This presentation deals with how to solve Poisson\u27s Equation for a two dimensional plate. The numbers can be represented as heat, charge density, and gravitational force. This is a generalization of Laplace\u27s Equation, which are steady state problems. In this problem, it is broken up into two simpler problems. For the first part, there are two ways of solving it since it is two dimensional. This presentation covers both ways of solving it using separation of variables, orthogonality, and partial differential equations. In addition, it shows how to solve the second part and finalize Poisson\u27s Equation

Braun, Ehren

Sward, Andrew P

English

Augustana College, Illinois: Augustana Digital Commons

Augustana CollegeAugustana Digital CommonsMathematics: Student Scholarship & CreativeWorks MathematicsSpring 2019Poisson's EquationEhren BraunAugustana College, Rock Island IllinoisAndrew P. SwardAugustana College, Rock Island IllinoisFollow this and additional works at: https://digitalcommons.augustana.edu/mathstudentPart of the Mathematics CommonsThis Presentation is brought to you for free and open access by the Mathematics at Augustana Digital Commons. It has been accepted for inclusion inMathematics: Student Scholarship & Creative Works by an authorized administrator of Augustana Digital Commons. For more information, pleasecontact digitalcommons@augustana.edu.Augustana Digital Commons CitationBraun, Ehren and Sward, Andrew P.. "Poisson's Equation" (2019). Mathematics: Student Scholarship & Creative Works.https://digitalcommons.augustana.edu/mathstudent/4Poisson’s Equation  By Ehren Braun Poisson’s Equation  ● Generalization of Laplace’s Equation: 𝚫𝚫u = 0  Poisson’s Equation  ● Generalization of Laplace’s Equation: 𝚫𝚫u = 0 ● Poisson’s Equation: 𝚫𝚫u = Q ○ Q represents sources in region Poisson’s Equation  ● Generalization of Laplace’s Equation: 𝚫𝚫u = 0 ● Poisson’s Equation: 𝚫𝚫u = Q ○ Q represents sources in region ● Sources: ○ Voltage ○ Heat ○ Gravity Poisson’s Equation  ● Generalization of Laplace’s Equation: 𝚫𝚫u = 0 ● Poisson’s Equation: 𝚫𝚫u = Q ○ Q represents sources in region ● Sources: ○ Voltage ○ Heat ○ Gravity ● Time-independent(Steady State) Poisson’s Equation  ● Generalization of Laplace’s Equation: 𝚫𝚫u = 0 ● Poisson’s Equation: 𝚫𝚫u = Q ○ Q represents sources in region ● Sources: ○ Voltage ○ Heat ○ Gravity ● Time-independent(Steady State) ● Geometry determines 𝚫𝚫 Poisson’s Equation Example  Poisson’s Equation Example  ● Rectangular Plate ○ Edges(boundary) given by u = 𝝰𝝰 ■ 𝝰𝝰 can vary  Poisson’s Equation Example  ● Rectangular Plate ○ Edges(boundary) given by u = 𝝰𝝰 ■ 𝝰𝝰 can vary ○ Rest  given by 𝚫𝚫u = Q Poisson’s Equation Example  ● Rectangular Plate ○ Edges(boundary) given by u = 𝝰𝝰 ■ 𝝰𝝰 can vary ○ Rest  given by 𝚫𝚫u = Q ● Nonhomogenous from Q and 𝝰𝝰 Poisson’s Equation Example  ● Rectangular Plate ○ Edges(boundary) given by u = 𝝰𝝰 ■ 𝝰𝝰 can vary ○ Rest  given by 𝚫𝚫u = Q ● Nonhomogenous from Q and 𝝰𝝰 ● Easier with homogenous components Solving 𝚫𝚫u = Q, u = 𝝰𝝰 on Boundary ● To Simplify: Break into two parts Solving 𝚫𝚫u = Q, u = 𝝰𝝰 on Boundary ● To Simplify: Break into two parts ○ Let u = u1+ u2 Solving 𝚫𝚫u = Q, u = 𝝰𝝰 on Boundary ● To Simplify: Break into two parts ○ Let u = u1+ u2 ○ 𝚫𝚫u1 = Q, u1 = 0 on boundary Solving 𝚫𝚫u = Q, u = 𝝰𝝰 on Boundary ● To Simplify: Break into two parts ○ Let u = u1+ u2 ○ 𝚫𝚫u1 = Q, u1 = 0 on boundary ○ 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on boundary Solving 𝚫𝚫u = Q, u = 𝝰𝝰 on Boundary ● To Simplify: Break into two parts ○ Let u = u1+ u2 ○ 𝚫𝚫u1 = Q, u1 = 0 on boundary ○ 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on boundary ● This satisfies 𝚫𝚫u = Q, u = 𝝰𝝰 on boundary Solving 𝚫𝚫u = Q, u = 𝝰𝝰 on Boundary ● To Simplify: Break into two parts ○ Let u = u1+ u2 ○ 𝚫𝚫u1 = Q, u1 = 0 on boundary ○ 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on boundary ● This satisfies 𝚫𝚫u = Q, u = 𝝰𝝰 on boundary ● Two “easier” problems to solve Solving 𝚫𝚫u = Q, u = 𝝰𝝰 on Boundary ● To Simplify: Break into two parts ○ Let u = u1+ u2 ○ 𝚫𝚫u1 = Q, u1 = 0 on boundary ○ 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on boundary ● This satisfies 𝚫𝚫u = Q, u = 𝝰𝝰 on boundary ● Two “easier” problems to solve ● Similar for other regions Solving 𝚫𝚫u1 = Q, u1 = 0 on Boundary Solving 𝚫𝚫u1 = Q, u1 = 0 on Boundary ● With homogeneous boundaries  Solving 𝚫𝚫u1 = Q, u1 = 0 on Boundary ● With homogeneous boundaries ● Implies eigenfunction expansion method  Solving 𝚫𝚫u1 = Q, u1 = 0 on Boundary ● With homogeneous boundaries ● Implies eigenfunction expansion method ● Two different ways of Expansion Solving 𝚫𝚫u1 = Q, u1 = 0 on Boundary ● With homogeneous boundaries ● Implies eigenfunction expansion method ● Two different ways of Expansion ○ Eigenfunctions related to 𝚫𝚫u1 = 0 Solving 𝚫𝚫u1 = Q, u1 = 0 on Boundary ● With homogeneous boundaries ● Implies eigenfunction expansion method ● Two different ways of Expansion ○ Eigenfunctions related to 𝚫𝚫u1 = 0 ○ Eigenfunctions related to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 Solving 𝚫𝚫u1 = Q, u1 = 0 on Boundary ● With homogeneous boundaries ● Implies eigenfunction expansion method ● Two different ways of Expansion ○ Eigenfunctions related to 𝚫𝚫u1 = 0 ○ Eigenfunctions related to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 ● Methods are different, but related ○ One-dimensional vs Two-dimensional One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 ● Relating to Laplace’s Equation: 𝚫𝚫u1 = 0 One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 ● Relating to Laplace’s Equation: 𝚫𝚫u1 = 0 ● Laplacian: u1xx+ u1yy = 0 One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 ● Relating to Laplace’s Equation: 𝚫𝚫u1 = 0 ● Laplacian: u1xx+ u1yy = 0 ● Separat ion of Variables: u1 = XY  One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 ● Relating to Laplace’s Equation: 𝚫𝚫u1 = 0 ● Laplacian: u1xx+ u1yy = 0 ● Separat ion of Variables: u1 = XY ● X″Y + XY″ = 0 One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 ● Relating to Laplace’s Equation: 𝚫𝚫u1 = 0 ● Laplacian: u1xx+ u1yy = 0 ● Separat ion of Variables: u1 = XY ● X″Y + XY″ = 0   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 ● Relating to Laplace’s Equation: 𝚫𝚫u1 = 0 ● Laplacian: u1xx+ u1yy = 0 ● Separat ion of Variables: u1 = XY ● X″Y + XY″ = 0  ● aa One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 ● Relating to Laplace’s Equation: 𝚫𝚫u1 = 0 ● Laplacian: u1xx+ u1yy = 0 ● Separat ion of Variables: u1 = XY ● X″Y + XY″ = 0  ● a One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 ● Relating to Laplace’s Equation: 𝚫𝚫u1 = 0 ● Laplacian: u1xx+ u1yy = 0 ● Separat ion of Variables: u1 = XY ● X″Y + XY″ = 0  ● aA ○ Note: Could subtract  Xs instead One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1      One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1      One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1  3 Cases:          ,          , Looking for Non-Trivial Solutions (Only λ > 0)   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1  3 Cases:          ,          , Looking for Non-Trivial Solutions (Only λ > 0)             :    One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1  3 Cases:          ,          , Looking for Non-Trivial Solutions (Only λ > 0)             :  Boundary Conditions =>                                   ,   ,   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1             Note:  One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1             Note:  From Xs,  λ > 0 One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1             Note:  From Xs,  λ > 0 One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1            Note:  From Xs,  λ > 0  Can be rewritten as: One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 Now that we have X and Y:  One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 Now that we have X and Y: u1= XnYn,   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 Now that we have X and Y: u1= XnYn,    For One-Dimensional:  One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 Now that we have X and Y: u1= XnYn,    For One-Dimensional: Now apply the laplacian  One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 Note: Q = Q(x, y) One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 Note: Q = Q(x, y) One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 Note: Q = Q(x, y) One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1      Now we just want only Ys Note: Q = Q(x, y) One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1       One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1       One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1       Now we have 3 cases(Orthogonality): m≠n, m=n≠0, m=n=0  One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1    One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1  Doing these cases, only m=n≠0 is nonzero =>   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1  Doing these cases, only m=n≠0 is nonzero =>   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 All there is left is bn(y)  Note:   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 All there is left is bn(y)  Note:   Solving for bn(y) is now just an ODE with IC  bn(0)=bn(H)=0   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 All there is left is bn(y)  Note:   Solving for bn(y) is now just an ODE with IC  bn(0)=bn(H)=0 For Nonhomogeneous ODEs: Variation of Parameters (yp= v1y1+ v2y2)   One -Dim e n sion a l Eig e n fu n c t ion s  fo r u1 All there is left is bn(y)  Note:   Solving for bn(y) is now just an ODE with IC  bn(0)=bn(H)=0 For Nonhomogeneous ODEs: Variation of Parameters (yp= v1y1+ v2y2)   Finishing One -Dim e n sion a l Eig e n fu n c t ion s   Finishing One -Dim e n sion a l Eig e n fu n c t ion s   We now have our solution:    Finishing One -Dim e n sion a l Eig e n fu n c t ion s   We now have our solution:   Where    Finishing One -Dim e n sion a l Eig e n fu n c t ion s   We now have our solution:   Where   Two-Dimensional Eigenfunctions are easier  Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1  ● Relating to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1  ● Relating to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 ● Laplacian: 𝝓𝝓XX+𝝓𝝓YY=-λ𝝓𝝓  Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1  ● Relating to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 ● Laplacian: 𝝓𝝓XX+𝝓𝝓YY=-λ𝝓𝝓 ● Separat ion of Variables: 