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Recently I am reading a textbook on P.D.E. Most of the time textbooks mainly deal with homogenous equations and boundary conditions.

I am curious how would one solve say, the heat equation with inhomogenous boundary conditions?

$u_t=u_{xx}$

$u(0,t)=b(t)$ (Dirichlet BC)

or $u_x(0,t)=b(t)$ (Neumann BC)

I read somewhere about a "subtraction method", where one lets $v(x,t)=u(x,t) - ...$ , but don't really understand it.

Sincere thanks for any help.

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There is also a method involving fourier transforms, if you know a little about them. –  Tom Oldfield Nov 23 '12 at 16:55
    
I have a similar type of experience in math.stackexchange.com/questions/172463. –  doraemonpaul Nov 23 '12 at 23:16
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3 Answers

up vote 2 down vote accepted

In fact the "subtraction method" you so called is a little trick that pointedly changing some of the conditions from inhomogeneous to become homogeneous by applying the variable transformation on the dependent variable only.

This little trick is briefly introduced in http://maths.swan.ac.uk/staff/vl/pdes-course.pdf#page=32 and http://maths.swan.ac.uk/staff/vl/pdes-course.pdf#page=38.

For example the PDEs with dependent variable $u$ and independent variables $x$ and $t$ , the little trick is as follows:

for pointedly changing $u(0,t)=f(t)$ and $u(a,t)=g(t)$ $(a\neq0)$ : Let $v(x,t)=u(x,t)-f(t)-\dfrac{x}{a}(g(t)-f(t))$

for pointedly changing $u(0,t)=f(t)$ and $u_x(a,t)=g(t)$ : Let $v(x,t)=u(x,t)-f(t)-xg(t)$

for pointedly changing $u_x(0,t)=f(t)$ and $u(a,t)=g(t)$ : Let $v(x,t)=u(x,t)-(x-a)f(t)-g(t)$

for pointedly changing $u_x(0,t)=f(t)$ and $u_x(a,t)=g(t)$ $(a\neq0)$ : Let $v(x,t)=u(x,t)-h(t)-xf(t)-\dfrac{x^2}{2a}(g(t)-f(t))$

The other types of pointedly changing examples leaves yourself to think.

Sometimes this little trick is a must to apply, for example in Boundaries in heat equation and Heat equation with an unknown diffusion coefficient, otherwise you will get the wrong conclusion that has no solution.

For this problem, you might let $v(x,t)=u(x,t)-b(t)-xb(t)$ , but since there are no other conditions exist, so you have not necessarily to apply this trick, just seckilling this problem by one of the these two approaches:

Approach $1$: separation of variables

Case $1$: $\text{Re}(t)\geq0$

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=X''(x)T(t)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s^2)e^{-ts^2}\\X(x)=\begin{cases}c_1(s^2)\sin xs+c_2(s^2)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s^2)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds$

$u_x(x,t)=C_1+\int_0^\infty sC_3(s^2)e^{-ts^2}\cos xs~ds-\int_0^\infty sC_4(s^2)e^{-ts^2}\sin xs~ds$

$u_x(0,t)=b(t)$ :

$C_1+\int_0^\infty sC_3(s^2)e^{-ts^2}~ds=b(t)$

$\int_0^\infty\dfrac{C_3(s^2)e^{-ts^2}}{2}d(s^2)=b(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{-ts}}{2}ds=b(t)-C_1$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_3(s)}{2}\biggr\}=b(t)-C_1$

$C_3(s)=2\mathcal{L}^{-1}_{t\to s}\{b(t)\}-2C_1\delta(s)$

$\therefore u(x,t)=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-2C_1\int_0^\infty\delta(s^2)e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\int_0^\infty\dfrac{\delta(s^2)e^{-ts^2}\sin xs}{s}d(s^2)+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\int_0^\infty\dfrac{\delta(s)e^{-ts}\sin x\sqrt{s}}{\sqrt{s}}ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}\dfrac{e^{-ts}\sin x\sqrt{s}}{\sqrt{s}}+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}\dfrac{\dfrac{xe^{-ts}\cos x\sqrt{s}}{2\sqrt{s}}-te^{-ts}\sin x\sqrt{s}}{\dfrac{1}{2\sqrt{s}}}+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1\lim\limits_{s\to 0}(xe^{-ts}\cos x\sqrt{s}-2t\sqrt{s}e^{-ts}\sin x\sqrt{s})+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds-C_1x+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+\int_0^\infty C_4(s^2)e^{-ts^2}\cos xs~ds$

$u(0,t)=b(t)$ :

$C_2+\int_0^\infty C_4(s^2)e^{-ts^2}~ds=b(t)$

$\int_0^\infty\dfrac{C_4(s^2)e^{-ts^2}}{2s}d(s^2)=b(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{-ts}}{2\sqrt{s}}ds=b(t)-C_2$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_4(s)}{2\sqrt{s}}\biggr\}=b(t)-C_2$

$C_4(s)=2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{b(t)\}-C_2\delta(\sqrt{s})$

$\therefore u(x,t)=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\cos xs~ds-C_2\int_0^\infty\delta(s)e^{-ts^2}\cos xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\cos xs~ds-C_2=2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\cos xs~ds$

Case $2$: $\text{Re}(t)\leq0$

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=X''(x)T(t)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=s^2\\X''(x)-s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s^2)e^{ts^2}\\X(x)=\begin{cases}c_1(s^2)\sinh xs+c_2(s^2)\cosh xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s^2)e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds$

