Silly String Derivation of the Wave Equation
Silly String Derivation of the Wave Equation
Silly String Derivation of the Wave Equation
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<strong>Silly</strong> <strong>String</strong>Our modeling presentation deals with <strong>the</strong> wave equation for a vibrating string. For ourmodel, we will use <strong>the</strong> motion <strong>of</strong> a string that has been plucked to represent this wave. The firstthing we will do is describe <strong>the</strong> vibrating string and <strong>the</strong> physics surrounding it. Next, we willderive <strong>the</strong> wave equation using basic modeling assumptions and a decomposition method <strong>of</strong>equation derivation. Finally, we will find <strong>the</strong> solution to <strong>the</strong> wave equation using partialdifferential equations, <strong>the</strong> multivariate chain rule, D'Alembert's solution, and infinite and finitestring cases.A string can be defined as a rigid body whose o<strong>the</strong>r dimensions are relatively small whencompared with its length. The string for our model will be stretched between two fixed pegs thatare separated by a distance L, where L will represent <strong>the</strong> length. The force that <strong>the</strong>se two pegsexert when pulling <strong>the</strong> string will be <strong>the</strong> tension. For our model, we will assume constanttension. Next, density can be defined as <strong>the</strong> ratio <strong>of</strong> an object's mass compared to its volume.For a string though, density will be <strong>the</strong> mass per unit length. Also we are assuming near constantdensity for our model.<strong>Derivation</strong> <strong>of</strong> <strong>the</strong> <strong>Wave</strong> <strong>Equation</strong>This part <strong>of</strong> <strong>the</strong> models presentation will show how to get <strong>the</strong> one-dimensional wave equation <strong>of</strong>a vibrating string. Some steps that will need to be done to calculate <strong>the</strong> equation are some basicassumptions, a quick review <strong>of</strong> Newton’s Second Law <strong>of</strong> Physics, prerequisite equations, and<strong>the</strong>n derive <strong>the</strong> equation.There are a couple <strong>of</strong> assumptions that are going to be need for <strong>the</strong> wave equation model.One will be to understand that for any point on <strong>the</strong> string from 0 to L will be traveling in atraverse motion, or perpendicular to <strong>the</strong> X-axis. The second assumption will be that we aregoing to assume that <strong>the</strong> density <strong>of</strong> <strong>the</strong> string is constant at all times. For this model will assumethat <strong>the</strong> density is equal to one. Ano<strong>the</strong>r assumption is that <strong>the</strong> initial deformation <strong>of</strong> <strong>the</strong> stringwill be small. The last assumption is that <strong>the</strong> tension for every point on <strong>the</strong> string will be tangentto <strong>the</strong> string at be constant, which will be equal to one also.A quick review is need about Newton’s Second Law <strong>of</strong> Physics, which we will beapplying to last part <strong>of</strong> are deriving <strong>the</strong> equation. Newton’s Second Law <strong>of</strong> Physics is equal to F= ma, which is equal to force, mass, and acceleration, respectively.Next for <strong>the</strong> preparation <strong>of</strong> <strong>the</strong> equation some prerequisite will need to be established.The figure below shows how <strong>the</strong> vector T equation was created:y = f(x)TθAngle <strong>of</strong> Inclination
