1

I am only starting my PDE course and I have problems solving this easy equation.

$$3 \frac{\partial u}{\partial y} + \frac{\partial ^2 u}{\partial x \partial y} = 0$$

Here's what I've tried:

$$\frac{\partial ^2 u}{\partial x \partial y} = \frac{\partial ^2 u}{\partial y \partial x}$$

So we can substitute: $z = \frac{\partial u}{\partial y} $

Now we have $3z + \frac{\partial z}{\partial x} =0 $

We get $z(x,y) = C(y)e^{-3x}$.

Is this reasoning correct? Because now it seems impossible to compute $u(x,y)$.

What should I do?

Hagrid
  • 2,581
  • 2
    Just integrate. $\frac{\partial u}{\partial y}\implies u=e^{-3x}\int C(y)dy+C_x(x)$. –  Oct 07 '14 at 12:59

1 Answers1

3

Hint $$u(x,y) = \int C(y) e^{-3x} \mathrm dy = e^{-3x} \tilde C_2(y) + \tilde C_1(x)$$ Now look at the original PDE to infer something about the $\tilde C_i$s $$u_y(x,y) = e^{-3x} \tilde C_2'(y)\\ u_x(x,y) = -3e^{-3x} \tilde C_2(y) + \tilde C_1'(x)\\ u_{yx}(x,y) = -3 e^{-3x} \tilde C_2'(y)\\ u_{xy}(x,y) = -3 e^{-3x} \tilde C_2'(y)$$ So you need $\tilde C_1 \in C^1(\mathbb R), \tilde C_2 \in C^1(\mathbb R)$ as the only requirements if no boundary conditions are imposed.

Example with given boundary conditions
If we impose the boundary conditions $u(x,0) = e^{-3x}, u_y(x,0) = 0\quad \forall x$ that translates to $$e^{-3x} = \tilde C_2(0) e^{-3x} + \tilde C_1(x) \qquad \forall x\\ 0 = e^{-3x} \tilde C_2'(0)$$ So we obtain $\tilde C_2'(0) = 0$ and $\tilde C_1(x) = e^{-3x}(1-\tilde C_2(0))$ so $$u(x,y) = e^{-3x}\underbrace{(1-\tilde C_2(0) + \tilde C_2(y))}_{=\tilde C_3(y)}$$ Verify that $\tilde C_3' = \tilde C_2'$ to see that we may chose $\tilde C_1 \equiv 0$. Thus the solution space is precisely given by $$u(x,y) = C(y) e^{-3x}$$ With $C\in C^1(\mathbb R), C(0) = 1, C'(0) = 0$. One possible choice is $C \equiv 1$ or $C \equiv \cos, C(y) = \frac1{1+y^2}$ would be a decaying solution.

AlexR
  • 24,905
  • I don't think there is anything to infer. Perhaps you are hinting at differentiability. –  Oct 07 '14 at 13:01
  • Thanks. I guess I just panicked. We need two initial conditions to know what $u$ is exactly. – Hagrid Oct 07 '14 at 13:01
  • @CareBear You can infer the class of $\tilde C_i$. I added that. – AlexR Oct 07 '14 at 13:06
  • Thank you very much. I just have one more question. If there are two initial conditions: $u(x, 0) = e^{-3x}$ and $u_y(x,0) = 0$, then what will $u$ be? I can see that then $e^{-3x}C_2(0) + C_1(x)=e^{-3x}, \ \ e^{-3x}C'_2(0)=0$, so $C'_2(0)=0$. Does that help me find the formula for $u$? – Hagrid Oct 07 '14 at 13:11
  • 2
    @Hagrid Yes, that means $C_2 = 1$ and $C_1 = 0$ if you look closely ;) – AlexR Oct 07 '14 at 13:38
  • @Hagrid Yes, that means $C_2 = 1$ and $C_1 = 0$ if you look closely ;) – AlexR Oct 07 '14 at 13:38
  • 1
    @AlexR Ok, I get it now. How do we know no other $C_i$s satisfy these equations? – Hagrid Oct 07 '14 at 14:49
  • 1
    @Hagrid I've elaborated on it and found that in fact $C_1 = 0$ and $C_2(0) = 1, C_2'(0) = 0$ are the only constraints, you still have infinite solutions, another example is $$u(x,y) = \cos(y) e^{-3x}$$ – AlexR Oct 08 '14 at 15:14