My lecture notes state the following:

When we were dealing with first order equations we saw that a differential operator of the form,

$$p\frac{\partial}{\partial{x}} + q\frac{\partial}{\partial{y}}$$

Led to the characteristic equations

$$\frac{dx}{dt} = p, \frac{dy}{dt} = q$$

The solution of these ODE in turn implied that the solution would be constant along characteristics of the form $qx − py = constant$.

Can someone please demonstrate this?

  • 2
    $\begingroup$ Are $p,q$ constants? Usually in the treatment of first order PDE one has $p=\frac{∂u}{∂x}$, $q=\frac{∂u}{∂y}$. $\endgroup$ – LutzL Aug 11 '18 at 6:49
  • $\begingroup$ @LutzL Yes, $p, q$ are constants. Sorry for not clarifying. $\endgroup$ – handler's handle Aug 11 '18 at 11:47

The wording of the question is doubtful about the notations $p$ and $q$ which are not conventional or ambiguously used. So, the answer below might be made obsolete if the OP rewrite the question on another form. Nevertheless, according to the present form of the question, my answer is :

When the differential operator $\quad p\frac{\partial}{\partial{x}} + q\frac{\partial}{\partial{y}}\quad$ is applied to an unknown function $\quad u(x,y)\quad$ this function is solution of the PDE : $$\left(p\frac{\partial }{\partial{x}} + q\frac{\partial }{\partial{y}}\right)u(x,y)=0$$ $$p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}=0$$ The differential of $u(x,y)$ is : $$du=\frac{\partial u}{\partial{x}}dx + \frac{\partial u}{\partial{y}}dy$$

Consider a curve (C) parametrically defined with parameter $t$ $$\quad\begin{cases} x=x(t)\\y=y(t)\end{cases} \quad ; \quad \quad\begin{cases} dx=\frac{dx}{dt}dt\\dy=\frac{dy}{dt}dt\end{cases}$$ along this curve : $$du=\frac{\partial u}{\partial{x}}\frac{dx}{dt}dt + \frac{\partial u}{\partial{y}}\frac{dy}{dt}dt$$ $$\frac{du}{dt}=\frac{\partial u}{\partial{x}}\frac{dx}{dt} + \frac{\partial u}{\partial{y}}\frac{dy}{dt}$$

Suppose now that (C) is no longer any curve, but a characteristic curve which equation is : $$qx-py=\text{constant}$$ where $p$ and $q$ are constant, or equivalently on parametric form :

$$\begin{cases} x=pt+x_0\\y=qt+y_0\end{cases}\quad \text{thus}\quad\begin{cases} \frac{dx}{dt}=p\\ \frac{dy}{dt}=q\end{cases}$$

Putting this into $\frac{du}{dt}=\frac{\partial u}{\partial{x}}\frac{dx}{dt} + \frac{\partial u}{\partial{y}}\frac{dy}{dt}$ $$\frac{du}{dt}=p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}$$ Comparing to the PDE : $p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}=0$ gives : $$\frac{du}{dt}=0$$ $$u(x,y)=\text{constant}$$ This proves that the function $u(x,y)$ solution of the PDE is constant along the characteristic curve $qx-py=$constant .

  • $\begingroup$ Thanks for the clear answer! Can you please clarify this part: Comparing to the PDE: $p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}=0$. What is meant by "compare" here? What operation was done? Did you subtract $p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}=0$ from $\frac{du}{dt}=p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}$? $\endgroup$ – handler's handle Aug 11 '18 at 11:49
  • 1
    $\begingroup$ It is straightforward : On one hand we have $$\frac{du}{dt}=p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}$$ and on the other hand : $$p\frac{\partial u}{\partial{x}} + q\frac{\partial u}{\partial{y}}=0$$ thus : $$\frac{du}{dt}=0$$ $\endgroup$ – JJacquelin Aug 11 '18 at 12:27
  • $\begingroup$ Ahh, of course. Thank you! $\endgroup$ – handler's handle Aug 11 '18 at 12:37
  • $\begingroup$ My apologies, I noticed something else when studying your answer: how did you get $du=\frac{\partial u}{\partial{x}}dx + \frac{\partial u}{\partial{y}}dy$ as the differential of $u(x,y)$? It looks a bit weird. $\endgroup$ – handler's handle Aug 12 '18 at 5:08
  • 1
    $\begingroup$ This is basic in differential calculus : $$df(x_1,x_2,...,x_k,...,x_n)=\sum_{k=1}^{n}\frac{\partial f}{\partial x_k}dx_k$$ See Eq.(*) in en.wikipedia.org/wiki/Differential_form#Differential_calculus . Note that the indices doesn't denote differentials, but are indices to distinguish different variables. In case of two variables $x_1=x\:,\: x_2=y\:,\: f(x_1,x_2)=u(x,y)$ corresponds to the formula that you are asking for. $\endgroup$ – JJacquelin Aug 12 '18 at 7:18

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.