MA-207 Differential Equations II Ronnie Sebastian Department of - PowerPoint PPT Presentation

In each case, we use method of separation of variables. Suppose v ( x, t ) = X ( x ) T ( t ) u t = k 2 u xx Substituting this in the Heat equation T ′ ( t ) X ( x ) = k 2 X ′′ ( x ) T ( t ) . We can now separate the variables: X ′′ ( x ) X ( x ) = T ′ ( t ) k 2 T ( t ) The equality is between a function of x and a function of t , so both must be constant, say − λ . We need to solve T ′ ( t ) = − k 2 λT ( t ) . X ′′ ( x ) + λX ( x ) = 0 and 17 / 51

Dirichlet boundary conditions u (0 , t ) = u ( L, t ) = 0 Initial-boundary value problem is u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 , t > 0 u ( x, 0) = f ( x ) , 0 ≤ x ≤ L The endpoints of the rod are maintained at temperature 0 at all time t . 18 / 51

Dirichlet boundary conditions u (0 , t ) = u ( L, t ) = 0 Initial-boundary value problem is u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 , t > 0 u ( x, 0) = f ( x ) , 0 ≤ x ≤ L The endpoints of the rod are maintained at temperature 0 at all time t . (The rod is isolated from the surroundings except at the endpoints from where heat will be lost to the surrounding.) Assuming the solution in the form v ( x, t ) = X ( x ) T ( t ) v (0 , t ) = X (0) T ( t ) = 0 and v ( L, t ) = X ( L ) T ( t ) = 0 we don’t want T to be identically zero, we get X (0) = 0 and X ( L ) = 0 . 18 / 51

We need to solve eigenvalue problem X ′′ ( x ) + λX ( x ) = 0 , X (0) = 0 , X ( L ) = 0 , ( ∗ ) T ′ ( t ) = − k 2 λT ( t ) = ⇒ T ( t ) = exp ( − k 2 λt ) and The eigenvalues of ( ∗ ) are λ n = n 2 π 2 L 2 with associated eigenfunctions X n = sin nπx L , n ≥ 1 . 19 / 51

We need to solve eigenvalue problem X ′′ ( x ) + λX ( x ) = 0 , X (0) = 0 , X ( L ) = 0 , ( ∗ ) T ′ ( t ) = − k 2 λT ( t ) = ⇒ T ( t ) = exp ( − k 2 λt ) and The eigenvalues of ( ∗ ) are λ n = n 2 π 2 L 2 with associated eigenfunctions X n = sin nπx L , n ≥ 1 . We get infinitely many solutions for IBVP, one for each n ≥ 1 v n ( x, t ) = T n ( t ) X n ( x ) � − n 2 π 2 k 2 � sin nπx = exp t L 2 L v n ( x, 0) = sin nπx Note L 19 / 51

Therefore � − n 2 π 2 k 2 � sin nπx v n ( x, t ) = exp t L 2 L satisfies the IBVP u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 t > 0 u ( x, 0) = sin nπx 0 ≤ x ≤ L L 20 / 51

Therefore � − n 2 π 2 k 2 � sin nπx v n ( x, t ) = exp t L 2 L satisfies the IBVP u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 t > 0 u ( x, 0) = sin nπx 0 ≤ x ≤ L L More generally, if α 1 , . . . , α m are constants and m � − n 2 π 2 k 2 � sin nπx � u m ( x, t ) = α n exp t L 2 L n =1 then u m ( x, t ) satisfies the IBVP with initial condition m α n sin nπx � u m ( x, 0) = L . n =1 20 / 51

Let us consider the formal series ∞ � − n 2 π 2 k 2 � sin nπx � u ( x, t ) = α n exp t L 2 L n =1 21 / 51

Let us consider the formal series ∞ � − n 2 π 2 k 2 � sin nπx � u ( x, t ) = α n exp t L 2 L n =1 Setting t = 0 we get ∞ α n sin nπx � u ( x, 0) = L n =1 To solve our IBVP we would like to have ∞ α n sin nπx � f ( x ) = 0 ≤ x ≤ L L n =1 Is it possible that f has such an expansion? 21 / 51

