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$$A_t-A_{xx} = \sin(\pi x)$$ $$A(0,t)=A(1,t)=0$$ $$A(x,t=0)=0$$ Find $A$.

I know I need to find the homogeneous and particular solutions. Im just not sure on this PDE.

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3 Answers 3

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The solution may be accomplished using a Laplace transform. Defining

$$\hat{A}(x,s) = \int_0^{\infty} dt \, A(x,t) \, e^{-s t}$$

and applying the initial condition, we get an ordinary differential equation in $x$:

$$\frac{d^2}{dx^2} \hat{A} - s \hat{A} = -\frac{1}{s} \sin{\pi x}$$

The zero boundary conditions in $x$ mean that the homogeneous solution is zero. The solution then takes the form $\hat{a}(x,s) = P \sin{\pi x}$. Plugging this into the equation, we get the solution

$$\hat{A}(x,s) = \frac{\sin{\pi x}}{s (\pi^2 + s)}$$

You can use partial fractions, or simply look up in a table of inverse LT's; the solution is

$$A(x,t) = \frac{1}{\pi^2} \sin{\pi x} \, (1-e^{-\pi^2 t})$$

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Since the non-homogeneity depends only on $x$, we can assume a solution of the form $A(x,t)=u(x,t)+\phi(x)$.

Substituting this into the PDE gives $$u_t-u_{xx}-\phi_{xx}=\sin(\pi x).$$ Choosing $\phi(x)$ such that $-\phi_{xx}=\sin(\pi x)$, means that $u$ only needs to satisfy a homogeneous PDE.

Note that the boundary conditions on $u$ will change with this assumed solution.

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You have to guess particular solution first. $$ A^p = B\sin \pi x \\ -A^p_{xx} = B\pi^2\sin \pi x = \sin \pi x \\ B = \frac 1{\pi^2} $$ so $$ A^p = \frac 1{\pi^2} \sin \pi x $$ General solution of inhomogeneous problem is a sum of general solution of homogeneous problem and particular solution. So $$ A = A^h + A^p $$ It'll be much easier if one solves homogeneous problem instead. So all you need to do is alter BCs as follows $$ A(0,t) = A^h(0,t)+A^p(0,t) = A^h(0,t)+0 = \fbox{$A^h(0,t)=0$} \\ A(1,t) = A^h(1,t)+A^p(1,t) = A^h(1,t)+0 = \fbox{$A^h(1,t)=0$} \\ A(x,0) = A^h(x,0)+A^p(x,0) = A^h(x,0)+\frac 1{\pi^2}\sin \pi x = 0 \Leftrightarrow \fbox{$A^h(x,0) = -\frac 1{\pi^2} \sin \pi x$} $$ So, now solve homogeneous heat equation with BCs provided above.

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