The Dirichlet Problem in a Two Dimensional Rectangle. Section 13.5

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1 The Dirichlet Prolem in Two Dimensionl Rectngle Section Dirichlet Prolem in Rectngle In these notes we will pply the method of seprtion of vriles to otin solutions to elliptic prolems in rectngle s n infinite sum involving Fourier coefficients, eigenvlues nd eigenvectors. These prolems represent the simplest cses consisting of the Dirichlet Prolem in -Dimensionl in rectngle. g y f 1 x Ω g 1 y f x As usul we will strt with simplest oundry conditions Dirichlet oundry conditions nd rectngulr region Ω. The most generl setup in this cse is to prescrie function on ech of the four sides of the rectngle s depicted in the figure. Thus we otin the prolem u xx x, y + u yy x, y =, x, y [, ] [, ], 1.1 u, y = g y, ux, = f x, u, y = g 1 y ux, = f 1 x Anlysis of this prolem would ecome rther messy ut the principle of superposition llows us to divide nd conquer. We cn write the solution to this prolem s sum of solutions to four simpler prolems The Principle of Superposition We note tht for liner prolem it is lwys possile to replce single hrd prolem y severl simpler prolems. More specificlly, we cn write the solution to hrd prolem s the sum of the solutions to severl simpler prolems. For exmple the solution u to 1.1 1

2 cn e otined s sum of the solutions to four simpler prolems u 1 xx x, y + u 1 yy x, y =, x, y [, ] [, ], 1. u 1, y = g y, u 1, y = u 1 x, =, u 1 x, = u xx x, y + u yy x, y =, x, y [, ] [, ], 1.3 u, y =, u, y = u x, =, u x, = f 1 x u 3 xx x, y + u 3 yy x, y =, x, y [, ] [, ], 1.4 u 3, y =, u 3, y = g 1 y u 3 x, =, u 3 x, = u 4 xx x, y + u 4 yy x, y =, x, y [, ] [, ], 1.5 u 4, y =, u 4, y = u 4 x, = f x, u 4 x, = Non-Zero Boundry Function of x To illustrte the method of seprtion of vriles pplied to these prolems let us consider the BVP in 1.3. In this cse the only non-zero oundry term occurs on the top of the ox when y = where we hve u x, = f 1 x. u = u = f 1 Ω u = µ n = u x, y = n = u =, λ n = µ n, ϕ n x = sinµ nx, n = 1,,, n sinh µ n y ϕ n x 1 sinh µ n f 1 xϕ n x dx.

3 As usul we look for simple solutions in the form u x, y = ϕxψy. Sustituting into 1.3 nd dividing oth sides y ϕxψy gives ψ y ψy = ϕ x ϕx Since the left side is independent of x nd the right side is independent of y, it follows tht the expression must e constnt: ψ y ψy = ϕ x ϕx Here ψ mens the derivtive of ψ with respect to y nd ϕ mens mens the derivtive of ϕ with respect to x. = λ. N.B. Notice tht we re using the negtive of the sign we used for λ in the het nd wve eqution. This is in keeping with stndrd prctice for the Dirichlet prolem. So in this cse our eigenvlues will e λ n = µ n insted of µ n. We seek to find ll possile constnts λ nd the corresponding nonzero functions ϕ nd ψ. We otin ϕ + λϕ =, ψ λψ =. Furthermore, the oundry conditions give ϕψy =, ϕψy = for ll y. Since ψy is not identiclly zero we otin the eigenvlue prolem ϕ x + λϕx =, ϕ =, ϕ =..1 We hve solved similr prolem mny times with the min difference eing tht now the eigenvlues re positive. µ n =, λ n = µ n, ϕ n x = sinµ nx, n = 1,,.. The generl solution of ψ y µ nψy = is then ψy = c 1 cosh µ n y + c sinh µ n y.3 where c 1 nd c re ritrry constnts. The oundry condition ψ = implies ψy = sinh µ n y. 3

4 So we look for u s n infinite sum ux, y = n sinh µ n y ϕ n x.4 The only thing remining is to somehow pick the constnts n so tht the initil condition ux, = f 1 x is stisfied. Setting y = in.4, we seek to otin { n } stisfying f 1 x = ux, = n sinh µ n ϕ n x. This is lmost Sine expnsion of the function f 1 x on the intervl,. In prticulr we otin or sinh µ n n = n = 1 sinh µ n f 1 xϕ n x dx f 1 xϕ n x dx..5 N.B. In order to otin the complete solution to the originl prolem 1.1 we would need to solve the three other similr prolems 1., 1.4, 1.5. Let us now consider n explicit exmple for f 1 x: Exmple.1. Consider the prolem 1.3 with { x x /, f 1 x = π x / x. For this exmple.4 ecomes ux, y = n sinh µ n y sinh µ n y ϕ n x, with ϕ n x = sinµ nx. In this cse we otin the following results for.5 The explicit integrtions re crried out elow [ / x ] x sinh n = x sin dx + x sin dx / [ = sin + = sin n π sin ] 4

