# Classifying (and solving) a linear, first order PDE

In my work, I've come up with a PDE of the following form:

$$\partial _xV(x,y) = ia \ (y A(x)) \left( V(x,y) - B(y) \right)$$

where: $a>0$, $x \in [0,L]$, $y \in [0,\infty]$. The $A(x)$ may be approximated by a step function ($1$ for $x < L/2$, $0$ otherwise), and $B$ is a known smooth function.

I've had very little experience with PDEs before and I can't seem to classify this equation. I'd appreciate any help with that as well as hints on how to proceed in solving it.

• It's not really a PDE. The only derivative is in the $x$ variable, so you can treat this an a first-order linear ODE with $x$ as the independent variable. Such ODEs can be solved using an integrating factor or variation of parameters. – Gyu Eun Lee Nov 15 '16 at 10:11
• @GyuEunLee But the $y$ dependence is important. I don't want to lose it, which I would if I just treat it like a parameter... – Spine Feast Nov 17 '16 at 19:39
• I think you can keep the $y$ dependence if you integrate correctly. When you integrate in $x$, you will get not an constant of integration, but an arbitrary function of $y$ that replaces the integrating constant. (e.g. $\int 2xy~dx = x^2y + \phi(y)$.) – Gyu Eun Lee Nov 18 '16 at 5:09
• Stupid question: what is $\,i\,$ in $\,ia$ ? Is it the complex unit? – Han de Bruijn Nov 22 '16 at 15:42
• yes, $V$ is complex valued – Spine Feast Nov 22 '16 at 16:18

$$\frac{\partial V(x,y)}{\partial x} = ia y A(x) \left( V(x,y) - B(y) \right) \tag 1$$ First, solve the associated homogeneous equation : $$\frac{\partial W(x,y)}{\partial x} = ia y A(x) W(x,y)$$ $$W(x,y)=f(y)e^{iay \int A(x)dx}$$ where $f(y)$ is any differentiable function.
Second, solve for $g(x)$ the inhomogeneous equation (1) with $V(x,y)=g(x)W(x,y)$ $$V=g(x)f(y)e^{iay \int A(x)dx} \quad\to\quad \frac{\partial V(x,y)}{\partial x}= g'(x)f(y)e^{iay \int A(x)dx}+g(x)iayA(x)e^{iay \int A(x)dx}$$ Putting it into eq.(1) leads to : $$g'(x)f(y)e^{iay \int A(x)dx}= iayA(x)B(y)$$ $$g'(x)=ia\frac{yB(y)}{f(y)}e^{-iay \int A(x)dx}$$ $$g(x)=ia \frac{yB(y)}{f(y)} \left( \int e^{-iay \int A(x)dx} dx +F(y)\right)$$ The general solution of eq.(1) is : $$V(x,y)=ia yB(y)e^{iay \int A(x)dx} \left( \int e^{-iay \int A(x)dx} dx +F(y)\right)$$ where $F(y)$ is any differentiable function.
Since $A(x)$ is a step function, $\int A(x)dx$ has a nice closed form and the above form of solution can be simplified. I suppose that you can take it from here.