Let us consider the formal series ∞ � − n 2 π 2 k 2 � sin nπx � u ( x, t ) = α n exp t L 2 L n =1 Setting t = 0 we get ∞ α n sin nπx � u ( x, 0) = L n =1 To solve our IBVP we would like to have ∞ α n sin nπx � f ( x ) = 0 ≤ x ≤ L L n =1 Is it possible that f has such an expansion? Given f on [0 , L ] , it has a Fourier sine series b n sin nπx � f ( x ) = L n ≥ 1 21 / 51

Definition The formal solution of IBVP u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L is ∞ � − n 2 π 2 k 2 � sin nπx � u ( x, t ) = α n exp t L 2 L n =1 where ∞ α n sin nπx � S ( x ) = L n =1 is the Fourier sine series of f on [0 , L ] i.e. � L α n = 2 f ( x ) sin nπx dx. L L 0 22 / 51

∞ � − n 2 π 2 k 2 � sin nπx � u ( x, t ) = α n exp t L 2 L n =1 We say u ( x, t ) is a formal solution, since the series for u ( x, t ) may NOT satisfy all the requirements of IBVP. When it does, we say it is an actual solution of IBVP. 23 / 51

∞ � − n 2 π 2 k 2 � sin nπx � u ( x, t ) = α n exp t L 2 L n =1 We say u ( x, t ) is a formal solution, since the series for u ( x, t ) may NOT satisfy all the requirements of IBVP. When it does, we say it is an actual solution of IBVP. Because of negative exponential in u ( x, t ) , the series in u ( x, t ) converges for all t > 0 . Each term in u ( x, t ) satisfies the heat equation and boundary condition. If u t and u xx can be obtained by differentiating the series term by term, once w.r.t. t and twice w.r.t. x for t > 0 , then u also satisfies these properties. If f ( x ) is continuous and piecewise smooth on [0 , L ] , then we can do it. Hence we get next result. 23 / 51

Theorem f ( x ) : continuous and piecewise smooth on [0 , L ] f (0) = f ( L ) = 0 � L ∞ α n sin nπx with α n = 2 f ( x ) sin nπx � S ( x ) = dx L L L 0 n =1 is Fourier sine series of f on [0 , L ] . Then the IBVP u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L has a solution ∞ � − n 2 π 2 k 2 � sin nπx � u ( x, t ) = α n exp t L 2 L n =1 Here u t and u xx can be obtained by term-wise differentiation for t > 0 . 24 / 51

Example Let f ( x ) = x ( x 2 − 3 Lx + 2 L 2 ) . Solve IBVP u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L 25 / 51

Example Let f ( x ) = x ( x 2 − 3 Lx + 2 L 2 ) . Solve IBVP u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L The Fourier sine expansion of f ( x ) is ∞ S ( x ) = 12 L 3 n 3 sin nπx 1 � L . π 3 n =1 25 / 51

Example Let f ( x ) = x ( x 2 − 3 Lx + 2 L 2 ) . Solve IBVP u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L The Fourier sine expansion of f ( x ) is ∞ S ( x ) = 12 L 3 n 3 sin nπx 1 � L . π 3 n =1 Therefore, the solution of IBVP is ∞ u ( x, t ) = 12 L 3 � − n 2 π 2 k 2 � 1 sin nπx � n 3 exp t L . π 3 L 2 n =1 � 25 / 51

Neumann boundary conditions Initial-boundary value problem is u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 , t > 0 u ( x, 0) = f ( x ) , 0 ≤ x ≤ L 26 / 51

Neumann boundary conditions Initial-boundary value problem is u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 , t > 0 u ( x, 0) = f ( x ) , 0 ≤ x ≤ L Assuming the solution in the form v ( x, t ) = X ( x ) T ( t ) v x (0 , t ) = X ′ (0) T ( t ) = 0 v x ( L, t ) = X ′ ( L ) T ( t ) = 0 and we don’t want T to be identically zero, we get X ′ (0) = 0 X ′ ( L ) = 0 . and 26 / 51

