4

Given $$u_{xx}-3u_{xy}+2u_{yy}=0$$

Can you apply the methods of characteristics to this problem? I was required to but don't know how to do it for second order PDEs.

with the boundary condition $$u(x,0)=-x^2, \frac{\partial u}{\partial y}(x,0)=0$$

Also in which range are these $x$ and $y$ valid?

EDIT:

Based on the comment we got $$c_1=y+x$$ and $$c_2=y+2x$$

The general solution is $$u(x,y)=f(y+x)+g(y+2x)$$

Now given the boundary condition $u(x,0)=-x^2$ and $u_y(x,0)=0$ I need to find the special solution.

So obviously $$u(x,0)=f(x)+g(2x)=-x^2$$

$$u_y(x,0)=f_y(x)+g_y(2x)=0\Rightarrow f(x,0)+g(x,0)=C=-x^2$$

But i don't know how to proceed..

Calvin Khor
  • 34,903
Tomy
  • 429

1 Answers1

3

Yes, you can use the method of characteristic to tackle these types of problems. The trick is you need to factor the differential operator. In this case we can write $$\left(\partial_{xx} + 2\partial_{yy} - 3\partial_{xy}\right)u = \left(\partial_x-\partial_y\right)\left(\partial_x -2\partial_y\right)u$$ You can now use the method of characteristics to solve each piece separately.

Update:

Here's how to find the solution. Following the above, we solve each first order hyperbolic pde separately. Let's take a look at the first one. We want to solve $$u_x - u_y = 0$$ Notice that this equation can be rewritten in the form $$\nabla u \cdot \begin{bmatrix} -1 \\ 1\end{bmatrix} = 0$$ implying that long lines with slope $\frac{dx}{dy} = -1$ the solution $u$ is constant. Solving this ODE for $x$ (i.e. using the method of characteristics) we find that $x(y) = -y + x_0$ where $x_0$ is the initial condition. Since $u$ is constant along these lines, we have $$u(x(y),y) = u(x_0,0) = -x_0^2$$ Since $x_0 = x(y)+y$, this tells us that the first solution is $u_1 = C_1(x+y)^2$. Using the same kind of argument for the second equation, we find that the second solution is $u_2 = C_2(x+\frac{y}{2})^2$. Since each of these two solutions solve the equation, the general solution is $$u_{\text{gen}} = C_1(x+y)^2 + C_2(x+ \frac{y}{2})^2$$ To get the particular solution, we use the fact that $u(x,0) = -x^2$ and $\frac{\partial u}{\partial y}(x,0) = 0$. Using these conditions we get the following system of equations for the two constants $$C_1 +C_2 = 0$$ $$2C_1 + C_2 = 0$$ Solving this system, we find that $C_1 = 1$ and $C_2 = -2$. Therefore, the particular solution is $$u = (x+y)^2 - 2(x + \frac{y}{2})^2.$$

ColeG97
  • 596
  • As in solving $(u_x-u_y)=0$ and $(u_x-2u_y)=0$ separately using the methods of characteristics? Thou the exercise does want me to give the characteristics "Find the characteristics and draw them in the x-y plane", Ouh wait, does that imply that the general solution of both these two pieces are the same? – Tomy Dec 12 '22 at 20:05
  • Ah its the combination of those, sry yeah ! – Tomy Dec 12 '22 at 20:06
  • Right, in this case you will have two sets of characteristic curves. They will be moving at different speeds. In this case, it might be useful to think about the $y$ variable as representing time. – ColeG97 Dec 12 '22 at 20:18
  • i made some progress, can you help me out with the special solution? I most of the time have difficulties even finding out the special solution of first order DEQs,...i find it somewhat not logical – Tomy Dec 15 '22 at 07:44
  • What do you not find logical? I can be of more help if you're more specific about what part of the problem you are stuck on. – ColeG97 Dec 15 '22 at 16:43
  • about how to find the special solution of the problem....I edited my post,,,spent some time on it...still can't figure it out... – Tomy Dec 15 '22 at 18:12
  • @Tomy I've updated my answer. Please accept it if you deem it worthy. I'm happy to answer follow up questions if you are still confused. – ColeG97 Dec 15 '22 at 22:30
  • Hey thank you so much for this...i didnt expect such a detailed answer! Thank you. And i feel bad even following up , in which range are x and y in the found solution valid? – Tomy Dec 16 '22 at 04:53