Solving the heat equation

Question

I have the heat equation $$\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2}$$ where $0\leq t<\infty$ and $0\leq x\leq1$

The Boundary conditions are $u(0,t)=0$ and $u(1,t)=1$ with initial condition $u(x,0)=x+sin(\pi x)$

I assume the solution is of the form $u=u(x,t)=X(x)T(t)$ I find the respective derivatives of $u$ and substitute them into the original heat equation to get $$\frac{T'}{T}(t)=\frac{X''}{X} (x)=k$$

I have two Ode's $$T'=KT$$ and $$X''=KX$$

I now look at three different cases.

Case 1. K=0 My solutions say $X(x)=1$ and $T(t)=1$. Where do they come from?

For case 2. I have $K=-p^2<0$ The solution I have is $$X(x)=Acos(px)+Bsin(px)$$ and then $X(x)=sin(\pi nx)$and $T(t)=e^{-\pi^2n^2t}$ How do I get this solution?

For case 3. I have $K=p^2>0$. This gives me $X(x)=Ae^{px}+Be^{-px}$ whic gives me $X(x)=0$ and $T(t)=e^{p^2 t}$ Where do these come from?

Answer 1

The rule of thumb is that you try to substitute appropriate general functions in your ODEs and thus determine the solution. For more detail see derivation of characteristic equation of a linear differential equation.

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As for the cases of different values of $K$, the answers can be derived using common sense.

First, we take a notion of the boundary and initial conditions. You have inhomogeneous Dirichlet BCs, therefore the general solution of your equation can be represented as a sum of the general solution of the same equation with homogeneous BCs and a particular solution satisfying inhomogeneous BCs. The separation of variables is applied to find the general solution of the equation with homogeneous BCs firs. See this webpage for details.

Since $u(x,t) = X(x)T(t)$, we have $$ u(0,t) = X(0)T(t) = 0 \implies X(0) = 0. $$

Second, the answers for the different cases of $K$ can be derived using common sense:

1. $K = 0 \implies $

   (1.a) $ T' = KT \implies T' = 0 \implies T(t) \equiv c - \text{ const.} $ Without loss of generality we can assume $c=1$, i.e. that $T(t) \equiv 1$.

   (1.b) $$ X'' = KX \iff X'' =0 \implies X(x) = ax + b, \quad a,b \in \mathbb{R}\\ X(x)T(0) = 1 \iff a\cdot 1 = 1 \implies a = 1 \implies X(x) = x $$

2. For simplicity, I am going to explain the solution of homogeneous Boundary Conditions problem, i.e. I assume $X(0)=X(1) = 0$. For generalization to inhomogeneous BC case see this link.

   $ K = -p^2 <0 \implies$

   (2.a) $ T' = KT \implies T' = -p^2 T \implies T(t) = ce^{-p^2 t}, \quad c \text{ is a const.}, $ since exponent is the only (nontrivial) function whose derivative equals the function itself multiplied by a constant: $$ \frac{d}{dt} \Big(c e^{-p^2 t} \Big) = -p^2 c e^{-p^2 t}. $$

   (2.b) $ X'' = KX \implies K'' = -p^2 K \implies X(x) = a \sin px + b \cos px, $ since sine and cosine are the only (nontrivial) functions whose second derivatives equal the functions themselves multiplied by a negative squared constant, e.g. $$ \frac{d^2}{dx^2} \Big(a \sin px + b \cos px \Big) = \frac{d}{dx} \Big(a\,p \cos px - b\, p \sin px \Big) = -a\, p^2 \sin px - b\,p^2 \cos px = -p^2 \big( a\sin px + b \cos px \big). $$ Applying (homogeneous) boundary conditions, we have $$ X(0) = 0 \implies a \sin 0 + b\cos 0 = a\cdot 0 + b \cdot 1 = 0 \implies b =0 ,\\ X(1) = 0 \implies a \sin p + 0 \cdot \cos pi = a\sin p = 0 \implies p = \pi n, n \in \mathbb{N}. $$ Therefore $u(x,t) = a \sin (\pi n x) e^{-\pi^2 n^2 t} .$ Applying initial conditions for the homogeneous problem, we get $$ u(x,0) =\sin \pi x \implies a \sin (\pi n x) e^{-\pi^2 n^2 0} = a \sin (\pi n x) = \sin \pi x \implies a=n=1, $$ so we finally write $$ u(x,t) = \sin (\pi n x) e^{-\pi^2 n^2 t} $$

3. Again, I assume $X(0)=X(1) = 0$. For generalization to inhomogeneous BC case see this link.

   $ K = p^2 >0 \implies$

   (3.a) $ T' = KT \implies T' = p^2 T \implies T(t) = ce^{p^2 t}, \ c - $const.

   (3.b) $ X'' = KX \implies X'' = p^2 X \implies X = a e^{px} + be^{-px}, \quad a,b - \text{ consts}$. This is because exponent is the only function satisfying $ X'' = KX$ ODE, because $$ \frac{d^2}{dx^2} \Big(a e^{px} + be^{-px} \Big) = \frac{d}{dx} \Big(a\,p e^{px} - b \, p e^{-px}\Big) = a\, p^2 e^{px} + b \, p^2e^{-px} = p^2 \big(a e^{px} + be^{-px} \big). $$ Apply BC: $$ X(0) = 0 \implies a e^{0} + be^{0} = 0 \implies a+b = 0 \implies a = -b \\ X(1) = 0 \implies a e^{p} + be^{-p} = 0 \implies a \big( e^{p} - e^{-p}\big) = 0 \implies e^{-p} = e^p \implies p=0, $$ so that $X(x) = a + b = 2a$, and thus $u(x,t) = 2a ce^{p^2 t } $.

   Apply IC for homogeneous problem: $$ X(x)T(0) = 0 \implies X(x) = 0 $$

   Applying inhomogeneous BC and IC you can show that $2 ac = 1 \implies T(t) = e^{p^2 t }$ for the case $K =p^2>0$.