# Method of characteristics hyperbolic PDE

Consider the hyperbolic PDE :

$$2u_{xx} + 8u_{xy} + 6u_{yy} = 0.$$

It can be shown using the method of characteristics that the above PDE has the following general solution:

$$u(x,y) = F(y-x) + G(3x-y).$$

I want to find a solution to the equation with the following general boundary conditions,

$$u(x,0) = g_0(x)$$ and $$u_y(x,0) = g_1(x)$$ for $$x \in \mathbb{R}.$$

My working thus far:

$$u_y(x,0) = F'(-x) - G'(3x) = g_1(x)\,\,\,\,\,\, (1)$$

$$u(x,0) = F(-x) + G(3x) = g_0(x) \,\,\,\, (2)$$

Differentiating (2), we get

$$-F'(-x) + 3G'(3x) = g_0'(x) \,\,\,\, (3)$$

Adding $$(1)$$ and $$(3),$$ we get

$$2G'(3x) = g_0'(x) + g_1(x) \,\,\,\, (4)$$

I want to integrate $$(4)$$ to get expression for $$G(3x)$$ but I don't know how to do it. Is this simply:

$$G(3x) = \frac{3}{2}\int g_1(x) dx + \frac{3}{2}g_0(x) + C.$$

Also, once $$G(3x)$$ is computed, how do I obtain expression for $$G(3x-y)?$$ I'm stuck here and need help.

Let we consider a system: $$u_y(x,0) = F'(-x) - G'(3x) = g_1(x)\,\,\,\,\,\, (1)$$ $$u(x,0) = F(-x) + G(3x) = g_0(x)\,\,\,\,\,\, (2)$$

From (2) we get $$F(x) = g_0(-x) - G(-3x)$$. Now we have a new form for solution of this problem $$u(x, y) = g_0(x-y)-G(3 x-3 y)+G(3 x-y)$$. Using condition $$u_y(x,0) = g_1(x)$$ we can define the function $$G$$.

$$2 G'(3 x)-g_0'(x)=g_1(x),\text{ or } 2G'(x)=g_1\left(\frac{x}3\right)+g_0'\left(\frac{x}3\right)$$

By integrating we obtain $$G(x) =C+ \frac{1}{2}\int\limits_0^x \left(g_1\left(\frac{z}3\right)+g_0'\left(\frac{z}3\right) dz\right)$$, where $$C$$ is some constant. And we can write $$F(x) = -C-\frac{1}{2} \int_0^{-3 x} \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3 }\right)\right) \, dz+g_0(-x)$$.

So, we have found the solution $$u(x, y) = -\frac{1}{2} \int_0^{3 x-3 y} \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3 }\right)\right) \, dz+\frac{1}{2} \int_0^{3 x-y} \left(g_0'\left(\frac{z}{3}\right)+g_1\left(\frac{z}{3 }\right)\right) \, dz+g_0(x-y).$$

• I'm not understanding how you got bounds in the equations. We start with indefinite integral. How did you get $0$ to $x$ in the first case and $0$ to $-3x$ in the second case? I'm not clear with this. Do you mind elaborating? Thank you.
– user871167
Mar 5, 2021 at 10:06
• We start from definite integral. I have integrated $G'(x)=\frac{1}2 \left(g_1\left(\frac{x}3\right)+g_0'\left(\frac{x}3\right)\right)$ from $0$ to $x$ in order to get $G$. If $G(x) = \int_0^x ... dz + C$, then $G(-3 x) = \int_0^{-3x} ... dz + C$ and $F(x) = g_0(x) - G(-3x) = g_0(x) -\int_0^{-3 x} ... dz - C$.
– jan
Mar 5, 2021 at 12:42
• Due to $x \in \mathbb{R}$, you can choose lower bound of integration as any real number, but upper will be always $x$.
– jan
Mar 5, 2021 at 12:49
• your answer is very clear but I don't understand how you can choose lower bound of integration as any real number ( and we are choosing $0$ here) ? What is the logic here? I haven't seen this kind of technique before. Can you provide any relevant reference or just explain it?
– user871167
Mar 6, 2021 at 1:31
• We have $G'(x) = ...$, $x \in \mathbb{R}$, so $G(x) = C+\int_c^x ... dz$, where $c$ can be any number from $\mathbb{R}$ (choose any). If we have $G'(x) = ...$, $x \in [a, b]$, then $G(x) = C+\int_c^x ... dz$, where $c$ is any number from $[a, b]$ (choose any).
– jan
Mar 6, 2021 at 7:46