Solving this problem by the method of characteristic curves we have to solve the ODE

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y}{x}$$

which gives us $$C = \ln(y/x)$$

where $C$ is constant.

Therefore, the solution should be

$$u(x,y) = f(\ln(y/x))$$

We can go further by noting that $e^C$ is also a constant which we can name $C_2$ giving us

$$C_2 = y/x$$

and the solution would give

$$u(x,y) = f(y/x)$$

Are both solutions valid?

  • $\begingroup$ All you are doing is rescaling how "quickly" your solution travels along the characteristics. Both solutions are essentially the same. With both formulations you may run into problems with things blowing up, but that is part of the differential equation, not your formulation $\endgroup$ – whpowell96 Dec 13 '18 at 5:17

They are the same solution but not the same function to express it. Using the same name for both functions is confusing. Call the second $g$, so $f(\ln(x/y))=g(x/y)$ For any chosen $f$ we have determined $g$ as $g=f\circ \ln$


I like to think of it that when:

$$C = \ln \left ( \frac{y}{x} \right ) \iff \frac{y}{x} = e^{C} = C_1$$

So you can just think of it as unwrapping the independent variables until you can't go any further.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.