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4.9: Steady state temperature and the Laplacian - Mathematics


Suppose we have an insulated wire, a plate, or a 3-dimensional object. We apply certain fixed temperatures on the ends of the wire, the edges of the plate, or on all sides of the 3-dimensional object. We wish to find out what is the steady state temperature distribution. That is, we wish to know what will be the temperature after long enough period of time.

We are really looking for a solution to the heat equation that is not dependent on time. Let us first solve the problem in one space variable. We are looking for a function (u) that satisfies

[u_t=ku_{xx},]

but such that (u_t=0) for all (x) and (t). Hence, we are looking for a function of (x) alone that satisfies (u_{xx}=0). It is easy to solve this equation by integration and we see that (u=Ax+B) for some constants (A) and (B).

Suppose we have an insulated wire, and we apply constant temperature (T_1) at one end (say where (x=0)) and (T_2) on the other end (at (x=L) where (L) is the length of the wire). Then our steady state solution is

[u(x)= dfrac{T_2-T_1}{L}x+T_1.]

This solution agrees with our common sense intuition with how the heat should be distributed in the wire. So in one dimension, the steady state solutions are basically just straight lines.

Things are more complicated in two or more space dimensions. Let us restrict to two space dimensions for simplicity. The heat equation in two space variables is

[ u_t=k(u_{xx}+u_{yy}), label{4.9.3}]

or more commonly written as ( u_t=k Delta u ) or ( u_t=k abla^2 u ). Here the ( Delta ) and ( abla^2 ) symbols mean ( dfrac{partial^2}{partial x^2}+ dfrac{partial^2}{partial y^2}). We will use ( Delta) from now on. The reason for using such a notation is that you can define ( Delta) to be the right thing for any number of space dimensions and then the heat equation is always (u_t=k Delta u). The operator ( Delta) is called the Laplacian.

OK, now that we have notation out of the way, let us see what does an equation for the steady state solution look like. We are looking for a solution to Equation ef{4.9.3} that does not depend on (t), or in other words (u_t=0). Hence we are looking for a function (u(x,y)) such that

[ Delta u = u_{xx}+u_{yy}=0.]

This equation is called the Laplace equation (aamed after the French mathematician Pierre-Simon, marquis de Laplace). Solutions to the Laplace equation are called harmonic functions and have many nice properties and applications far beyond the steady state heat problem.

Harmonic functions in two variables are no longer just linear (plane graphs). For example, you can check that the functions (x^2-y^2) and (xy) are harmonic. However, if you remember your multi-variable calculus we note that if (u_{xx}) is positive, (u) is concave up in the (x) direction, then (u_{yy}) must be negative and (u) must be concave down in the (y) direction. Therefore, a harmonic function can never have any “hilltop” or “valley” on the graph. This observation is consistent with our intuitive idea of steady state heat distribution; the hottest or coldest spot will not be inside.

Commonly the Laplace equation is part of a so-called Dirichlet problem ( named after the German mathematician Johann Peter Gustav Lejeune Dirichlet). That is, we have a region in the (xy)-plane and we specify certain values along the boundaries of the region. We then try to find a solution (u) defined on this region such that (u) agrees with the values we specified on the boundary.

For simplicity, we consider a rectangular region. Also for simplicity we specify boundary values to be zero at 3 of the four edges and only specify an arbitrary function at one edge. As we still have the principle of superposition, we can use this simpler solution to derive the general solution for arbitrary boundary values by solving 4 different problems, one for each edge, and adding those solutions together. This setup is left as an exercise.

We wish to solve the following problem. Let (h) and (w) be the height and width of our rectangle, with one corner at the origin and lying in the first quadrant.

[ Delta u=0,]

[ u(0,y)=0 ~~{ m{for}~} 0

[ u(x,h)=0 ~~{ m{for}~} 0

[u(w,y)=0 ~~{ m{for}~} 0

[u(x,0)=f(x) ~~{ m{for}~} 0

The method we apply is separation of variables. Again, we will come up with enough building-block solutions satisfying all the homogeneous boundary conditions (all conditions except (4.9.9)). We notice that superposition still works for the equation and all the homogeneous conditions. Therefore, we can use the Fourier series for (f(x)) to solve the problem as before.

We try (u(x,y)=X(x)Y(y)). We plug (u) into the equation to get

[X''Y+XY''=0.]

We put the (X)s on one side and the (Y)s on the other to get

[ - dfrac{X''}{X}= dfrac{Y''}{Y}.]

The left hand side only depends on (x) and the right hand side only depends on (y). Therefore, there is some constant ( lambda ) such that ( lambda = dfrac{-X''}{X}= dfrac{Y''}{Y}). And we get two equations

[ X'' + lambda X = 0, Y'' - lambda Y=0.]

Furthermore, the homogeneous boundary conditions imply that (X(0)=X(w)=0) and (Y(h)=0). Taking the equation for (X) we have already seen that we have a nontrivial solution if and only if ( lambda= lambda_n = dfrac{n^2 pi^2}{w^2}) and the solution is a multiple of

[ X_n(x) = sin left( dfrac{n pi}{w}x ight). ]

For these given ( lambda_n ), the general solution for (Y) (one for each (n)) is

[Y_n(y)=A_n cosh left( dfrac{n pi}{w}y ight) + B_n sinh left( dfrac{n pi}{w}y ight). ]

We only have one condition on (Y_n) and hence we can pick one of (A_n) or (B_n) to be something convenient. It will be useful to have (Y_n(0)=1), so we let (A_n=1). Setting (Y_n(h)=0) and solving for (B_n) we get that

[ B_n = dfrac{ - cosh left( dfrac{n pi h}{w} ight)}{ sinh left( dfrac{n pi h}{w} ight)}.]

After we plug the (A_n) and (B_n) we into (4.9.14) and simplify, we find

[ Y_n(y) = dfrac{ sinh left( dfrac{n pi (h-y)}{w} ight)}{ sinh left( dfrac{n pi h}{w} ight)}.]

We define (u_n(x,y)=X_n(x)Y_n(y)). And note that (u_n) satisfies (4.9.5)–(4.9.8).

Observe that

[ u_n(x,0)= X_n(x)Y_n(0) = sin left( dfrac{n pi}{w}x ight). ]

Suppose

[ f(x)= sum_{n=1}^{ infty}b_n sin left( dfrac{n pi x}{w} ight).]

Then we get a solution of (4.9.5)–(4.9.9) of the following form.

[ u(x,y)= sum_{n=1}^{ infty}b_n u_n(x,y) = sum_{n=1}^{ infty}b_n sin left( dfrac{n pi }{w}x ight) left( dfrac{ sinh left( dfrac{n pi (h-y)}{w} ight)}{ sinh left( dfrac{n pi h}{w} ight)} ight).]

As (u_n) satisfies Equation ef{4.9.5}– ef{4.9.8}) and any linear combination (finite or infinite) of (u_n) must also satisfy (4.9.5)–(4.9.8), we see that (u) must satisfy Equations ef{4.9.5}– ef{4.9.8}. By plugging in (y=0) it is easy to see that (u) satisfies (4.9.9) as well.

