Explicit calculation of strong solution on linear parabolic equation
aa r X i v : . [ m a t h . A P ] N ov Explicit calculation of strong solution on linear parabolicequation ∗ Xiaoping Fang † Youjun Deng ‡ Jing Li § Abstract
In this paper, we give the existence and uniqueness of the strong solution of one dimen-sional linear parabolic equation with mixed boundary conditions. The boundary conditionscan be any kind of mixed Dirichlet, Neumann and Robin boundary conditions. We use theextension method to get the unique solution. Furthermore, the method can also be easilyimplemented as a numerical method. Some simple examples are presented.
Keywords: Linear parabolic equation, Existence, Strong solution, Extension, Mixed boundarycondition
We consider the existence and uniqueness of the strong solution of the following problem u t = ku xx + F ( x, t ) ( x, t ) ∈ Ω T ,u ( x,
0) = µ ( x ) x ∈ (0 , l ) ,u x (0 , t ) = 0 (or u (0 , t ) = 0) t ∈ (0 , T ] , − ku x ( l, t ) = ν ( u ( l, t ) − T ( t )) t ∈ (0 , T ] , (1.1)where Ω T := { < x < l, < t ≤ T } , k > ν >
0. For the sake of convenience we assume that F ( x, t ), µ ( x ) and T ( t ) are all analytic functions in space and time variables, and we shall seethat the solution to (1.1) is also analytic.The mathematical model (1.1) arises in various physical and engineering settings, in particularin hydrology [1], material sciences [12], heat transfer [13] and transport problems [17]. In the caseof heat transfer, the term F ( x, t ) stands for the heat source. In the mean time the Robin boundarycondition in (1.1) is quite physically meaningful. Suppose that the end x = l of the rod is exposedto air or other fluid with temperature T ( t ), then u ( l, t ) − T ( t ) is the temperature differencebetween the rod and its surroundings at x = l . By the Newton’s law of cooling, which states thatthe rate of change of the temperature of an object is proportional to the difference between itsown temperature and the ambient temperature (i.e., the temperature of its surroundings), we getthat − ku x ( l, t ) = ν [ u ( l, t ) − T ( t )]. The constant ν > Neumann-Robin boundary problem if boundary condition u x (0 , t ) = 0 is used and Dirichlet-Robin boundary problem if u (0 , t ) = 0 is used.Not only doest the direct heat conduction problem attract lots of researchers’ attentions, butalso the inverse problem. The inverse problems of the recovery of unknown sources in (1.1) fromoverspecified boundary data or final overdetermination have attracted many mathematicians fordecades since these problems are of paramount importance in heat conduction processes (see, ∗ Fang is supported by NSF grants No. NSFC70921001 and No. NSFC71210003. Deng and Li are supported byNSF grants No. NSFC11301040, † Postdoctoral, Management Science and Engineering Postdoctoral Mobile Station, School of Business; Schoolof Mathematics and Statistics, Central South University, Changsha, Hunan 410083, P. R. China. Email: [email protected] ‡ Corresponding author. School of Mathematics and Statistics, Central South University, Changsha, Hunan410083, P. R. China. Email: [email protected], dengyijun [email protected] § Department of Mathematics, Changsha University of Science and Technology, Changsha, Hunan 410004,P.R.China. Email: [email protected] F ( x, t ) and T ( t ) with final overspecified data u ( x, T ) has been studied in [6].The uniqueness of the solution to (1.1) is clear and easily proved. In the mean time, theexistence of the weak solution of similar form of (1.1) but with different boundary conditions, hasbeen proved long time ago by Riesz representation theorem (see e.g. [8–10, 14, 16]). The existenceof the strong solution has also been proved by using the maximum principle and comparisonprinciple for the weak solution in Dirichlet boundary problem (cf. [15]). However, the existenceof the strong solution of (1.1) is still not quite clear. Here we mainly consider the strong solutionof (1.1). More importantly, we shall give a direct method to solve the strong solution instead ofusing the numerical methods such as finite difference method, etc. We know that the solution of(1.1) without boundary condition would be u ( x, t ) = Z ∞−∞ √ kπt e − ( x − y )24 kt µ ( y ) dy + Z t Z ∞−∞ p kπ ( t − s ) e − ( x − y )24 k ( t − s ) F ( y, s ) dyds. (1.2)Concerning on the initial condition there hold u ( x,
0) = lim t → Z ∞−∞ √ kπt e − ( x − y )24 kt µ ( y ) dy = µ ( x ) u x ( x,
0) = − lim t → Z ∞−∞ x − y kt √ kπt e − ( x − y )24 kt µ ( y ) dy = µ ′ ( x ) . In fact, if we introduce the well-known Dirac function for δ ( x ) = √ kπt e − x kt while t = 0, we caneasily deduce the above results.In this paper we explore the extension method to get the strong solution to (1.1). The methodcan also be used in other boundary conditions such as both Neumann boundaries or Dirichletboundaries, Neumann and Dirichlet mixed boundaries, etc. We present the main theorem in this paper.