$u_x(x,t)=C_1+\int_0^\infty sC_3(s^2)e^{ts^2}\cosh xs~ds+\int_0^\infty sC_4(s^2)e^{ts^2}\sinh xs~ds$

$u_x(0,t)=b(t)$ :

$C_1+\int_0^\infty sC_3(s^2)e^{ts^2}~ds=b(t)$

$\int_0^\infty\dfrac{C_3(s^2)e^{ts^2}}{2}d(s^2)=b(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{ts}}{2}ds=b(t)-C_1$

$\int_0^\infty\dfrac{C_3(s)e^{-ts}}{2}ds=b(-t)-C_1$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_3(s)}{2}\biggr\}=b(-t)-C_1$

$C_3(s)=2\mathcal{L}^{-1}_{t\to s}\{b(-t)\}-2C_1\delta(s)$

$\therefore u(x,t)=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-2C_1\int_0^\infty\delta(s^2)e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\int_0^\infty\dfrac{\delta(s^2)e^{ts^2}\sinh xs}{s}d(s^2)+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\int_0^\infty\dfrac{\delta(s)e^{ts}\sinh x\sqrt{s}}{\sqrt{s}}ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}\dfrac{e^{ts}\sinh x\sqrt{s}}{\sqrt{s}}+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}\dfrac{\dfrac{xe^{ts}\cosh x\sqrt{s}}{2\sqrt{s}}+te^{ts}\sinh x\sqrt{s}}{\dfrac{1}{2\sqrt{s}}}+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1\lim\limits_{s\to 0}(xe^{ts}\cosh x\sqrt{s}+2t\sqrt{s}e^{ts}\sinh x\sqrt{s})+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_1x+C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds-C_1x+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+\int_0^\infty C_4(s^2)e^{ts^2}\cosh xs~ds$

$u(0,t)=b(t)$ :

$C_2+\int_0^\infty C_4(s^2)e^{ts^2}~ds=b(t)$

$\int_0^\infty\dfrac{C_4(s^2)e^{ts^2}}{2s}d(s^2)=b(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{ts}}{2\sqrt{s}}ds=b(t)-C_2$

$\int_0^\infty\dfrac{C_4(s)e^{-ts}}{2\sqrt{s}}ds=b(-t)-C_2$

$\mathcal{L}_{s\to t}\biggl\{\dfrac{C_4(s)}{2\sqrt{s}}\biggr\}=b(-t)-C_2$

$C_4(s)=2\sqrt{s}\mathcal{L}^{-1}_{t\to s}\{b(-t)\}-C_2\delta(\sqrt{s})$

$\therefore u(x,t)=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\cosh xs~ds-C_2\int_0^\infty\delta(s)e^{ts^2}\cosh xs~ds=C_2+2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\cosh xs~ds-C_2=2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\cosh xs~ds$

Hence $u(x,t)=\begin{cases}2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\sin xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(t)\}e^{-ts^2}\cos xs~ds&\text{when Re}(t)\geq0\\2\int_0^\infty\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\sinh xs~ds+2\int_0^\infty s\mathcal{L}^{-1}_{t\to s^2}\{b(-t)\}e^{ts^2}\cosh xs~ds&\text{when Re}(t)\leq0\end{cases}$

Approach $2$: power series method

Similar to PDE - solution with power series:

Let $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^n}{n!}\dfrac{\partial^nu(0,t)}{\partial x^n}$ ,

Then $u(x,t)=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^{2n}u(0,t)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(0,t)}{\partial x^{2n+1}}=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^nu(0,t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^nu_x(0,t)}{\partial t^n}=\sum\limits_{n=0}^\infty\dfrac{x^{2n}}{(2n)!}\dfrac{\partial^nb(t)}{\partial t^n}+\sum\limits_{n=0}^\infty\dfrac{x^{2n+1}}{(2n+1)!}\dfrac{\partial^nb(t)}{\partial t^n}$

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The idea is essentially the same as the idea behind solving inhomogeneous ODEs by finding a particular integral and a complementary function. The way I know to do it is, I think, what you are calling the subtraction method.

I notice that you have only given one boundary condition for each type of condition, generally you would need two boundary conditions to get a unique solution to your problem, just thought I'd mention that. I will refer to boundary conditions plural, since this gives a unique solutions but I imagine the same method works for just one boundary condition, you will just have a series of non-unique solutions.

You first find a function $u'$ called the particular solution which solves the equation and satisfies your inhomogeneous BCs. You then solve the same equation for a new function $\hat u := u - u' $ with the now homogeneous BCs. (Bear in mind that when you do this your initial conditions for $\hat u$ will be different to those of the original $u$)

Finding a particular solution $u'$ is not always easy, depending on the form of $b(t)$, and in fact I'm not sure that there is a way to do it for general BCs. However, one good way to do it (when your boundary conditions $b(t)$ are independent of time, i.e. there are points with a fixed amount of the quantity that is diffusing) is to find a "steady state solution" which doesn't vary with time, and can be thought of as the long term behaviour of the general solution, where all short-term behaviour has faded away (in terms of diffusion, this would be the limiting behaviour when the quantity in question has diffused to be spread over the space in an equilibrium determined by our boundary conditions).

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Here is a different approach, called the method of eigenfunction expansions.

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