Example (PageIndex{1}):

Suppose that we take (w=h= pi) and we let (f(x)= pi). We compute the sine series for the function ( pi)(we will get the square wave). We find that for (0

[ f(x)= sum_ {underset{n~ { m{odd}} }{n=1}}^{ infty} dfrac{4}{n} sin(n x).]

Therefore the solution (u(x,y)), see Figure 4.22, to the corresponding Dirichlet problem is given as

[ u(x,y)= sum_ {underset{n~ { m{odd}} }{n=1}}^{ infty} dfrac{4}{n} sin(n x) left( dfrac{ sinh(n( pi-y)) }{ sinh(n pi) } ight).]

Figure 4.22: Steady state temperature of a square plate with three sides held at zero and one side held at ( pi).

This scenario corresponds to the steady state temperature on a square plate of width ( pi) with 3 sides held at 0 degrees and one side held at ( pi) degrees. If we have arbitrary initial data on all sides, then we solve four problems, each using one piece of nonhomogeneous data. Then we use the principle of superposition to add up all four solutions to have a solution to the original problem.

A different way to visualize solutions of the Laplace equation is to take a wire and bend it so that it corresponds to the graph of the temperature above the boundary of your region. Cut a rubber sheet in the shape of your region—a square in our case—and stretch it fixing the edges of the sheet to the wire. The rubber sheet is a good approximation of the graph of the solution to the Laplace equation with the given boundary data.


4.9: Steady state temperature and the Laplacian - Mathematics

Complete the plucked string problem to get equation 4.0

Here we start with the solution given in 4.8

Where (f elax (x) ) represents the initial position (shape) of the string.

First need to define (f elax (x) ), from diagram we see that from (x=0) to (x=L/2) the slope is (frac =frac <2h>) hence from equation of line we get ( y=frac <2h>x)

From (x=L/2) to (x=L), slope is (-frac <2h>), so (y=h-frac <2h>left ( x-frac <2> ight ) =h-frac <2h>x+h=2h-frac <2h>x= 2left ( h-frac ight ))

so we have [ f elax (x) =left < egin [c]frac <2h>x & & 0leq xleq frac <2> 2left (h-frac ight ) & & frac <2><xleq L end ight . ]

so from ( elax (1) ) we get, after applying inner product w.r.t. (sin left (frac ight ) )

Looking at few values of n to see the pattern

Notice that we have terms for only odd n.

Now, substituting the above in the general solution given in equation 4.7 in book, which is

egin y & = frac <8h>>left (sin left (frac ight ) cos left (frac ight ) +0+-frac <1><9>sin left (frac <3pi x> ight ) cos left (frac <3pi vt> ight ) +0+frac <1><25>sin left ( frac <5pi x> ight ) cos left (frac <5pi vt> ight ) +. ight ) y & =frac <8h>>left (sin left (frac ight ) cos left (frac ight ) -frac <1><9>sin left (frac <3pi x> ight ) cos left (frac <3pi vt> ight ) +frac <1><25>sin left ( frac <5pi x> ight ) cos left (frac <5pi vt> ight ) +. ight ) end

The above is the result we are asked to show.

4.9.3 chapter 13, problem 4.2. Mary Boas, second edition

A string of length L has zero initial velocity and a displacement (y_<0>left ( x ight ) ) as shown. Find the displacement as a function of x and t.

The PDE that governs this problem is the wave equation ( abla ^<2>y=frac <1>>frac y>>)

The candidate solutions are

[ y=left < egin [c] sin (kx),sin (omega t) sin (kx),cos (omega t) cos (kx),sin (omega t) cos (kx),cos (omega t) end ight . ]

where (omega =kv) and (k=frac <2pi >) where (lambda ) is the wave length

Now we discard solutions that contains (cos kx) since the string is fixed at (x=0).

[ y=left < egin [c]sin (kx),sin (omega t) sin (kx),cos (omega t) end ight . ]

Now, (y=0) at (x=L) then from (sin kx=0) or (sin kL=0 )we need (k=frac )

Applying initial conditions, which says that at time (t=0), velocity is zero.

Hence from above, after taking (frac ), we get

For the above to be zero at (t=0) then we discard first solution above with (cos t) in it. Hence final general solution is now

A general solution is a linear combination of the above solutions, hence

To find (b_), we apply the second initial condition, which is (y=y_<0>=f elax (x) )

(Notice that we use two initial conditions, i.e. at time t=0 we are looking at speed and position, this is because we started with a PDE with (frac y>>) in it, which is a second order in t.)

To find (f elax (x)) from diagram, we see that for (0leq xleq frac <4>), (y=xfrac =frac <4h>x)

Do the inner product on both sides of equation (2) w.r.t. (sin frac x)

Looking at few values of (b_)

egin b_ & =frac <8h><1^<2>pi ^<2>>left (2sin left (frac <4> ight ) -sin frac <2> ight ) ,frac <8h><2^<2>pi ^<2>>left (2sin left ( frac <2pi ><4> ight ) -sin frac <2pi ><2> ight ) ,frac <8h><3^<2>pi ^<2>>left (2sin left (frac <3pi ><4> ight ) -sin frac <3pi ><2> ight ) ,dots & =frac <8h>>left [ left (2sin left (frac <4> ight ) -sin frac <2> ight ) ,frac <1><2^<2>>left (2sin frac <2pi ><4>-sin frac <2pi ><2> ight ) ,frac <1><3^<2>>left (2sin frac <3pi ><4>-sin frac <3pi ><2> ight ) ,dots ight ] & =frac <8h>>left [ frac <1>>left < left (2sin left ( frac <4> ight ) -sin frac <2> ight ) ,left (2sin frac <2pi ><4>-sin frac <2pi ><2> ight ) ,left (2sin frac <3pi ><4>-sin frac <3pi ><2> ight ) ,dots ight > ight ] & =frac <8h>> left [ frac <1>>left (2sin left (frac <4> ight ) -sin frac <2> ight ) ight ] end

Hence from equation (1) above, we get

Where [ B_=frac <1>>left (2 sin left (frac <4> ight ) -sin frac <2> ight ) ]

The above is the result required to show.

4.9.4 chapter 13, problem 4.6. Mary Boas, second edition

A string of length L is initially stretched straight, its ends are fixed for all time t. At time t=0 its points are given the velocity (V elax (x) =left (frac ight ) _ )as shown in diagram below. Determine the shape of the string at time t.

The PDE that governs this problem is the wave equation ( abla ^<2>y=frac <1>>frac y>>)

The candidate solutions are

[ y=left < egin [c] sin (kx),sin (omega t) sin (kx),cos (omega t) cos (kx),sin (omega t) cos (kx),cos (omega t) end ight . ]

Where (omega =kv) and (k=frac <2pi >) where (lambda ) is the wave length

Now we discard solutions that contains (cos kx) since the string is fixed at (x=0).

