Section 12.6: Non-homogeneous Problems
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1 Section 12.6: Non-homogeneous Problems 1 Introduction Up to this point all the problems we have considered are we what we call homogeneous problems. This means that for an interval < x < l the problems were of the form u t (x, t = u xx (x, t, B (u =, B 1 (u = u(x, = f(x In contrast, in Section we are concerned with some non-homogeneous cases: u t (x, t = u xx (x, t + R(x, B (u = γ, B 1 (u = γ 1 u(x, = f(x where γ j are constants and R is a function of x but not t. I would lie to do the case where it is a function of t but the boo does not and there is no time. Here we have used the notation B j (u to indicate our usual boundary conditions B (u = α u x (, t + α 1 u(, t, B 1 (u = β u x (l, t + β 1 u(l, t. Specifically then for Dirichlet boundary conditions we have B (u = u(, t, B 1 (u = u(l, t and for Neumann conditions we have B (u = u x (, t, B 1 (u = u x (l, t. 1.1 Non-Homogeneous Equation, Homogeneous Dirichlet BCs We first show how to solve a non-homogeneous heat problem with homogeneous Dirichlet boundary conditions u t (x, t = u xx (x, t + R(x, < x < l, t > (1 u(, t =, u(l, t = u(x, = f(x 1
2 Let us recall from all our examples involving Dirichlet BC the Sturm-Liouville problem gives λ n = µ 2 n, µ n = nπ 2 l, ϕ n(x = l sin(µ nx. To solve this problem we loo for a function ψ(x so that the change of dependent variables u(x, t = v(x, t + ψ(x transforms the non-homogeneous problem into a homogeneous problem. In particular we have R = u t u xx = (v + ψ t (v + ψ xx = v t v xx ψ. So if we want v t v xx = then we need ψ = 1 R. In addition we want ψ to satisfy the BCs so we loo for ψ satisfying: ψ (x = 1 f(x, < x < l, ψ( =, ψ(l =. Notice this is a non-homogeneous second order constant coefficient boundary value problem. At t = we have v(x, = u(x, ψ(x = f(x ψ(x so the heat problem for v has a different initial condition. Thus, in order to solve (1 we need to solve the following: and then ψ (x = 1 f(x, < x < l, ψ( =, ψ(l =. v t (x, t = v xx (x, t, to obtain v(x, =, v(l, t = v(x, = f(x ψ(x v (x u(x, t = v(x, t + ψ(x. We already now how to solve the heat problem: v(x, t = c n e λnt ϕ n (x, c n = v (xϕ n (x dx. 2
3 Before we turn to finding ψ let us mae a very important remar. Remar 1.1. The solution v approaches zero as t goes to infinity which means that u converges to a non-constant steady state, i.e., the solution consists of two parts which are often referred to as the transient and the steady state. u(x, t = v(x, t + ψ(x ψ(x as t. This is a very important property of the heat equation. Solutions tend to a equilibrium solution., i.e., a solution for which u t (x, t = which means a solution of = u xx + R(x, u(x, t =, u(l, t =. Example 1.1. Find the steady state solution for the heat problem u t (x, t = u xx (x, t 6x, < x < 1, t > u(, t =, u(1, t = u(x, = ϕ(x As described in the remar the steady state problem is obtained by setting u t solving the non-homogeneous BVP = and ψ (x = 6x, < x < 1, ψ( =, ψ(1 = For this problem we apply the techniques from an elementary ODE class. Namely, we now that the general solution is the sum of the general solution of the homogenous problem ψ h and any particular solution ψ p. The general solution of the homogeneous problem ψ (x = is ψ h (x = c 1 x + c 2 and it is clear that ψ p (x = x 3 is a particular solution. N.B. Remember we learned two methods to find a particular solution: Undetermined Coefficients, Variation of Parameters. So we have ψ(x = ψ h (x + ψ p (x = c 1 x + c 2 + x 3. Now we try to find c 1 and c 2 so that the boundary conditions are satisfied. We need = ψ( = c 2, and = ψ(1 = c which implies c 1 = 1 and ψ(x = x 3 x. Thus for every initial condition ϕ(x the solution u(x, t to this forced heat problem satisfies lim u(x, t = ψ(x. t More generally we can give a formula for the solution of the steady state problem ψ = 1 R, ψ( =, ψ(l = 3
