I came across a partial differential equation (IVP PDE) that I would like to solve:
$$\{u_y+\cos(ky)u_x=ax^2 | u(x,0)=0\}$$
This should be a quasilinear PDE, and is in the format of a Cauchy Problem, in the form of:
$$au_x+bu_y=c$$
Such that a, b, and c are constants.
In my particular case, I have:
$$
\left\{
\begin{array}{c}
a=\cos(ky) \\
b=1 \\
c=ax^2
\end{array}
\right.
$$
Using the Lagrange-Carpit Equations: $$\frac{dx}{\cos(ky)}=\frac{dy}{1}=\frac{du}{ax^2}$$
Rearranging gives: $$\frac{dx}{dy}=\cos(ky)$$
$$dy=\frac{du}{ax^2}$$
I then canclulated $dx-dy$:
$$du=(ax^2\cos(ky)+1)dy-dx$$
Integrate to obtain $u(x,y,a,k)$:
$$u(x,y,a,k)=\frac{1}{k}ax^2\sin(ky)+y-x+C_1$$
I applied the initial condition of $u(x,0)=0$ and made $x=0$, which gave $C_1=0$.
The question is thus finalised.
This however does not look correct. Am I doing the right thing?
I would also like to see what the characteristic curves look like, what are these? Are they essentially the terms $a, b, c$ I obtained above? Or rather, I simply differentiate those three variables in terms of $s$. Essentially calculating $\frac{dy}{ds}$, $\frac{dx}{ds}$ and $\frac{dz}{ds}$.
Best Answer
The integration you performed is not correct as $x$ has dependence on $y$ (in fact, the first equation tell us how they are related).
Nevertheless, we can integrate the equations. Further, we get an explicit expression for $u$. I take as starting point these equations:
$$\frac{dx}{\cos(ky)}=\frac{dy}{1}=\frac{du}{ax^2}$$
From them:
$dx=dy\cos(ky)$ and $\dfrac{ax^2dx}{\cos(ky)}=du$ From the first one,
$x+c_1=\dfrac{1}{k}\sin(ky)$ or $c_1=\dfrac{1}{k}\sin(ky)-x$
From here, we can write the cosine as function of $x$ to integrate the second one:
$\sqrt{1-k^2(x+c_1)^2}=\cos(ky)$
Substituting into the second one
$\dfrac{ax^2dx}{\sqrt{1-k^2(x+c_1)^2}}=du$
And integrating:
$$u+c_2=\dfrac{a\left((2k^2c_1^2+1)\arcsin(k(x+c_1))+k(3c_1-x)\sqrt{1-k^2(x+c_1)^2}\right)}{2k^3}$$
Eliminating $c_1$
$$u+c_2=\dfrac{a\left(\left(2k^2\left(\dfrac{1}{k}\sin(ky)-x\right)^2+1\right)y+\left(\dfrac{3}{k}\sin(ky)-4x\right)\cos(ky)\right)}{2k^2}$$
At last, considering that $c_2=f(c_1)$ with $f$ a single argument differentiable function, the general solution is:
$$u(x,y)=\dfrac{a\left(\left(2k^2\left(\dfrac{1}{k}\sin(ky)-x\right)^2+1\right)y+\left(\dfrac{3}{k}\sin(ky)-4x\right)\cos(ky)\right)}{2k^2}-f\left(\dfrac{1}{k}\sin(ky)-x\right)$$
Now, the boundary conditions $u(x,0)=0$ impose some restriction for $f$:
$$u(x,0)=\dfrac{-2ax}{k^2}-f(-x)=0$$
So $f(x)=\dfrac{2ax}{k^2}$ and
$$u(x,y)=\dfrac{a\left(\left(2k^2\left(\dfrac{1}{k}\sin(ky)-x\right)^2+1\right)y+\left(\dfrac{3}{k}\sin(ky)-4x\right)\cos(ky)\right)}{2k^2}-\dfrac{2a}{k^2}\left(\dfrac{1}{k}\sin(ky)-x\right)$$
$$u(x,y)=\dfrac{a}{2k^2}\left(\left(2k^2\left(\dfrac{1}{k}\sin(ky)-x\right)^2+1\right)y+\left(\dfrac{3}{k}\sin(ky)-4x\right)\cos(ky)-\left(\dfrac{1}{k}\sin(ky)-x\right)4\right)$$