Let u(x, y) be the solution of the Cauchy problem

uu_x + u_y = 0, x ∈ ℝ, y > 0,

u(x, 0) = x, x ∈ ℝ.

Which of the following is the value of u(2, 3)?

Concept:

Let Pp + Qq = R be a PDE where P, Q, R are functions of x, y, z then by Lagrange's method

\(\frac{dx}{P}\) = \(\frac{dy}{Q}\) = \(\frac{du}{R}\)

Explanation:

Given 

uux + uy = 0, x ∈ ℝ, y > 0,

u(x, 0) = x, x ∈ ℝ.

Using Lagrange's method

\(\frac{dx}{P}\) = \(\frac{dy}{Q}\) = \(\frac{du}{R}\)

⇒ \(\frac{dx}{u}\) = \(\frac{dy}{1}\) = \(\frac{du}{0}\)

Solving this we get 

u = c₁...(i)

and putting u = c1 we get from first two terms

\(\frac{dx}{c_1}\) = \(\frac{dy}{1}\)  

dx = c₁dy
⇒ x - c₁y = c₂

⇒ x - uy = c₂...(ii)

 From (i) and (ii) we get the general solution as

u = ϕ(x - uy)

Using u(x, 0) = x we get

x = ϕ(x)  so ϕ(x - uy) = x - uy

Hence solution is

u = x - uy ⇒  u(1 + y) = x ⇒ u = \(\frac{x}{1+y}\)

Therefore u(2, 3) = \(\frac{2}{4}=\frac12\)

Option (3) is correct