-tivial solution exists only fo the case (pove that fo the othe two cases, the solution is identically zeo) u(x, y) =(c 1 e λx + c e λx ) (c 3 cos λy + c 4 sin λy) Now, applying the bounday conditions at y =andy = b on (c 3 cos λy + c 4 sin λy) gives c 3 =andsinλb =. Theefoe λ n = nπ,,,. ( the othe n ae omitted because, n = gives tivial solution and negative n only epeats the existing eigenfunctions with a minus b sign). Similaly, using the zeo bounday condition at x = on (c 1 e λx + c e λx )gives c 1 = c. Now, using the supeposition pinciple, the solution can be witten as u(x, y) = whee A n. A n (eλnx e λnx )sinλ n y = A n sinh λ n x sin λ n y

3 Lectue Notes MA Diffeential Equations 113 Finally, the values of A n can be computed using the non-zeo bounday condition at x = a in the following way: At x = a we have u(a, y) = A n sinh λ n a sin λ n y = f(y) The above is a Fouie sine seies, theefoe, A n sinh λ n a = f(y) sinλ n ydy o b b f(y) sinλ n ydy b sinh λ n a The solution of the given poblem, afte substituting the values of λ n,is u(x, y) = A n sinh nπx sin nπy b b Note : b b sinh nπa b b f(y) sin nπy b (1.7) dy (1.8) 1. The convegence of the seies in the final solution, unde cetain conditions on f and f, is not included in the pesent Lectue notes.. If the bounday conditions in the above poblem ae modified to u(,y) = u(a, y) =,u(x, ) = and u(x, b) =f(x) then the solution of the coesponding poblem is u(x, y) = A n sin nπx nπy sinh (1.9) a a Numeical Example a sinh nπb a a f(x) sin nπx a dx (1.1) The faces of a thin squae plate of length 4cm ae pefectly insulated (to avoid any atmospheical effects). Find the tempeatue distibution on the plate if the side at y =4iskeptat o C and all the othe thee sides ae kept o C. The solution of the poblem, fom the discussion given above, is u(x, y) = A n sin nπx nπy sinh sinh nπ 4 sin nπx 4 dx = 4 nπ sinh nπ (1 cos nπ)

4 114 Sanyasiaju V S S Yedida syedida@iitm.ac.in u(x, y) = 8 π 1 (n 1)πx (n 1)πy sin sinh (n 1)π sinh (n 1)π Poblems to Wokout 1. The faces of a thin squae plate of length cm ae pefectly insulated. Find the tempeatue distibution on the plate if the side at y =iskeptatsinπx and all the othe thee sides ae kept o C.. Solve u x + u y =, (x, y) Ω whee Ω = (,π) (,π) satisfying the bounday conditions u(x, ) = u(x, π) =along x π and u(,y)=,u(π, y) =1 along <y<π. 3. The faces of a thin squae plate of length cm ae pefectly insulated. Find the tempeatue distibution on the plate if u =atx =andx = a, andthe othe two sides ae insulated. 4. The faces of a thin squae plate of unit length ae pefectly insulated. Find the tempeatue distibution on the plate if the uppe and lowe sides of the plate ae insulated, left side is kept at o C and the ight side is kept at f(y) o C 5. Find the steady state tempeatue in a ectangula plate bounded x =,x=1, y= and y = π. Theedgesx =andx = 1 ae insulated and the tempeatue along y =iscosπx and along y = π is. 6. Neumann poblem : Solve u x + u y =, (x, y) Ω whee Ω = (,a) (,b) with bounday conditions u y (x, ) = u y (x, b) =,u x (,y)=andu x (a, y) = f(y) 1.4 Laplacian in Pola Coodinates Taking the tansfomation we have x = cos θ, = x + y, y = sin θ θ =tan 1 y x Theefoe, x =cosθ, y =sinθ, θ x = sin θ, θ y = cos θ

5 Lectue Notes MA Diffeential Equations 115 x = cosθ sin θ θ = sinθ y + cos θ θ = cosθ (cos θ x sin θ ( + sin θ θ + sin θ ) θ )(cos θ θ sin θ sin θ = sinθ (sin θ y + cos θ ( cos θ + θ cos θ θ cos θ ) θ )(sin θ θ +cosθ + cos θ ) θ θ sin θ ) θ x + u = u y + 1 θ + 1 θ = (1.11) Taking R()T (θ) and substituting in the pola fom of the Laplace equation gives the odinay diffeential equations d R d + dr d kr = d R dθ + kt Fo k = λ,,λ (k is negative, zeo and positive), the solution u is espectively. (c 1 cos λ log + c sin λ log ) (c 3 e λθ + c 4 e λθ ) = (c 1 log + c ) (c 3 θ + c 4 ) = (c 1 λ + c λ ) (c 3 cos λθ + c 4 sin λθ) 1.5 Diichlet Inteio Poblem + 1 θ + 1 =, θ π, < a θ subjected to the bounday conditions u(a, θ) =f(θ) fo θ π. Solution : Since fo Diichlet inteio poblem, = is also a pat of the domain at which log is not defines, theefoe, the equied solution can be obtained only fom k = λ,thatis, (c 1 λ + c λ ) (c 3 cos λθ + c 4 sin λθ)

