1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ You use the method of characteristics for the inhomogeneous component too. If you parameterise the curve by $t$ you have \begin{align} \frac{d}{dt} u(x(t), y(t)) &= u_{x} \frac{dx}{dt} + u_{y} \frac{dy}{dt} & &\text{(chain rule)} \\ &= u_{x} + u_{y} x & &\text{(equating to the PDE)} \\ &= e^{x+y} \end{align} and eliminating the parameterisation gives $$\frac{dx}{1} = \frac{dy}{x} = \frac{du}{e^{x+y}}$$ You already solved across the first equality, now you need to solve another to get an explicit characteristic containing $u$ (hint: note that $e^{x + y} = e^{x + c + x^{2}/2}$). $\endgroup$ Apr 22, 2023 at 15:58
  • $\begingroup$ @MatthewCassell I was not aware of this parametrization, and it seems that this is the "Charpit-Lagrange characteristic ODEs" that JJacquelin has mentioned. I will look this up. Thank you. $\endgroup$
    – Joshua Woo
    Apr 23, 2023 at 1:53

1 Answer 1

1
$\begingroup$

$$u_x+xu_y=e^{x+y}$$ 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.

$\endgroup$
1
  • $\begingroup$ I checked the results, and it was indeed correct. It seems that the method of characteristics normally refers to the Charpit-Lagrange characteristic ODEs that you mentioned. I will look up this. Thank you for your answer! $\endgroup$
    – Joshua Woo
    Apr 23, 2023 at 1:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .