Dirichlet boundary conditions: Getting some solutions For each λ n we consider the equation in the t variable T ′′ ( t ) + k 2 λT ( t ) = 0 Thus, for each λ n we get a solution for T given by � knπ + β n L � knπ � � T n ( t ) = α n cos L t knπ sin L t , 5 / 40
Dirichlet boundary conditions: Getting some solutions For each λ n we consider the equation in the t variable T ′′ ( t ) + k 2 λT ( t ) = 0 Thus, for each λ n we get a solution for T given by � knπ + β n L � knπ � � T n ( t ) = α n cos L t knπ sin L t , where α n and β n are real numbers. 5 / 40
Dirichlet boundary conditions: Getting some solutions For each λ n we consider the equation in the t variable T ′′ ( t ) + k 2 λT ( t ) = 0 Thus, for each λ n we get a solution for T given by � knπ + β n L � knπ � � T n ( t ) = α n cos L t knπ sin L t , where α n and β n are real numbers. Thus, we get a solution for each n ≥ 1 5 / 40
Dirichlet boundary conditions: Getting some solutions For each λ n we consider the equation in the t variable T ′′ ( t ) + k 2 λT ( t ) = 0 Thus, for each λ n we get a solution for T given by � knπ + β n L � knπ � � T n ( t ) = α n cos L t knπ sin L t , where α n and β n are real numbers. Thus, we get a solution for each n ≥ 1 � knπ + β n L � knπ sin nπx � � �� u n ( x, t ) = T n ( t ) X n ( x ) = α n cos knπ sin L t L t L 5 / 40
Dirichlet boundary conditions: Formal solution From the above we conclude that one possible solution of the wave equation with boundary conditions is, 6 / 40
Dirichlet boundary conditions: Formal solution From the above we conclude that one possible solution of the wave equation with boundary conditions is, � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos L t knπ sin L t L . n ≥ 1 6 / 40
Dirichlet boundary conditions: Formal solution From the above we conclude that one possible solution of the wave equation with boundary conditions is, � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos L t knπ sin L t L . n ≥ 1 This function satisfies 6 / 40
Dirichlet boundary conditions: Formal solution From the above we conclude that one possible solution of the wave equation with boundary conditions is, � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos L t knπ sin L t L . n ≥ 1 This function satisfies α n sin nπx � u ( x, 0) = and L n ≥ 1 β n sin nπx � u t ( x, 0) = L . n ≥ 1 6 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by 7 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n sin nπx � f ( x ) = and L n ≥ 1 β n sin nπx � g ( x ) = L . n ≥ 1 7 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n sin nπx � f ( x ) = and L n ≥ 1 β n sin nπx � g ( x ) = L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. 7 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n sin nπx � f ( x ) = and L n ≥ 1 β n sin nπx � g ( x ) = L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. Definition Consider the wave equation with initial and boundary values given by 7 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n sin nπx � f ( x ) = and L n ≥ 1 β n sin nπx � g ( x ) = L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. Definition Consider the wave equation with initial and boundary values given by u tt = k 2 u xx 0 < x < L, t > 0 7 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n sin nπx � f ( x ) = and L n ≥ 1 β n sin nπx � g ( x ) = L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. Definition Consider the wave equation with initial and boundary values given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 7 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n sin nπx � f ( x ) = and L n ≥ 1 β n sin nπx � g ( x ) = L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. Definition Consider the wave equation with initial and boundary values given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L 7 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n sin nπx � f ( x ) = and L n ≥ 1 β n sin nπx � g ( x ) = L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. Definition Consider the wave equation with initial and boundary values given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L 7 / 40