[ y=left < egin [c]sin (kx),sin (omega t) sin (kx),cos (omega t) end ight . ]

Now, (y=0) at (x=L) then from (sin kx=0) or (sin kL=0 )we need (k=frac )

Applying initial conditions, which says that at time (t=0), velocity is given by (V elax (x) )

Hence from above, after taking (frac ), we get

For the above we discard velocity solution above with (sin t) in it since that will give zero velocity at time t=0, which is not the case here. Hence we discard y solution with (cos t) in it, then the final general solution for y is now

A general solution is a linear combination of the above solutions, hence

To find (b_), we apply the velocity initial condition. Hence differentiate equation (1) and set t=0, we have

Now to find (V_). From diagram, we see that for (0leq xleq frac <2>-w), (V_=0)

Do the inner product on both sides of equation (2) w.r.t. (sin frac x)

But (cos left (a+b ight ) =cos elax (a) cos elax (b) -sin elax (a) sin elax (b) )

and (cos left (a-b ight ) =cos elax (a) cos elax (b) +sin elax (a) sin elax (b) )

egin b_ & =-frac <2hL>pi ^<2>v>left [ cos left (a+b ight ) -cos left (a-b ight ) ight ] & =-frac <2hL>pi ^<2>v>left [ cos elax (a) cos left ( b ight ) -sin elax (a) sin elax (b) -left < cos left ( a ight ) cos elax (b) +sin elax (a) sin elax (b) ight > ight ] & =-frac <2hL>pi ^<2>v>left [ cos elax (a) cos left ( b ight ) -sin elax (a) sin elax (b) -cos elax (a) cos elax (b) -sin elax (a) sin elax (b) ight ] & =-frac <2hL>pi ^<2>v>left [ -sin elax (a) sin left ( b ight ) -sin elax (a) sin elax (b) ight ] & =frac <4hL>pi ^<2>v>sin elax (a) sin elax (b) & =frac <4hL>pi ^<2>v>sin left (frac <2> ight ) sin left ( frac ight ) end

For even (n), the term (sin left (frac <2> ight ) ) is zero. For (n) odd (sin left (frac <2> ight ) =1) when (n=1,5,9,dots ) and (sin left (frac <2> ight ) =-1) when (n=3,7,11,dots )Hence

[ b_=A elax (n) frac <4hL>pi ^<2>v>sin left (frac ight )qquad n=1,3,5,7,dots ]

And (A elax (n) ) is a function which returns (1) when (n=1,5,9. ) and returns (-1) when (n=3,7,11,dots )

Hence now we have (b_) we can substitute in (1)

Which is the general solution. Looking at few expanded terms in the series we get

Which is the result required.

4.9.5 chapter 13, problem 5.1. Mary Boas, second edition

Compute numerically the coefficients ( c_=frac <200><>J_<1>left ( k_ ight ) >) for the first 3 terms of the series (u=^> c_J_<0>left (k_r ight ) e^<-k_z>) for the steady state temp. in a solid semi-infinite cylinder when (u=0) at (r=1) and (u=100) at (z=0.) find (u) at (r=1/2,z=1)

Here, we are looking at the solution for temp. inside a semi-infinite cylinder. This solution is for the case of a uniform temp. distribution on the boundary (z=0) is given by (u) equation shown above. Note that in the expression (c_=frac <200><>J_<1>left (k_ ight ) >), the (k_) are the zeros of (J_<0>) not (J_<1>.)

Need to find (c_<1>,c_<2>,c_<3>) where (c_<1>=frac <200>J_<1>left (k_<1> ight ) >)

To find (k_<1>) and (J_<1>left (k_<1> ight ) ) I used mathematica.

I plotted (J_<0> elax (x) ) to see where the zeros are located first

So I see there is a zero near 2,5, and 9. I use mathematica to find these:

Now I need to find (J_<1>left (k_ ight ) ). This is the result for 3 terms:

Hence, now the (c_) terms can be found:

egin u & =c_<1>J_<0>left (k_<1>r ight ) e^<-k_<1>z>+c_<2>J_ <0>left (k_<2>r ight ) e^<-k_<2>z>+c_<3>J_<0>left (k_<3>r ight ) e^<-k_<3>z> & =c_<1>J_<0>left (k_<1>frac <1><2> ight ) e^<-k_<1>>+c_<2>J_<0>left ( k_<2>frac <1><2> ight ) e^<-k_<2>>+c_<3>J_<0>left (k_<3>frac <1><2> ight )e^<-k_<3>> & =left (160.30 ight ) J_<0>left (2.404frac <1><2> ight ) e^<-2.404>-left (106.56 ight ) J_<0>left (5.52frac <1><2> ight ) e^<-5.52>+left (85.635 ight ) J_<0>left (8.65frac <1><2> ight ) e^<-8.65> & =left (160.30 ight ) J_<0>left (1.202 ight ) e^<-2.404>-left (106.56 ight ) J_<0>left (2.76 ight ) e^<-5.52>+left (85.635 ight ) J_<0>left (4.325 ight ) e^ <-8.65>end

Mathematica was used to evaluate (J_<0>) values above.

egin u & =left (160.30 ight ) left (0.67 ight ) e^<-2.404>-left (106.56 ight ) left (-0.168 ight ) e^<^<-5.52>>+left (85.635 ight ) left (-0.356 ight ) e^<-8.65> u & =9.704,3+7.171,3 imes 10^<-2>-5.3389 imes 10^<-3> u & =9.7707 ext end

4.9.6 chapter 13, problem 5.2. Mary Boas, second edition

Find the solution for the steady state temp. distribution in a solid semi-infinite cylinder if the boundary temp. are (u=0) at (r=1) and (u=y=rsin heta ) at (z=0).

The candidate solutions are given by the solution to the Laplace equation in cylindrical coordinates which are

[ u=left < egin [c]J_left (k r ight ) sin left (n heta ight ) e^ <-k z>& & (1) & & J_left (k r ight ) cos left (n heta ight ) e^ <-k z>& & (2) end ight . ]

Where (k) is a zero of (J_) (This is because we have used the B.C. of (u=0) at (r=1) to determine that the (k^s) have to be the zeros of (J_)) when deriving the above solutions. See book page 560.

From boundary conditions we want (u=rsin heta ) when (z=0), hence we need to keep the solution (1) above, with (n=1). Hence a solution is [ u=J_<1>left (k r ight ) sin left ( heta ight ) e^<-k z>qquad (3) ]

A general solution is a linear series combinations (eigenfunctions) of (3), each eigenfunction for each of the zeros of (J_<1>). Call these zeros (k_)

[ u=sum _^c_ J_<1>left (k_ r ight ) sin left ( heta ight ) e^<-k_ z>qquad (4) ]

We now apply B.C. at (z=0) to find (c_). From (4) when (z=0) [ rsin heta =sum _^c_ J_<1>left (k_ r ight ) sin left ( heta ight ) qquad (5) ]

We use (5) to find (c_) and then substitute into (4) to obtain the final solution.

