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[SOLVED] Heat PDE

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Feb 1, 2012
1,673
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
Feb 5, 2012
1,621
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
Feb 1, 2012
1,673
My guess is the initial condition since $f$ is usually denoted as an arbitrary IC.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Feb 1, 2012
1,673
That is the series with the eigenvalues obtain from Desmos of the first 10 positive of $\tan x = \frac{1}{x}$.