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Given $$ux+x^2u_y=0$$

a) Determine the characteristics and the general solution.

b) Determine the range of $y$ where we need to fix the boundary condition $u(0,y)$ such that the solution is determined everywhere on the unit square $Q=(0,1)\times(0,1) \subset \Bbb R^2$.

c) Find the special solution to the PDE satisfying the boundary condition $u(2,y)=5\sin(y)$.

d) find the solution to the non-homogeneous problem $$u_x+x^2u_y=2x^2$$ with the boundary condition $u(2,y)=5\sin(y)$.

So far I got:

a) The characteristics: $\frac{dy}{dx}=x^2$. The ODE has the solution $y= \frac{x^3}{3}+C \Rightarrow u(x,y)=f(y-\frac{x^3}{3})$

b) Here I'm a bit confused. Basically, $u(0,y) \subset Q=(0,1)\times(0,1) \subset \Bbb R^2$. If I say $y \in (0,1)$, then it's fulfilled, but this seems to be too easy so I think it's wrong.

c) By continuing from a) obviously, $u(2,y)=5\sin(y)=f(y-\frac{8}{3})=5\sin(y)$. Here I believe I made a mistake somewhere because it doesn't seem to be possible for such a function $f$ to exist.

d) How would you solve a non-homo PDE? Same way as for ODE? And before I start doing something here I think it's best I get some guidance for c) first:) thanks

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  • $\begingroup$ For c, why can such a function not exist? Let $\alpha = y - 8/3$ to see that $f(\alpha) = 5 \sin(\alpha + 8/3)$ - this makes it clear to see how the function works explicitly. Replace $\alpha$ with $y - x^3/3$ and you will get the solution. $\endgroup$
    – VeDAN
    Dec 12, 2022 at 18:03
  • $\begingroup$ I'm always a bit confused with this part, getting particular solution, much easier with ODEs...but in this case if i plug $\alpha$ back in i get $u(x,y)=5sin(y)$. I pluged in the derivatives back and its not correct..I know i made a mistake somewhere please tell me where... $\endgroup$ Dec 12, 2022 at 18:36

1 Answer 1

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The idea behind characteristics (in simple terms) is to introduce some new variables, $\xi$ and $\eta$ say, so that you transform a PDE (or system of PDEs) into a linear ODE. In this case, we have $\eta = y - x^3/3$ and $\xi = x$, which you have identified - note, this transformation is valid as the Jacobian is well-defined. Letting $u(x,y) = w(x(\xi, \eta), y(\xi, \eta))$ one gets \begin{equation} \cfrac{\partial w}{\partial \xi} = 0 \implies w = f(\eta), \end{equation} and in $(x,y)$ space this is $u(x,y) = f(y - x^3/3)$. This agrees with what you got. $$\\$$ For part c, I do not see the issue. Namely, we are given $u(2,y) = 5 \sin(y) \implies f(y - 8/3) = 5 \sin(y)$. This is giving us all the information we need about the function $f$. Let us take $\alpha = y - 8/3$ and so $f(\alpha) = 5 \sin(\alpha + 8/3)$. Therefore, the particular solution is given by \begin{equation} u(x,y) = f(y - x^3/3) = 5 \sin(y - x^3/3 + 8/3) \end{equation} where have set $\alpha = y -x^3/3$. It should be clear to see that computing the derivatives of such a function will certainly satisfy the PDE. $$\\$$ For part d, the only thing that changes is the ODE we must solve in $(\xi, \eta)$ space. Namely, this time we will have (recalling $\xi = x$), \begin{equation} \cfrac{\partial w}{\partial \xi} = 2 \xi^2 \implies w = \cfrac{2 \xi^3}{3} + g(\eta) \end{equation} and therefore in $(x,y)$ space this gives us the solution \begin{equation} u(x,y) = \cfrac{2 x^3}{3} + g(y - x^3/3). \end{equation} Now as before, apply the given condition and the particular solution should fall out quickly. If you want more detail please let me know :)

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  • $\begingroup$ Thank you very much, this was very clear (i never saw that variable transformation thing, once i figured that out the rest was crystal:)) Would you have any idea for b) thou) $\endgroup$ Dec 13, 2022 at 7:45
  • $\begingroup$ No problem :) Hmm, I will think on it - super busy today with PhD but at first glance, I think that your answer is correct. $\endgroup$
    – VeDAN
    Dec 13, 2022 at 10:37

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