Toronto Math Forum

Second condition: $u_t|_{t=x^2/2}$

$$ 2x = \phi'(x+x^2/2) - \psi'(x-x^2/2) $$
$$ \implies x^2 = \phi(x+x^2/2) - \psi(x-x^2/2) $$

I'm not following your implication here. It seems we have:

$$u_{x^2/2} = \phi_{x^2/2} - \psy_{x^2/2) = 2x$$

So don't we need to integrate with respect to $x^2/2$, in which case:

$$\phi(x+x^2/2) - \psi(x+x^2/2) = x^3 + C$$ 

Also, is this C truly constant?

Edit: Not sure what's wrong with my script for it not to display.

I just want to check if I did it right before writing out details:

u(t,x) = 3xt + xt² for all (t,x)

Jaisen, you arrived to the equality I maked by red. And it is not just an equality, it must be an identity, that is, fulfilled for all $x$ and $y$.

What are $f$ and $g$ here by  your definition? And which of these functions make that equation to be an identity?

Ok, picking up from where I erred:

$$ x = f_{yy} = g_{yy} = 0 \implies f_{y} = c_{1} $$ and $$ g_{y} = c_{2} $$

Concluding:

$$  u(x,y)=\frac{x^2y^2}{2} + {c_{1}xy}+c_{3}x+{c_{2}y}+c_{4} $$

But the interesting question: Why this solution includes just 4 arbitrary constants rather than arbitrary functions of one variable?

Perhaps a hint? Are we supposed to know this from the solution or prior to arriving at the solution? The solution indicates this is a linear(?) PDE so perhaps it is related to that fact. Alternatively, perhaps I can ask, as this is a second order PDE, and our only constraints are related to the second partial derivatives, does that mean we need constants to determine the initial values and boundaries?

For (14)

$$u_{xx}=y^2 \implies u=\frac{x^2y^2}{2} + xf(y)+g(y) \implies u_{y}=x^2y+xf_{y}+g_{y} \implies u_{yy}= x^2 + xf_{yy} + g_{yy} = -x^2 $$

So we have:

$$2x^2 + xf_{yy} + g_{yy}= 0 $$

Suppose x = 0, therefore

$$g_{yy} = 0 \implies g_{y} = c \implies g(y)=cy+d \implies u(x,y) = u(0,y) = cy + d $$

Suppose x not equal to 0

When we solve the quadratic we arrive at imaginary values for x. I'm not sure if I ought to continue.

Here is my attempt at a solution, perhaps someone can review it?