Theorem 2.1
Suppose µ ( x ) , F ( x, t ) and T ( t ) are all analytic functions. Then the solution to(1.1) has a unique solution. In particular, if F ( x, t ) = 0 and T ( t ) = P N d i t i then(1) Neumann-Robin boundary condition ( u x (0 , t ) = 0 ). The solution u to (1.1) has the form u = N X i =0 a i i X j =0 c j C j i x i − j (4 kt ) j + ∞ X n =1 b n e − σ n kt cos( σ n x ) + C T (2.1) where C T = d − kν µ ′ ( l ) − µ ( l ) and the function µ ( x ) is defined by µ ( x ) = P Ni =0 a i x i . Thecoefficients a i , i = 0 , , . . . , N are determined by l ( l − kν ) · · · l N − ( l − kNν )0 4 kc · · · l N − ( lC N c − kν C N c )4 k ... ... . . . ... · · · · · · l ( lC N − N c N − − kν C N − N c N )(4 k ) N − · · · c N (4 k ) N a a ... a N − a N = dd ... d N − d N (2.2) where d = kν µ ′ ( l ) + µ ( l ) and c j = (2 j − j − · · · j j = 0 , , . . . , N. The coefficients b n , n = 1 , , . . . , ∞ are b n = 2 ννl + k sin ( σ n l ) Z l ( µ ( x ) − µ ( x ) − C T ) cos( σ n x ) dx with σ n > , kσ n tan( σ n l ) = ν .
2) Dirichlet-Robin boundary condition ( u (0 , t ) = 0 ). The solution u to (1.1) has the form u ( x, t ) = N X i =0 ˜ a i i X j =0 c j C j i +1 x i − j +1 (4 kt ) j + ∞ X n =1 ˜ b n e − σ n kt sin(˜ σ n x ) + ˜ C T (2.3) where ˜ C T = d − kν ˜ µ ′ ( l ) − ˜ µ ( l ) . The function ˜ µ ( x ) is defined by ˜ µ ( x ) = P Ni =0 ˜ a i x i +1 and ˜ a i , i = 1 , , . . . , N are determined by l − kν l ( l − kν ) · · · l N ( l − k (2 N +1) ν )0 ( lC c − kν c )4 k · · · l N − ( lC N +1 c − kν C N +1 c )4 k ... ... . . . ... · · · · · · l ( lC N − N +1 c N − − kν C N − N +1 c N )(4 k ) N − · · · ( lC N N +1 c N − − kν )(4 k ) N ˜ a ˜ a ... ˜ a N − ˜ a N = ˜ dd ... d N − d N (2.4) where ˜ d = kν ˜ µ ′ ( l ) + ˜ µ ( l ) . The coefficients ˜ b n , n = 1 , , . . . , ∞ are given by ˜ b n = 2 ννl + k cos (˜ σ n l ) Z l ( µ ( x ) − ˜ µ ( x ) − ˜ C T ) sin(˜ σ n x ) dx where σ n > satisfy k ˜ σ n = − ν tan(˜ σ n l ) . In what follows, we shall prove the Theorem by extension method introduced in this paper. Weshall consider the Neumann-Robin problem and Dirichlet-Robin problem, separately.
