# (nonhomogeneous) Partial differential Equation (partial solution check)

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

• 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. Dec 12, 2022 at 18:03
• 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... Dec 12, 2022 at 18:36

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 $$$$\cfrac{\partial w}{\partial \xi} = 0 \implies w = f(\eta),$$$$ and in $$(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 $$$$u(x,y) = f(y - x^3/3) = 5 \sin(y - x^3/3 + 8/3)$$$$ where have set $$\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 $$\xi = x$$), $$$$\cfrac{\partial w}{\partial \xi} = 2 \xi^2 \implies w = \cfrac{2 \xi^3}{3} + g(\eta)$$$$ and therefore in $$(x,y)$$ space this gives us the solution $$$$u(x,y) = \cfrac{2 x^3}{3} + g(y - x^3/3).$$$$ Now as before, apply the given condition and the particular solution should fall out quickly. If you want more detail please let me know :)