Differential equations – Partial differential equations – Heat equation

The heat conduction equation is: \(\alpha^2 u_{xx}=u_t\) for \(0 < x < L\) and \(t>0\).

Here \(u=u(x,t)\) is a function of two variables (the position variable \(x\) and the time variable \(t\)). This describes for example the temperature in a metal rod of length \(L > 0\) of which the thickness can be neglected. The positive constant \(\alpha^2\) is called the thermal diffusivity and depends on the thermal conductivity properties of the rod. At time \(t=0\) there is a certain initial temperature distribution in the rod, described by the function \(f(x)\) for \(0\leq x\leq L\). The rod is isolated over its full length such that heat trasnfer is only possible at the two ends of the rod at \(x=0\) and at \(x=L\)). The temperature might be both positive and negative; for instance, a certain value above or below a fixed value (the zero level).

If at both ends a heat source is attached with a fixed temperature, then this temperature can be chosen as the zero level. This leads to the initial-boundary-value problem with homogeneous boundary conditions:

\[\left\{\begin{array}{l}\alpha^2u_{xx}=u_t,\quad 0 < x < L,\quad t>0\\[2.5mm] u(0,t)=0,\quad u(L,t)=0,\quad t > 0\\[2.5mm] u(x,0)=f(x),\quad 0\leq x\leq L.\end{array}\right.\]

We use the method of separation of variables: let \(u(x,t)=X(x)T(t)\), then we have:

\[\alpha^2u_{xx}=u_t\quad\Longleftrightarrow\quad \alpha^2X''(x)T(t)=X(x)T'(t).\]

For \(u(x,t)=X(x)T(t)\neq 0\) we now have (divide by \(X(x)T(t)\)): \(\displaystyle\frac{X''(x)}{X(x)}=\frac{1}{\alpha^2}\cdot\frac{T'(t)}{T(t)}\). The left-hand side only depends on \(x\) (and not on \(t\)), while the right-hand side only depends on \(t\) (and not on \(x\)). Therefore these can only be equal for all \(x\in(0,L)\) and \(t>0\) if they are constant. Hence:

\[\frac{X''(x)}{X(x)}=\frac{1}{\alpha^2}\cdot\frac{T'(t)}{T(t)}=\sigma\quad(\text{separation constant}).\]

This implies:

\[X''(x)-\sigma X(x)=0\quad\text{and}\quad T'(t)-\sigma\alpha^2 T(t)=0.\]

These are ordinary differential equations for \(X(x)\) and for \(T(t)\). Using the boundary conditions we have (with \(T(t)\neq 0\)):

\[u(0,t)=0:\quad X(0)T(t)=0\quad\Longrightarrow\quad X(0)=0\quad\quad\text{and}\quad\quad u(L,t)=0:\quad X(L)T(t)=0\quad\Longrightarrow\quad X(L)=0.\]

This leads to the homogeneous boundary-value problem

\[\left\{\begin{array}{l}X''(x)-\sigma X(x)=0,\quad 0 < x < L\\[2.5mm]X(0)=0,\quad X(L)=0.\end{array}\right.\]

We distinguish three possibilities: \(\sigma=0\), \(\sigma > 0\) and \(\sigma < 0\):

• \(\sigma=0\): \(X''(x)=0\;\Longrightarrow\;X(x)=a_1x+a_2\). Using \(X(0)=0\) and \(X(L)=0\) we then have: \(a_2=0\) and \(a_1L+a_2=0\). Hence: \(a_1=a_2=0\) (the trivial solution).


• \(\sigma=\mu^2 > 0\): \(X''(x)-\mu^2X(x)=0\;\Longrightarrow\;X(x)=b_1\cosh(\mu x)+b_2\sinh(\mu x)\). Using \(X(0)=0\) and \(X(L)=0\) we then have: \(b_1=0\) and \(b_1\cosh(\mu L)+b_2\sinh(\mu L)=0\). Hence: \(b_1=b_2=0\) (the trivial solution), since \(\mu L\neq 0\) (and therefore: \(\sinh(\mu L)\neq 0\)).


• \(\sigma=-\mu^2 < 0\): \(X''(x)+\mu^2X(x)=0\;\Longrightarrow\;X(x)=c_1\cos(\mu x)+c_2\sin(\mu x)\). Using \(X(0)=0\) and \(X(L)=0\) we then have: \(c_1=0\) and \(c_1\cos(\mu L)+c_2\sin(\mu L)=0\). So there are nontrivial solutions if \(\sin(\mu L)=0\), so if \(\mu L=\pm n\pi\) with \(n\in\{1,2,3,\ldots\}\).

Hence there are only nontrivial solutions for \(\sigma_n=-\mu_n^2 < 0\) with \(\mu_n=\displaystyle\frac{n\pi}{L},\;n=1,2,3,\ldots\).

This leads to the eigenvalues \(\displaystyle\sigma_n=-\frac{n^2\pi^2}{L^2}\;n=1,2,3,\ldots\) and the eigenfunctions \(X_n(x)=\displaystyle\sin\left(\frac{n\pi x}{L}\right),\;n=1,2,3,\ldots\).

For \(T(t)\) we then have: \(T_n'(t)-\sigma_n\alpha^2 T_n(t)=0\) with eigenfunctions \(T_n(t)=\displaystyle\exp\left(-\frac{n^2\pi^2\alpha^2t}{L^2}\right)\).

Then we have: \(u_n(x,t)=X_n(x)T_n(t)=\displaystyle\sin\left(\frac{n\pi x}{L}\right)e^{-\frac{n^2\pi^2\alpha^2t}{L^2}},\;n=1,2,3,\ldots\). Now we use the super position principle:

\[u(x,t)=\sum_{n=1}^{\infty}c_nu_n(x,t)=\sum_{n=1}^{\infty}c_n\sin\left(\frac{n\pi x}{L}\right)e^{-\frac{n^2\pi^2\alpha^2t}{L^2}}.\]

Finally the initial value implies for \(0\leq x\leq L\):

\[u(x,0)=f(x)\quad\Longleftrightarrow\quad\sum_{n=1}^{\infty}c_n\sin\left(\frac{n\pi x}{L}\right)=f(x).\]

This is a Fourier sine series for \(f(x)\). Using the Euler-Fourier formulas we then have:

\[c_n=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{n\pi x}{L}\right)\,dx,\quad n=1,2,3,\ldots.\]

Other heat conduction problems

1) If at the two ends heat sources are attached with different temperatures, say \(T_1\) and \(T_2\) with \(T_1\neq T_2\), then we have:

\[\left\{\begin{array}{l}\alpha^2u_{xx}=u_t,\quad 0 < x < L,\quad t>0\\[2.5mm] u(0,t)=T_1,\quad u(L,t)=T_2,\quad t > 0\\[2.5mm] u(x,0)=f(x),\quad 0\leq x\leq L.\end{array}\right.\]

Now we have nonhomogeneous boundary conditions if \(T_1\neq 0\) or \(T_2\neq 0\)>.

Suppose that \(u(x,t)=v(x)+w(x,t)\) with \(v(x)=\displaystyle\frac{T_2-T_1}{L}x+T_1\), then we have: \(v(0)=T_1\) and \(v(L)=T_2\) and therefore

\[\left\{\begin{array}{l}\alpha^2w_{xx}=w_t,\quad 0 < x < L,\quad t>0\\[2.5mm] w(0,t)=0,\quad w(L,t)=0,\quad t\geq 0\\[2.5mm] w(x,0)=f(x)-w(x),\quad 0\leq x\leq L.\end{array}\right.\]

This is again the same initial-boundary-value problem as above with homogeneous boundary conditions.

2) If both ends are isolated, then there also no heat transfer at these ends:

\[\left\{\begin{array}{l}\alpha^2u_{xx}=u_t,\quad 0 < x < L,\quad t>0\\[2.5mm] u_x(0,t)=0,\quad u_x(L,t)=0,\quad t > 0\\[2.5mm] u(x,0)=f(x),\quad 0\leq x\leq L.\end{array}\right.\]

In that case the method of separation of variables leads through \(u(x,t)=X(x)T(t)\neq 0\) to \(T'(t)-\sigma\alpha^2 T(t)=0\) and the boundary-value problem

\[\left\{\begin{array}{l}X''(x)-\sigma X(x)=0,\quad 0 < x < L\\[2.5mm]X'(0)=0,\quad X'(L)=0.\end{array}\right.\]

We distinguish three possibilities: \(\sigma=0\), \(\sigma > 0\) and \(\sigma < 0\):

• \(\sigma=0\): \(X''(x)=0\;\Longrightarrow\;X(x)=a_1x+a_2\) and therefore \(X'(x)=a_1\). Using \(X'(0)=0\) and \(X'(L)=0\) we then have: \(a_1=0\) and \(a_2\in\mathbb{R}\) arbitrary. Hence: \(\sigma_0=0\) is an eigenvalue with eigenfunction \(X_0(x)=1\).