For the existence of the strong solution of (1.1) with Neumann-Robin boundary condition we firstconsider the following problem u t = ku xx ( x, t ) ∈ Ω T ,u ( x,
0) = µ ( x ) x ∈ (0 , l ) ,u x (0 , t ) = 0 , − ku x ( l, t ) = ν ( u ( l, t ) − T ( t )) t ∈ (0 , T ] , (2.5)where T ( t ) = T ( t ) − T (0) + kν µ ′ ( l ) + µ ( l ) and µ ( x ) is an analytic function to be constructed.We mention that in the construction of T ( t ), compatibility condition on ( x = l, t = 0) is satisfiedautomatically. We consider the solution in the following form u ( x, t ) = Z ∞−∞ √ kπt e − ( x − y )24 kt µ ( y ) dy (2.6)then we have u x ( x, t ) = − Z ∞−∞ x − y kt √ kπt e − ( x − y )24 kt µ ( y ) dy. (2.7)The strategy we use in this paper is to construct the function µ ( x ), x ∈ (0 , l ) and then make anextension to R to cope with the boundary conditions. For the boundary condition u x (0 , t ) = 0,we extend µ ( x ) to be an even function in R , that is µ ( x ) = µ ( − x ). Clearly we have u x (0 , t ) = 12 kt √ kπt Z ∞−∞ xe − x kt µ ( x ) dx = 0 . Concerning with the boundary condition at x = l we define g ( t ) = 1 √ kπt Z ∞−∞ (cid:18) − l − y νt (cid:19) e − ( l − y )24 kt µ ( y ) dy (2.8)We note that g ( t ) ∈ C ∞ ((0 , T )). We suppose that µ ( x ) has the following form in R µ ( x ) = N X i =0 a i x i . (2.9)3ubstituting (2.9) into (2.8) we have g ( t ) = 1 √ kπt Z ∞−∞ (cid:18) − l − y νt (cid:19) e − ( l − y )24 kt N X i =0 a i y i dy = 1 √ kπt N X i =0 a i Z ∞−∞ (cid:16) − y νt (cid:17) e − y kt ( l − y ) i dy = 1 √ π N X i =0 a i Z ∞−∞ e − y ( l − √ kty ) i dy − √ kt νt √ π N X i =0 a i Z ∞−∞ ye − y ( l − √ kty ) i dy, By elementary calculation we obtain Z ∞−∞ e − y ( l − √ kty ) i dy = i X j =0 C j i l i − j (2 j − j − · · · j √ π (4 kt ) j and Z ∞−∞ ye − y ( l − √ kty ) i dy = − i X j =1 C j − i l i − j +1 (2 j − j − · · · j √ π (4 kt ) j − . Thus by setting c j = (2 j − j − ··· j , with c = 1 we have g ( t ) = N X i =0 a i l i + N X i =1 a i i X j =1 C j i l i − j c j (4 kt ) j − νt N X i =1 a i i X j =1 C j − i l i − j +1 c j (4 kt ) j = a + N X i =1 a i l i − ( l − kiν ) + N X i =1 a i i X j =1 C j i l i − j c j (4 kt ) j − kν N X i =2 a i i − X j =1 C j +12 i l i − j − c j +1 (4 kt ) j = a + N X i =1 a i l i − ( l − kiν ) + N X j =1 a j c j + n X i = j +1 a i l i − j − (cid:18) lC j i c j − kν C j +12 i c j +1 (cid:19) (4 kt ) j We see from above that g ( t ) is a polynomial of t thus if there is a selection of a i , i = 1 , , · · · suchthat g ( t ) = T ( t ) = T ( t ) − T (0) + kν µ ′ ( l ) + µ ( l )then the boundary conditions in (2.5) are satisfied. Since T ( t ) is analytic we suppose T ( t ) = P Ni =0 d i t i , then T ( t ) = kν µ ′ ( l ) + µ ( l ) + P Ni =1 d i t i and letting ˜ d := kν µ ′ ( l ) + µ ( l ) the coefficients ofthe function µ ( x ) can be calculated by solving the linear equation (2.2). We see from (2.2) thatthe