To find (c_) from (5), take the inner product of each side with respect to (rJ_<1>left (k_ r ight ) ) from (r=0) to (r=1)

egin int _<0>^<1>rsin heta left [ rJ_<1>left (k_ r ight ) ight ] dr & =sum _^c_ left (int _<0>^<1>J_<1>left (k_ r ight ) sin left ( heta ight ) left [ rJ_<1>left (k_ r ight ) ight ] dr ight ) sin heta int _<0>^<1>r^<2>J_<1>left (k_ r ight ) dr & =sum _^c_ sin left ( heta ight ) left (int _<0>^<1>J_<1>left (k_ r ight ) left [ rJ_<1>left (k_ r ight ) ight ] dr ight ) end

Dividing each side by (sin heta )

[ int _<0>^<1>r^<2>J_<1>left (k_ r ight ) dr=sum _^c_ left (int _<0>^<1>J_<1>left (k_ r ight ) left [ rJ_<1>left ( k_ r ight ) ight ] dr ight ) ]

From orthogonality of Bessel function, we know that

[ int _<0>^<1>J_

left (k_ r ight ) r J_

left (k_ r ight ) dr=0 ]

If (m eq u). Hence in above equation all terms on the right drop except for one when (u=m). We get

[ int _<0>^<1>r^<2>J_<1>left (k_ r ight ) dr= c_ int _<0>^<1>r J_<1>left (k_ r ight ) J_<1>left (k_ r ight ) dr ]

[ c_=frac ^<1>r^<2>J_<1>left (k_ r ight ) dr>^<1>r J_<1>left (k_ r ight ) J_<1>left (k_ r ight ) dr>qquad (6) ]

The integral in the denominator above is found from equation 19.10 in text on page 523 which gives [ int _<0>^<1>r J_<1>left (k_ r ight ) J_<1>left (k_ r ight ) dr=frac <1><2>left [ J_<2>left (k_ ight ) ight ] ^<2>qquad (7) ]

Now, we need to find the integral of the numerator in equation (6).

Using equation 15.1 in text, page 514, which says

[ frac left [ x^

J_

elax (x) ight ] =x^

J_left ( x ight ) ]

Putting (p=2) above, and letting (x=k_r) gives

egin frac <1><>>frac left [ left (k_r ight ) ^<2>J_<2>left ( k_r ight ) ight ] & =left (k_r ight ) ^<2>J_<1>left ( k_r ight ) frac <1><>>frac left [ k_^<2>r^<2>J_<2>left (k_r ight ) ight ] & =k_^<2>r^<2>J_<1>left (k_r ight ) frac <1><>>frac left [ r^<2>J_<2>left (k_r ight ) ight ] & =r^<2>J_<1>left (k_r ight ) end

Integrating each side w.r.t (r) from (0dots 1)

egin frac <1><>>int _<0>^<1>frac left [ r^<2>J_<2>left ( k_r ight ) ight ] dr & =int _<0>^<1>r^<2>J_<1>left (k_r ight ) dr frac <1><>> left [ r^<2>J_<2>left (k_r ight ) ight ] _<0>^ <1>& =int _<0>^<1>r^<2>J_<1>left (k_r ight ), dr frac <1><>> left [ J_<2>left (k_ ight ) -0 ight ] & =int _<0>^<1>r^<2>J_<1>left (k_r ight ) dr frac <1><>> J_<2>left (k_ ight ) & =int _<0>^<1>r^<2>J_<1>left ( k_r ight ),drqquad (8) end

Substituting (7) and (8) into (6)

egin c_ & =frac ^<1>r^<2>J_<1>left (k_ r ight ) dr>^<1>r J_<1>left (k_ r ight ) J_<1>left (k_ r ight ) dr> & =frac <>> J_<2>left (k_ ight ) ><2>left [ J_<2>left (k_ ight ) ight ] ^<2> > & =frac <2 ><> J_<2>left (k_ ight )> end

Substituting this into (4) above, we get

egin u & =sum _^c_ J_<1>left (k_ r ight ) sin left ( heta ight ) e^<-k_ z> u & =sum _^frac <2 ><> J_<2>left (k_ ight ) > J_<1>left (k_ r ight ) sin left ( heta ight ) e^<-k_ z> end

The above is the result we are asked to show.

4.9.7 chapter 13, problem 5.4. Mary Boass, second edition

A flat circular plate of radius 1 is initially at temp. 100(^<0>). From (t=0) on, the circumference of the plate is held at 0(^<0>.) Find the time-dependent temp distribution (uleft (r, heta ,t ight ) )

First convert heat equation from Cartesian coordinates to polar.

First need to express Laplacian operator ( abla ^<2>) in polar coordinates: egin x &= rcos heta y &=rsin heta end

From geometry, we also know that egin r &=sqrt +y^<2>> heta &=arctan frac end

The above 2 relations imply

Hence we can express the above, using equations (A) and (B) as follows:

Squaring each sides of (1) gives

Notice that when manipulating of differential operators, (xfrac eq frac x). Similarly

Squaring each side of (3) gives

Adding equation (2) and (4) and carry cancellations

Now that we have the Laplacian in polar coordinates, we can solve the problem by applying separation of variables on the heat PDE expressed in polar coordinates.

Let solution (uleft (r, heta ,t ight ) ) be a linear combination of functions each depends on only (r, heta ,) or (t)

egin uleft (r, heta ,t ight ) =R elax (r) Theta left ( heta ight ) T elax (t) ag <6>end

Substitute (6) in (5). First evaluate the various derivatives:

egin frac >>u+frac <1>frac u+frac <1>>frac >>u & =frac <1>>frac u Theta left ( heta ight ) T elax (t) frac >>Rleft ( r ight ) +frac <1>Theta left ( heta ight ) T elax (t) frac R elax (r) +frac <1>>R elax (r) Tleft ( t ight ) frac >>Theta left ( heta ight ) & =frac <1>>R elax (r) Theta left ( heta ight ) frac

T elax (t) end

Divide by (R elax (r) Theta left ( heta ight ) T elax (t) )

We notice that the RHS depends only on (t) and the LHS depends only on (r, heta ) and they equal to each others, hence they both must be constant. Let this constant be (-k^<2>)

equation (7) is a linear first order ODE with constant coeff. (frac

T elax (t) =-alpha ^<2>T elax (t) k^<2>) or (frac
=-alpha ^<2>k^<2>dt)

Integrating to solve gives egin int frac

&=int -alpha ^<2>k^<2>dt ln T elax (t) &=-alpha ^<2>k^<2>t end

oregin T elax (t) =e^<-alpha ^<2>k^<2>t> ag <9>end Looking at equation (8). Multiply each sides by (r^<2>) we getegin frac >left [ frac >>Rleft ( r ight ) +frac <1>frac R elax (r) ight ] +frac <1>frac >>Theta left ( heta ight ) & =-r^<2>k^<2> onumber frac >left [ frac >>Rleft ( r ight ) +frac <1>frac R elax (r) ight ] +r^<2>k+frac <1>frac >>Theta left ( heta ight ) & =0 onumber r^<2>left (frac < 1>left [ frac >>R elax (r) +frac <1>frac R elax (r) ight ] +k^<2> ight ) +frac <1>frac >>Theta left ( heta ight ) & =0 ag <10>end

The second term depends only on ( heta ) and the first term depends only on (r) and they are equal, hence they must be both constant. Let this constant be (-n^<2>) hence

