# [SOLVED]Heat PDE

#### dwsmith

##### Well-known member
I have already solved the main portions.
I have
$$T(x,t) = \sum_{n = 1}^{\infty}A_n\cos\lambda_n x\exp(-\lambda_n^2t)$$
The eigenvalues are determined by
$$\tan\lambda_n = \frac{1}{\lambda_n}$$
The initial condition is $T(x,0) =1$.
For the particular case of $f(x) = 1$, numerically determine the series coefficients $A_n$ and construct a series representation for $T(x,t)$.
How do I do this?
$$A_n = 2\int_0^1\cos\lambda_n xdx$$

#### dwsmith

##### Well-known member
I have already solved the main portions.
I have
$$T(x,t) = \sum_{n = 1}^{\infty}A_n\cos\lambda_n x\exp(-\lambda_n^2t)$$
The eigenvalues are determined by
$$\tan\lambda_n = \frac{1}{\lambda_n}$$
The initial condition is $T(x,0) =1$.
For the particular case of $f(x) = 1$, numerically determine the series coefficients $A_n$ and construct a series representation for $T(x,t)$.
How do I do this?
$$A_n = 2\int_0^1\cos\lambda_n xdx$$
\begin{alignat*}{3} T(x,t) & = & 1.7624\cos(0.86x)e^{-0.86^2t} - 0.1638\cos(3.426x)e^{-3.426^2t} + 0.476\cos(6.437x)e^{6.437^2t}\\ & - & 0.0218\cos(9.529x)e^{-9.529^2t} + 0.0124\cos(12.645x)e^{-12.645^2t} - 0.0080\cos(15.771x)e^{-15.771^2t}\\ & + & 0.0055\cos(18.902x)e^{-18.902^2t} - 0.0041\cos(22.036x)e^{-22.036^2t} + 0.0031\cos(25.172x)e^{-25.172^2t}\\ & - & 0.0025\cos(38.31x)e^{-28.31^2t} \end{alignat*}

#### Sudharaka

##### Well-known member
MHB Math Helper
I have already solved the main portions.
I have
$$T(x,t) = \sum_{n = 1}^{\infty}A_n\cos\lambda_n x\exp(-\lambda_n^2t)$$
The eigenvalues are determined by
$$\tan\lambda_n = \frac{1}{\lambda_n}$$
The initial condition is $T(x,0) =1$.
For the particular case of $f(x) = 1$, numerically determine the series coefficients $A_n$ and construct a series representation for $T(x,t)$.
How do I do this?
$$A_n = 2\int_0^1\cos\lambda_n xdx$$
Hi dwsmith,

Can you please clarify as to what $$f$$ is?

Kind Regards,
Sudharaka.

#### dwsmith

##### Well-known member
My guess is the initial condition since $f$ is usually denoted as an arbitrary IC.

#### Sudharaka

##### Well-known member
MHB Math Helper
How do I do this?
$$A_n = 2\int_0^1\cos\lambda_n xdx$$
$A_n=2\int_0^1\cos\lambda_n xdx=\left.\frac{2\sin\lambda_n x}{\lambda_n}\right|_{0}^{1}=\frac{2\sin\lambda_n}{\lambda_n}$

\begin{alignat*}{3} T(x,t) & = & 1.7624\cos(0.86x)e^{-0.86^2t} - 0.1638\cos(3.426x)e^{-3.426^2t} + 0.476\cos(6.437x)e^{6.437^2t}\\ & - & 0.0218\cos(9.529x)e^{-9.529^2t} + 0.0124\cos(12.645x)e^{-12.645^2t} - 0.0080\cos(15.771x)e^{-15.771^2t}\\ & + & 0.0055\cos(18.902x)e^{-18.902^2t} - 0.0041\cos(22.036x)e^{-22.036^2t} + 0.0031\cos(25.172x)e^{-25.172^2t}\\ & - & 0.0025\cos(38.31x)e^{-28.31^2t} \end{alignat*}
I don't understand why you wrote this. Did you obtain this by simplifying the series mentioned in your first post?

#### dwsmith

##### Well-known member
That is the series with the eigenvalues obtain from Desmos of the first 10 positive of $\tan x = \frac{1}{x}$.