4 using the method of variation of parameters. First the general solution of the homogeneous problem ψ = is ψ h = c 1 + c 2 x. So we next need to find a particular solution y p and then the general solution is y = y h + y p. The method of Variation of Parameters is used to give a particular solution y p for a problem y + P (xy + Q(xy = R(x. It requires having a pair of linearly independent solutions of the homogeneous problem, y 1 and y 2. The formula is t y 2 (sr(s t y 1 (sr(s y p (x = y 1 (x ds + y 2 (x ds, W (s W (s [ ] y1 (s y 2 (s W (s = det y 1(s y 2(s In our case we tae y 1 = 1 and y 2 = x and notice that the right hand side has 9 1/r(x so we have [ ] 1 x W (s = det = 1. 1 y p (x = 1 ( sr(s ds x R(s ds. Then the general solution is ψ(x = 1 ( sr(s ds x Next we apply the BCs to find c 1 and c 2. = ψ( = c 1 c 1. Next = ψ(l = 1 ( sr(s ds l This implies c 2 = 1 ( sr(s ds + l l so we have ψ(x = 1 ( sr(s ds x R(s ds + x ( l R(s ds + c 1 + c 2 x. R(s ds + c 2 l R(s ds sr(s ds + l This can be considerably simplified by factoring out an l to obtain ψ(x = 1 ( l sr(s ds xl R(s ds x sr(s ds + xl l R(s ds. R(s ds. in the denominator then combining the first and third integrals and the second and fourth integrals. We have l sr(s ds x sr(s ds = (l x 4 sr(s ds + x x sr(s ds
5 and xl R(s ds + xl Combining these results we have ψ(x = 1 ( (l x l R(s ds = xl s R(s ds + x x x R(s ds. (l sr(s ds. Collecting these results we have the following results. has solution ψ(x = 1 l ψ (x = 1 R(x, < x < l, ψ( =, ψ(l =. ( (l x s R(s ds + x x (l sr(s ds. 1.2 Non-homogeneous Dirichlet Boundary Conditions In this section we consider forcing through Dirichlet boundary conditions u t (x, t = u xx (x, t, < x < l, t > (2 u(, t = γ, u(l, t = γ 1 u(x, = f(x In order to obtain a continuous solution we also need to impose the compatibility conditions f( = γ, f(l = γ 1. Our method to solve this problem is to transform it to a homogeneous problem lie we did in the previous section. In order to do this we introduce the function h(x = γ + x l (γ 1 γ. Then we introduce a new function v(x, t by w(x, t = u(x, t h(x. Our goal is to see what problem w(x, t satisfies. To this end we note that h xx (x = h t (x = 5
6 We see that and w t w xx = (u(x, t h(x t (u(x, t h(x xx = w(, t = u(, t h( = γ γ =, w(l, t = u(l, t h(l = γ 1 γ 1 =. ( w(x, = u(x, h(x = f(x γ + x l (γ 1 γ w (x. (3 Collecting this information we find that w(x, t satisfies w t (x, t = w xx (x, t, < x < l, t > (4 w(, t =, w(l, t = w(x, = w (x So we can apply our earlier results to obtain a formula for w(x, t. Once we do this (see below we can obtain the desired solution from u(x, t = w(x, t + h(x. Then for the initial condition we compute w (x = b n ϕ n (x. (5 where c n = ϕ n (xw (x dx. (6 Combining these results we obtain and finally w(x, t = c n e λnt ϕ n (x (7 u(x, t = w(x, t + h(x. Notice in this case, just as in the case of the non-homogeneous equation, that as t all the exponential terms in the sum tend to zero and we have lim u(x, t = h(x. t This represents a nonzero and non constant steady state temperature profile. 6