This is a second order linear ODE with constant coeff. Solution is

egin Theta left ( heta ight ) =left < egin [c]sin n heta & & & & cos n heta & & end ight . ag <11>end

egin r^<2>left (frac < 1>left [ frac >>R elax (r) +frac <1>frac R elax (r) ight ] +k^<2> ight ) -n^ <2>& =0 onumber frac < r^<2>>left [ frac >>Rleft ( r ight ) +frac <1>frac R elax (r) ight ] +r^<2>k^<2>-n^ <2>& =0 onumber r^<2>left [ frac >>R elax (r) +frac <1>frac R elax (r) ight ] +left (r^<2>k^<2>-n^<2> ight ) Rleft ( r ight ) & =0 onumber r^<2>frac >>R elax (r) +rfrac R elax (r) +left (r^<2>k^<2>-n^<2> ight ) R elax (r) & =0 ag <12>end

Equation (12) is the Bessel D.E., its solutions are (J_left (kr ight ) ) and (N_left (kr ight ) ) . As described on book on page 560, we can not use the (N_left (kr ight ) ) solution since plate contains the origin and (N_ elax (0) ) is not defined. So we use solution (Rleft ( r ight ) =J_left (kr ight ) .) From boundary conditions, we want solution to be zero at (r=1), hence we want (J_ elax (k) =0), hence the k’s are the zeros of (J_)

Putting these solutions together, we get from (6)

egin uleft (r, heta ,t ight ) & =R elax (r) Theta left ( heta ight ) T elax (t) & =left < egin [c]J_left (kr ight ) sin n heta e^<-alpha ^<2>k^<2>t> & & & & J_left (kr ight ) cos n heta e^<-alpha ^<2>k^<2>t> & & end ight . end

From symmetry of plate, the solution can not depend on the angle ( heta ), hence let (n=0) and so as not to get (u=0), we must pick the solution with (cos n heta ) term. Hence our solution now is

The general solution is a linear combination of this eigenfunction for all zeros of (J_<0>), hence

We find (c_) by using initial condition. When (t=0) (,) temp. was 100(^ <0>) hence

Applying inner product w.r.t. (rJ_<0>left (k_r ight ) ) from (0dots 1)

egin int _<0>^<1>100 rJ_<0>left (k_r ight ) dr & =int _<0>^<1>left ( sum _^c_ J_<0>left (k_r ight ) ight ) rJ_<0>left ( k_r ight ) dr 100int _<0>^<1> rJ_<0>left (k_r ight ) dr & =sum _^c_int _<0>^<1>J_<0>left (k_r ight ) rJ_<0>left (k_r ight ) dr end

From orthogonality of (J_<0>left (k_r ight ) )and (J_<0>left ( k_r ight ) ), all terms drop expect when (m=u)

[ 100int _<0>^<1> rJ_<0>left (k_r ight ) dr= c_int _<0>^<1> r left [ J_<0>left (k_r ight ) ight ] ^<2> dr ]

From here we can follow the book on page 561 to get [ c_=frac <200><>J_<1>left (k_ ight ) >]

Substitute this in equation 13

Notice that final solution does not depend on ( heta )

4.9.8 chapter 13, problem 5.11. Mary Boas, second edition

Solve egin rfrac left (rfrac ight ) &=n^<2>R frac left (r^<2>frac ight ) &=lleft (l+1 ight ) R end

First equation, use power series method.

egin R & =a_<0>r^+a_<1>r^+a_<2>r^+a_<3>r^+a_<4>r^+cdots -n^<2>R & =-n^<2>a_<0>r^-n^<2>a_<1>r^-n^<2>a_<2>r^-n^<2>a_<3>r^-n^<2>a_<4>r^-cdots frac & =s a_<0>r^+left (s+1 ight ) a_<1>r^+left ( s+2 ight ) a_<2>r^+left (s+3 ight ) a_<3>r^+ cdots rfrac & =s a_<0>r^+left (s+1 ight ) a_<1>r^+left ( s+2 ight ) a_<2>r^+left (s+3 ight ) a_<3>r^+ cdots frac R>> & = left (s-1 ight ) s a_<0>r^+sleft ( s+1 ight ) a_<1>r^+left (s+1 ight ) left (s+2 ight ) a_<2>r^+left (s+2 ight ) left (s+3 ight ) a_<3>r^+ cdots r^<2>frac R>> & = left (s-1 ight ) s a_<0>r^+sleft ( s+1 ight ) a_<1>r^+left (s+1 ight ) left (s+2 ight ) a_<2>r^+left (s+2 ight ) left (s+3 ight ) a_<3>r^+ cdots end

(r^) (r^) (r^) (r^)
(-n^<2>R) (-n^<2>a_<0>) (-n^<2>a_<1>) (-n^<2>a_<2>) (-n^<2> a_)
(rfrac ) (s a_<0>) (left (s+1 ight ) a_<1>) (left ( s+2 ight ) a_<2>) ( left (s+m ight ) a_)
(r^<2>frac R>>) ( left (s-1 ight ) s a_<0>) (sleft ( s+1 ight ) a_<1>) (left (s+1 ight ) left (s+2 ight ) a_<2>) (left (s+m-1 ight ) left (s+m ight ) a_)

Hence, from first column we see , and since (a_<0>) ( eq 0) we solve for (s)

egin -n^<2>a_<0>+s a_<0>+left (s-1 ight ) s a_ <0>& =0 a_<0>left (-n^<2>+s+left (s-1 ight ) s ight ) & =0 -n^<2>+s+left (s-1 ight ) s & =0 -n^<2>+ s^ <2>& =0 s & =pm n end

We see from second column, (a_<1>left (-n^<2>+left (s+1 ight ) +s^<2>+s ight ) =0) or (a_<1>left (-s^<2>+2s+1+s^<2> ight ) =0), hence (a_<1>left (2s+1 ight ) =0)

For (a_<1> eq 0) then (s=-frac <1><2>), this means (n) is not an integer since (s=pm n). hence (a_<1>) must be zero.

The same applies to all (a_) , (m>0) Hence solution contains only (a_<0>)

For some constant (a_<0>). This solution is when (n eq 0)

(r^) (r^) (r^) (r^)
(-n^<2>R) (0) (0) (0) (0)
(rfrac ) (s a_<0>) (left (s+1 ight ) a_<1>) (left ( s+2 ight ) a_<2>) ( left (s+m ight ) a_)
(r^<2>frac R>>) ( left (s-1 ight ) s a_<0>) (sleft ( s+1 ight ) a_<1>) (left (s+1 ight ) left (s+2 ight ) a_<2>) (left (s+m-1 ight ) left (s+m ight ) a_)

And all other (a^s) are zero. Hence (R=a_<0>) or (R) is constant.

egin frac left (r^<2>frac ight ) & =lleft (l+1 ight ) R r^<2>frac R>>+2rfrac -lleft (l+1 ight ) R & =0 end

egin R & =a_<0>r^+a_<1>r^+a_<2>r^+a_<3>r^+a_<4>r^+cdots -lleft (l+1 ight ) R & =-lleft (l+1 ight ) a_<0>r^-lleft ( l+1 ight ) a_<1>r^-lleft (l+1 ight ) a_<2>r^-lleft ( l+1 ight ) a_<3>r^-lleft (l+1 ight ) a_<4>r^-cdots frac & =s a_<0>r^+left (s+1 ight ) a_<1>r^+left ( s+2 ight ) a_<2>r^+left (s+3 ight ) a_<3>r^+ cdots 2rfrac & =2s a_<0>r^+2left (s+1 ight ) a_<1>r^+2left (s+2 ight ) a_<2>r^+2left (s+3 ight ) a_<3>r^+ cdots frac R>> & = left (s-1 ight ) s a_<0>r^+sleft ( s+1 ight ) a_<1>r^+left (s+1 ight ) left (s+2 ight ) a_<2>r^+left (s+2 ight ) left (s+3 ight ) a_<3>r^+ cdots r^<2>frac R>> & = left (s-1 ight ) s a_<0>r^+sleft ( s+1 ight ) a_<1>r^+left (s+1 ight ) left (s+2 ight ) a_<2>r^+left (s+2 ight ) left (s+3 ight ) a_<3>r^+ cdots end

