# Can someone help me fill in the gaps of a Cauchy PDE problem?

I was given the following problem to solve:

Ux + Uy = U^2
x=0, U = e^(-y^2)


This is how I was taught to solve it:

dx/dt = 1
dy/dt = 1
y = x + C1

x = 0, let y = S
C1 = S - x
C1 = S
x = y - S


The PDE becomes:

du/dt = U^2
du/dt = dx/dt * du/dx = 1 * du/dx = du/dx
du/dx = U^2


Then this is the part I don't understand, this is the conclusion that was made:

du/dx = U^2
U = 1/(c2 - x)


What integral is done in order to get that equation? I am entirely lost, please help, and thank you!

The basic idea is that $$u_x \frac{dx}{dt} + u_y \frac{dy}{dt} = \frac{d}{dt} u(x(t),y(t))$$. So you pick a path, called a characteristic, so that $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$ line up with the LHS of the equation, and then along this path you have $$\frac{du}{dt}=u^2$$.

This is a separable ODE which can be solved into an implicit form by integration:

$$\int du \, u^{-2} = -u^{-1} = \int dt = t+c.$$

(Note that I abuse notation a little bit here.)

You can now solve that for $$u$$ explicitly, obtaining

$$u=-(t+c)^{-1}.$$

This gives $$u$$ as a function of the "artificial" variables $$(t,c)$$, whereas you want it as a function of $$(x,y)$$. The variable $$c$$ determines which characteristic you are looking at, whereas $$t$$ determines where along that characteristic you are looking. You obtain $$c$$ by using the initial data at the point on the characteristic where the initial data is specified.

There is some slack in the way the characteristics are parameterized, which means we can make one arbitrary decision about them. I will arbitrarily decide that on each characteristic, when $$t=0$$, $$x=0$$. With that established, our calculation above yields $$-c^{-1}=u(0,y)$$. This eliminates the "artificial" variable $$c$$. Do you see how to eliminate the "artificial" variable $$t$$?

• Would you be able to show how this relates to my problem above and the exact steps that would be taken? Commented Oct 6, 2022 at 18:34
• @SamMoldenha I showed most of the steps in detail.
– Ian
Commented Oct 6, 2022 at 18:41
• Thank you so much! Commented Oct 6, 2022 at 18:42