Dirichlet boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n sin nπx � f ( x ) = and L n ≥ 1 β n sin nπx � g ( x ) = L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. Definition Consider the wave equation with initial and boundary values given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L 7 / 40
Dirichlet boundary conditions: Formal solution Definition (continued) The formal solution of the above problem is 8 / 40
Dirichlet boundary conditions: Formal solution Definition (continued) The formal solution of the above problem is � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos knπ sin L t L t L . n ≥ 1 8 / 40
Dirichlet boundary conditions: Formal solution Definition (continued) The formal solution of the above problem is � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos knπ sin L t L t L . n ≥ 1 where 8 / 40
Dirichlet boundary conditions: Formal solution Definition (continued) The formal solution of the above problem is � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos knπ sin L t L t L . n ≥ 1 where � L α n = 2 f ( x ) sin nπx and dx L L 0 � L β n = 2 g ( x ) sin nπx dx. L L 0 8 / 40
Dirichlet boundary conditions: Formal solution Definition (continued) The formal solution of the above problem is � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos knπ sin L t L t L . n ≥ 1 where � L α n = 2 f ( x ) sin nπx and dx L L 0 � L β n = 2 g ( x ) sin nπx dx. L L 0 We say u ( x, t ) is a formal solution, since the series for u ( x, t ) may NOT make sense, or it may not make sense to differentiate it term wise. 8 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by 9 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 9 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 9 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L 9 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L 9 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L has an actual solution, which is given by 9 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L has an actual solution, which is given by � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos L t knπ sin L t L . n ≥ 1 9 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L has an actual solution, which is given by � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos L t knπ sin L t L . n ≥ 1 where 9 / 40
Dirichlet boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] such that f (0) = f ( L ) = 0 . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L has an actual solution, which is given by � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos L t knπ sin L t L . n ≥ 1 where � L � L α n = 2 f ( x ) sin nπx β n = 2 g ( x ) sin nπx dx and dx. L L L L 0 0 9 / 40
Dirichlet boundary conditions: Example Example Consider the wave equation with initial and boundary value given by 10 / 40
Dirichlet boundary conditions: Example Example Consider the wave equation with initial and boundary value given by u tt = 5 u xx 0 < x < 1 , t > 0 10 / 40
Dirichlet boundary conditions: Example Example Consider the wave equation with initial and boundary value given by u tt = 5 u xx 0 < x < 1 , t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 10 / 40
Dirichlet boundary conditions: Example Example Consider the wave equation with initial and boundary value given by u tt = 5 u xx 0 < x < 1 , t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = sin πx + 3 sin 5 πx 0 ≤ x ≤ 1 10 / 40
Dirichlet boundary conditions: Example Example Consider the wave equation with initial and boundary value given by u tt = 5 u xx 0 < x < 1 , t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = sin πx + 3 sin 5 πx 0 ≤ x ≤ 1 u t ( x, 0) = sin 5 πx − 26 sin 9 πx 0 ≤ x ≤ 1 10 / 40
Dirichlet boundary conditions: Example Example Consider the wave equation with initial and boundary value given by u tt = 5 u xx 0 < x < 1 , t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = sin πx + 3 sin 5 πx 0 ≤ x ≤ 1 u t ( x, 0) = sin 5 πx − 26 sin 9 πx 0 ≤ x ≤ 1 Since both f and g are given by their Fourier series in the above example, it is clear that α 1 = 1 β 1 = 0 α 5 = 3 β 5 = 1 α 9 = 0 β 9 = − 26 10 / 40
Dirichlet boundary conditions: Example Example Consider the wave equation with initial and boundary value given by u tt = 5 u xx 0 < x < 1 , t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = sin πx + 3 sin 5 πx 0 ≤ x ≤ 1 u t ( x, 0) = sin 5 πx − 26 sin 9 πx 0 ≤ x ≤ 1 Since both f and g are given by their Fourier series in the above example, it is clear that α 1 = 1 β 1 = 0 α 5 = 3 β 5 = 1 α 9 = 0 β 9 = − 26 10 / 40
Dirichlet boundary conditions: Example Example (continued) Thus, the solution to the above problem is given by √ √ u ( x, t ) = cos( 5 πt ) sin( πx ) + (3 cos( 5 πt )+ √ √ 1 5 πt )) sin(5 πx ) + − 26 5 sin( 5 sin( 9 πt ) sin(9 πx ) √ √ 5 π 9 π 11 / 40
Dirichlet boundary conditions: Formal solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u (0 , t ) = u ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L has an actual solution, which is given by � knπ + β n L � knπ sin nπx � � �� � u ( x, t ) = α n cos L t knπ sin L t L . n ≥ 1 where � L � L α n = 2 f ( x ) sin nπx β n = 2 g ( x ) sin nπx dx and dx. L L L L 0 0 12 / 40
Neumann boundary condition Consider the following differential equation u tt = k 2 u xx , 0 < x < L, t > 0 , We wish to find solutions of the above PDE which satisfy the following initial and boundary conditions The initial conditions are u ( x, 0) = f ( x ) and u t ( x, 0) = g ( x ) . The (Neumann) boundary conditions are u x (0 , t ) = u x ( L, t ) = 0 . 13 / 40
Neumann boundary conditions: Getting some solutions We will use the method of separation of variables to first find some solutions to the wave equation with boundary conditions. That is, we forget about the initial conditions for now. 14 / 40
Neumann boundary conditions: Getting some solutions We will use the method of separation of variables to first find some solutions to the wave equation with boundary conditions. That is, we forget about the initial conditions for now. Suppose u ( x, t ) = X ( x ) T ( t ) u tt = k 2 u xx Substituting this in wave equation X ( x ) T ′′ ( t ) = k 2 X ′′ ( x ) T ( t ) . 14 / 40
Neumann boundary conditions: Getting some solutions We will use the method of separation of variables to first find some solutions to the wave equation with boundary conditions. That is, we forget about the initial conditions for now. Suppose u ( x, t ) = X ( x ) T ( t ) u tt = k 2 u xx Substituting this in wave equation X ( x ) T ′′ ( t ) = 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 . 14 / 40
Neumann boundary conditions: Getting some solutions Thus, we get the conditions T ′′ ( t ) + k 2 λT ( t ) = 0 . X ′′ ( x ) + λX ( x ) = 0 and 15 / 40
Neumann boundary conditions: Getting some solutions Thus, we get the conditions T ′′ ( t ) + k 2 λT ( t ) = 0 . X ′′ ( x ) + λX ( x ) = 0 and We also have the boundary conditions u x (0 , t ) = X ′ (0) T ( t ) = 0 u x ( L, t ) = X ′ ( L ) T ( t ) = 0 . and Since we don’t want T to be identically zero, we get X ′ (0) = 0 X ′ ( L ) = 0 . and 15 / 40
