Initial value problem $xyz_x + x^2z_y = x^2+y^2-yz$ with initial curve

Question

Problem
Solve the initial value problem in the domain $x<0$:
$$
xyz_x + x^2z_y = x^2+y^2-yz
$$
where $z=x$ on the line $y=1$.
Attempt
Edit: As JJacquelin pointed out in the answers, my first integral $u_2$ is incorrect which would explain the problem. It should be $x(y-z)$.
I solved for two functionally independent first integrals $u_1(x,y):=x^2-y^2+1$ and $u_2(x,y,z):=dfrac{y-z}{x}$ of the vector field $V(x,y,z)=(xy,x^2,x^2+y^2-yz)$. We can parameterize the initial curve as follows: $x=t, y=1, z=t$ where $t<0$.
Evaluating $u_1$ and $u_2$ on the initial curve, we get
$$
begin{align}
&U_1 = t^2\
&U_2 = dfrac{1-t}{t}
end{align}
$$
Here is where I have some confusion. I know we need to eliminate $t$, but there seems to be a subtlety about how to go about it. I am tempted to write $t=pmsqrt{U_1}$ so that $U_2=dfrac{1mpsqrt{U_1}}{pmsqrt{U_1}}$. Then, we can write
$$
dfrac{y-z}{x}=dfrac{1mpsqrt{x^2-y^2+1}}{pmsqrt{x^2-y^2+1}}
$$
When $y=1$, we must have $x=z$, so this determines the choice of signs above as:
$$
dfrac{y-z}{x}=dfrac{1-sqrt{x^2-y^2+1}}{sqrt{x^2-y^2+1}}
$$
Solving for $z$, we then have
$$
z=x+y-dfrac{x}{sqrt{x^2-y^2+1}}.
$$
But when I plug this back into the PDE, it fails to satisfy the equation. My suspicion for what went wrong is how I eliminated $t$. Certainly, I can also get rid of $t$ by writing $U_1U_2^2=(1pmsqrt{U_1})^2$. If we follow the same procedure, we get a more complicated formula for $z$ with more terms. What is the correct way to proceed, and where did I go wrong in my attempt?

JJacquelin · Answer

This is not a direct answer to your question because I am not quite certain of your first integrals $u_1$ and $u_2$. You should edit the detailed calculus so that one could check it. I think that might be $u_2=x(y-z)$ instead of $frac{y-z}{x}$ .
Nevertheless I hope that the comparison with the characteristic equations below would help to clarify.
$$xyz_x+x^2z_y=x^2+y^2-yz$$
The Charpit-Lagrange system of characteristic ODEs is :
$$frac{dx}{xy}=frac{dy}{x^2}=frac{dz}{x^2+y^2-yz}$$
A first characteristic equation comes from $frac{dx}{xy}=frac{dy}{x^2}$ which solving leads to :
$$x^2-y^2=c_1$$
A second characteristic equation comes from $frac{dy}{x^2}=frac{dz}{x^2+y^2-yz}$ which solving leads to :
$frac{dy}{c_1+y^2}=frac{dz}{(c_1+y^2)+y^2-yz}quadimpliesquadfrac{dz}{dy} =frac{c_1+2y^2-yz}{c_1+y^2}$ which is a linear first order ODE easy to solve.
$$(z-y)sqrt{c_1+y^2}=c_2$$
$$(z-y)x=c_2$$
The general solution of the PDE expessed on the form of implicite equation $c_2=F(c_1)$ is :
$$(z-y)x=F(x^2-y^2)$$
with arbitrary function $F$, to be determined later according to the specified condition.
$$boxed{z(x,y)=y+frac{1}{x}F(x^2-y^2)}$$
Condition :
$$z(x,1)=x=1+frac{1}{x}F(x^2-1)quadimpliesquad F(x^2-1)=x^2-x$$
Let $X=x^2-1quad;quad x=pmsqrt{X+1}$
$$F(X)=X+1pmsqrt{X+1}$$
The function $F$ is determined. We put it into the above general solution where $X=x^2-y^2$.
$$z(x,y)=y+frac{1}{x}left(x^2-y^2+1pmsqrt{x^2-y^2+1} right) $$
The sign is determined so that $z(x,1)=x$
$$boxed{z(x,y)=begin{cases}
x+y+frac{1}{x}left(1-y^2-sqrt{x^2-y^2+1} right) quadtext{if }x>0 \
x+y+frac{1}{x}left(1-y^2+sqrt{x^2-y^2+1} right) quadtext{if }x<0
end{cases}}$$

Initial value problem $xyz_x + x^2z_y = x^2+y^2-yz$ with initial curve

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