T= 1+1dydx2i+1+dydxdydx2jTNext a small chunk <strong>of</strong> <strong>the</strong> string is needed to calculate <strong>the</strong> one-dimensional wave equation. Asshow in <strong>the</strong> diagram below:Horizontal∆SForcesVerticalForcesSince all points on <strong>the</strong> string are going in a vertical motion <strong>the</strong> horizontal force <strong>of</strong> <strong>the</strong> vector Tcan be dropped. Hence, from all <strong>the</strong> information calculated so far <strong>the</strong> following equation can beestablished:T ∂u∂x ∂u1+ ∂x( x + ∆x,t)( x + ∆x,t)2−∂u∂x ∂u1+ ∂x( x,t)2((x,t) ∂u∂uAlso since <strong>the</strong> initial deformation <strong>of</strong> <strong>the</strong> string was small <strong>the</strong> ( x + ∆x,t)and <strong>the</strong> ((x,t)will∂x∂xget small, and since <strong>the</strong>re squared <strong>the</strong>y will go to zero faster. Plus, <strong>the</strong> tension T and <strong>the</strong> densityρ is equal to one, <strong>the</strong>y can be dropped by <strong>the</strong> equation. This will <strong>the</strong>n give us <strong>the</strong> equation:∂u∂x∂u∂x( x + ∆xt) − ( x,t) = ∆s( x,t)∂,22∂tuNet forcemassacceleration
Hence Newton’s Second Law <strong>of</strong> Physics is show here. The last thing that is needed to be doneis divide both sides by ∆x, which will <strong>the</strong>n equal <strong>the</strong> one-dimensional wave equation:∂2∂xu2( )∂ ux, t = ( x,t)∂t22<strong>Derivation</strong> <strong>of</strong> two-dimensional wave equationThe next part <strong>of</strong> our modeling is coming up with <strong>the</strong> basis for our two-dimensional waveequation. The model is based on a second order homogeneous partial differential equation in <strong>the</strong>form <strong>of</strong>:222∂ y ∂ y ∂ y ∂y∂yA + B + C + D + E + Fy = 022∂x∂x∂y∂t∂x∂tThere are three different classifications <strong>of</strong> partial differential equations. Using <strong>the</strong> equationabove if ∆ = B*B – AC, and ∆>0 <strong>the</strong> equation is Hyperbolic, If ∆
That gives:F = f +2 2( u,v)= uv uLooking at a Variable Dependency Diagram you will see that:∂F∂f∂g∂f∂u= +∂x∂u∂x∂v∂xWhich is equal tox(( v 2 + 2u)(y cos( x))+ (2uv)eWith substitution this is broken down tox 2x x( e ) + 2ysin( x)(y cos( x))+ 2( y sin( x)e eThat shows <strong>the</strong> multivariable chain rule for <strong>the</strong> first derivative. The multivariable chain rule for<strong>the</strong> second derivative is very close to that so we will use <strong>the</strong> same explanation to show it.The partial differential equation that we are going to work with is set up in <strong>the</strong> following matter:ξ = x – t and η = x + tSo <strong>the</strong>refore ξ + η = 2x and η - ξ = 2tNow using <strong>the</strong> Multi-Variable Chain Rule which we just established we have∂u∂u∂u= − +∂t∂ξ ∂ηSo <strong>the</strong>refore <strong>the</strong> partial <strong>of</strong> that is equal to2∂ u ∂ ∂u∂u= [ − + ]2∂t∂t∂ξ ∂ηUsing algebra to simplify this you get2∂ u2∂t=2∂ u2∂ξ2∂ u− 2 =∂η∂ξ2∂ u2∂ηAnd now with respect to x∂u∂u∂u= − +∂x∂ξ ∂ηSo <strong>the</strong>refore <strong>the</strong> partial <strong>of</strong> this equation is equal to2∂ u ∂ ∂u∂u= [ − + ]2∂t∂t∂ξ ∂ηUsing algebra to simplify this equation you get2∂ u2∂x=2∂ u2∂ξ2∂ u+ 2 =∂η∂ξ2∂ u2∂η