6 116 Sanyasiaju V S S Yedida syedida@iitm.ac.in Futhe, the solution must be peiodic with peiod π, theefoe c 3 cos λθ + c 4 sin λθ = c 3 cos λ(θ +π)+c 4 sin λ(θ +π) c 3 (cos λθ cos λ(θ +π)) + c 4 (sin λθ sin λ(θ +π)) = sinλπ (c 3 sin(λθ + λπ)+c 4 cos(λθ + λπ)) = Theefoe, sin λπ =, λπ = nπ λ = n, n =, 1,,. Using the supeposition pinciple, the solution can be witten as (a n n + b n n ) (c n cos nθ + d n sin nθ) n= Futhe since the solution must be finite at = implies d n must be zeo (fo oute poblem wheein the domain is defined ove >1, c n hastobezeotomakethe solution finite). Using d n = and enaming the constants will give n (A n cos nθ + B n sin nθ) n= Now using the given bounday condition gives f(θ) = a n (A n cos nθ + B n sin nθ) n= which is a full Fouie seies hence the coefficients ae A = 1 π f(θ) dθ π 1 π f(θ)cosnθ dθ a n π B n = 1 π f(θ)sinnθ dθ a n π Numeical Example + 1 θ + 1 =, θ π, < 1 θ subjected to the bounday conditions u(1,θ)=1cos θ fo θ π. Solution : The solution is n (A n cos nθ + B n sin nθ) n=

7 Lectue Notes MA Diffeential Equations 117 with the coefficients A = 1 π π B n = π π π π 5(1 + cos θ) dθ =5 5(1 + cos θ)cosnθ dθ = (n ) & A =5 5(1 + cos θ)sinnθ dθ = Theefoe, 5+5 cos θ = 5(1+ cos θ) Poblems to Wokout u conditions f(θ) =sin 3 θ { θ π. f(θ) = <θ< π π <θ< 3π { θ π <θ< 3. f(θ) = θ <θ<π + 1 =, θ π, < 1 subjected to the bounday θ θ 1.6 Neumann Inteio Poblem + 1 θ + 1 =, θ π, < a θ subjected to the bounday conditions u u (a, θ) = (a, θ) =f(θ) fo θ π. n Solution : Following the same computations given in Diichlet inteio poblem, we get (since no change in any of the conditions until the application of the bounday conditions at = a) n (A n cos nθ + B n sin nθ) n= u n = u = n n 1 (A n cos nθ + B n sin nθ) n= Now using the given bounday condition gives f(θ) = na n 1 (A n cos nθ + B n sin nθ)

8 118 Sanyasiaju V S S Yedida syedida@iitm.ac.in which is once again a full Fouie seies hence the coefficients can be witten as B n = 1 na n 1 π 1 na n 1 π π π f(θ)cosnθ dθ f(θ)sinnθ dθ Theefoe, the solution of the Neumann inteio poblem is A + n (A n cos nθ + B n sin nθ) (1.1) whee B n = 1 1 π f(θ)cosnθ dθ (1.13) na n 1 π 1 1 π f(θ)sinnθ dθ (1.14) na n 1 π Notice that, in this case, the solution can diffe by an abitay constant A. 1.7 Semicicula Domain + 1 θ + 1 =, θ π, < a θ subjected to the bounday conditions u(a, θ) = f(θ) fo θ π, u(, ) = u(, π) =. Solution : We have (c 1 λ + c λ ) (c 3 cos λθ + c 4 sin λθ) Applying the conditions at θ =andπ gives c 3 =andλ = n fo n =1,,. Using the supeposition pinciple, the solution can be witten as (a n n + b n n )sinnθ Futhe since the solution must be finite at = implies d n must be zeo. Using d n = and enaming the constants will give A n n sin nθ

9 Lectue Notes MA Diffeential Equations 119 Now using the given bounday condition gives f(θ) = A n a n sin nθ n= which is a full Fouie sine seies hence the coefficients ae 1 π f(θ)sinnθ dθ a n π Numeical Example + 1 θ + 1 =, θ π, < 1 θ satisfying the bounday conditions u(, ) = u(, π) =,u(1, θ)= 1θ(π θ) fo θ π. Solution : The solution is A n n sin nθ 1 π 1θ(π θ)sinnθ dθ = 4 (cos nπ 1) a n π πn an 1.7. Poblems to Wokout θ + 1 =, θ π, < 1 θ satisfying the conditions u(, ) = u(, π) = and (a) f(θ) = 1 sin 3 θ (b) f(θ) =T,wheeT is a constant. u(, θ) is a function satisfying + 1 θ + 1 θ = in an semicicula annulus defined by θ π, a < < b. If its value along the bounday = a is θ( π θ) and is zeo on the emaining pat of the bounday then pove that ( b )4n ( b )4n sin(4n )θ π ( a b )4n ( b a )4n (n 1) 3