matrix is an upper triangle matrix and it is nonsingular since the elements in the diagonal arenonzero. Furthermore, if l = kν , the matrix is even more simple (the elements in the secondarydiagonal of the matrix are zero). Thus for any analytic function T ( t ) we can construct a function µ ( x ) such that g ( t ) = T ( t ).Suppose we have solved problem (2.5) and the solution is u ( x, t ), u ( x, t ) = Z ∞−∞ √ kπt e − ( x − y )24 kt µ ( y ) dy = N X i =0 a i √ kπt Z ∞−∞ e − ( x − y )24 kt y i dy = N X i =0 a i √ kπt i X j =0 C j i x i − j Z ∞−∞ y j e − y kt dy = N X i =0 a i i X j =0 c j C j i x i − j (4 kt ) j (2.10)then the solution to (1.1) is u ( x, t ) = u ( x, t ) + u ( x, t ), where u ( x, t ) is the solution to u t = ku xx ( x, t ) ∈ Ω T ,u ( x,
0) = µ ( x ) − µ ( x ) x ∈ (0 , l ) ,u x (0 , t ) = 0 , − ku x ( l, t ) = ν ( u ( l, t ) − C T ) t ∈ (0 , T ] , (2.11)4here C T = T (0) − kν µ ′ ( l ) − µ ( l ). Denote ¯ u ( x, t ) = u ( x, t ) − C T then problem (2.11) turns to ¯ u t = k ¯ u xx ( x, t ) ∈ Ω T , ¯ u ( x,
0) = µ ( x ) − µ ( x ) − C T x ∈ (0 , l ) , ¯ u x (0 , t ) = 0 , − k ¯ u x ( l, t ) = ν ¯ u ( l, t ) t ∈ (0 , T ] . (2.12)We can thus use the well-known variational separation method (see, e.g. [11]), which is lookingfor simple solutions in the form ¯ u ( x, t ) = X ( x ) T ( t ) . By some elementary analysis, we get the solution form to (2.11) u ( x, t ) = ∞ X n =1 b n e − σ n kt cos( σ n x ) + C T , (2.13)with σ n > , kσ n tan( σ n l ) = ν . The choice of σ n is increasing as n increases and we have σ n → nπl as n → ∞ . It is easy to verify that (2.13) satisfies the boundary conditions. The only problemremaining here is to seek for the constants b n such that ∞ X n =1 b n cos( σ n x ) = µ ( x ) − µ ( x ) − C T . (2.14)As we know that { cos( nπl x ) } is an orthonormal basis in L ([0 , l ]). We can also prove that { cos( σ n x ) } is orthogonal in L ([0 , kσ n tan( σ n l ) = ν we have Z l cos( σ n x ) cos( σ m x ) dx = 1 σ n sin( σ n x ) cos( σ m x ) (cid:12)(cid:12)(cid:12) l + σ m σ n Z l sin( σ n x ) sin( σ m x ) dx = 1 σ n ( σ n sin( σ n l ) cos( σ m l ) − σ m sin( σ m l ) cos( σ n l ))+ σ m σ n Z l cos( σ n x ) cos( σ m x ) dx = σ m σ n Z l cos( σ n x ) cos( σ m x ) dx. Furthermore, by Robin boundary condition we know that u ( x,
0) can not be nonzero constantfunctions. Thus we have b n = R l ( µ ( x ) − µ ( x ) − C T ) cos( σ n x ) dx R l cos ( σ n x ) dx = 2 ννl + k sin ( σ n l ) Z l ( µ ( x ) − µ ( x ) − C T ) cos( σ n x ) dx. We should mention that in order to make the summation in the left hand side of (2.14) converge,some continuous conditions on µ ( x ) should be given. However, we do not give further discussionon this because it can be similarly inherited from the Fourier series theory (see, e.g., [11]).From the analysis above, we conclude that the solution to (1.1) with Neumann-Robin