Simple Heat Transfer Problem

Partial Differential Equations (PDEs)

an equation containing derivatives of a function of two or more variables

a second order PDE contains no derivatives higher than 2

Linear second-order partial differential equations form:

Au_xx + 2Bu_xy + Cu_yy + Eu_x + Fu_y + Gu = H

examples second order PDEs

4u_xx + 6xyu_xy = 0

πx_xy + x²u_yy = sin (xy)

not-linear second order PDEs

u²_xx + u_yy = 0

Difference Equations

• approximating derivative with a difference
• reducing h (interval) can improve approximation

f’(x) ≈ f (x + h/2) – f (x - h/2)

h

• second derivative

f (x + h/2 – h/2) – f (x + h/2 – h/2) f (x – h/2 + h/2) – f (x – h/2 – h/2)

f”(x) ≈                             h                                                           h                          _       

h

               ≈     f (x +h) – f (x) – ((f (x) – f (x – h))

                                                h²

≈    f (x + h) – 2f (x) + f (x – h )

                                                  h²

Steady State Heat (voltage) Distribution

• Poisson equation

u_xx + u_yy = f (x,y),  0 ≤ x ≤ a,  0 ≤ y ≤ b

• Boundary Value Problem

u (x,0) = G₁ (x)  and  u (x,b) = G₂ (x)  0 ≤ x ≤ a

u (o,y) = G₃ (y)  and  u (a,y) = G₄ (x)  0 ≤ y ≤ b

Deriving Finite Difference Equations

• finite difference in the x variable

u_xx (x_i,y_j) ≈  u (x_i + h,y_j) – 2u (x_i + y_j) + u (x_i - h,y_j)

                       h²

                                                          ≈  u_{i + 1,j} – 2u_i,j + u_{i – 1,j}

                                                                                h²

• Similarly, deriving finite difference in the y variable with k sized intervals

u_yy (x_i,y_j)  ≈  u_{i,j + 1} – 2u_i,j + u_{i,j – 1}

                                                                      k²

• Substituting Finite Differences into Poisson Equation

u_{i + 1,j} – 2u_i,j + u_{i – 1,j}   + u_{i,j + 1} – 2u_i,j + u_{i,j – 1} = f (x_i, y_j)

                                                       h² k²

• Assuming h ( interval size in the x) =k (interval size in the y)

w_{i,j =} λ² (w_{i + 1,j} + w_{i- 1,j} + w_{i,j + 1} + w _{i,j – 1} – k²f_i,j

2 (1 + λ²)

• Each point on the grid is updated as an average of its four neighbors

w_{i,j =} (w_{i + 1,j} + w_{i - 1,j} + w_{i,j + 1} + w _{i,j – 1}

4