Theorem The BVP for the one-space dimensional heat equation, infinitely. where k > 0, L > 0 are constants, has infinitely. #, c n R.

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1 The IBVP: Neumann Conditions. Theorem 7... The BVP for the one-space dimensional heat equation, U KIU act o o Welt t u = k 2 xu, BC: x u(t, ) =, x u(t, ) =, here k >, > are constants, has infinitely act x u(t, x) = c 2 +! c n e many solutions, c n R. nπ k( )2t cos n=1 get Cu e KCET't cost cue K2 Furthermore, for every continuous function f on [, ] satisfying fco FE O f () = f () = problem above that also satisfies the initial UCO X infinitely, there is a unique solution u of the boundary value initial condition fax u(, x) = f(x). 2 to Neumann Be This solution u is given by the expression above, here the coefficients c n are $ f(x) cos q to fcx coscn DX c n = 2 Remarks: (a) This is an Initial-Boundary (b) The boundary conditions are called Neumann dx, n =, 1, 2,. Value Problem (IBVP). boundary conditions. Remark: The physical meaning of the initial-boundary conditions is simple. (1) The boundary conditions is to keep the heat flux at the sides of the bar is constant Initial constant. (2) The initial condition is the initial temperature on the hole bar. Remark: One can use Dirichlet on the other side. This is called a mixed Boundary Neumann initial Dirichlet mixed heat flux temperatures conditions on one side and Neumann boundary condition. separation of Neumann variables Remark: The proof is based on the separation of variables method.

2 1 Proof of the Theorem: First look for simple solutions of the heat equation given by ZU k EU NBC UCE separation of o o variables Simple solutions u(t, x) = v(t) (x). So e look for solutions having the variables separated into to functions. Introduce this particular function in the heat equation, v(t) (x) = k v(t) (x) 1 t.is v(t) k v(t) = (x) (x), here e used the notation v = dv/dt and = d/dx. The separation of variables in the IIgzCvctscxi function u implies a separation of variables in the heat equation. The left hand side in the last equation above depends only on t and the right hand side depends only on x. The only possible solution is that both sides are equal the same constant, call it λ. So e end up ith to equations 1 v(t) k v(t) = λ, and (x) (x) = λ. v λ (t) = c λ e kλt, c λ = v λ (). The second equation leads to an eigenfunction problem for once boundary conditions are provided. These boundary conditions come from the heat equation boundary conditions, x u(t, ) = v(t) () = for all t! x u(t, ) = v(t) () = for all t! () = () =. So e need to solve the folloing BVP for ; + λ =, () = () =. 2 4 act x vets as 1 o kdxvctt'd V as k Vitt ZW u 1 1 o kv The equation on the left is first order and simple to solve. The solution depends on λ, Ivas co tuk KEI o vi In iffy t W WA 2 constant TER k l I I i This is an eigenfunction problem, hich has solutions only for λ >, because in that case ICE ka Ct unction problem the asociated characteristic polynomial has complex roots. If e rite λ = µ 2, for µ >, i J qq.my e get the general solution c UCO arbitrary l ta ou s s (x) = c 1 cos(µx) + c 2 sin(µx) only for X o M2 W O r TM e o s

3 1 Proof of the Theorem: First look for simple solutions of the heat equation given by ZU k EU NBC UCE separation of o o variables Simple solutions u(t, x) = v(t) (x). So e look for solutions having the variables separated into to functions. Introduce this particular function in the heat equation, v(t) (x) = k v(t) (x) 1 t.is v(t) k v(t) = (x) (x), here e used the notation v = dv/dt and = d/dx. The separation of variables in the IIgzCvctscxi function u implies a separation of variables in the heat equation. The left hand side in the last equation above depends only on t and the right hand side depends only on x. The only possible solution is that both sides are equal the same constant, call it λ. So e end up ith to equations 1 v(t) k v(t) = λ, and (x) (x) = λ. v λ (t) = c λ e kλt, c λ = v λ (). The second equation leads to an eigenfunction problem for once boundary conditions are provided. These boundary conditions come from the heat equation boundary conditions, x u(t, ) = v(t) () = for all t! x u(t, ) = v(t) () = for all t! () = () =. So e need to solve the folloing BVP for ; + λ =, () = () =. 2 4 act x vets as 1 o kdxvctt'd V as k Vitt ZW u 1 1 o kv The equation on the left is first order and simple to solve. The solution depends on λ, Ivas co tuk KEI o vi In iffy t W WA 2 constant TER k l I I i This is an eigenfunction problem, hich has solutions only for λ >, because in that case ICE ka Ct unction problem the asociated characteristic polynomial has complex roots. If e rite λ = µ 2, for µ >, i J qq.my e get the general solution c UCO arbitrary l ta ou s s (x) = c 1 cos(µx) + c 2 sin(µx) only for X o M2 W O r TM e o s