boundaryconditions has the form (2.1) if F ( x, t ) = 0. The strong solution to (1.1) with F ( x, t ) not identicallyzero can be solved similarly as discussed. Followed the idea above, we only need to extend F ( x, t )to be an even function in the whole x -axis, that is F ( − x, t ) = F ( x, t ) then same method can beused to get the solution of (1.1). In this case, the boundary values are u ( l, t ) = Z ∞−∞ √ kπt e − ( l − y )24 kt µ ( y ) dy + Z t Z ∞−∞ p kπ ( t − s ) e − ( l − y )24 k ( t − s ) F ( y, s ) dyds,u x ( l, t ) = − Z ∞−∞ l − y kt √ kπt e − ( l − y )24 kt µ ( y ) dy − Z t Z ∞−∞ l − y k ( t − s ) p kπ ( t − s ) e − ( l − y )24 k ( t − s ) F ( y, s ) dyds. We can actually treat T ( t ) as T ( t ) = T ( t ) − Z t Z ∞−∞ p kπ ( t − s ) e − ( l − y )24 k ( t − s ) F ( y, s ) dyds − Z t Z ∞−∞ l − y ν ( t − s ) p kπ ( t − s ) e − ( l − y )24 k ( t − s ) F ( y, s ) dyds − d + kν ˜ µ ′ ( l ) + ˜ µ ( l )(2.15)and do some same analysis to get the solution of (1.1).5 .2 Dirichlet-Robin boundary condition To solve the strong solution of (1.1) with Dirichlet-Robin boundary condition, we shall first con-sider the solution to u t = ku xx ( x, t ) ∈ Ω T ,u ( x,
0) = ˜ µ ( x ) x ∈ (0 , l ) ,u (0 , t ) = 0 , − ku x ( l, t ) = ν ( u ( l, t ) − T ( t )) t ∈ (0 , T ] . (2.16) T ( t ) = T ( t ) − d + kν ˜ µ ′ ( l ) + ˜ µ ( l ). In order to satisfy the Dirichlet boundary condition the searchof ˜ µ ( x ) is now for an odd function of the following form˜ µ ( x ) = N X i =0 ˜ a i x i +1 , (2.17)and by using the Robin boundary we compute g ( t ) g ( t ) = 1 √ kπt Z ∞−∞ (cid:18) − l − y νt (cid:19) e − ( l − y )24 kt N X i =0 a i y i +1 dy = 1 √ π N X i =0 a i Z ∞−∞ e − y ( l − √ kty ) i +1 dy − √ kt νt √ π N X i =0 a i Z ∞−∞ ye − y ( l − √ kty ) i +1 dy = N X j =0 N X i = j a i l i − j (cid:18) lC j i +1 c j − kν C j +12 i +1 c j +1 (cid:19) (4 kt ) j . By setting g ( t ) = T ( t ), we thus get the linear equation (2.4). The matrix in (2.4) is also an uppertriangle matrix with the diagonal elements nonzero if l = kν . In the case l = kν we can actuallyset ˜ µ ( x ) = P n +1 i =0 a i x i +1 with a arbitrarily given, since elements of the secondary diagonal ofthe matrix (2.4) are totally nonzero. In fact, we know if u ( x, t ) is the solution of u t = u xx then u ( x, t ) + cx with arbitrary constant c is also the solution of u t = u xx . Furthermore, the boundaryconditions are also satisfied. Then the solution to (2.16) is u ( x, t ) = Z ∞−∞ √ kπt e − ( x − y )24 kt µ ( y ) dy = n X i =0 a i √ kπt Z ∞−∞ e − ( x − y )24 kt y i +1 dy = n X i =0 a i i X j =0 c j C j i +1 x i − j +1 (4 kt ) j (2.18)Next we need further solve u t = ku xx ( x, t ) ∈ Ω T ,u ( x,
