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PDE with extern function

Markov

Member
Feb 1, 2012
149
Consider the equation

$\begin{aligned} & {{u}_{t}}=K{{u}_{xx}}+g(t),\text{ }0<x<L,\text{ }t>0, \\
& {{u}_{x}}(0,t)={{u}_{x}}(L,t)=0,\text{ }t>0 \\
& u(x,0)=f(x), \\
\end{aligned}
$

a) Show that $v=u-G(t)$ satisfies the initial value boundary problem where $G(t)$ is the primitive of $g(t)$ with $g(0)=0$

b) Find the solution when $g(t)=\cos wt$ and $f(x)=0.$

Attempts:

Okay so I have $g(t)=\displaystyle\int_0^t G(t)\,ds,$ is that how do I need to proceed?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Consider the equation

$\begin{aligned} & {{u}_{t}}=K{{u}_{xx}}+g(t),\text{ }0<x<L,\text{ }t>0, \\
& {{u}_{x}}(0,t)={{u}_{x}}(L,t)=0,\text{ }t>0 \\
& u(x,0)=f(x), \\
\end{aligned}
$

a) Show that $v=u-G(t)$ satisfies the initial value boundary problem where $G(t)$ is the primitive of $g(t)$ with $g(0)=0$

b) Find the solution when $g(t)=\cos wt$ and $f(x)=0.$

Attempts:

Okay so I have $g(t)=\displaystyle\int_0^t G(t)\,ds,$ is that how do I need to proceed?
Hi Markov, :)

"Primitive" means the anti-derivative(Refer this). Therefore, \(G'(t)=g(t)\).

\[v=u-G(t)\]

\[\Rightarrow v_{t}=u_{t}-g(t)\mbox{ and }v_{xx}=u_{xx}\]

Substituting these in the original equation we get, \(v_{t}=Kv_{xx}+2g(t)\). Therefore \(v=u-G(t)\) is not a solution of the given partial differential equation.

Kind Regards,
Sudharaka.