Neumann boundary conditions Initial-boundary value problem is u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 , t > 0 u ( x, 0) = f ( x ) , 0 ≤ x ≤ L Assuming the solution in the form v ( x, t ) = X ( x ) T ( t ) v x (0 , t ) = X ′ (0) T ( t ) = 0 v x ( L, t ) = X ′ ( L ) T ( t ) = 0 and we don’t want T to be identically zero, we get X ′ (0) = 0 X ′ ( L ) = 0 . and We need to solve eigenvalue problem X ′′ ( x ) + λX ( x ) = 0 , X ′ (0) = 0 , X ′ ( L ) = 0 , ( ∗ ) T ′ ( t ) = − k 2 λT ( t ) = ⇒ T ( t ) = exp ( − k 2 λt ) and 26 / 51

The eigenvalues of ( ∗ ) are λ n = n 2 π 2 L 2 with associated eigenfunctions X n = cos nπx L , n ≥ 0 . 27 / 51

The eigenvalues of ( ∗ ) are λ n = n 2 π 2 L 2 with associated eigenfunctions X n = cos nπx L , n ≥ 0 . We get infinitely many solutions for IBVP, one for each n ≥ 0 v n ( x, t ) = T n ( t ) X n ( x ) � − n 2 π 2 k 2 � cos nπx = exp t L 2 L 27 / 51

The eigenvalues of ( ∗ ) are λ n = n 2 π 2 L 2 with associated eigenfunctions X n = cos nπx L , n ≥ 0 . We get infinitely many solutions for IBVP, one for each n ≥ 0 v n ( x, t ) = T n ( t ) X n ( x ) � − n 2 π 2 k 2 � cos nπx = exp t L 2 L v n ( x, 0) = cos nπx Note L 27 / 51

The eigenvalues of ( ∗ ) are λ n = n 2 π 2 L 2 with associated eigenfunctions X n = cos nπx L , n ≥ 0 . We get infinitely many solutions for IBVP, one for each n ≥ 0 v n ( x, t ) = T n ( t ) X n ( x ) � − n 2 π 2 k 2 � cos nπx = exp t L 2 L v n ( x, 0) = cos nπx Note L Therefore � − n 2 π 2 k 2 � cos nπx v n ( x, t ) = exp t L 2 L 27 / 51

satisfies the IBVP u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 t > 0 u ( x, 0) = cos nπx 0 ≤ x ≤ L L 28 / 51

satisfies the IBVP u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 t > 0 u ( x, 0) = cos nπx 0 ≤ x ≤ L L More generally, if α 0 , . . . , α m are constants and m � − n 2 π 2 k 2 � cos nπx � u m ( x, t ) = α n exp t L 2 L n =0 then u m ( x, t ) satisfies the IBVP with initial condition m α n cos nπx � u m ( x, 0) = L . n =0 28 / 51

Let us consider the formal series ∞ � − n 2 π 2 k 2 � cos nπx � u ( x, t ) = α n exp t L 2 L n =0 29 / 51

Let us consider the formal series ∞ � − n 2 π 2 k 2 � cos nπx � u ( x, t ) = α n exp t L 2 L n =0 Setting t = 0 we get ∞ α n cos nπx � u ( x, 0) = L n =0 To solve our IBVP we would like to have ∞ α n cos nπx � f ( x ) = 0 ≤ x ≤ L L n =0 Is it possible that f has such an expansion? 29 / 51

Let us consider the formal series ∞ � − n 2 π 2 k 2 � cos nπx � u ( x, t ) = α n exp t L 2 L n =0 Setting t = 0 we get ∞ α n cos nπx � u ( x, 0) = L n =0 To solve our IBVP we would like to have ∞ α n cos nπx � f ( x ) = 0 ≤ x ≤ L L n =0 Is it possible that f has such an expansion? Given f on [0 , L ] , it has a Fourier cosine series a n cos nπx � f ( x ) = L n ≥ 0 29 / 51