5 Which implies n = sin. n π sinh µ n Then we rrive t the solution ux, y = 4 π sin sinh µ n y sin µ n π n x. sinh µ n To otin the ove formuls we needed to crryout two integrtion y prts. We do them seprtely in the following. nd / / x sin x sin Thus we hve x dx = = x / x x cos dx x / cos = / cos = cos = cos / / x cos dx cos x / sin sin x dx = x x / cos dx = x x cos sinh µ n n = = / cos = cos = cos + [ sin / /. x dx x cos dx / x sin / sin 5 +. sin 1 x cos ] dx

6 or finlly, n = sin. n π sinh µ n 3 Non-Zero Boundry Function of y As nother exmple illustrting the method of seprtion of vriles pplied to these prolems let us consider the BVP in 1.4. In this cse the only non-zero oundry term occurs on the right hnd side of the ox when x = where we hve u 3, y = g 1 y. µ n = u 3 x, y = n =, λ n = µ n, ϕ n y = sinµ ny, n = 1,, n sinh µ n x sin µ n y 1 sinh µ n g 1 yϕ n y dy. To otin this formul we proceed s usul nd look for simple solutions in the form u 3 x, y = ϕyψx. The min difference here is tht in this cse we interchnge the roles of x nd y since we will wnt to do Fourier series in y this time insted of x. Sustituting into 1.4 nd dividing oth sides y ϕyψx gives ψ x ψx = ϕ y ϕy Since the left side is independent of y nd the right side is independent of x, it follows tht the expression must e constnt: ψ x ψx = ϕ y ϕy Here ψ mens the derivtive of ψ with respect to x nd ϕ mens mens the derivtive of ϕ with respect to y. We seek to find ll possile constnts λ nd the corresponding nonzero functions ϕ nd ψ. We otin Furthermore, the oundry conditions give = λ. ϕ y + λϕy =, ψ x λψx =. ϕψx =, ϕψx = for ll x. 6

7 Since ψx is not identiclly zero we otin the eigenvlue prolem ϕ y + λϕy =, ϕ =, ϕ =. 3.1 Once gin we note the min difference now is tht the eigenvlues λ n re positive. µ n =, λ n = µ n, ϕ n y = sinµ ny, n = 1,,. 3. The generl solution of ψ x µ nψx = is then ψy = c 1 cosh µ n x + c sinh µ n x 3.3 where c 1 nd c re ritrry constnts. The oundry condition ψ = implies ψx = sinh µ n x. So we look for u s n infinite sum ux, y = n sinh µ n x ϕ n y 3.4 The only remining prt is to find the n so tht the initil condition u, y = g 1 y is stisfied. Setting x = in.4, we seek to otin { n } stisfying g 1 y = u, y = n sinh µ n sin µ n y. This is lmost Sine expnsion of the function g 1 y on the intervl,. In prticulr we otin sinh µ n n = g 1 yϕ n y dy The Lplce Eqution with other Boundry Conditions Next we consider slightly different prolem involving mixture of Dirichlet nd Neumnn oundry conditions. To simplify the prolem it we set = π nd keep ny numer. 7

8 Nmely we consider u xx x, y + u yy x, y =, x, y [, π] [, ], 4.1 u, y =, u x π, y = u y x, =, ux, = f 1 x u = f 1 Ω u = u x = u x = π Look for simple solutions in the form ux, y = ϕxψy. Sustituting into 4.1 nd dividing oth sides y ϕxψy gives ψ y ψy = ϕ x ϕx Since the left side is independent of x nd the right side is independent of y, it follows tht the expression must e constnt: ψ y ψy = ϕ x ϕx Here ψ mens the derivtive of ψ with respect to y nd ϕ mens mens the derivtive of ϕ with respect to x. We seek to find ll possile constnts λ nd the corresponding nonzero functions ϕ nd ψ. We otin Furthermore, the oundry conditions give = λ. ϕ + λϕ =, ψ λψ =. ϕψy =, ϕ πψy = for ll y. Since ψy is not identiclly zero we otin the desired eigenvlue prolem ϕ x + λϕx =, ϕ =, ϕ π =. 4. µ n = n 1, λ n = µ n, ϕ n x = π sinµ nx, n = 1,,

9 The generl solution of ψ µ nψ = is ψy = c 1 cosh µ n y + c sinh µ n y 4.4 where c 1 nd c re ritrry constnts. The oundry condition ψ = implies ψy = cosh µ n y. So we look for u s n infinite sum ux, y = n cosh µ n y ϕ n x. 4.5 Finlly we need to find the constnts n so tht f 1 x = ux, = n cosh µ n ϕ n x. As usul we otin n expnsion of the function f 1 x on the intervl, π in the form π cosh µ n n = f 1 x sinµ n x dx. 4.6 π 9

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