# Constants and superposition principle in PDEs

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.

• 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? Oct 11, 2020 at 18:25
• The indication on my paper clearly states I have to use a separation of variables methods. Oct 11, 2020 at 18:39
• 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 $f(t)=1+\sin(\sqrt{t})$. Oct 11, 2020 at 18:44
• Hi, thank you for helping me. I'm not quite sure what you're trying to tell me. Would It be possible to clarify? Oct 11, 2020 at 19:26
• 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. Oct 11, 2020 at 20:03

$$x \frac{\partial u}{\partial x} -\frac 12 y\frac { \partial u}{\partial y} = 0$$ 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 $$Y=xy^2$$

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