Neumann boundary conditions: Getting some solutions Thus, we get the conditions T ′′ ( t ) + k 2 λT ( t ) = 0 . X ′′ ( x ) + λX ( x ) = 0 and We also have the boundary conditions u x (0 , t ) = X ′ (0) T ( t ) = 0 u x ( L, t ) = X ′ ( L ) T ( t ) = 0 . and Since we don’t want T to be identically zero, we get X ′ (0) = 0 X ′ ( L ) = 0 . and First let us solve the eigenvalue problem X ′′ ( x ) + λX ( x ) = 0 X ′ (0) = X ′ ( L ) = 0 , Recall from the section on eigenvalue problems, that we need that λ ≥ 0 . The solutions to this problem are given by λ n = n 2 π 2 X n = cos nπx n ≥ 0 L , n ≥ 0 . L 2 15 / 40
Neumann boundary conditions: Getting some solutions For each λ n we consider the equation in the t variable T ′′ ( t ) + k 2 λ n T ( t ) = 0 For n = 0 we get T 0 ( t ) = β 0 t + α 0 16 / 40
Neumann boundary conditions: Getting some solutions For each λ n we consider the equation in the t variable T ′′ ( t ) + k 2 λ n T ( t ) = 0 For n = 0 we get T 0 ( t ) = β 0 t + α 0 For each n ≥ 1 we get a solution for T given by � knπ + β n L � knπ � � T n ( t ) = α n cos L t knπ sin L t , where α n and β n are real numbers. 16 / 40
Neumann boundary conditions: Getting some solutions For each λ n we consider the equation in the t variable T ′′ ( t ) + k 2 λ n T ( t ) = 0 For n = 0 we get T 0 ( t ) = β 0 t + α 0 For each n ≥ 1 we get a solution for T given by � knπ + β n L � knπ � � T n ( t ) = α n cos L t knπ sin L t , where α n and β n are real numbers. Thus, we get a solution for each n ≥ 1 � knπ + β n L � knπ cos nπx � � �� u n ( x, t ) = T n ( t ) X n ( x ) = α n cos knπ sin L t L t L 16 / 40
Neumann boundary conditions: Formal solution For n = 0 we get u 0 ( x, t ) = T 0 ( t ) X 0 ( x ) = β 0 t + α 0 From the above we conclude that one possible solution of the wave equation with boundary conditions is, 17 / 40
Neumann boundary conditions: Formal solution For n = 0 we get u 0 ( x, t ) = T 0 ( t ) X 0 ( x ) = β 0 t + α 0 From the above we conclude that one possible solution of the wave equation with boundary conditions is, � knπ + β n L � knπ cos nπx � � �� � u ( x, t ) = β 0 t + α 0 + α n cos L t knπ sin L t L . n ≥ 1 17 / 40
Neumann boundary conditions: Formal solution For n = 0 we get u 0 ( x, t ) = T 0 ( t ) X 0 ( x ) = β 0 t + α 0 From the above we conclude that one possible solution of the wave equation with boundary conditions is, � knπ + β n L � knπ cos nπx � � �� � u ( x, t ) = β 0 t + α 0 + α n cos L t knπ sin L t L . n ≥ 1 This function satisfies 17 / 40
Neumann boundary conditions: Formal solution For n = 0 we get u 0 ( x, t ) = T 0 ( t ) X 0 ( x ) = β 0 t + α 0 From the above we conclude that one possible solution of the wave equation with boundary conditions is, � knπ + β n L � knπ cos nπx � � �� � u ( x, t ) = β 0 t + α 0 + α n cos L t knπ sin L t L . n ≥ 1 This function satisfies α n cos nπx � u ( x, 0) = α 0 + and L n ≥ 1 β n cos nπx � u t ( x, 0) = β 0 + L . n ≥ 1 17 / 40
Neumann boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n cos nπx � f ( x ) = α 0 + and L n ≥ 1 β n cos nπx � g ( x ) = β 0 + L . n ≥ 1 18 / 40
Neumann boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n cos nπx � f ( x ) = α 0 + and L n ≥ 1 β n cos nπx � g ( x ) = β 0 + L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. 18 / 40
Neumann boundary conditions: Formal solution Thus, if f ( x ) and g ( x ) have Fourier expansions given by α n cos nπx � f ( x ) = α 0 + and L n ≥ 1 β n cos nπx � g ( x ) = β 0 + L . n ≥ 1 then we will have solved our wave equation with the given boundary and initial conditions. Definition Consider the wave equation with initial and boundary values given by u tt = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = u x ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L 18 / 40