Definition The formal solution of IBVP u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L is ∞ � − n 2 π 2 k 2 � cos nπx � u ( x, t ) = α n exp t L 2 L n =0 where ∞ α n cos nπx � S ( x ) = L n =0 is the Fourier sine series of f on [0 , L ] i.e. � L � L α 0 = 1 α n = 2 f ( x ) cos nπx f ( x ) dx dx. L L L 0 0 30 / 51

∞ � − n 2 π 2 k 2 � cos nπx � u ( x, t ) = α n exp t L 2 L n =0 We say u ( x, t ) is a formal solution, since the series for u ( x, t ) may NOT satisfy all the requirements of IBVP. When it does, we say it is an actual solution of IBVP. 31 / 51

∞ � − n 2 π 2 k 2 � cos nπx � u ( x, t ) = α n exp t L 2 L n =0 We say u ( x, t ) is a formal solution, since the series for u ( x, t ) may NOT satisfy all the requirements of IBVP. When it does, we say it is an actual solution of IBVP. Because of negative exponential in u ( x, t ) , the series in u ( x, t ) converges for all t > 0 . Each term in u ( x, t ) satisfies the heat equation and boundary condition. If u t and u xx can be obtained by differentiating the series term by term, once w.r.t. t and twice w.r.t. x for t > 0 , then u also satisfies these properties. If f ( x ) is continuous and piecewise smooth on [0 , L ] , then we can do it. Hence we get next result. 31 / 51

Theorem f ( x ) is continuous, piecewise smooth on [0 , L ] ; f ′ (0) = f ′ ( L ) = 0 . ∞ α n cos nπx � S ( x ) = with L n =1 � L � L α 0 = 1 α n = 2 f ( x ) cos nπx f ( x ) dx dx L L L 0 0 is Fourier sine series of f on [0 , L ] . Then the IBVP u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L has a solution ∞ � − n 2 π 2 k 2 � cos nπx � u ( x, t ) = α n exp t L 2 L n =0 Here u t and u xx can be obtained by term-wise differentiation for t > 0 . 32 / 51

Example Let f ( x ) = x on [0 , L ] . Solve IBVP u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L 33 / 51

Example Let f ( x ) = x on [0 , L ] . Solve IBVP u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L The Fourier cosine expansion of f ( x ) is ∞ C ( x ) = L 2 − 4 L (2 n − 1) 2 cos (2 n − 1) πx 1 � . π 2 L n =1 33 / 51

Example Let f ( x ) = x on [0 , L ] . Solve IBVP u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L The Fourier cosine expansion of f ( x ) is ∞ C ( x ) = L 2 − 4 L (2 n − 1) 2 cos (2 n − 1) πx 1 � . π 2 L n =1 Therefore, the solution of IBVP is u ( x, t ) = ∞ � − (2 n − 1) 2 π 2 k 2 � L 2 − 4 L 1 cos (2 n − 1) nπx � (2 n − 1) 2 exp t . π 2 L 2 L n =1 33 / 51

Definition (Formal solution for Dirichlet boundary ) The formal solution of IBVP u t = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = 0 t > 0 u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L is ∞ � − n 2 π 2 k 2 � sin nπx � u ( x, t ) = α n exp t L 2 L n =1 where ∞ α n sin nπx � S ( x ) = L n =1 is the Fourier sine series of f on [0 , L ] i.e. � L α n = 2 f ( x ) sin nπx dx. L L 0 34 / 51

Definition (Formal solution for Neumann boundary condition) The formal solution of IBVP u t = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = 0 t > 0 u x ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L is ∞ � − n 2 π 2 k 2 � cos nπx � u ( x, t ) = α n exp t L 2 L n =0 where ∞ α n cos nπx � S ( x ) = L n =0 is the Fourier cosine series of f on [0 , L ] i.e. � L � L α 0 = 1 α n = 2 f ( x ) cos nπx f ( x ) dx dx. L L L 0 0 35 / 51

Non homogeneous PDE: Dirichlet boundary condition Let us now consider the following PDE u t − k 2 u xx = F ( x, t ) 0 < x < L, t > 0 u (0 , t ) = f 1 ( t ) t > 0 u ( L, t ) = f 2 ( t ) t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L How do we solve this? 36 / 51

