The following partial differential equation is defined for u:u (x, y)  \(\frac{{\partial u}}{{\partial y}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}};y \ge 0; ~ {x_1} \le x \le {x_2}\)

The set auxiliary conditions necessary to solve the equation uniquely, is 

Calculation:

Given:

\(\frac{{\partial v}}{{\partial y}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}};\) y ≥ 0; x₁ ≤ x ≤ x₂

∵ y ≥ 0 ⇒ It can be replaced with ‘t’.

\(\therefore \frac{{\partial v}}{{\partial t}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}}\)

This is a 1-D Heat equation. It measures temperature distribution in a uniform rod.

The general solution is u = f(x, t)

u = (c₁ cos px + c₂ sin px) \(\left( {{c_3}{e^{ - {c^2}{p^2}t}}} \right)\)

Auxiliary solutions include both initial and boundary conditions.

1) Number of initial conditions = Highest order of time derivative in partial differential = 1

2) The number of boundary conditions:

\(\frac{{\partial v}}{{\partial t}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}}\) ; To solve this partial differential equation, it needs to be integrated twice that will introduce two arbitrary constants.

Hence 2 boundary conditions and 1 initial condition are required to solve this Partial differential equation.