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**TT1 / Re: TT1-P2**

« **on:**October 19, 2016, 10:54:06 PM »

My solution is:

(a)

$$u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$

As:

$$u = {1 \over 2}( - {e^{ - {{{{(x + t)}^2}} \over 2}}} - {e^{ - {{{{(x - t)}^2}} \over 2}}}) + \underbrace {{1 \over {2c}}\int_0^t {\int_{x - t + t'}^{x + t - t'} {({y^2} - 1)} } {e^{ - {{{y^2}} \over 2}}}dydt'}_{(3)}$$

The inner integral of $(3)$ yields:

$$\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}} - {e^{ - {{{y^2}} \over 2}}}dy = \underbrace {\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}}dy}_{(4)} - \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy$$

$$(4) = \int_{x - t + t'}^{x + t - t'} y {e^{ - {{{y^2}} \over 2}}}ydy = \left. { - {e^{ - {{{y^2}} \over 2}}}y} \right|_{y = x - t + t'}^{y = x + t - t'} + \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy$$

Therefore

$$\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}} - {e^{ - {{{y^2}} \over 2}}}dy = \left. { - {e^{ - {{{y^2}} \over 2}}}y} \right|_{y = x - t + t'}^{y = x + t - t'} + \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy - \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy = (x - t + t'){e^{ - {{{{(x - t + t')}^2}} \over 2}}} - (x + t - t'){e^{ - {{{{(x + t - t')}^2}} \over 2}}}$$

so $(3)$ becomes:

$${1 \over 2}\int_0^t {(x - t + t'){e^{ - {{{{(x - t + t')}^2}} \over 2}}} - (x + t - t'){e^{ - {{{{(x + t - t')}^2}} \over 2}}}} dt'\matrix{

{} & {} \cr

} (c = 1)$$

$$ = {1 \over 2}\int_{x - t}^x {z{e^{ - {{{z^2}} \over 2}}}} dz + {1 \over 2}\int_{x + t}^x {z{e^{ - {{{z^2}} \over 2}}}} dz$$

$$ = - {e^{ - {{{x^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x - t)}^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x + t)}^2}} \over 2}}}$$

Finally,

$$u = {1 \over 2}( - {e^{ - {{{{(x + t)}^2}} \over 2}}} - {e^{ - {{{{(x - t)}^2}} \over 2}}})- {e^{ - {{{x^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x - t)}^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x + t)}^2}} \over 2}}} $$

$$ \Rightarrow u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$

(b) Since u does not depend on t, so we basically get the same equation:

$$\mathop {\lim }\limits_{t \to \infty } = u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$

(a)

$$u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$

As:

$$u = {1 \over 2}( - {e^{ - {{{{(x + t)}^2}} \over 2}}} - {e^{ - {{{{(x - t)}^2}} \over 2}}}) + \underbrace {{1 \over {2c}}\int_0^t {\int_{x - t + t'}^{x + t - t'} {({y^2} - 1)} } {e^{ - {{{y^2}} \over 2}}}dydt'}_{(3)}$$

The inner integral of $(3)$ yields:

$$\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}} - {e^{ - {{{y^2}} \over 2}}}dy = \underbrace {\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}}dy}_{(4)} - \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy$$

$$(4) = \int_{x - t + t'}^{x + t - t'} y {e^{ - {{{y^2}} \over 2}}}ydy = \left. { - {e^{ - {{{y^2}} \over 2}}}y} \right|_{y = x - t + t'}^{y = x + t - t'} + \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy$$

Therefore

$$\int_{x - t + t'}^{x + t - t'} {{y^2}} {e^{ - {{{y^2}} \over 2}}} - {e^{ - {{{y^2}} \over 2}}}dy = \left. { - {e^{ - {{{y^2}} \over 2}}}y} \right|_{y = x - t + t'}^{y = x + t - t'} + \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy - \int_{x - t + t'}^{x + t - t'} {{e^{ - {{{y^2}} \over 2}}}} dy = (x - t + t'){e^{ - {{{{(x - t + t')}^2}} \over 2}}} - (x + t - t'){e^{ - {{{{(x + t - t')}^2}} \over 2}}}$$

so $(3)$ becomes:

$${1 \over 2}\int_0^t {(x - t + t'){e^{ - {{{{(x - t + t')}^2}} \over 2}}} - (x + t - t'){e^{ - {{{{(x + t - t')}^2}} \over 2}}}} dt'\matrix{

{} & {} \cr

} (c = 1)$$

$$ = {1 \over 2}\int_{x - t}^x {z{e^{ - {{{z^2}} \over 2}}}} dz + {1 \over 2}\int_{x + t}^x {z{e^{ - {{{z^2}} \over 2}}}} dz$$

$$ = - {e^{ - {{{x^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x - t)}^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x + t)}^2}} \over 2}}}$$

Finally,

$$u = {1 \over 2}( - {e^{ - {{{{(x + t)}^2}} \over 2}}} - {e^{ - {{{{(x - t)}^2}} \over 2}}})- {e^{ - {{{x^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x - t)}^2}} \over 2}}} + {1 \over 2}{e^{ - {{{{(x + t)}^2}} \over 2}}} $$

$$ \Rightarrow u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$

(b) Since u does not depend on t, so we basically get the same equation:

$$\mathop {\lim }\limits_{t \to \infty } = u(x,t) = - {e^{ - {{{x^2}} \over 2}}}$$