7 1.3 More General Non-homogeneous Boundary Conditions In this section we consider forcing through Neumann boundary conditions. u t (x, t = u xx (x, t, < x < l, t > (8 u(, t = γ, u x (l, t + 1 u(l, t = γ 1 u(x, = f(x The primary difference between this problem and that considered in the previous section is that we need a different function h. In order to find an appropriate function h let us examine the properties we desire. We want a function that satisfies the following conditions: h t =, h xx =, h( = γ, h x (l + 1 h(l = γ 1. We see that this first suggests h not depend on t and since h xx = we would have h(x = c 1 x + c 2. Applying the boundary conditions we would have γ = h( = c 2, γ 1 = h (l + 1 h(l = c 1 (1 + 1 l + 1 c 2. This implies So we have c 2 = γ, c 1 = (γ 1 1 γ (1 + 1 l. h(x = (γ 1 1 γ (1 + 1 l x + γ. Next we mae a change of variables u(x, t = w(x, t + h(x and find that w(x, t satisfies w t (x, t = w xx (x, t, < x < l, t > (9 w x (, t =, w x (l, t + 1 w(l, t = w(x, = w (x = f(x h(x. Notice once again that the initial condition w (x = f(x + h(x which is not the original initial condition f(x. Having found w(x, t we then obtain u(x, t = w(x, t + h(x. The only thing that remains is to solve for w and we do this using an eigenfunction expansion and see a solution to the problem in (9 using separation of variables. Thus we see a solution in the form w(x, t = c n e λnt ϕ n (x. In this case, due to the boundary conditions, we have the Sturm-Liouville problem ϕ = λϕ, ϕ( =, ϕ (l + 1 ϕ(l =. 7
8 We can readily show that the eigenvalues are negative since λ ϕ 2 = ϕ (xϕ(x dx = ϕ 2 + ϕ (xϕ(x l = ϕ 2 1 ϕ(l 2. So λ = µ 2 and ϕ(x = a cos(µx + b sin(µx. To satisfy the first boundary condition we need = ϕ( = a a =, ϕ(x = b sin(µx. Then we need which implies = ϕ (l + 1 ϕ(l = bµ cos(µl + 1 b sin(µl, tan(µl = µ 1. This equation has infinitely solutions which diverge to infinity but we cannot solve for them in closed form. We denote the associated eigenvalues and normalized eigenfunctions by µ n, λ n = µ 2 n, ϕ n (x = κ n sin(µ n x. Then the solution of our heat problem for w is w(x, t = c n e λnt ϕ n (x, where c n = w (xϕ n (x dx. Now, once again, since the λ n < (and tend to minus infinity we have sup w(x, t t. x [,l] Therefore we have u(x, t = w(x, t + h(x t h(x. 1.4 Heat Equation with Conduction and Convection Another variation on the heat equation is to add extra terms that correspond to heat conduction and convection. u t (x, t = ( u xx (x, t 2au(x, t x + bu(x, t, < x < l, t > (1 u(, t =, u(l, t =, (11 (12 u(x, = f(x. (13 There are many different ways to approach this problem. One such method would be to apply separation of variable directly. The dissadvantange to this is that one gets a more complicated ode for ϕ(x and there is a more difficult analysis of the eigenvalues and eigenvectors. We will tae a different approach and define v(x, t via u(x, t = e ax+βt v(x, t, β = (b a 2. (14 8
9 Thus we have and we can compute v(x, t = e (ax+βt u(x, t Furthermore and v t v xx = e (ax+βt ( βu + u t [ e (ax+βt ( au + u x ] x = e (ax+βt {( βu + u t [ a( au + u x + ( au x + u xx ]} = e (ax+βt [ u t (u xx 2au x + a 2 u + βu ] Therefore, v(x, t is the solution of = e (ax+βt [ u t (u xx 2au x + a 2 u + (b a 2 u ] = e (ax+βt [u t (u xx 2au x + bu] =. v(, t = e βt u(, t =, v(l, t = e (al+βt u(l, t = We have eigenvalues and eigenfunctions ( nπ µ n = l v(x, = e ax u(x, = e ax f(x. v t (x, t = v xx (x, t v(, t =, v(l, t = v(x, = e ax f(x., λ n = µ 2 n, ϕ n (x = and we obtain the solution to this problem as v(x, t = ( nπ b n e λnt sin l x with b n = 2 l sin (µ nx e ax f(xϕ n (x dx. Finally our solution to (1-(13 can be written as u(x, t = e ax+βt ( nπ b n e λnt sin l x. 9
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