Given this partial differential equation: $$ x \frac{\partial u}{\partial x} -\frac 12 y\frac { \partial u}{\partial y} = 0$$

I would like to first find the general solution and then apply the boundary condition: $u(1,y) = 1+ \sin y$ .

I have used a separation of variables technique, and have gotten two ODE's : $$\frac 1X dX = \frac \lambda x dx$$ and $$\frac 1Y dY = \frac{2\lambda}y dy $$

So one solution will be: $u(x,y) = Cx^\lambda y^{2 \lambda} $. I am confused when using the superposition which states that I can sum up all the solutions because lambda can take any value (even a complex number). I don't understand if this means I have to sum over lambda like so: $$u(x,y) = \sum C_{\lambda} x^\lambda y^{2\lambda } $$ or integrate because lambda is a technically a continuous variable: $$u(x,y) = \int_{-\infty}^{+\infty} C_{\lambda} x^\lambda y^{2\lambda } d \lambda$$ I feel like using the sum would be easier to solve the boundary condition problem as I could equate the summand to the summand in the Taylor series for $1+ \sin x $ . If the integral is the right way to proceed, how do I go about evaluating this integral, and how will I be able to solve the BCP from there? Any help would be great.

  • 1
    $\begingroup$ Are you sure that you want to go via a separation approach and not via the Lagrange equations for the characteristic curves $xy^2=c$ where $u(x,y)=f(xy^2)$ is constant? $\endgroup$ Oct 11, 2020 at 18:25
  • $\begingroup$ The indication on my paper clearly states I have to use a separation of variables methods. $\endgroup$ Oct 11, 2020 at 18:39
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    $\begingroup$ But even in your form $u(x,y)$ is a function of $xy^2$, $u(x,y)=f(xy^2)$ with $f(t)=\sum C_λt^λ$ or the same in the integral version, and $f(t)=1+\sin(\sqrt{t})$. $\endgroup$ Oct 11, 2020 at 18:44
  • $\begingroup$ Hi, thank you for helping me. I'm not quite sure what you're trying to tell me. Would It be possible to clarify? $\endgroup$ Oct 11, 2020 at 19:26
  • $\begingroup$ If you look at your basis solutions, they can all be combined to $x^λy^{2λ}=(xy^2)^λ$. So in the end, $u(x,y)$ is not a function of two variables, but only a function of one variable $t=xy^2$. What ever you do to compute the coefficient sequence or function $C_λ$ and the corresponding functional expression, the result still satisfies $u(x,y)=u(sign(x),\sqrt{|x|}y)$. So you can as well take a shortcut and directly insert the initial condition. $\endgroup$ Oct 11, 2020 at 20:03

1 Answer 1


$$ x \frac{\partial u}{\partial x} -\frac 12 y\frac { \partial u}{\partial y} = 0$$ In order to make it more understandable we will compare the results of two different approachs.

The method of characteristics leads to this general solution : $$u(x,y)=F(x\,y^2)$$ $F$ is an arbitrary function.

The method of separation of variables leads to solutions on this forms with $t=x\, y^{2 }$ : $$u(x,y) = \sum_{\lambda} C_{\lambda} x^\lambda y^{2\lambda } =\sum_{\lambda} C_{\lambda} t^\lambda$$ $C_\lambda$ are arbitrary constants.

or : $$u(x,y) = \int C(\lambda) x^\lambda y^{2\lambda } d \lambda=\int C(\lambda) t^\lambda d \lambda$$ $C(\lambda)$ is an arbitrary function.

Since $C_{\lambda}$ are arbitrary constants and since $C(\lambda)$ is arbitrary thus $\sum C_{\lambda} t^\lambda$ or $\int C(\lambda) t^\lambda d \lambda$ are arbitrary functions of $t$ say $F(t)$. $$u(x,y)=F(t)=F(xy^2)$$ The two methods lead to equivalent forms of the same general solution.

CONDITION : $u(1,y)=1+\sin(y)$ $$u(1,y)=1+\sin(y)=F(y^2)$$ Let $Y=y^2$ $$1+\sin(\sqrt{Y})=F(Y)$$ Now the function $F$ is known. We put it into the above general solution where $Y=xy^2$

$u(x,y)=1+\sin(\sqrt{xy^2})$ $$u(x,y)=1+\sin(y\sqrt{x})$$


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