# PDE with extern function

#### Markov

##### Member
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
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.