Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was thinking about the following problem :

Find out which of the following option(s) is/are correct?

The differential equation $xu_{x}+yu_{y}=2u$ satisfying the initial condition $y=xg(x),u=f(x)$, with

(a)$f(x)=2x,g(x)=1$ has no solution,

(b)$f(x)=2x^2,g(x)=1$ has infinite number of solutions,

(c)$f(x)=x^3,g(x)=x$ has a unique solution,

(d)$f(x)=x^4,g(x)=x$ has a unique solution.

My Attempt:

Using lagrange's method, we see that $\frac{dx}{x}=\frac{dy}{y}=\frac{du}{2u} $ gives $y/x=c_1,$ and $y=\sqrt uc_2$ ,where $c_1,c_2$ being constants. Now we have to apply to the given initial conditions and check the given options.Here ,I observe that $y=xg(x)=x.1=x$,[as $g(x)=1$] and $u=f(x)=2x^2$But,this relation does not satisfy $xu_{x}+yu_{y}=2u$ and so the choice"(a)$f(x)=2x,g(x)=1$ has no solution," is right. Now I am stuck here and could not progress further.Am i going in the right direction? Please help.Thanks in advance for your time.

share|cite|improve this question
How exactly is $y = x g(x)$ an initial condition? – Alex M. Aug 7 '15 at 13:35

The PDE is first-order linear. We can think of it as a prescription for the directional derivative of $u$. Specifically, $(x,y)\cdot \nabla u=2u$ tells us something about the derivative of $u$ in the direction straight away from the origin. Let's fix a half-line $\{(ta,tb):t>0\}$ where $(a,b)\in\mathbb R^2\setminus \{(0,0)\}$. Along this half-line our PDE will become an ODE, which is obviously a big help. Indeed, the function $v(t)=u(ta,tb)$ satisfies $$v'(t)=au_x(ta,tb)+bu_y(ta,tb) = 2t^{-1}u(ta,tb) = 2t^{-1}v(t)$$ This is a separable ODE with solution $v(t)=Ct^2$, or more specifically $u(ta,tb)=t^2 u(a,b)$. Thus, any differentiable function $u$ which grows quadratically on every half-line from the origin ($u(ta,tb)=t^2 u(a,b)$, $t>0$) solves the PDE. The upshot is that once $u$ is given at one point $(a,b)$, it is automatically prescribed on the rest of the half-line.

Great. Let's see how this structure of $u$ squares with the given conditions. (a) says than on the line $y=x$ the function must grow linearly. No way. (b) says $u=2x^2$ on the line $y=x$. No problem, we can arrange that and still have a lot of freedom on the rest of the plane. (c) and (d) prescribe values of $u$ on the parabola $y=x^2$. This parabola meets each half-line at most once, so these conditions are neither inconsistent as in (a) nor redundant as in (b). They do leave us freedom to do whatever we want in the bottom half-plane, because the parabola does not go there.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.