I am currently trying to solve a problem I have already solved, but am trying to solve it the way our professor solved it. The PDE is given by:

$$yu_x-2xyu_y=2xu, \>\>\>\>\>\>\>u=y^3 \text{ when } x=0, 1\leq y\leq 2$$

Now, I found the characteristic equations (noting that $u(0,y)=z(y)$): $$\dot x(s)=y, \>\dot y(s)=-2xy, \>\dot z(s)=2xz$$ I multiplied $\dot x(s)$ by $2x$ and added it to $\dot y(s)$ to get: $$2x\dot x+\dot y=0$$ Using Separation of Variables Method I got: $$x^2+y=y_0, \>\>\>\>\Longrightarrow\>\>\>\> y=y_0-x^2$$ Where $y_0$ is based off of the initial data. Then , my professor says: $$\frac{\partial x}{\partial s}=y_0-x^2$$ And claims it is easy to solve the above, and then the ODE: $$\dot z(s)=2xz$$ is trivial to solve. My question is, how is the above easy? I tried Separation of Variables which led to: $$x=-\sqrt{y_0}\frac{1+e^{2\sqrt{y_0} s}}{1-e^{2\sqrt{y_0}s}}$$ I don't see how thats easy. I know the method where you solve for: $$\frac{\dot z(s)}{z(s)}=-\frac{\dot y(s)}{y(s)}$$ and separate the variables to solve for this, but I want to know how to work it out this way. If anyone could help it would be appreciated!


Solving the ODE $x'(s)=y_0-x^2(s)$ is 'easy' in a course of PDEs, because the variables can easily be separated. Of course the integration can sometimes be tricky but here it is a standard integral. We have

$$s+C=\int ds +C = \int \frac{dx}{y_0-x^2}=\frac{1}{\sqrt{y_0}} \tanh^{-1} \left( \frac{x}{\sqrt{y_0}} \right)$$

and thus $$x(s)=\sqrt{y_0} \tanh(\sqrt{y_0} s) \left(=\sqrt{y_0} \frac{e^{2\sqrt{y_0}s}-1}{e^{2\sqrt{y_0}s} +1} \right).$$

Then the ODE $z'(s)=2xz=2\sqrt{y_0} \tanh(\sqrt{y_0}s)z$ is again separable since

$$2 \underbrace{\int \sqrt{y_0} \tanh(\sqrt{y_0}s) ds}_{\ln(\cosh(\sqrt{y_0} s)}+C=\int \frac{dz}{z}=\ln(z)$$

and thus


I wouldn't call it trivial but in a course of PDEs these steps can of course be left out as an exercise.


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.