2.1.27 \(x u_x+y u_y=n u\) Example 3.5.2 in Lokenath Debnath

problem number 27

Added June 2, 2019.

From example 3.5.2, page 211 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y)\)

\begin{align*} x u_x+y u_y&=n u \end{align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde =  x*D[u[x, y], x] + y*D[u[x, y], y] ==n*u[x, y]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{u(x,y)\to x^n c_1\left (\frac {y}{x}\right )\right \}\right \}\]

Maple ✓

restart; 
pde :=x*diff(u(x,y),x)+y*diff(u(x,y),y)=n*u(x,y); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y))),output='realtime'));
 

\[u \left (x , y\right ) = f_{1} \left (\frac {y}{x}\right ) x^{n}\]

Hand solution

Solve

\[ xu_{x}+yu_{y}=nu \]

Using the Lagrange-charpit method

\[ \frac {dx}{x}=\frac {dy}{y}=\frac {du}{nu}\]

The first pair of equations gives

\[ x=C_{1}y \]

And \(\frac {dx}{x}=\frac {du}{nu}\) gives

\begin{align*} \ln x & =\frac {1}{n}\ln u+C_{2}\\ x & =C_{2}u^{\frac {1}{n}}\\ x^{n} & =C_{3}u \end{align*}

Since \(C_{3}=G\left ( C_{1}\right ) \) then \(\frac {x^{n}}{u}=G\left ( \frac {x}{y}\right ) \) or \(u=x^{n}G^{-1}\left ( \frac {x}{y}\right ) \). Let \(G^{-1}=F\). Then the solution

\[ u\left ( x,y\right ) =x^{n}F\left ( \frac {x}{y}\right ) \]

___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________