Non homogeneous PDE: Dirichlet boundary condition Let us now consider the following PDE u t − k 2 u xx = F ( x, t ) 0 < x < L, t > 0 u (0 , t ) = f 1 ( t ) t > 0 u ( L, t ) = f 2 ( t ) t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L How do we solve this? Let us first make the substitution z ( x, t ) = u ( x, t ) − (1 − x L ) f 1 ( t ) − x Lf 2 ( t ) Then clearly z t − k 2 z xx = G ( x, t ) z (0 , t ) = 0 z ( L, t ) = 0 z ( x, 0) = g ( x ) 36 / 51

Non homogeneous PDE: Dirichlet boundary condition It is clear that we would have solved for u iff we have solved for z . In view of this observation, let us try and solve the problem for z . 37 / 51

Non homogeneous PDE: Dirichlet boundary condition It is clear that we would have solved for u iff we have solved for z . In view of this observation, let us try and solve the problem for z . By observing the boundary conditions, we guess that we should try and look for a solution of the type Z n ( t ) sin( nπx � z ( x, t ) = L ) n ≥ 1 37 / 51

Non homogeneous PDE: Dirichlet boundary condition It is clear that we would have solved for u iff we have solved for z . In view of this observation, let us try and solve the problem for z . By observing the boundary conditions, we guess that we should try and look for a solution of the type Z n ( t ) sin( nπx � z ( x, t ) = L ) n ≥ 1 Differentiating the above term by term we get that is satisfies the equation n ( t ) + k 2 n 2 π 2 sin( nπx � z t − k 2 z xx = Z ′ � � Z n ( t ) L ) L 2 n ≥ 1 37 / 51

Non homogeneous PDE: Dirichlet boundary condition It is clear that we would have solved for u iff we have solved for z . In view of this observation, let us try and solve the problem for z . By observing the boundary conditions, we guess that we should try and look for a solution of the type Z n ( t ) sin( nπx � z ( x, t ) = L ) n ≥ 1 Differentiating the above term by term we get that is satisfies the equation n ( t ) + k 2 n 2 π 2 sin( nπx � z t − k 2 z xx = Z ′ � � Z n ( t ) L ) L 2 n ≥ 1 Let us write G n ( t ) sin( nπx � G ( x, t ) = L ) n ≥ 1 37 / 51

Non homogeneous PDE: Dirichlet boundary condition Thus, if we need z t − k 2 z xx = G ( x, t ) then we should have that n ( t ) + k 2 n 2 π 2 G n ( t ) = Z ′ Z n ( t ) ( ∗ ) L 2 38 / 51

Non homogeneous PDE: Dirichlet boundary condition Thus, if we need z t − k 2 z xx = G ( x, t ) then we should have that n ( t ) + k 2 n 2 π 2 G n ( t ) = Z ′ Z n ( t ) ( ∗ ) L 2 We also need that z ( x, 0) = g ( x ) . 38 / 51

Non homogeneous PDE: Dirichlet boundary condition Thus, if we need z t − k 2 z xx = G ( x, t ) then we should have that n ( t ) + k 2 n 2 π 2 G n ( t ) = Z ′ Z n ( t ) ( ∗ ) L 2 We also need that z ( x, 0) = g ( x ) . If b n sin nπx � g ( x ) = L n ≥ 1 then we should have that Z n (0) = b n (!) Clearly, there is a unique solution to the differential equation ( ∗ ) with initial condition (!) . 38 / 51

Non homogeneous PDE: Dirichlet boundary condition The solution to the above equation is given by � t Z n ( t ) = Ce − k 2 n 2 π 2 t + e − k 2 n 2 π 2 k 2 n 2 π 2 t s ds G n ( s ) e L 2 L 2 L 2 0 We can find the constant using the initial condition. 39 / 51