0) = µ ( x ) − ˜ µ ( x ) x ∈ (0 , l ) ,u (0 , t ) = 0 , − ku x ( l, t ) = ν ( u ( l, t ) − ˜ C T ) t ∈ (0 , T ] , (2.19)where ˜ C T = T (0) − kν ˜ µ ′ ( l ) − ˜ µ ( l ). Similarly, we can get the solution to (2.19), denoted by u ( x, t ) u ( x, t ) = ∞ X n =1 ˜ b n e − ˜ σ n kt sin(˜ σ n x ) + ˜ C T , where ˜ σ n > k ˜ σ n = − ν tan(˜ σ n l ) and˜ b n = R l (˜ µ ( x ) − ˜ µ ( x ) − ˜ C T ) sin(˜ σ n x ) dx R l sin (˜ σ n x ) dx = 2 ννl + k cos (˜ σ n l ) Z l (˜ µ ( x ) − ˜ µ ( x ) − C T ) sin(˜ σ n x ) dx. In this section we study the solution of the following problem u t = ku xx + f ( x ) ,u ( x,
0) = µ ( x ) u x (0 , t ) = 0 , u x (1 , t ) = 0 . (3.1)6irstly, by separating variables we have the solution of (3.1) as follows: u ( x, t ) = ∞ X n =0 " e − n π kt a n cos( nπx ) + 1 − e − n π kt n π k b n cos( nπx ) , (3.2)where for n = 0, a = R µ ( x ) dx , b = R f ( x ) dx , and for n > a n = 2 R µ ( x ) cos( nπx ) dx , b n = 2 R f ( x ) cos( nπx ) dx . (3.2) makes no sense for n = 0. However we can define1 − e − n π kt n π | n =0 := lim n → − e − n π kt n π k = t, to make it meaningful at n = 0. Secondly, by using the extension method in this paper, we candeduce a different solution form of (3.1) as follows: u ( x, t ) = Z ∞−∞ √ kπt e − ( x − y )24 kt µ ( y ) dy + Z t Z ∞−∞ √ kπs e − ( x − y )24 ks f ( y ) dyds, (3.3)where µ ( x ) and f ( x ) are extended to the following form by Fourier expansion µ ( x ) = Z µ ( y ) dy + 2 ∞ X n =1 cos( nπx ) Z µ ( y ) cos( nπy ) dy,f ( x ) = Z f ( y ) dy + 2 ∞ X n =1 cos( nπx ) Z f ( y ) cos( nπy ) dy. The boundary conditions are satisfied automatically by this kind of extension. The solutions (3.3)and (3.2) have totally different forms while the solution of (3.1) is unique. Thus the two differentsolution forms must be the same function. To explain this in detail, we first verify that the valuesare the same at x = 0. By (3.2) we obtain u (0 , t ) = ∞ X n =0 " e − n π kt a n + 1 − e − n π kt n π k b n . On the other hand by (3.3) we have u (0 , t ) = Z ∞−∞ √ kπt e − y kt ∞ X n =0 a n cos( nπy ) dy + Z t Z ∞−∞ √ kπs e − y ks ∞ X n =0 b n cos( nπy ) dyds = ∞ X n =0 a n Z ∞−∞ cos( nπy ) 1 √ kπt e − y kt dy + ∞ X n =0 b n Z t Z ∞−∞ cos( nπy ) 1 √ kπs e − y ks dyds. Comparing the two forms above, it is enough to verify the following
Lemma 3.1
There holds the following Z ∞−∞ cos ( nπy ) 1 √ kπt e − y kt dy = e − n π kt . (3.4) Proof . By Taylor expansion of cos( nπy ) we have Z ∞−∞ cos( nπy ) 1 √ kπt e − y kt dy = Z ∞−∞ ∞ X i =0 ( − i ( nπy ) i (2 i )! 1 √ kπt e − y kt dy = ∞ X i =0 ( − i ( nπ ) i (2 i )! Z ∞−∞ y i √ kπt e − y kt dy = ∞ X i =0 ( − i ( nπ ) i (2 i )! (4 kt ) i i Y j =1 j − ∞ X i =0 ( − i ( nπ ) i i ! ( kt ) i = e − n π kt . Z ∞−∞ cos( nπy ) 1 √ kπt e − ( x − y )24 kt dy = Z ∞−∞ cos( nπ ( x − y )) 1 √ kπt e − y kt dy = Z ∞−∞ [cos( nπx ) cos( nπy ) + sin( nπx ) sin( nπy )] 1 √ kπt e − y kt dy = cos( nπx ) Z ∞−∞ cos( nπy ) 1 √ kπt e − y kt dy = cos( nπx ) e − n π kt . (cid:3) This Lemma verifies that (3.2) and (3.3) are the same solution to (3.1) but with two differentsolution forms .
