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Solve with the use of Duhamel’s Principle

$$ U_t-4U_{xx}=e^t\sin\frac{x}{2}-\sin t,\quad 0\leq x\leq \pi, t\geq0 $$ $$ U(0,t)=\cos t,\quad U_x(\pi,t)=0 $$ $$ U(x,0)=f(x)=1 $$

I know that the Duhamel's Principle can only be used when the Initial Condition $U(x,0)=0$ In this case $U(x,0)=f(x)=1$

Does the Duhamel Principle still apply?

I tried bringing the question down to

$$ V_t=4U_{xx} $$ $$ V(0,t;s)=\cos t\quad V(\pi,t;s)=0 $$ $$ V(x,s;s)=h(x,s)=1 $$

I am not sure on how to go on after this

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1 Answer 1

up vote 0 down vote accepted

Let $U=V+\cos t$ ,

Then $U_x=V_x$

$U_{xx}=V_{xx}$

$U_t=V_t-\sin t$

$\therefore V_t-\sin t-4V_{xx}=e^t\sin\dfrac{x}{2}-\sin t$

$V_t-4V_{xx}=e^t\sin\dfrac{x}{2}$

with $V(0,t)=0$ , $V_x(\pi,t)=0$ and $V(x,0)=0$

So Duhamel's Principle can apply.

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