Non homogeneous PDE: Dirichlet boundary condition The solution to the above equation is given by � t Z n ( t ) = Ce − k 2 n 2 π 2 t + e − k 2 n 2 π 2 k 2 n 2 π 2 t s ds G n ( s ) e L 2 L 2 L 2 0 We can find the constant using the initial condition. Thus, we let Z n ( t ) be this unique solution, then the series Z n ( t ) sin( nπx � z ( x, t ) = L ) n ≥ 1 solves our non homogeneous PDE with Dirichlet boundary conditions for z . 39 / 51

Non homogeneous PDE: Dirichlet boundary condition Example Let us now consider the following PDE u t − u xx = e t 0 < x < 1 , t > 0 u (0 , t ) = 0 t > 0 u (1 , t ) = 0 t > 0 u ( x, 0) = x ( x − 1) 0 ≤ x ≤ 1 40 / 51

Non homogeneous PDE: Dirichlet boundary condition Example Let us now consider the following PDE u t − u xx = e t 0 < x < 1 , t > 0 u (0 , t ) = 0 t > 0 u (1 , t ) = 0 t > 0 u ( x, 0) = x ( x − 1) 0 ≤ x ≤ 1 From the boundary conditions u (0 , t ) = u (1 , t ) = 0 it is clear that we should look for solution in terms of Fourier sine series. 40 / 51

Non homogeneous PDE: Dirichlet boundary condition Example Let us now consider the following PDE u t − u xx = e t 0 < x < 1 , t > 0 u (0 , t ) = 0 t > 0 u (1 , t ) = 0 t > 0 u ( x, 0) = x ( x − 1) 0 ≤ x ≤ 1 From the boundary conditions u (0 , t ) = u (1 , t ) = 0 it is clear that we should look for solution in terms of Fourier sine series. The Fourier sine series of F ( x, t ) is given by (for n ≥ 1 ) � 1 F n ( t ) = 2 F ( x, t ) sin nπx dx 0 � 1 e t sin nπx dx = 2 0 = 2(1 − ( − 1) n ) e t nπ 40 / 51

Non homogeneous PDE: Dirichlet boundary condition Example (continued ...) Thus, the Fourier series for e t is given by 2(1 − ( − 1) n ) e t = e t sin nπx � nπ n ≥ 1 41 / 51

Non homogeneous PDE: Dirichlet boundary condition Example (continued ...) Thus, the Fourier series for e t is given by 2(1 − ( − 1) n ) e t = e t sin nπx � nπ n ≥ 1 The Fourier sine series for f ( x ) = x ( x − 1) is given by 4(( − 1) n − 1) � x ( x − 1) = sin nπx ( nπ ) 3 n ≥ 1 41 / 51

Non homogeneous PDE: Dirichlet boundary condition Example (continued ...) Thus, the Fourier series for e t is given by 2(1 − ( − 1) n ) e t = e t sin nπx � nπ n ≥ 1 The Fourier sine series for f ( x ) = x ( x − 1) is given by 4(( − 1) n − 1) � x ( x − 1) = sin nπx ( nπ ) 3 n ≥ 1 Substitute u ( x, t ) = � n ≥ 1 u n ( t ) sin nπx into the equation u t − u xx = e t 2(1 − ( − 1) n ) e t sin nπx � � u ′ n ( t ) + n 2 π 2 u n ( t ) � � sin nπx = nπ n ≥ 1 n ≥ 1 41 / 51

Non homogeneous PDE: Dirichlet boundary condition Example (continued ...) Thus, for n ≥ 1 and even we get n ( t ) + n 2 π 2 u n ( t ) = 0 u ′ that is, u n ( t ) = C n e − n 2 π 2 t 42 / 51

Non homogeneous PDE: Dirichlet boundary condition Example (continued ...) Thus, for n ≥ 1 and even we get n ( t ) + n 2 π 2 u n ( t ) = 0 u ′ that is, u n ( t ) = C n e − n 2 π 2 t If n ≥ 1 and even, we have that the Fourier coefficient of x ( x − 1) is 0. Thus, when we put u n (0) = 0 we get C n = 0 . For n ≥ 1 odd we get n ( t ) + n 2 π 2 u n ( t ) = 4 nπe t u ′ that is, � t 4 u n ( t ) = e − n 2 π 2 t nπe s e n 2 π 2 s ds + C n e − n 2 π 2 t 0 42 / 51