(r^) (r^) (r^) (r^)
(-n^<2>R) (-lleft (l+1 ight ) a_<0>) (-lleft (l+1 ight ) a_<1>) (-lleft (l+1 ight ) a_<2>) (-lleft (l+1 ight ) a_)
(2rfrac ) (2s a_<0>) (2left (s+1 ight ) a_<1>) (2left ( s+2 ight ) a_<2>) ( 2left (s+m ight ) a_)
(r^<2>frac R>>) ( left (s-1 ight ) s a_<0>) (sleft ( s+1 ight ) a_<1>) (left (s+1 ight ) left (s+2 ight ) a_<2>) (left (s+m-1 ight ) left (s+m ight ) a_)

From first column: egin -lleft (l+1 ight ) a_<0>+2s a_<0>+left (s-1 ight ) s a_<0>&=0 a_<0>left (-lleft (l+1 ight ) +2s +left (s-1 ight ) s ight ) &=0 -lleft (l+1 ight ) +2s +left (s-1 ight ) s&=0 -lleft (l+1 ight ) +s +s^<2>&=0 left (s-l ight ) left (s-left (-l-1 ight ) ight ) &=0 end

We also see that all other (a^s) will be zero, since recursive formula has only (a_) in it and no other (a). Hence


ME3291: The Governing Equation for the Temperature Distribution with Time on a 2D: Numerical Methods for Engineers Assignment, NUS,, Singapore

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Assignment Details

(1). The governing equation for the temperature distribution with time on a 2D square plate measuring 1 unit by1 unit is

Subjected to the Dirichlet boundary conditions for T provided in Fig.1. You are to obtain the following:

(a) The temperature contour plot on the square plate with time, say at t=0.01, 0.1 and at steady state. (You can provide contours at other times too to depict the convergence of the results at steady state.) Take the initial condition at t=0 as T=0.0 for the whole domain.

(b) Separately, program and compute for the Laplace Equation

and obtain the solution for comparison to the steady-state solution in (a).

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For the above, you have to show clearly how you treat the Dirichlet boundary conditions, provide a listing of your program, and other pertinent workings. The various contour plots can be carried out using the Techp1ot or any other suitable software. (On matrix inversion, you have the choice to use the direct method like Gauss Elimination or indirect iterative methods.)

(2) The same governing equation for the temperature distribution with time on a 2D square plate measuring 1 unit by 1 unit is given as

In this case, the boundary conditions are given as the Dirichlet type for 3 sides of the plate and reflected as follows (Fig. 2),

and the Neumann boundary condition for

Obtain the temperature contour plot on the square plate with time, say at=0.01, 0.1 and at steady state. (You can provide contours at other times too to depict the convergence of the results at steady state.)

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Notes on Diffy Qs: Differential Equations for Engineers

Notes on Diffy Qs is an introductory textbook on differential equations for engineering students. In approach it is close to (and internally cross-referenced with) books by Edwards and Penney (Differential Equations and Boundary Value Problems) and Boyce and DiPrima (Elementary Differential Equations and Boundary Value Problems). Lebl&rsquos intent is to provide a comparable book in hardcopy at a low cost (or freely available here for download). The hardcopy is entirely in black-and-white whereas the downloadable version has several color illustrations.

The text is intended to support a one-semester or two-quarter course, and could be stretched to two semesters with some supplementary material. The contents are quite standard for a course that aims to introduce both ordinary and partial differential equations. The major topics are first and higher order ordinary differential equations (ODEs), systems of ODEs, then Fourier series and partial differential equations (PDEs), eigenvalue problems (including Sturm-Liouville problems), the Laplace transform and power series methods. Linear algebra is introduced as needed, but is kept to a minimum.

If one considers what knowledge and skills an engineering student should take from an introductory course on differential equations, one might include:

  • Recognizing common differential equations (e.g., variations of harmonic oscillator equation, heat equation, wave equation)
  • Understanding basic techniques for solving ODEs and PDEs
  • Creating awareness of issues involving existence and uniqueness
  • Learning rudiments of qualitative behavior of solutions.
  • Getting some experience with numerical solutions.

The current text does reasonably well by these criteria. One might wish for some more depth on qualitative methods, and a clearer message that most differential equations do not have closed form solutions.

The writing is plain but clear throughout and the style is relaxed and conversational. Exercises are plentiful but not particularly distinctive. There are plenty of examples, all carefully worked out. Where examples involve applications, these are all (unsurprisingly) in physics or engineering. Yet it might not be a bad idea for engineers to see at least a few applications outside their field.

Overall, this is a very competent &mdash but not especially inspired &ndash introduction to differential equations, one which is sensitive to the needs of engineering students in future coursework.

Bill Satzer ([email protected]) is a senior intellectual property scientist at 3M Company, having previously been a lab manager at 3M for composites and electromagnetic materials. His training is in dynamical systems and particularly celestial mechanics his current interests are broadly in applied mathematics and the teaching of mathematics.

Introduction
0.1 Notes about these notes
0.2 Introduction to differential equations

1 First order ODEs
1.1 Integrals as solutions
1.2 Slope fields
1.3 Separable equations
1.4 Linear equations and the integrating factor
1.5 Substitution
1.6 Autonomous equations
1.7 Numerical methods: Euler&rsquos method

2 Higher order linear ODEs
2.1 Second order linear ODEs
2.2 Constant coefficient second order linear ODEs
2.3 Higher order linear ODEs
2.4 Mechanical vibrations
2.5 Nonhomogeneous equations
2.6 Forced oscillations and resonance

3 Systems of ODEs
3.1 Introduction to systems of ODEs
3.2 Matrices and linear systems
3.3 Linear systems of ODEs
3.4 Eigenvalue method
3.5 Two dimensional systems and their vector fields
3.6 Second order systems and applications
3.7 Multiple eigenvalues
3.8 Matrix exponentials
3.9 Nonhomogeneous systems

4 Fourier series and PDEs
4.1 Boundary value problems
4.2 The trigonometric series
4.3 More on the Fourier series
4.4 Sine and cosine series
4.5 Applications of Fourier series
4.6 PDEs, separation of variables, and the heat equation
4.7 One dimensional wave equation
4.8 D&rsquoAlembert solution of the wave equation
4.9 Steady state temperature and the Laplacian
4.10 Dirichlet problem in the circle and the Poisson kernel

