Solve one dimensional heat equation

The steady-state heat equation for a plate is Laplace's equation: ∂²u/∂x² + ∂²u/∂y² = 0.

Derivation. Take a thin rectangular plate of thickness δz, density ρ, specific heat s and thermal conductivity K. The faces are insulated, so heat flows only in the xy-plane. Consider a small element with sides δx and δy, and let u(x, y, t) be the temperature.

Heat flow rate across a face is −K × (area) × (temperature gradient normal to the face).

        ↓ heat flow in y
     ┌────────────┐
 →   │  δx × δy   │  →   heat flow in x
     └────────────┘
        ↓

• Along x. Heat entering through the left face: −K δy δz (∂u/∂x)ₓ. Heat leaving through the right face: −K δy δz (∂u/∂x)ₓ₊δₓ. Net gain = K δy δz [(∂u/∂x)ₓ₊δₓ − (∂u/∂x)ₓ] ≈ K δx δy δz ∂²u/∂x².
• Along y. In the same way the net gain = K δx δy δz ∂²u/∂y².

Total net gain per unit time = K δx δy δz (u_xx + u_yy).

This equals the rate at which the heat content increases, ρ s δx δy δz ∂u/∂t. Hence

∂u/∂t = c² (∂²u/∂x² + ∂²u/∂y²), c² = K/(ρs)

Steady state. In the steady state the temperature does not change with time, so ∂u/∂t = 0:

∂²u/∂x² + ∂²u/∂y² = 0