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##### Chapter 3 / Re: Chapter 3.1 Theorem 4

« Last post by**Victor Ivrii**on

*February 27, 2022, 07:52:52 PM*»

Indeed, to be corrected. Thanks

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Indeed, to be corrected. Thanks

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In the online text book, chapter 3.1 theorem 4, we have the formula for inhomogeneous heat equations. For the second integral in the formula, since it represents the homogeneous part of the heat equation and the heat equation is linear, should the range be from $-\infty$ to $\infty$ instead of $0$ to $\infty$? The equation we are solving have the range of $-\infty < x < \infty$ and $t > 0$ as usual.

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how did we derive 2 from 1 and 4 from 3

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Hi Professor,

I think one coefficient is wrong in the solution.

In this problem the solution to the inhomogeneous equation with homogeneous initial condition should start with coefficient $\frac{1}{2c} = \frac{1}{6}$. After cancelling the 36 in the $f(x,t)$, we should arrive at $6\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$ instead of $3\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$.

Can you help me confirm if I missed something or the textbook has a typo?

I think one coefficient is wrong in the solution.

In this problem the solution to the inhomogeneous equation with homogeneous initial condition should start with coefficient $\frac{1}{2c} = \frac{1}{6}$. After cancelling the 36 in the $f(x,t)$, we should arrive at $6\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$ instead of $3\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$.

Can you help me confirm if I missed something or the textbook has a typo?

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Quote

But how does this qualify us to replace the indefinite integral with the definite one?

Did you take Calculus I? Then you

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I understand that I can show $\phi'(\xi)=0$ by:

1) when t = 0, $\xi = x+ct = x = x-ct = \eta$

2) $u_\xi = u_t \frac{dt}{d\xi} + u_x \frac{dx}{d\xi}$ by chain rule.

3)By initial condition: $u_t|_{t=0} = u_x|_{t=0}$ = 0, we must have $u_\xi=0$

4) $u_\xi = \phi'(x)$ hence $\phi'(\xi)=0$

But how does this qualify us to replace the indefinite integral with the definite one?

1) when t = 0, $\xi = x+ct = x = x-ct = \eta$

2) $u_\xi = u_t \frac{dt}{d\xi} + u_x \frac{dx}{d\xi}$ by chain rule.

3)By initial condition: $u_t|_{t=0} = u_x|_{t=0}$ = 0, we must have $u_\xi=0$

4) $u_\xi = \phi'(x)$ hence $\phi'(\xi)=0$

But how does this qualify us to replace the indefinite integral with the definite one?

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We integrate from $t=0$ because for $t=0$ initial conditions are done. $-1<t $ is a domain where $f(x,t)$ is defined

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Reproduce formula correctly (there are several errors) and think about explanation why $\phi'(\xi)=0$.

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In this problem, $-1 < t < -\infty$, when applying D' Alembert's formula, why we integrate from 0 to t instead of from -1 to t?

Thanks in advance!

Thanks in advance!

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In text book 2.4, we had one line in the derivation: $\tilde{u_{\xi}}=-\frac{1}{4c^2} \int^\xi f(\xi,\eta')d\eta'=-\frac{1}{4c^2} \int_\xi^\eta\tilde f(\xi,\eta')d\eta' + \phi'(\xi)$.

Why can we replace the definite integral with the indefinite integral? Why we choose $\xi$ as lower limit and $\eta$ as upper limit?

Why can we replace the definite integral with the indefinite integral? Why we choose $\xi$ as lower limit and $\eta$ as upper limit?