5 Eigenvalue problems
5.1 Sturm-Liouville problems
5.2 Application of eigenfunction series
5.3 Steady periodic solutions

6 The Laplace transform
6.1 The Laplace transform
6.2 Transforms of derivatives and ODEs
6.3 Convolution
6.4 Dirac delta and impulse response

7 Power series methods
7.1 Power series
7.2 Series solutions of linear second order ODEs
7.3 Singular points and the method of Frobenius


GENERALIZED SCHRODINGER EQUATIONS FOR SHIFTS, WIDTHS, AND WAVE FUNCTIONS OF ATOMIC AND MOLECULAR STATES IN DENSE MATTER

4 DISCRETE ATOMIC EIGENFUNCTIONS OF THE REDUCED DENSITY MATRIX

The thermal equilibrium one-atom reduced density matrix ρ(X, X') is defined by

where the notation is the same as in Sec. 1 (X for atomic coordinates, Y for those of the medium) and ψi are the orthonormal states of the full Hamiltonian (1.1) with eigenvalues Ei. ρ(X, X') is a hermitian kernel and hence possesses orthonormal eigenfunctions ψα and eigenvalues λα satisfying

In cases where there exist bound states of the atom immersed in the medium, the eigenvalue spectrum <>α> will contain both a discrete and continuum part, and the medium-perturbed atomic wave functions may be defined as the discrete eigenfunctions ψα.

This definition of perturbed bound-state wave functions has been used, for example, in the theory of superconductivity, where the Cooper pair wave function is an eigenfunction of the two-electron reduced density matrix 11 . It has the advantage of being a fundamental approach but also several disadvantages:(a) the reduced density matrix of a system as complicated as a partially-ionized plasma can only be evaluated in crude approximation(b) the λα, being real, do not incorporate information regarding finite-lifetime effects.


A new mixed-FEM for steady-state natural convection models allowing conservation of momentum and thermal energy

In this work we present a new mixed finite element method for a class of steady-state natural convection models describing the behavior of non-isothermal incompressible fluids subject to a heat source. Our approach is based on the introduction of a modified pseudostress tensor depending on the pressure, and the diffusive and convective terms of the Navier–Stokes equations for the fluid and a vector unknown involving the temperature, its gradient and the velocity. The introduction of these further unknowns lead to a mixed formulation where the aforementioned pseudostress tensor and vector unknown, together with the velocity and the temperature, are the main unknowns of the system. Then the associated Galerkin scheme can be defined by employing Raviart–Thomas elements of degree k for the pseudostress tensor and the vector unknown, and discontinuous piece-wise polynomial elements of degree k for the velocity and temperature. With this choice of spaces, both, momentum and thermal energy, are conserved if the external forces belong to the velocity and temperature discrete spaces, respectively, which constitutes one of the main feature of our approach. We prove unique solvability for both, the continuous and discrete problems and provide the corresponding convergence analysis. Further variables of interest, such as the fluid pressure, the fluid vorticity, the fluid velocity gradient, and the heat-flux can be easily approximated as a simple postprocess of the finite element solutions with the same rate of convergence. Finally, several numerical results illustrating the performance of the method are provided.

This is a preview of subscription content, access via your institution.


1 Condition (1.11) is natural, since it determines the phase: Ω(t)=<q(t)>0>.

One contribution of 15 to a theme issue ‘Free boundary problems and related topics’.

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6.7 Traveling Waves

Now we turn our attention to a new PDE: the traveling wave equation [ u_t + v u_x = 0. ] In this equation (u(t,x)) is the height of a wave at time (t) and spatial location (x) . The parameter (v) is the velocity of the wave. Imagine this as sending a single solitary wave pulsing down a taught rope or as sending a single pulse of light down a fiber optic cable.

Exercise 6.55 Consider the PDE (u_t + v u_x = 0) . There is a very easy way to get an analytic solution to this traveling wave equation. If we have the initial condition (u(0,x) = f(x) = e^<-(x-4)^2>) then we claim that (u(t,x) = f(x-vt)) is an analytic solution to the PDE. More explicitely, we are claiming that [ u(t,x) = e^ <-(x-vt-4)^2>] is the analytic solution to the PDE. Let’s prove this.

  1. Take the (t) derivative of (u(t,x)) .
  2. Take the (x) derivative of (u(t,x)) .
  3. The PDE claims that (u_t + vu_x = 0) . Verify that this equal sign is indeed true.

The traveling wave equation (u_t + vu_x = 0) has a very nice analytic solution which we can always find. Therefore there is no need to ever find a numerical solution – we can just write down the analytic solution if we are given the initial condition. As it turns out though the numerical solutions exhibit some very interesting behavior.

Exercise 6.58 Consider the traveling wave equation (u_t + vu_x = 0) with initial condition (u(0,x) = f(x)) for some given function (f) and boundary condition (u(t,0) = 0) . To build a numerical solution we will again adopt the notation (U_i^n) for the approximation to (u(t,x)) at the point (t=t_n) and (x=x_i) .

  1. Write an approximation of (u_t) using (U_i^) and (U_i^n) .
  2. Write an approximation of (u_x) using (U_^n) and (U_i^n) .
  3. Substitute your answers from parts (a) and (b) into the traveling wave equation and solve for (U_i^) . This is our first finite difference scheme for the traveling wave equation.
  4. Write Python code to get the finite difference approximation of the solution to the PDE. Plot your finite difference solution on top of the analytic solution for (f(x) = e^<-(x-4)^2>) . What do you notice? Can you stabilize this method by changing the values of (Delta t) and (Delta x) like with did with the heat and wave equations?

The finite difference scheme that you built in the previous exercise is called the downwind scheme for the traveling wave equation. Figure 6.12 shows the finite difference stencil for this scheme. We call this scheme “downwind” since we expect the wave to travel from left to right and we can think of a fictitious wind blowing the solution from left to right. Notice that we are using information from “downwind” of the point at the new time step.

Figure 6.12: The finite difference stencil for the 1D downwind scheme on the traveling wave equation.

Figure 6.13 shows the finite difference stencil for the upwind scheme. We call this scheme “up” since we expect the wave to travel from left to right and we can think of a fictitious wind blowing the solution from left to right. Notice that we are using information from “upwind” of the point at the new time step.

Figure 6.13: The finite difference stencil for the 1D downwind scheme on the traveling wave equation.

Exercise 6.61 Complete the following sentences:

  1. In the downwind finite difference scheme for the traveling wave equation, the approximate solution moves at the correct speed, but …
  2. In the upwind finite difference scheme for the traveling wave equation, the approximate solution moves at the correct speed, but …

Exercise 6.62 Neither the downwind nor the upwind solutions for the traveling wave equation are satisfactory. They completely miss the interesting dynamics of the analytic solution to the PDE. Some ideas for stabilizing the finite difference solution for the traveling wave equation are as follows. Implement each of these ideas and discuss pros and cons of each. Also draw a finite difference stencil for each of these methods.

