Solving linear partial differential equation in a particular way

Question

This problem came up when I was studying my final exam for my PDE class.

Find the solution to the first-order PDE $$u_x+xu_y=e^{x+y}$$ with the initial conditions $u(0, y)=y$.

I have only learnt two ways about find solutions to a linear PDE. The first method is by transforming the coordinates such that the equation becomes more simple, and the second method is using the method of characteristics(or this is what my professor calls it). It seems that there is some more general version of the "method of characteristics" that I am using like in this question. It seems that they introduce some dummy variable, and change it to a system of ODE's, but up to my knowledge, the method of characteristics is the method of finding the characteristic curves, in this case, which is the family of curves satisfying $\frac{dy}{dx}=\frac{x}{1}$ so that the function must be constant on these curves. We have $\frac{dy}{dx}=x$, $y-\frac{x^2}{2}=c$ in this case, so the homogeneous solution is given as $f(y-\frac{x^2}{2})$. All I need to find now is a particular solution to the non homogeneous equation, and I would add to the homogeneous solution. I have tried functions of the form $f(x)e^{x+y}, f(y)e^{x+y}, (ax+by)e^{x+y}$, but they did not work. I have also tried transforming the coordinates to $$\xi= y-\frac{1}{2}x^2, \quad \eta=x+y, x, y, f(x), f(y) (\text{and several variants})$$ but they did not work.

So, my question is, is there an elagant way to find either a particular solution to the non homogeneous solution, or a elegant transformation that makes the equation much simple?

Please tell me if some parts needs clarification, or edit. Thank you in advance.

Answer 1

$$u_x+xu_y=e^{x+y}$$ What you did seems correct. $f(y-\frac12 x^2)$ is the homogeneous solution.

Coming back to the Charpit-Lagrange characteristic ODEs : $$\frac{dx}{1}=\frac{dy}{x}=\frac{du}{e^{x+y}},$$ a first characteristic equation comes from solving $\quad \frac{dx}{1}=\frac{dy}{x}\quad$ which gives the equation that you already found : $$y-\frac12 x^2=c_1$$ A second characteristic equation comes from solving $\quad \frac{dx}{1}=\frac{du}{e^{x+y}}\quad$ on the first characteristic curve $y=\frac12 x^2+c_1$ : $$\frac{dx}{1}=\frac{du}{e^{x+(\frac12 x^2+c_1)}}$$ $$\frac{du}{dx}=e^{x+\frac12 x^2+c_1}=e^{\frac12 (x+1)^2-\frac12+c_1}$$ This isn't simple because the closed form involves the special function erfi : https://mathworld.wolfram.com/Erfi.html $$u=\sqrt{\frac{\pi}{2}}e^{c_1-\frac12}\text{ erfi}\left(\frac{x+1}{\sqrt{2}} \right)+c_2$$ $$u=\sqrt{\frac{\pi}{2}}e^{y-\frac12 x^2-\frac12}\text{ erfi}\left(\frac{x+1}{\sqrt{2}}\right)+c_2$$ A particular solution of the PDE is $\sqrt{\frac{\pi}{2}}e^{y-\frac12 x^2-\frac12}\text{ erfi}\left(\frac{x+1}{\sqrt{2}}\right)$. Then the general solution is : $$\boxed{u(x,y)=\sqrt{\frac{\pi}{2}}e^{y-\frac12 x^2-\frac12}\text{ erfi}\left(\frac{x+1}{\sqrt{2}}\right)+f(y-\frac12 x^2)}$$ The arbitrary function $f$ has to be determined according to the condition $u(0,y)=y$. $$u(0,y)=\sqrt{\frac{\pi}{2}}e^{y-\frac12}\text{ erfi}\left(\frac{1}{\sqrt{2}}\right)+f(y)=y$$ $$f(y)=y-\sqrt{\frac{\pi}{2}}e^{y-\frac12}\text{ erfi}\left(\frac{1}{\sqrt{2}}\right)$$ The function $f$ is known. We put it into the above general solution of the PDE where the argument is $y-\frac12 x^2$ : $$u(x,y)=\sqrt{\frac{\pi}{2}}e^{y-\frac12 x^2-\frac12}\text{ erfi}\left(\frac{x+1}{\sqrt{2}}\right)+(y-\frac12 x^2)-\sqrt{\frac{\pi}{2}}e^{(y-\frac12 x^2)-\frac12}\text{ erfi}\left(\frac{1}{\sqrt{2}}\right)$$

$$u(x,y)=\sqrt{\frac{\pi}{2}}e^{y-\frac12 x^2-\frac12}\left(\text{ erfi}\left(\frac{x+1}{\sqrt{2}}\right)-\text{ erfi}\left(\frac{1}{\sqrt{2}}\right)\right)+y-\frac12 x^2$$

To be checked. The result seems rather complicated. I wonder if there is no mistake in the calculus.