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 $dx-dy$: $$du=(ax^2cos(ky)+1)dy-dx$$ Integrate to obtain $u(x,y,a,k)$: $$u(x,y,a,k)=\frac{1}{k}ax^2sin(ky)+y-x+C_1$$ 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 $\frac{dz}{ds}$.


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:


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:


Substituting into the second one


And integrating:


Eliminating $c_1$


At last, considering that $c_2=f(c_1)$ with $f$ a single argument differentiable function, the general solution is:


Now, the boundary conditions $u(x,0)=0$ impose some restriction for $f$:


So $f(x)=\dfrac{2ax}{k^2}$ and




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