• \(\sigma=\mu^2 > 0\): \(X''(x)-\mu^2X(x)=0\;\Longrightarrow\;X(x)=b_1\cosh(\mu x)+b_2\sinh(\mu x)\) and therefore \(X'(x)=\mu b_1\sinh(\mu x)+\mu b_2\cosh(\mu x)\). Using \(X'(0)=0\) and \(X'(L)=0\) we then have: \(\mu b_2=0\) and \(\mu b_1\sinh(\mu L)+\mu b_2\sinh(\mu L)=0\). Hence: \(b_1=b_2=0\) (the trivial solution), since \(\mu L\neq 0\) (and therefore: \(\sinh(\mu L)\neq 0\)).


• \(\sigma=-\mu^2 < 0\): \(X''(x)+\mu^2X(x)=0\;\Longrightarrow\;X(x)=c_1\cos(\mu x)+c_2\sin(\mu x)\) and therefore \(X'(x)=-\mu c_1\sin(\mu x)+\mu c_2\cos(\mu x)\). Using \(X'(0)=0\) and \(X'(L)=0\) we then have: \(\mu c_2=0\) and \(-\mu c_1\sin(\mu L)+\mu c_2\cos(\mu L)=0\). So there also nontrivial solutions if \(\sin(\mu L)=0\), so if \(\mu L=\pm n\pi\) with \(n\in\{1,2,3,\ldots\}\). This leads to the eigenvalues \(\sigma_n=-\displaystyle\frac{n^2\pi^2}{L^2},\;n=1,2,3,\ldots\) and the eigenfunctions \(X_n(x)=\displaystyle\cos\left(\frac{n\pi x}{L}\right),\;n=1,2,3,\ldots\).

Now we have:

\[u(x,t)=\frac{c_0}{2}+\sum_{n=1}^{\infty}c_n\cos\left(\frac{n\pi x}{L}\right)e^{-\frac{n^2\pi^2\alpha^2t}{L^2}}.\]

Using the initial condition we have for \(0\leq x\leq L\):

\[u(x,0)=f(x)\quad\Longleftrightarrow\quad\frac{c_0}{2}+\sum_{n=1}^{\infty}c_n\cos\left(\frac{n\pi x}{L}\right)=f(x).\]

This is a Fourier cosine series for \(f(x)\). Using the Euler-Fourier formulas we then have:

\[c_0=\frac{2}{L}\int_0^Lf(x)\,dx\quad\text{and}\quad c_n=\frac{2}{L}\int_0^Lf(x)\cos\left(\frac{n\pi x}{L}\right)\,dx,\quad n=1,2,3,\ldots.\]

Other combination are also possible; a heat source at one end and the other one isolated for instance.

The two-dimensional heat conduction equation is: \(\alpha^2\left(u_{xx}+u_{yy}\right)=u_t\) with \(u=u(x,y,t)\).

For this we can also use the method of separation of variables: let \(u(x,y,t)=X(x)Y(y)T(t)\), then we have:

\[\alpha^2\left(u_{xx}+u_{yy}\right)=u_t\quad\Longleftrightarrow\quad \alpha^2\left(X''(x)Y(y)T(t)+X(x)Y''(y)T(t)\right)=X(x)Y(y)T'(t)\]

For \(u(x,y,t)=X(x)Y(y)T(t)\neq 0\) we now have (divide by \(X(x)Y(y)T(t)\)): \(\displaystyle\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=\frac{1}{\alpha^2}\cdot\frac{T'(t)}{T(t)}\). The left-hand side only depends on \(x\) and \(y\) (and not on \(t\)), while the right-hand side only depends on \(t\) (and not on \(x\) and \(y\)). Therefore these can only be equal for all \(x\) and \(y\) in the domain and \(t>0\) if they are constant. Hence:

\[\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=\frac{1}{\alpha^2}\cdot\frac{T'(t)}{T(t)}=\sigma\quad(\text{separation constant}).\]

This implies: \(T'(t)-\sigma\alpha^2 T(t)=0\) and \(\displaystyle\frac{X''(x)}{X(x)}=\sigma-\frac{Y''(y)}{Y(y)}\). The left-hand side only depends on \(x\) (and not on \(y\)), while the right-hand side only depends on \(y\) (and not on \(x\)). Therefore these can only be equal for all \(x\) and \(y\) in the domain if they are constant, so:

\[\frac{X''(x)}{X(x)}=\sigma-\frac{Y''(y)}{Y(y)}=\tau\quad(\text{separation constant}).\]

Hence:

\[X''(x)-\tau X(x)=0,\quad Y''(y)-(\sigma-\tau)Y(y)=0\quad\text{and}\quad T'(t)-\sigma\alpha^2 T(t)=0.\]

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Last modified on August 23, 2021