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### Messages - Jaisen Kuhle

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1
##### Web Bonus Problems / Re: Web bonus problem -- Week 2
« on: January 14, 2018, 10:39:07 PM »
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.

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##### Web Bonus Problems / Re: Web bonus problem -- Week 2
« on: January 14, 2018, 06:08:52 PM »
Hint: What is the general solution of (\ref{A})?

For (1) it is $$u(t,x)= \phi(x+t) + \psi(x-t)$$

For the Cauchy Problem it is:

$$u(t,x)=\frac{1}{2}\bigl[g(x+t)+g(x-t)\bigr]+ \frac{1}{2}\int_{x-t}^{x+t} h(y)\,dy$$

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##### Web Bonus Problems / Re: Web bonus problem -- Week 2
« on: January 12, 2018, 02:52:56 PM »
I just want to check if I did it right before writing out details:

u(t,x) = 3xt + xt2 for all (t,x)

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##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 09:55:35 AM »
Wrong again. What are $f$ and $g$?

Functions dependent on the variable y but constant with respect to x. Other than that, I'm unsure.

5
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 09:38:09 AM »
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:

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

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##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 07, 2018, 08:01:51 AM »
Quote from: Jaisen Kuhle
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.

I agree with your first part. Now suppose x is not identically $0$, what function of $f_{yy}(y)$ multiply to $x$ will give you such a quadratic in $x$?
Also the matter seems not to be with possible complex values, but that in this case the quadratic formula gives a relation $x=F(y).$

I'm not sure I follow. If x is not equal to zero, the quadratic would be satisfied if $f_{yy}$ and $g_{yy}$ are constant. So we end up with:

$$u(x,y)=\frac{x^2y^2}{2} + \frac{c_{1}xy^2}{2}+c_{3}xy+c_{5}x+\frac{c_{2}y^2}{2}+{c_{4}y}+c_{6}$$

7
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 10:24:31 PM »
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?

8
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 10:24:00 PM »
oops.

9
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 10:16:07 PM »
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.

10
##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 09:30:19 AM »
Ok, typed here are my steps:

$u_{x} = f(y)$
$u(x,y) = xf(y) + g(y)$
$u_{y} = x f_{y} + g_{y}$, but
$u_{yy} = 0$ which implies
$f_{y} = c_{1}$ and $g_{y} = c_{2}$ which in turn implies
$f(y) = c_{1}y + c_{3}$ and $g(y) = c_{2}y + c_{4}$, concluding:

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

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##### Web Bonus Problems / Re: Web Bonus Problem –– Week 1
« on: January 06, 2018, 09:03:29 AM »
Here is my attempt at a solution, perhaps someone can review it?

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