4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $$ \left\{ \matrix{ u_{\,x} + x\,u_{\,y} = 1 \hfill \cr u(3,y) = y^{\,2} \hfill \cr} \right. $$

$\endgroup$
3
$\begingroup$

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!

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.