Now if you substitute2 2∂ u ∂ u=2 2∂x∂tYou end up getting that2∂ u4 = 0∂η∂ξSo we finally come up with2∂ u= 0∂η∂ξWhen ξ = x - t and η = x + tD’Alambert’s SolutionD’Alambart’s solution is a general solution to <strong>the</strong> one-dimensional wave equation found earlier.In order to find D’Alambert’s solution, two substitutions that we will be making will be thatη = x + tξ = x – tIn order to find D’Alambert’s solution we need to find <strong>the</strong> second partial <strong>of</strong> U with respect totime and second partial <strong>of</strong> U with respect to x, which were also found earlier. Substituting <strong>the</strong>second partial <strong>of</strong> U with respect to time and x back into <strong>the</strong> one-dimension wave equation gives22 2∂ u ∂ u ∂ u− 2 +22∂ε∂ndε∂n22 2∂ u ∂ u ∂ u= + 2 +22∂ε∂ndε∂nCanceling like terms and simplifying gives2∂ u= 0∂η∂εIf we integrate with respect to Ada, we get <strong>the</strong> partial <strong>of</strong> U with respect to Xi.∂u = k (ε )∂εIf we integrate <strong>the</strong>n with respect to Xi, and simplify, we getu ( ε,η)= k(ε ) + c(η)
Relabeling in more conventional terms givesu ( ε,η)= f ( η)+ g(ε)Then substituting gives us D’Alambart’s Solution.u( x,t)= f ( x + t)+ g(x − t)Infinite <strong>String</strong> ProblemNext, we will find a solution to <strong>the</strong> infinite string problem. First we set some reasonable initialconditions.∂u( x,0)= 0∂tu( x,0)= φ(x)Next find <strong>the</strong> partial <strong>of</strong> U with respect to time∂u∂tand,=f '(x + t)− g'( x − t)∂u( x,0)= f '( x)− g'( x)∂tso,∂u( x,0)= f '( x)− g'( x)= 0∂tand also,u( x,t)= f ( x + t)+ g(x − t)remembering initial conditionsu( x,0)= f ( x)+ g(x)= φ(x)which gives us
f ( x)+ g(x)= φ(x)Now using <strong>the</strong> previous two equations, we have to solve for f and g. When we do solve for f andg, we get,c +φ(x)f ( x)=( x2) − cg(x)2= φ 2Now to find <strong>the</strong> solution to <strong>the</strong> Infinite <strong>String</strong> Problem, we need to plug f and g back into U, weget,φ(x + t)+ φ(x − t)U ( x,t)=Which is <strong>the</strong> solution to <strong>the</strong> Infinite <strong>String</strong> Problem.Infinite <strong>String</strong> ProblemNext we will find a solution to <strong>the</strong> finite string problem. In this example we will consider astring with endpoints 0 and L. For <strong>the</strong> finite string problem we will start out with <strong>the</strong> two initialconditions from <strong>the</strong> infinite string problem plus two o<strong>the</strong>rs.u( x,0)= φ(x)∂u( x,0)= 0∂tu( 0, t)= 0u( L,t)= 0Using <strong>the</strong> solution to <strong>the</strong> infinite string problem, we can considerφ(t)+ φ(−t)u( 0, t)== 02φ(L + t)+ φ(L − t)u( L,t)== 02Now we can simplify <strong>the</strong>se two equations by multiplying each side by two. Now take <strong>the</strong>equation u (0,t) and let b stand for t. Now through manipulation, we get <strong>the</strong> equationφ(b)= −φ( −b)
Then substitute L-t back in for b and come up with this equation.φ( L − t)= −φ( t − L)Which equalsφ( L + t)−φ(t − L)= 0Manipulating this equation gives usφ(x + 2L)−φ(x)= 0Which is <strong>the</strong> solution to <strong>the</strong> finite string problem. Now this is a periodic function with a period<strong>of</strong> 2L. Because this function is periodic, it is <strong>of</strong>ten modeled using a series <strong>of</strong> sine and cosine's,which is more commonly known as <strong>the</strong> Fourier series.
<strong>Silly</strong> <strong>String</strong>By: Michael JanssenJeffery RogersNathaniel SmithZebulon ZielieMay 17, 2000
References:S.L. Sobolev. Partial Differential <strong>Equation</strong>s <strong>of</strong> Ma<strong>the</strong>matical Physics.Steve Deckelman. Math Models II Pr<strong>of</strong>essor