In this section we shall present some concrete examples to show how to get the strong solution to(1.1). We set l = 1, k = 1 / ν = 1 /
2, then the problem (1.1) becomes u t = u xx + F ( x, t ) ( x, t ) ∈ Ω T ,u ( x,
0) = µ ( x ) x ∈ (0 , l ) ,u x (0 , t ) = 0 , − u x (1 , t ) = 2( u (1 , t ) − T ( t )) t ∈ (0 , T ] , (4.1)and the linear equation in (2.2) turns to · · · − n c · · · C n c − C n c ... ... . . . ...0 · · · c n − C n − n c n − − C n − n c n · · · c n a a ... a n − a n = dd ... d n − d n . (4.2)Since c n = n − c n − we get C n − n c n − − C n − n c n = 0. For n = 4, the matrix appearing in thelinear equation (4.2) reads − − −
30 0 . − . −
280 0 0 .
75 0 − .
50 0 0 1 .
875 00 0 0 0 6 . In what follows, we show how to get the strong solution in four different choice of F ( x, t ), T ( t )and µ ( t ). Ex1: F ( x, t ) = 2 t + 1, T ( t ) = t + t .From (2.15), by unitary extension, we obtain T ( t ) = 0 and by (4.2) we have a = a = a =0. Thus µ ( x ) = 0. In fact, in this case, problem (1.1) is u t = u xx + 2 t + 1 ( x, t ) ∈ [0 , × [0 , T ] ,u ( x,
0) = µ ( x ) x ∈ (0 , ,u x (0 , t ) = 0 , − u x (1 , t ) = 2( u (1 , t ) − t − t ) t ∈ (0 , T ] , which can be solved by first setting ˜ u ( x, t ) = u ( x, t ) − t − t and then solving the homogeneousproblem ˜ u t = ˜ u xx ( x, t ) ∈ [0 , × [0 , T ] , ˜ u ( x,
0) = µ ( x ) x ∈ (0 , , ˜ u x (0 , t ) = 0 , − ˜ u x ( l, t ) = 2˜ u ( l, t ) t ∈ (0 , T ] . x2: F ( x, t ) = 3 t + 2 t , T ( t ) = t + t + t + 5, µ ( x ) = 2 x + 1.From (2.15), by unitary extension, we have T ( t ) = t +5. Thus a = 5, a = 2. µ ( x ) = 2 x +5.By (1.2), (2.10) and (2.15), the solution u ( x, t ) = a + a ( x + t )+ t + t = 2 x + t + t + t +5is the solution of u t = u xx + 3 t + 2 t ( x, t ) ∈ [0 , × [0 , T ] ,u ( x,
0) = 2 x + 5 x ∈ (0 , ,u x (0 , t ) = 0 , − u x (1 , t ) = 2( u (1 , t ) − t − t − t − t ∈ (0 , T ] , And we need further solve u t = u xx ( x, t ) ∈ [0 , × [0 , T ] ,u ( x,
0) = − x ∈ (0 , ,u x (0 , t ) = 0 , − u x (1 , t ) = 2( u (1 , t ) + 4) t ∈ (0 , T ] , to get the solution u ( x, t ) = −
4. Finally we have u ( x, t ) = u ( x, t ) + u ( x, t ) = 2 x + t + t + t + 1.On the other hand, if we use the zero extension outside [-1,1] then T ( t ) = (1 − erf (1))( t + t ) + t + 5. Thus a = 5 . a = 3 . a = 0 . a = 0 . µ ( x ) =0 . x + 0 . x + 3 . x + 5 . µ (1) = 8 . µ ′ (1) = 7 . u ( x, t ) = 0 . x + 0 . x + 3 . x + 5 . . . x + 0 . x ) t + 0 . x t + t + t is the solution of u t = u xx + 3 t + 2 t ( x, t ) ∈ [0 , × [0 , T ] ,u ( x,