Neumann boundary conditions: Formal solution Definition (continued) The formal solution of the above problem is u ( x, t ) = β 0 t + α 0 + � knπ + β n L � knπ cos nπx � � �� � α n cos L t knπ sin L t L . n ≥ 1 where � L � L α 0 = 1 α n = 2 f ( x ) cos nπx f ( x ) dx and dx L L L 0 0 � L � L β 0 = 1 β n = 2 g ( x ) cos nπx g ( x ) dx dx. L L L 0 0 We say u ( x, t ) is a formal solution, since the series for u ( x, t ) may NOT make sense, or it may not make sense to differentiate it term wise. 19 / 40
Neumann boundary conditions: Actual solution Theorem Let f and g be continuous and piecewise smooth functions on [0 , L ] . Then the problem given by u tt = k 2 u xx 0 < x < L, t > 0 u x (0 , t ) = u x ( L, t ) = 0 t > 0 u ( x, 0) = f ( x ) 0 ≤ x ≤ L u t ( x, 0) = g ( x ) 0 ≤ x ≤ L has an actual solution, which is given by u ( x, t ) = β 0 t + α 0 + � knπ + β n L � knπ cos nπx � � �� � α n cos L t knπ sin L t L . n ≥ 1 where � L � L α 0 = 1 α n = 2 f ( x ) cos nπx f ( x ) dx dx and L L L 0 0 � L � L β 0 = 1 β n = 2 g ( x ) cos nπx g ( x ) dx dx. L L L 0 0 20 / 40
Neumann boundary conditions: Example Example Consider the wave equation with initial and boundary value given by u tt = 5 u xx 0 < x < 1 , t > 0 u x (0 , t ) = u x ( L, t ) = 0 t > 0 u ( x, 0) = 34 + cos πx + 3 cos 5 πx 0 ≤ x ≤ 1 u t ( x, 0) = 23 + cos 5 πx − 26 cos 9 πx 0 ≤ x ≤ 1 21 / 40
Neumann boundary conditions: Example Example Consider the wave equation with initial and boundary value given by u tt = 5 u xx 0 < x < 1 , t > 0 u x (0 , t ) = u x ( L, t ) = 0 t > 0 u ( x, 0) = 34 + cos πx + 3 cos 5 πx 0 ≤ x ≤ 1 u t ( x, 0) = 23 + cos 5 πx − 26 cos 9 πx 0 ≤ x ≤ 1 Since both f and g are given by their Fourier series in the above example, it is clear that α 0 = 34 β 0 = 23 α 1 = 1 β 1 = 0 α 5 = 3 β 5 = 1 α 9 = 0 β 9 = − 26 21 / 40
Neumann boundary conditions: Example Example (continued) Thus, the solution to the above problem is given by √ u ( x, t ) = 23 t + 34 + cos( 5 πt ) cos( πx ) √ √ 1 + (3 cos( 5 πt ) + √ 5 sin( 5 πt )) cos(5 πx ) 5 π √ − 26 5 sin( 9 πt ) cos(9 πx ) √ 9 π 22 / 40
Non homogeneous PDE: Dirichlet boundary condition Let us now consider the following PDE u tt − 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 u t ( x, 0) = g ( x ) 0 ≤ x ≤ L How do we solve this? 23 / 40
Non homogeneous PDE: Dirichlet boundary condition Let us now consider the following PDE u tt − 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 u t ( x, 0) = g ( 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 tt − k 2 z xx = G ( x, t ) z (0 , t ) = 0 z ( L, t ) = 0 z ( x, 0) = v ( x ) z t ( x, 0) = w ( x ) 23 / 40
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 . 24 / 40
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 24 / 40
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 tt − k 2 z xx = Z ′′ � � Z n ( t ) L ) L 2 n ≥ 1 24 / 40
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 tt − 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 24 / 40
Non homogeneous PDE: Dirichlet boundary condition Thus, if we need z tt − 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 25 / 40
Non homogeneous PDE: Dirichlet boundary condition Thus, if we need z tt − 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) = v ( x ) and z t ( x, 0) = w ( x ) . 25 / 40
Non homogeneous PDE: Dirichlet boundary condition Thus, if we need z tt − 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) = v ( x ) and z t ( x, 0) = w ( x ) . If b n sin nπx c n sin nπx � � v ( x ) = w ( x ) = L L n ≥ 1 n ≥ 1 then we should have that Z ′ Z n (0) = b n n (0) = c n (!) Clearly, there is a unique solution to the differential equation ( ∗ ) with initial condition (!) . 25 / 40
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