Problem with solving $u_x+xu_y=1$ using method of characteristics

Question

I got an exercise in my PDE class which I'm struggling to solve.

Solve following eq using the method of characteristics $$u_x(x,y)+xu_y(x,y) = 1 \qquad (x,y) \in \mathbb{R}^2$$ $$u(3,y) = y^2 \qquad y \in \mathbb{R}$$

My approach was :
To find characteristics solve $(x'(t),y'(t)) = (1,x(t)) $
So I got $\quad x(t) = t+x_o; \quad y(t) = \frac{1}{2}t^2+x_0t+y_0 $
Now we want our characteristics to start at a curve where we know the value of $u$, hence start at $\Gamma = \{ (3,s) : s \in \mathbb{R} \}$.
We get $\quad x_0 = 3 ; \quad y_0 = s$

Now $u'(x(t),y(t)) = u_x(x(t),y(t)) + x(t)u_y(x(t),y(t)) = 1$ hence $u(x(t),y(t)) = t+ u_0 $ where $u_0 = s^2$
So we get $u((t+3),(\frac{1}{2}t^2+3t+s)) = s^2 +t$

I couldn't find an easy way to calculate the equation for $u$. This is the point where I started wondering, if everything was alright.

My approach to solve this would be using polynomial division but I think thats not the point of the exercise.

Answer 1

Your solution $$ u\left( {t + 3,\;t^2 /2 + 3t + s} \right) = s^2 + t $$ is correct.
You just have to complete it by putting $$ \left\{ \matrix{ x = t + 3 \hfill \cr y = {{t^{\,2} } \over 2} + 3t + s = {1 \over 2}t\left( {t + 6} \right) + s = {1 \over 2}\left( {x - 3} \right)\left( {x + 3} \right) + s \hfill \cr} \right. $$ and invert it (the trick is to convert $t$ in $x$, without passing through the square root ..) to obtain $$ \left\{ \matrix{ t = x - 3 \hfill \cr s = y - {1 \over 2}\left( {x - 3} \right)\left( {x + 3} \right) = y - {1 \over 2}x^{\,2} + {9 \over 2} \hfill \cr} \right. $$ and thus $$ s^{\,2} + t = \left( {y - {1 \over 2}x^{\,2} + {9 \over 2}} \right)^{\,2} + \left( {x - 3} \right) = u\left( {x,y} \right) $$

You can easily countercheck that you get $$ \left\{ \matrix{ u_{\,x} = 1 + x^{\,3} - 2xy - 9x = 1 + x\left( {x^{\,2} - 2y - 9} \right) \hfill \cr u_{\,y} = - \left( {x^{\,2} - 2y - 9} \right) \hfill \cr} \right. $$ which respect the given conditions $$ \left\{ \matrix{ u_{\,x} + x\,u_{\,y} = 1 \hfill \cr u(3,y) = y^{\,2} \hfill \cr} \right. $$

Answer 2

First, we can drop $x_0$ as it doesn't entail the graph of the curves to change, then apply the initial conditions.

$$\quad x(t) = t; \quad y(t) = \frac{1}{2}t^2+y_0; \quad u(t)=t+u_0$$

$$y= \frac{1}{2}x^2+y_0; \quad u=x+u_0$$

Now $u(3,y)=y^2$, so is $3+u_0=\left(\frac{1}{2}3^2+y_0\right)^2$, bringing us the desired relation between $y_0$ and $u_0$... to get rid of them!

$y_0=y-\frac{1}{2}x^2; \quad u_0=\left(\frac{1}{2}3^2+y_0\right)^2-3$

$u=x+\left(\frac{1}{2}3^2-\frac{1}{2}x^2+y\right)^2-3$

Answer 3

It's inhomogeneous so we need two parts: the homogeneous solution and a particular solution.

For the homogeneous part, we use $v$ as the variable. With characteristics you are solving $\frac{dy}{dx} = \frac{x}{1}$, i.e. $y = \frac{1}{2}x^2 + C$. That's because $v$ must be constant on these characteristic curves $(1,x)$.

Now this implies that $v$ relies only on the value of $C$. Hence $v(x,y) = f(C)$. Since $C = y - \frac{1}{2}x^2$, so we get $$v(x,y) = f\left(y - \frac{1}{2}x^2\right).$$ Now with $v(3,y) = y^2$, we get $f(y-4.5) = y^2$. Substituting $t = y-4.5$ we arrive at $$f(t) = (t+4.5)^2.$$ Hence we have $$v(x,y) = \left(\left(y - \frac{1}{2}x^2\right) + \frac{9}{2}\right)^2.$$

For the particular solution, $v_0(x,y) = x$ fits the bill.

Hence the solution is $$u(x,y) = v_0(x,y) + v(x,y) = x + \left(\left(y - \frac{1}{2}x^2\right) + \frac{9}{2}\right)^2.$$