Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
up vote 1 down vote accepted

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

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

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.