0) = 0 . x + 0 . x + 3 . x + 5 . x ∈ (0 , ,u x (0 , t ) = 0 , − u x (1 , t ) = 2( u (1 , t ) − t − t − t − . t ∈ (0 , T ] , and we need further solve u t = u xx ( x, t ) ∈ [0 , × [0 , T ] ,u ( x,
0) = − . x − . x − . x − . x ∈ (0 , ,u x (0 , t ) = 0 , − u x (1 , t ) = 2( u (1 , t ) + 7 . t ∈ (0 , T ] . By elementary calculation we know the solution is approximately u ( x, t ) = − . x − . x − . x − . − (0 . . x + 0 . x ) t − . x t . Finally we also get the solution u ( x, t ) = u ( x, t ) + u ( x, t ) = 2 x + 1 + t + t + t . Ex3: F ( x, t ) = x (5 t + 2), T ( t ) = 3 t + 1, µ ( x ) = x + 3 x + 1.If we do not change F ( x, t ) in the whole x -axis, then from (2.15) we obtain T ( t ) = 1 +3 t − t − t . Using (4.2) we obtain a = − , a = , a = − , a = − . So µ ( x ) = − x − x + x − and µ (1) = , µ ′ (1) = . By (2.10) we have u ( x, t ) = − x − x + 83 x − − t ( 53 x −
43 )be the solution of u t = u xx + x (5 t + 2) ( x, t ) ∈ [0 , × [0 , T ] ,u ( x,
0) = − x − x + x − x ∈ (0 , ,u x (0 , t ) = 0 , − u x (1 , t ) = 2( u (1 , t ) − t − ) t ∈ (0 , T ] , and we need further solve u t = u xx ( x, t ) ∈ [0 , × [0 , T ] ,u ( x,
0) = x + x + 3 x + x + x ∈ (0 , ,u x (0 , t ) = 0 , − u x (1 , t ) = 2( u (1 , t ) + ) t ∈ (0 , T ] . Conclusions
We have shown the existence and uniqueness of the strong solution (or the classical solution)for linear parabolic equation with mixed boundary conditions. Although we only analyzed the
Neumann-Robin boundary problem and the
Dirichlet-Robin boudnary problem, the other bound-ary condition problems can also be solved similarly. Furthermore, the extension method used inthis paper can also be easily implemented as a numerical method for solving the linear parabolicequation and the efficiency of the method is quite competitive. Even if T ( t ) is not infinitely dif-ferentiable in [0 , l ] the numerical solution can also be quite accurate if T ( t ) can be approximatedwell with certain degrees of polynomials. Future works would focus on the numerical comparisonbetween the method in this paper and other numerical methods such as finite difference and finiteelement method. The method can also be extended to two dimensional or higher dimensionallinear parabolic equations in the future. References [1] J. Bear, Dynamics of Fluids in Porous Media, Elsevier, New York, 1972.[2] J.V. Beck, B. Blackwell, S.R. Chair,
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