Non homogeneous PDE: Dirichlet boundary condition Example (continued ...) If n ≥ 1 and odd, we have the Fourier coefficient of x ( x − 1) is − 8 ( nπ ) 3 . Thus, we get − 8 u n (0) = C n = ( nπ ) 3 Thus, the solution we are looking for is � t 4 � e − (2 n +1) 2 π 2 t (2 n + 1) πe s e (2 n +1) 2 π 2 s ds + � u ( x, t ) = 0 n ≥ 0 − 8 ((2 n + 1) π ) 3 e − n 2 π 2 t � sin(2 n + 1) πx 43 / 51

Non homogeneous PDE: Neumann boundary condition Let us now consider the following PDE u t − k 2 u xx = F ( x, t ) 0 < x < L, t > 0 u x (0 , t ) = f 1 ( t ) t > 0 u x ( L, t ) = f 2 ( t ) t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L How do we solve this? 44 / 51

Non homogeneous PDE: Neumann boundary condition Let us now consider the following PDE u t − k 2 u xx = F ( x, t ) 0 < x < L, t > 0 u x (0 , t ) = f 1 ( t ) t > 0 u x ( L, t ) = f 2 ( t ) t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L How do we solve this?Let us first make the substitution z ( x, t ) = u ( x, t ) − ( x − x 2 2 L ) f 1 ( t ) − x 2 2 Lf 2 ( t ) Then clearly z t − k 2 z xx = G ( x, t ) z x (0 , t ) = 0 z x ( L, t ) = 0 z ( x, 0) = g ( x ) 44 / 51

Non homogeneous PDE: Neumann boundary condition It is clear that we would have solved for u iff we have solved for z . In view of this observation, let us try and solve the problem for z . 45 / 51

Non homogeneous PDE: Neumann boundary condition It is clear that we would have solved for u iff we have solved for z . In view of this observation, let us try and solve the problem for z . By observing the boundary conditions, we guess that we should try and look for a solution of the type Z n ( t ) cos( nπx � z ( x, t ) = L ) n ≥ 0 45 / 51

Non homogeneous PDE: Neumann boundary condition It is clear that we would have solved for u iff we have solved for z . In view of this observation, let us try and solve the problem for z . By observing the boundary conditions, we guess that we should try and look for a solution of the type Z n ( t ) cos( nπx � z ( x, t ) = L ) n ≥ 0 Differentiating the above term by term we get that is satisfies the equation n ( t ) + k 2 n 2 π 2 cos( nπx � z t − k 2 z xx = Z ′ � � Z n ( t ) L ) L 2 n ≥ 0 45 / 51

Non homogeneous PDE: Neumann boundary condition It is clear that we would have solved for u iff we have solved for z . In view of this observation, let us try and solve the problem for z . By observing the boundary conditions, we guess that we should try and look for a solution of the type Z n ( t ) cos( nπx � z ( x, t ) = L ) n ≥ 0 Differentiating the above term by term we get that is satisfies the equation n ( t ) + k 2 n 2 π 2 cos( nπx � z t − k 2 z xx = Z ′ � � Z n ( t ) L ) L 2 n ≥ 0 Let us write G n ( t ) cos( nπx � G ( x, t ) = L ) n ≥ 0 45 / 51

Non homogeneous PDE: Neumann boundary condition Thus, if we need z t − k 2 z xx = G ( x, t ) then we should have that n ( t ) + k 2 n 2 π 2 G n ( t ) = Z ′ Z n ( t ) ( ∗ ) L 2 46 / 51

Non homogeneous PDE: Neumann boundary condition Thus, if we need z t − k 2 z xx = G ( x, t ) then we should have that n ( t ) + k 2 n 2 π 2 G n ( t ) = Z ′ Z n ( t ) ( ∗ ) L 2 We also need that z ( x, 0) = g ( x ) . 46 / 51

