- #1

timjones007

- 10

- 0

u

_{t}= ku

_{x}

_{x}

-infinity < x < infinity and 0 < t < infinity

with u(x,0)= x

^{2}and u

_{x}

_{x}

_{x}(x,0)= 0

first i showed that u

_{x}

_{x}

_{x}(x,t) solves the equation (easy part)

the next step is to conclude that u(x,t) must be of the form A(t)x

^{2}+ B(t)x + C(t).

i tried to do this by integrating u

_{x}

_{x}

_{x}(x,t) with respect to x and i got u

_{x}

_{x}(x,t) + A(t). then i solved for u

_{x}

_{x}(x,t), integrated it with respect to x, and so on until i got u(x,t) = (triple integral of u

_{x}

_{x}

_{x}(x,t) with respect to x) + A(t)x

^{2}/2 - B(t)x - C(t).

It seems like the only way to get rid of the u

_{x}

_{x}

_{x}(x,t) is to say that it is identically equal to 0 since u

_{x}

_{x}

_{x}(x,0)= 0. Then i can get the desired form of the solution. But, can you say that u

_{x}

_{x}

_{x}(x,t) is identically equal to 0?

Once I show that A(t)x

^{2}+ B(t)x + C(t) is the form of the equation, I'm supposed to use the initial conditions to find A(t), B(t), and C(t). But, the initial conditions will only tell me what A(0), B(0), and C(0) are.That's another problem

and lastly, this may have little to do with the problem but if u(x,0)= x

^{2}then does it mean that u

_{x}(x,0)= 2x just by differentiating both sides with respect to x or are there special conditions that must be satisfied?