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What is the general solution form u(x,y)u(x,y) for the PDE 5ux+2uy=05u_x+2u_y=0?

The general solution is u(x,y)=f(y−(2/5)x)u(x,y) = f(y - (2/5)x), where f is an arbitrary differentiable function of one variable. This form arises because u is constant along the characteristic curves y=(2/5)x+cy = (2/5)x + c, so u depends only on the invariant c=y−(2/5)xc = y - (2/5)x.

Conditions

  • u is constant along the characteristic curves.
  • f is an arbitrary function of one variable.

Reasoning, step by step

  1. Identify the characteristic curves as y=(2/5)x+cy = (2/5)x + c.
  2. Rearrange to isolate the constant: c=y−(2/5)xc = y - (2/5)x.
  3. Since u is constant along each curve, u must be a function of c.
  4. Write the general solution as u(x,y)=f(c)=f(y−(2/5)x)u(x,y) = f(c) = f(y - (2/5)x).

Example

The board states: 'Along each such curve (characterised by c), u is a constant. We deduce u(x,y)=f(c)=f(y−2/5x)u(x,y) = f(c) = f(y - 2/5 x) where f is a function of one variable.'

Common misconceptions

  • Thinking f is a specific function like sine or cosine before applying boundary conditions.
  • Confusing the argument of f with x or y directly.

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