# Trouble in Checking PDE Boundary Solution Problem

I'm doing this problem with some other students, but it seems that our solution doesn't work?

We have the partial differential equation $\dfrac{\partial^2 u}{\partial x^2} -2 \dfrac{\partial^2 u}{\partial x \, \partial y} - 3\dfrac{\partial^2 u}{\partial y^2} = 0$.

We found the general solution to be $u(x, y) = F_1(x - y) + F_2(3x + y)$, where $F_1$ and $F_2$ are arbitrary functions.

We want a solution to the equation with the general boundary conditions $u(x,0) = g_0(x)$ and $u_y(x,0) = g_1(x)$.

$$u(x,0) = g_0(x): u(x, 0) = F_1(x) + F_2(3x) = g_0(x)$$

$$u_y(x,0) = g_1(x): u_y(x, 0) = -F_1'(x) + F_2'(3x) = g_1(x)$$

Our work is as follows:

So we have $F_1(x) + F_2(3x) = g_0(x)$ and $-F_1'(x) + F_2'(3x) = g_1(x)$.

Integration of $-F_1'(x) + F_2'(3x) = g_1(x)$:

$-F_1(x) + \frac13 F_2(3x) = \int g_1(x)dx+c_1$

Adding this to $F_1(x) + F_2(3x) = g_0(x)$:

$F_2(3x) =\frac34 \left(g_0(x)+\int g_1(x)dx +c_1\right)$

And since $F_1(x) + F_2(3x) = g_0(x) \rightarrow F_1(x) = g_0(x) - F_2(3x)$:

$$F_1(x) = g_0(x) - \frac{3}{4} \int g_1(x) \ dx - \frac{3}{4} g_0(x)$$

And so we check our work:

$$u(x,y) = F_1(x - y) + F_2(3x + y) = \\ = g_0(x - y) - \frac{3}{4} \int g_1 (x - y) dx - \frac{3}{4} g_0(x - y) + \frac{3}{4} \int g_1 (x - y) dx + \frac{3}{4} g_0 (x - y)$$

$$= g_0(x - y)$$

$$u_y(x, y) = - g_0(x - y)$$

$$u_y(x, 0) = - g_0(x) \not= g_1(x)$$

If I am not mistaken, we should have $u_y(x, 0) = g_1(x)$ here?

I would greatly appreciate it if people could please take the time to clear this up for us.

• Other than using the limits of integration as Julian did you can just do it as you wanted to do it. You just need to be careful at the very moment when you replace $x$ for $x-y$ and $3x$ for $3x+y$. The first substitution is clear and correct. The second would correctly be $x\rightarrow x + \frac{y}{3}$. – Diger Aug 18 '18 at 10:41
• @Diger I'm still not getting the correct values for $u_y(x, 0)$. Writing comment to Julian's post now. Would appreciate help :( – The Pointer Aug 18 '18 at 10:43
• There is not much to say. The solution for $F_2$ reads $$F_2(3x+y)=\frac{3}{4} \left\{ g_0\left(x+\frac{y}{3}\right) + \int g_1\left(x+\frac{y}{3}\right) + c_1 \right\} \, .$$ Adding to $F_1(x-y)$ gives the propper boundary conditions. – Diger Aug 18 '18 at 10:44

Your problemms come from the integration of the equation $-F'_1(x)+F_2(3\,x)=g_1(x)$. You should write $$-F_1(x)+\frac13\,F_2(3\,x)=\int_0^xg_1(t)\,dt.$$ The choice of $0$ as the lower limit of integration is arbitrary. Then \begin{align} F_1(x)&=\frac14\,g_0(x)-\frac34\int_0^xg_1(t)\,dt\\ F_2(x)&=\frac34\,g_0\Bigl(\frac x3\Bigr)+\frac34\int_0^{x/3}g_1(t)\,dt \end{align}
• Julian, thank you for this. But how do I check $u(x, y) = F_1(x - y) + G(3x + y)$? Notice that you have $F_1(x)$ and $F_2(x)$, but we have $x - y$ and $3x + y$ in $F_1(x - y) + G(3x + y)$? – The Pointer Aug 18 '18 at 10:11
• Substitute $x\to x-y$ in $F_1$ and $x\to 3\,x+y$ in $F_2$ (including the upper integration limit). Use the fundamental theorem of calculus for the derivatives of the integral terms. – Julián Aguirre Aug 18 '18 at 10:14
• I get $u_y(x, y) = - \dfrac{1}{4} g'_0 (x - y) - \dfrac{3}{4} g_1 \left( x - y \right) + \dfrac{1}{4} g'_0 \left( \dfrac{3x + y}{3} \right) + \dfrac{3}{4} g_1 \left( \dfrac{3x + y}{3} \right)$. So $u_y(x, 0) = - \dfrac{1}{4} g'_0 (x) - \dfrac{3}{4} g_1 \left( x \right) + \dfrac{1}{4} g'_0 \left( x \right) + \dfrac{3}{4} g_1 \left( x \right) \not= g_1(x)$ – The Pointer Aug 18 '18 at 10:45
• You are forgetting to use the chain rule when derivating with respect to $y$ in the upper limit of integration. There should be a factor of $-1$ in the second term, and $1/3$ in the last two. – Julián Aguirre Aug 18 '18 at 10:50
• $$\frac{d}{dx}\int_a^xf(t)\,dt=f(x)$$ and $$\frac{d}{dx}\int_a^{g(x)}f(t)\,dt=f(g(x))\,g'(x)$$ – Julián Aguirre Aug 18 '18 at 10:55