Question

(a) Find the solution to Laplace's equation on a disk with boundary condition u(1,0) = 5 + sin(40). (You do not need to derive the general solution to the polar Laplace's equation.) (b) Verify that the solution to (a) satisfies the mean value property. (Hint: compare the average value of u(r, ) on the boundary r=1 to the value of u(r,() at r = 0.) (c) Find the minimum and maximum of the solution to (a) and verify they occur on the boundary.

Answer 1

+ + B =- a'u 1 ou 1 au ama you or = 0. assume solution of the form uír, ) = R(r)7() pa aga 16.,y (r, 8) = R"(r)7(2), , (r,)= R'(r)(),600 (1,2)= R(r)T"(), pde =R" +- R'T + 2+LRT + ŽRT"=0 Multiplying both sides by ra, we can rewrite this as (~2R" + rR')T =-T"R Dividing both sides by RT pR" +R pa R" +rR 7" = 2 =rR"+rR'- AR=0,7" + AT = 0 R 7 т Strum liouville problem is 7" (0) + AT () = 0 periodic boundary conditions: uír, -1)=u(r,), (, - ) = 4(r, 17) = R(r)T(-1) = R(r)T(T), R(r)T"(-1) = R(r)T() 7(-1) = 7(), T"(-)=T"() consider the de 7"(©) + AT () = 0, 7(-7)=7 (T), T"(-1) = T () let 2 = m? T"()+AT()=0=T(0) +mºT () = 0 T(O)= a cos(mo) +b sin (me) T(-1)=77) =acos(mm)-b sin (MA) = a cos(MIT)+b sin (MA) :: sin (mA) = 0 = MIT = n1 =m=n T"()=-am sin (mo)+bmcos (mo) T'(-1) =T"(T) =amsin (mA)+bmcos(MT) =-amsin (MIT) +bm cos(MT) =sin(MT) = 0 = MT= NT=m=» eigen values are = nº corresponding eigen functions are cos(nx) or sin (nx) for 2 = 0 T"(?) + AT ()= 0 =T"(7)=0=T(T) = A+B8 T(-1) =T(T), T'(-1) =T'(7)=7. ()= A for the eigen value 1 = 0, corresponding eigen function is T. (O)=1 consider the de pR" +rR'-AR=0=R" +rR'- n°R=0 This is an ordinary differential equation it is called an Euler equation. let R(r) = p.* Differentiating twice and substituting into the equation, we get ala-1);* + am* - »*-*=0=2-22 = 0 This is the characteristic equation for the Euler equation arĚn. R(r) = 4,7*+By+*when r →0, we have p-* → but solution is bounded, so R(r) = 4**. When 2 = 0 then R(r) = 4, +Boln r. when r → 0, we have inr →-00, But solution is bounded, so R (r) = 4 Solution is lv(5, 6) = A +Š((43") cos(n@)+(C_7" )sin (10) boundary conditions: u(1, ) = f ( ) = 5+sin (40) 2 (1,6)= 4*+Š(4,60s() +C, sin (18)=+ sin(40) >> A = 5, 4, = 0, CA = 1, C =0 for n74 : u(1,2)=5+* sin(48)

(b) the average value of a harmonic function over a disc is equal to its value at the centre. 1 u(0,0) = | u(R, EdA, (R =1) 27TR A R,0) 1 = 2(Ssin (46)}de 107 = 5 27 uír, 7) = 5+r* sin(40) =u(0,0) = 5 (C) u(r, )= 5+r* sin (48), 0 Sr 51,-1 SOST maximum value of r* sin(49) is 1 ocuurs at r = 1, 0= 4 minimum value of ** sin(49) is - 1 ocuurs at r=1, 8= 4 :: maximum and minimum occurs on the boundary of circle. ग ग