𝝓𝝓 = XY  Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1  ● Relating to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 ● Laplacian: 𝝓𝝓XX+𝝓𝝓YY=-λ𝝓𝝓 ● Separat ion of Variables: 𝝓𝝓 = XY ● X″Y+XY″=-λXY  Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1  ● Relating to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 ● Laplacian: 𝝓𝝓XX+𝝓𝝓YY=-λ𝝓𝝓 ● Separat ion of Variables: 𝝓𝝓 = XY ● X″Y+XY″=-λXY ● A  Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1  ● Relating to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 ● Laplacian: 𝝓𝝓XX+𝝓𝝓YY=-λ𝝓𝝓 ● Separat ion of Variables: 𝝓𝝓 = XY ● X″Y+XY″=-λXY ● A ● Each term should be constant  (λ not  dependent)  Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1  ● Relating to 𝚫𝚫𝝓𝝓 + λ𝝓𝝓 = 0 ● Laplacian: 𝝓𝝓XX+𝝓𝝓YY=-λ𝝓𝝓 ● Separat ion of Variables: 𝝓𝝓 = XY ● X″Y+XY″=-λXY ● A ● Each term should be constant  (λ not  dependent) ● Let λ=λx+λy  Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1 Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1     Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1    Similar to previous 3 cases and result:     Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1    Similar to previous 3 cases and result:  Boundary Conditions =>    Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1    Similar to previous 3 cases and result:  Boundary Conditions =>                          =>    Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1 Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1 Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1    Since                                                    , Two -Dim e n sion a l Eig e n fu n c t ion s  fo r u 1    Since                                                    , substitute => Finishing Two -Dim e n sion a l Eig e n fu n c t ion s         Finishing Two -Dim e n sion a l Eig e n fu n c t ion s  All we have left is coefficients      Finishing Two -Dim e n sion a l Eig e n fu n c t ion s  All we have left is coefficients Orthogonality(3 cases): s≠n/t≠m, s=n≠0/t=m≠0, s=n=0/t=m=0        Finishing Two -Dim e n sion a l Eig e n fu n c t ion s  All we have left is coefficients Orthogonality(3 cases): s≠n/t≠m, s=n≠0/t=m≠0, s=n=0/t=m=0 Now for both      and       Finishing Two -Dim e n sion a l Eig e n fu n c t ion s  All we have left is coefficients Orthogonality(3 cases): s≠n/t≠m, s=n≠0/t=m≠0, s=n=0/t=m=0 Now for both      and       Finishing Two -Dim e n sion a l Eig e n fu n c t ion s  All we have left is coefficients Orthogonality(3 cases): s≠n/t≠m, s=n≠0/t=m≠0, s=n=0/t=m=0 Now for both      and       Finishing Two -Dim e n sion a l Eig e n fu n c t ion s  All we have left is coefficients Orthogonality(3 cases): s≠n/t≠m, s=n≠0/t=m≠0, s=n=0/t=m=0 Now for both      and       Finishing Solutions to u 1 Finishing Solutions to u 1 ● We now have our solutions for u1  Finishing Solutions to u 1 ● We now have our solutions for u1 ● Both One and Two-Dimensional Finishing Solutions to u 1 ● We now have our solutions for u1 ● Both One and Two-Dimensional ● u = u1+u2  Finishing Solutions to u 1 ● We now have our solutions for u1 ● Both One and Two-Dimensional ● u = u1+u2 ● u2 left  Finishing Solutions to u 1 ● We now have our solutions for u1 ● Both One and Two-Dimensional ● u = u1+u2 ● u2 left  ● Thankfully, similar Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary ● To identify different edges Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary ● To identify different edges ● To deal with nonhomogeneous boundary: Let u2(x,y) = v(x,y) + w(x,y)  Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary ● To identify different edges ● To deal with nonhomogeneous boundary: Let u2(x,y) = v(x,y) + w(x,y) ● Where v(x, y) represents boundary  and w(x,y) = 0 on boundary  Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary ● u2(x,y) = v(x,y) + w(x,y) Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary ● u2(x,y) = v(x,y) + w(x,y) ● 𝚫𝚫u2= u2xx+ u2yy  Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary ● u2(x,y) = v(x,y) + w(x,y) ● 𝚫𝚫u2= u2xx+ u2yy ● u2xx= vxx+ wxx   ● u2yy= vyy+ wyy Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary ● u2(x,y) = v(x,y) + w(x,y) ● 𝚫𝚫u2= u2xx+ u2yy ● u2xx= vxx+ wxx   ● u2yy= vyy+ wyy ● 𝚫𝚫u2= wxx+ wyy+ vxx+ vyy Solving 𝚫𝚫u2 = 0, u2 = 𝝰𝝰 on Boundary ● u2(x,y) = v(x,y) + w(x,y) ● 𝚫𝚫u2= u2xx+ u2yy ● u2xx= vxx+ wxx   ● u2yy= vyy+ wyy ● 𝚫𝚫u2= wxx+ wyy+ vxx+ vyy ● vxx= vyy= 0 Solving v(x,y)  vxx= vyy= 0    Note: 𝛂𝛂1,3= 𝛂𝛂(y)1,3, 𝛂𝛂2,4= 𝛂𝛂(x)2,4   Solving v(x,y)  vxx= vyy= 0    Note: 𝛂𝛂1,3= 𝛂𝛂(y)1,3, 𝛂𝛂2,4= 𝛂𝛂(x)2,4 v(x,y)xx= 0 v(x,y)x= f(y) v(x,y) = f(y)x + g(y)  Solving v(x,y)  vxx= vyy= 0    Note: 𝛂𝛂1,3= 𝛂𝛂(y)1,3, 𝛂𝛂2,4= 𝛂𝛂(x)2,4 v(x,y)xx= 0 v(x,y)x= f(y) v(x,y) = f(y)x + g(y) v(0,y) = g(y) = 𝛂𝛂3  Solving v(x,y)  vxx= vyy= 0    Note: 𝛂𝛂1,3= 𝛂𝛂(y)1,3, 𝛂𝛂2,4= 𝛂𝛂(x)2,4 v(x,y)xx= 0 v(x,y)x= f(y) v(x,y) = f(y)x + g(y) v(0,y) = g(y) = 𝛂𝛂3 v(L,y) = f(y)L+ 𝛂𝛂3 = 𝛂𝛂1  Solving v(x,y)  vxx= vyy= 0    Note: 𝛂𝛂1,3= 𝛂𝛂(y)1,3, 𝛂𝛂2,4= 𝛂𝛂(x)2,4 v(x,y)xx= 0 v(x,y)x= f(y) v(x,y) = f(y)x + g(y) v(0,y) = g(y) = 𝛂𝛂3 v(L,y) = f(y)L+ 𝛂𝛂3 = 𝛂𝛂1  Solving v(x,y)  Similarly for vyy:  Solving v(x,y)  Similarly for vyy: v(x,y)yy= 0 v(x,y)y= h(x) v(x,y) = h(x)y + q(x)  Solving v(x,y)  Similarly for vyy: v(x,y)yy= 0 v(x,y)y= h(x) v(x,y) = h(x)y + q(x) v(x,0) = q(x) = 𝛂𝛂4 Solving v(x,y)  Similarly for vyy: v(x,y)yy= 0 v(x,y)y= h(x) v(x,y) = h(x)y + q(x) v(x,0) = q(x) = 𝛂𝛂4 v(H,y) = h(x)H+ 𝛂𝛂4 = 𝛂𝛂2   Solving v(x,y)  Similarly for vyy: v(x,y)yy= 0 v(x,y)y= h(x) v(x,y) = h(x)y + q(x) v(x,0) = q(x) = 𝛂𝛂4 v(H,y) = h(x)H+ 𝛂𝛂4 = 𝛂𝛂2   Solving v(x,y)  Similarly for vyy: v(x,y)yy= 0 v(x,y)y= h(x) v(x,y) = h(x)y + q(x) v(x,0) = q(x) = 𝛂𝛂4 v(H,y) = h(x)H+ 𝛂𝛂4 = 𝛂𝛂2  Add solut ion of vxx: Solving w(x,y)  ● 𝚫𝚫w = wxx+ wyy= 0, w(x,y)=0 on boundary   Solving w(x,y)  ● 𝚫𝚫w = wxx+ wyy= 0, w(x,y)=0 on boundary   Solving w(x,y)  ● 𝚫𝚫w = wxx+ wyy= 0, w(x,y)=0 on boundary ● Reminder: Time-independent   Solving w(x,y)  ● 𝚫𝚫w = wxx+ wyy= 0, w(x,y)=0 on boundary ● Reminder: Time-independent ● The plate is 0 everywhere   Solving w(x,y)  ● 𝚫𝚫w = wxx+ wyy= 0, w(x,y)=0 on boundary ● Reminder: Time-independent ● The plate is 0 everywhere ● w(x,y) = 0  Finishing u2   ● u2(x,y) = v(x,y) + w(x,y)  Finishing u2   ● u2(x,y) = v(x,y) + w(x,y) ● Substitution: u2(x,y) = ● We now have the solutions to u1 and u2 Concluding Poisson’s Equation  𝚫𝚫u = Q, u = 𝝰𝝰 on boundary u = u1+ u2=            +  Where   For one-dimensional  Concluding Poisson’s Equation  𝚫𝚫u = Q, u = 𝝰𝝰 on boundary u =               +   Where  For two-dimensional  Finishing u2   ● u2(x,y) = v(x,y) + w(x,y) ● Substitution: u2(x,y) =  In Summary  ● Purpose of Poisson’s Equation ● Solved Poisson’s Equation ● Nonhomogeneous Internal and boundaries ● One and Two dimensional ways ● Separation of Variables ● Orthogonality  

Poisson\u27s Equation

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Poisson\u27s Equation

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