4 Initial condition UCO X f X f Cx coz t n If e prescribe the c n e get a solution u that at t = is given by the previous formula. Is Cn ar n it the converse true? The anser is yes. Given f(x) = u(, x), here f () = f () =, e can find all the coefficients c n. Here is ho: Given f on [, ], extend it to the domain f a Exp feuencx [, ] as an even function, f even (x) = f(x) and f even ( x) = f(x), x [, ] We get that f even it is continuous on [, ]. So f even has a Fourier series expansion. Since f even is even, the Fourier series is a cosine series f even (x) = a 2 +! a n cos Thi m and the coefficients are given by the formula a n = 1 n=1 $ $ E t NE on an f even (x) cos dx = 2 y A f(x) cos fox ea cx dx, n =, 1, 2,. d Cn Z J feueu.co cos dx Since f even (x) = f(x) for x [, ], then c n = a n. This establishes the Theorem. f x fauencx on o fcn zjofcxjcos.ci ITn o i z ftp.eozt cne kce5tcoscne D8

5 1 Example 7..2: Find the solution to the initial-boundary value problem t u = 2 xu, t >, x [, ], ith initial and boundary conditions given by 7 x % IC: u(, x) = 2, &, x %, ', 2 BC: ( u (t, ) =, u (t, ) =. Solution: We look for simple solutions of the form u(t, x) = v(t) (x), Separation of variables F (x) dv dt (t) = v(t) d2 v(t) (x) dx2 v(t) = (x) (x) = λ. simpleso, thesoya equations foron v and are p p act VCt WH v(t) = λ v(t), (x) + λ (x) =. The solution for v depends on λ, and is given by to ua a v λ (t) = c λ e λt, c λ = v λ (). Next e turn to the the equation for, and e solve the BVP (x) + λ (x) =, ith BC () = () =. I i u icons This is an eigenfunction problem for and λ. This problem has solution only for λ >, since only in that case the characteristic polynomial has complex roots. et λ = µ 2, then ya F 2 Cv logo fcx on o I p(r) = r 2 + µ 2 = r ± = ±µ i. The general solution of the differential equation is Its derivative is I IE HIEI n (x) = c 1 cos(µx) + c 2 sin(µx). v a eigenfunction problem seat 7.1 (x) = µ c 1 sin(µx) + µ c 2 cos(µx). uco's The first boundary conditions on implies n = () = µ c 2, c 2 = (x) = c 1 cos(µx)..gxsj.ec in g

6 co Ct E co 2 2 UnCts en e c t e act 14 The second boundary condition on implies coz Ii on e 5tasCng = () = µ c 1 sin(µ), c 1 =, sin(µ) =. B Then, µ n = nπ, that is, µ n = nπ. Choosing c 2 = 1, e conclude, I C UCO X f Cx T f! nπ 2,! nπx λ n = n (x) = cos, n = 1, 2,. Using the values of λ n found above in the formula for v λ e get fch v n (t) = c n e EE ( nπ )2t t II c cos, c n I = v n (). x Therefore, e get u(t, x) = c 2 + nπ ( c n e )2t cos n=1 here e have added the trivial constant solution ritten as c /2. The initial condition is E coz t on conf 7 x ( Cn 2, ), E if.name f(x) = u(, x) = x (,, j 2* I.fcx We extend f to [, ] as an even function 7 x ( 2, ), f even (x) = x ( 2,, 2* feuencx on,! nπx 7 x (,. 2) on fix cos ax Since f rmeven is even, its Fourier expansion is a cosine series T 2, t TE Ex Cn Zz 7 The coefficient a is given by a = 2 + f even (x) = a 2 +! nπx a n cos. cos f(x) dx = 2 + n=1 ok /2 7 dx = a = 7. co A ok G E E

7 Cn 717,4 sin E n a 21 iffy Ian ing No the coefficients a n for n! 1 are given by sin Singha a n = 2 = 2!! /2 f(x) cos 7 cos $ $$ = 2 7 nπ sin = 2 7 nπ sin cuts sin = 7 2 nπ sin(nπ). dx nπ 2 dx /2 15 ucti 5 E sine yei mn But for n = 2k e have that sin(2kπ/2) = sin(kπ) =, hile for n = 2k 1 e have that sin((2k 1)π/2) = ( 1) k 1. Therefore a 2k =, a 2k 1 = 7 2( 1)k, k = 1, 2,. (2k 1)π We then obtain the Fourier series expansion of f even, he f even (x) = % k=1 7 2( 1)k (2k 1)πx (2k 1)π cos But the function f has exactly the same Fourier expansion on [, ], hich means that c = 7, c 2k =, c (2k 1) = 7 2( 1)k (2k 1)π. So the solution of the initial-boundary value problem for the heat equation is u(t, x) = % k=1 2( 1) k (2k 1)π (2k 1)π e ( ) 2t cos (2k 1)πx.