The BVP $x \phi_x - 2 \phi_y = 0$, $\phi_x(x,0) = x \cos(x^2)$, $x > 0$

Question

Solve the following homogeneous first order PDEs subject to the given boundary condition:

$$x\phi_{x}-2\phi_{y}=0, \quad \phi_{x}(x,0)=x\cos(x^2), \quad x>0$$

My attempt using method of characteristic: Integrating $\frac{dx}{x}= \frac{dy}{-2}$ on both sides gives $y=-2\ln x+\alpha$. The characteristic curve is now given: $\zeta = y+2\ln x$. I know that $\phi(x,y)$ is a function $f(\zeta)$ but I don't know how to apply the boundary condition. Would appreciate anyone who can help me solve this problem.

Answer 1

The family of characteristic curves is indeed $$ y+2\ln(x)=C, \quad (C \in \mathbf R), $$ and the main staterment of the method of characteristics for first-order semilinear equations implies then that the general solutution is $$ \varphi(x,y)=F(y+2\ln(x)) $$ where $F$ is a continuously differentiable function (I'd recommend my book on PDEs to consult the version of the method of characteristics for f.-o. semilinear equations). Now to solve the boundary value problem you've got to find out what a/the function $F$ above is.

Deduce from your boundary conditions that $$ F(2 \ln(x))=\frac{\sin(x^2)}2+C $$ and then that $$ F(x)=\frac{\sin(e^x)}2+C. $$ Then you'll get the family of solutions $$ \phi(x,y)= \frac{\sin(e^y x^2)}2+C $$ satisfying the problem in question.