Non homogeneous PDE: Neumann boundary condition Thus, if we need z t − k 2 z xx = G ( x, t ) then we should have that n ( t ) + k 2 n 2 π 2 G n ( t ) = Z ′ Z n ( t ) ( ∗ ) L 2 We also need that z ( x, 0) = g ( x ) . If b n cos nπx � g ( x ) = L n ≥ 0 then we should have that Z n (0) = b n (!) Clearly, there is a unique solution to the differential equation ( ∗ ) with initial condition (!) . 46 / 51

Non homogeneous PDE: Neumann boundary condition The solution to the above equation is given by � t Z n ( t ) = Ce − k 2 n 2 π 2 t + e − k 2 n 2 π 2 k 2 n 2 π 2 t s ds G n ( s ) e L 2 L 2 L 2 0 We can find the constant using the initial condition. 47 / 51

Non homogeneous PDE: Neumann boundary condition The solution to the above equation is given by � t Z n ( t ) = Ce − k 2 n 2 π 2 t + e − k 2 n 2 π 2 k 2 n 2 π 2 t s ds G n ( s ) e L 2 L 2 L 2 0 We can find the constant using the initial condition. Thus, we let Z n ( t ) be this unique solution, then the series Z n ( t ) cos( nπx � z ( x, t ) = L ) n ≥ 0 solves our non homogeneous PDE with Dirichlet boundary conditions for z . 47 / 51

Non homogeneous PDE: Neumann boundary condition Example Let us now consider the following PDE u t − u xx = e t 0 < x < 1 , t > 0 u x (0 , t ) = 0 t > 0 u x (1 , t ) = 0 t > 0 u ( x, 0) = x ( x − 1) 0 ≤ x ≤ 1 48 / 51

Non homogeneous PDE: Neumann boundary condition Example Let us now consider the following PDE u t − u xx = e t 0 < x < 1 , t > 0 u x (0 , t ) = 0 t > 0 u x (1 , t ) = 0 t > 0 u ( x, 0) = x ( x − 1) 0 ≤ x ≤ 1 From the boundary conditions u x (0 , t ) = u x (1 , t ) = 0 it is clear that we should look for solution in terms of Fourier cosine series. 48 / 51

Non homogeneous PDE: Neumann boundary condition Example Let us now consider the following PDE u t − u xx = e t 0 < x < 1 , t > 0 u x (0 , t ) = 0 t > 0 u x (1 , t ) = 0 t > 0 u ( x, 0) = x ( x − 1) 0 ≤ x ≤ 1 From the boundary conditions u x (0 , t ) = u x (1 , t ) = 0 it is clear that we should look for solution in terms of Fourier cosine series. The Fourier cosine series of F ( x, t ) is given by (for n ≥ 0 ) � 1 � 1 e t dx = e t F 0 ( t ) = F ( x, t ) dx = 0 0 � 1 � 1 e t cos nπx dx = 0 F n ( t ) = 2 F ( x, t ) cos nπx dx = 2 n > 0 0 0 48 / 51

Non homogeneous PDE: Neumann boundary condition Example (continued ...) Thus, the Fourier series for e t is simply e t . 49 / 51

Non homogeneous PDE: Neumann boundary condition Example (continued ...) Thus, the Fourier series for e t is simply e t . The Fourier cosine series for f ( x ) = x ( x − 1) is given by 2(( − 1) n + 1) x ( x − 1) = − 1 � 6 + cos nπx ( nπ ) 2 n ≥ 1 49 / 51

Non homogeneous PDE: Neumann boundary condition Example (continued ...) Thus, the Fourier series for e t is simply e t . The Fourier cosine series for f ( x ) = x ( x − 1) is given by 2(( − 1) n + 1) x ( x − 1) = − 1 � 6 + cos nπx ( nπ ) 2 n ≥ 1 Substitute u ( x, t ) = � n ≥ 0 u n ( t ) cos nπx into the equation u t − u xx = e t � n ( t ) + n 2 π 2 u n ( t ) u ′ cos nπx = e t � � n ≥ 0 49 / 51