  1. Perhaps one of the issues is that we are using first-order methods to approximate (u_t) and (u_x) . What if we used a second-order approximation for these first derivatives [ u_t approx frac <>- U_i^><2Delta t>quad ext < and >quad u_x approx frac<>^n - U_^n><2Delta x>? ] Solve for (U_i^) and implement this method. This is called the leapfrog method.
  2. For this next method let’s stick with the second-order approximation of (u_x) but we’ll do something clever for (u_t) . For the time derivative we originally used [ u_t approx frac <>- U_i^n>] what happens if we replace (U_i^n) with the average value from the two surrounding spatial points [ U_i^n approx frac<1><2>left( U_^n + U_^n ight)? ] This would make our approximation of the time derivative [ u_t approx frac <>- frac<1><2>left( U_^n + U_^n ight)>. ] Solve this modified finite difference equation for (U_i^) and implement this method. This is called the Lax-Friedrichs method.
  3. Finally we’ll do something very clever (and very counter intuitive). What if we inserted some artificial diffusion into the problem? You know from your work with the heat equation that diffusion spreads a solution out. The downwind scheme seemed to have the issue that it was bunching up at the beginning and end of the wave, so artificial diffusion might smooth this out. The Lax-Wendroff method does exactly that: take a regular Euler-type step in time [ u_t approx frac <>- U_i^n>, ] use a second-order centered difference scheme in space to approximate (u_x) [ u_x approx frac<>^n - U_^n><2Delta x>, ] but add on the term [ left( frac<2Delta x^2> ight) left( U_^n - 2 U_i^n + U_^n ight) ] to the right-hand side of the equation. Notice that this new term is a scalar multiple of the second-order approximation of the second derivative (u_) . Solve this equation for (U_i^) and implement the Lax-Wendroff method.

Table of contents

1. Ordinary Differential Equations

1.3 Complete Solutions of Linear Equations

1.4 The Linear Differential Equation of First Order

1.5 Linear Differential Equations with Constant Coefficients

1.6 The Equidimensional Linear Differential Equation

1.7 Properties of Linear Operators

1.8 Simultaneous Linear Differential Equations

1.9 particular Solutions by Variation of Parameters

1.11 Determination of Constants

1.12 Special Solvable Types of Nonlinear Equations

2.1 An introductory Example

2.2 Definition and Existence of Laplace Transforms

2.3 Properties of Laplace Transforms

2.7 Use of Table of Transforms

2.8 Applications to Linear Differential Equations with Constant Coefficients

3. Numerical Methods for Solving Ordinary Differential Equations

3.4 The Modified Adams Method

3.7 Extrapolation with Differences

4. Series Solutions of Differential Equations: Special Functions

4.1 Properties of Power Series

4.3 Singular Points of Linear Second-Order Differential Equations

4.4 The Method of Frobenius

4.5 Treatment of Exceptional Cases

4.6 Example of an Exceptional Case

4.7 A Particular Class of Equations

4.9 Properties of Bessel Functions

4.10 Differential Equations Satisfied by Bessel Functions

4.11 Ber and Bei Functions

4.13 The Hypergeometric Function

4.14 Series Solutions Valid for Large Values of x

5. Boundary-Value Problems and Characteristic-Function Representations

5.4 Buckling of Long Columns Under Axial Loads

5.5 The Method of Stodola and Vianello

5.6 Orthogonality of Characteristic Functions

5.7 Expansion of Arbitrary Functions in Series of Orthogonal Functions

5.8 Boundary-Value Problems Involving Nonhomogeneous Differential Equations

5.9 Convergence of the Method of Stodola and Vianello

5.10 Fourier Sine Series and Cosine Series

5.11 Complete Fourier Series

5.12 Term-by-Term Differentiation of Fourier Series

6.1 Elementary Properties of Vectors

6.2 The Scalar Product of Two Vectors

6.3 The Vector Product of Two Vectors

6.5 Differentiation of Vectors

6.6 Geometry of a Space Curve

6.9 Differentiation Formulas

6.11 The Potential Function

6.13 Interpretation of Divergence. The Divergence Theorem

6.15 Interpretation of Curl. Laplace's Equation

6.17 Orthogonal Curvilinear Coordinates

6.18 Special Coordinate Systems

6.19 Application to Two-Dimensional Incompressible Fluid Flow

6.20 Compressible Ideal Fluid Flow

7. Topics in Higher-Dimensional Calculus

7.1 Partial Differentiation. Chain Rules

7.2 Implicit Functions. Jacobian Determinants

7.4 Jacobians and Curvilinear Coordinates. Change of Variables in Integrals

7.7 Constraints and Lagrange Multipliers

7.8 Calculus of Variations

7.9 Differentiation of Integrals Involving a Parameter

7.10 Newton's Iterative Method

8. Partial Differential Equations

8.1 Definitions and Examples

8.2 The Quasi-Linear Equation of First Order

8.3 Special Devices. Initial Conditions

8.4 Linear and Quasi-Linear Equations of Second Order

8.5 Special Linear Equations of Second Order, with Constant Coefficients

8.6 Other Linear Equations

8.7 Characteristics of Linear First-Order Equations

8.8 Characteristics of Linear Second-Order Equations

8.9 Singular Curves on Integral Surfaces

8.10 Remarks on Linear Second-Order Initial-Value Problems

8.11 The Characteristics of a Particular Quasi-Linear Problem

9. Solutions of Partial Differential Equations of Mathematical Physics

9.3 Steady-State Temperature Distribution in a Rectangular Plate

9.4 Steady-State Temperature Distribution in a Circular Annulus

9.6 Axisymmetrical Temperature Distribution in a Solid Sphere

9.7 Temperature Distribution in a Rectangular Parallelepiped

9.8 Ideal Fluid Flow about a Sphere

9.9 The Wave Equation. Vibration of a Circular Membrane

9.10 The Heat-Flow Equation. Heat Flow in a Rod

9.11 Duhamel's Superposition Integral

9.13 The Pulsating Cylinder

9.14 Examples of the Use of Fourier Integrals

9.15 Laplace Transform Methods

9.16 Application of the Laplace Transform to the Telegraph Equations for a Long Line

9.17 Nonhomogeneous Conditions. The Method of Variation of Parameters

9.18 Formulation of Problems

9.19 Supersonic Flow of ldeal Compressible Fluid Past an Obstacle

10. Functions of a Complex Variable

10.1 Introduction. The Complex Variable

10.2 Elementary Functions of a Complex Variable

10.3 Other Elementary Functions

10.4 Analytic Functions of a Complex Variable

10.5 Line Integrals of Complex Functions

10.6 Cauchy's Integral Formula

10.9 Singularities of Analytic Functions

10.10 Singularities at Infinity

10.11 Significance of Singularities

10.13 Evaluation of Real Definite Integrals

10.14 Theorems on Limiting Contours

10.16 Integrals Involving Branch Points

11. Applications of Analytic Function Theory

11.2 Inversion of Laplace Transforms

11.3 Inversion of Laplace Transforms with Branch Points. The Loop Integral


Attendance and accomodations for excused absences

Attendance to lectures is recommended, but does not directly influence the course grade.

For absences from required activities (exams and homework), legitimate excuses include illness, family emergency, religious holidays, and obligations to officially-recognized university groups, including sports teams. See the policy in the Student Code, Article 1, Part 5. This policy requires that "For excused absences, &hellip [t]he student must make arrangements with the instructor to make up missed work expeditiously." So please let the instructor know as soon as possible if you will miss a test or homework assignment.


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