I need to solve this semi-linear PDE:
$u_x - 3u_y = \sin (y) + \cos (x)$
The initial condition provided is:
$ u (t,t)= t^2$
I need to use the Characteristic Method. I learned the method from this video.
I have reached an answer. However, I am not sure if it is wright. I am insecure in two different moments of my solution.
My intermediate steps are:
First constant: $c_1= y + 3x $
Second constant: $c_2= x\cos y -\cos x-u$
The second constant above is my first source of insecurity.
I needed to solve this equality as an ODE:
$\frac{dx}{1}= \frac{du}{\sin x + \cos y}$
I did so with this approach:
$\int (\sin x + \cos y) dx = \int 1 du $
Using integral properties: $(\int \sin x dx) + (\int \cos y dx) = \int 1 du $
Which results in: $c_2= x\cos y -\cos x-u$
I am not sure if this is right. I don't know why, but I am affraid I can't treat \cos y as a constant on the second integral.
Supposing this was correct, I tried to go on. So I used an arbitrary function G to make the relation between both constants. Hence, $c_2 =G(c_1) $ and we have that:
$ x\cos y -\cos x-u = G(y + 3x) $
With the initial condition we have:
$G(4x) = x\cos y -\cos x- x^2$
This result above is my second source of insecurity.
Before, I was dealing with clear arbitraty functions like $G(x)$ or $G(y)$. Now, the result is different. I did some manipulation to put it on the more casual form. So:
$G(x) = \frac{x}{4}\cos y -\cos\frac{x}{4}- \frac{x^2}{16}$
After the definition of $G(x)$ above , I inputed the value of $c_1$ , having:
$G(y + 3x) = \frac{y+3x}{4}\cos y -\cos\frac{y+3x}{4} - \frac{(y+3x)^2}{16} $.
Finally, solving for $u$:
$u(x,y) = -\frac{y+3x}{4}\cos y +\cos\frac{y+3x}{4} + \frac{(y+3x)^2}{16} + x\cos y -\cos x $
Is this right?
If I did something wrong, what was it?
Thanks in advance!
$$u_x - 3u_y = \sin (y) + \cos (x),\qquad (1)\\ u(t,t)=t^2 \qquad\qquad\qquad\qquad\quad\ (2)$$ Solution of $u_x - 3u_y =0$ is $$u_h=G(y+3x).$$ Particular solution of $(1)$ is $$u_p=\sin{(x)}+\frac{\cos{(y)}}{3}.$$ Then general solution of $(1)$ is $$u=u_h+u_p=G(y+3x)+\sin{(x)}+\frac{\cos{(y)}}{3}.$$ From $(2)$ we get $$G(4t) +\sin{(t)}+\frac{\cos{(t)}}{3}={{t}^{2}}.$$ Then $$G(t)=\frac{t^2}{16}-\sin\left(\frac{t}{4}\right)-\frac{\cos{(\frac{t}{4})}}{3}.$$
Solution of problem $(1), (2)$ is $$u=\frac{{{\left( y+3 x\right) }^{2}}}{16}-\sin{\left( \frac{y+3 x}{4}\right) }-\frac{\cos{\left( \frac{y+3 x}{4}\right) }}{3}+\sin{(x)}+\frac{\cos{(y)}}{3}$$