Question

Solution of the PDE \(u_x+yu_y=0\) with the initial condition \(u(0,y)=y^3\) is ____.

• A. \(u(x,y)=y^3e^{-5x}\)
• B. \(u(x,y)=y^3e^{-4x}\)
• C. \(u(x,y)=y^3e^{-3x}\)
• D. \(u(x,y)=y^3e^{-2x}\)

Hint:

Use Lagrange's auxiliary equations \(\frac{dx}{1}=\frac{dy}{y}=\frac{du}{0}\) to find the invariant along each characteristic.

Answer 1

The Correct Option is C

Step 1: The given PDE \(u_x + y u_y = 0\) is a linear first order PDE of the form \(Pu_x + Qu_y = R\) with \(P=1\), \(Q=y\), \(R=0\). By Lagrange's method, the auxiliary equations are

\[\frac{dx}{1}=\frac{dy}{y}=\frac{du}{0}\]

Step 2: From \(du = 0\), we get \(u = c_1\), a constant along each characteristic curve.

Step 3: From \(\frac{dx}{1}=\frac{dy}{y}\), separate variables: \(\frac{dy}{y} = dx\). Integrating gives \(\ln y = x + k\), so

\[y e^{-x} = c_2\]

Step 4: The general solution of the PDE is \(u = F(c_2) = F(y e^{-x})\) for an arbitrary function \(F\).

Step 5: Apply the initial condition \(u(0,y) = y^3\). At \(x=0\), \(c_2 = y\), so \(F(y) = y^3\), which means \(F(s) = s^3\) for any argument \(s\).

Step 6: Substitute \(s = y e^{-x}\) back into \(F\):

\[u(x,y) = (y e^{-x})^3 = y^3 e^{-3x}\] \[\boxed{u(x,y) = y^3 e^{-3x}}\]