What is Partial Differential Equations?

What is Partial Differential Equations?

Again by myself

Content

• Introduction

• Wave Equation and Example

• Modeling : Vibrating String, Wave Equations

– Step One :Separation of Variables Use Fourier Series

– Step Two :Satisfying the Boundary Conditions

– Step Three : Solution of the entire problem. Fourier Series

• Laplace’s Equation for PDE

– Example

Introduction

• Partial Differential Equations (PDE) are a type of differential equation, i.e., a relation involving an unknown function (or functions) of several independent variables and their partial derivatives with respect to those variables.

• We can see this phenomena such as sound or heat, electrostatics, electrodynamics, fluid flow, and elasticity.

Wave Equation

• An equation involving one or more partial derivatives of a function of two or more independent variables called a partial differential equation.

• The order of the highest derivative is called the order of the equation.

• Any PDE is linear if it is of the first degree in the dependent variable and its partial derivatives.

• If each term of such an equation contains either the dependent variable or one of its partial derivatives, the equation is said to be homogeneous. Otherwise, it is said to be nonhomogeneous.

Let see what is PDE in non-sentence manner

𝑑4𝑢

𝑑𝑥4+ 𝑢2 = 𝑥

Is a differential equation for the function U(x) depending on a single variable x only, but

𝜕𝑢

𝜕𝑡−

𝜕2𝑢

𝜕𝑥2−

𝜕2𝑢

𝜕𝑦2+ 𝑢 = 0

Is a differential equation involving a function u(t,x,y) of three variables

Wave Equations – The Problem

1. I want to place a string and stretch it into length L


1. I want to place a string and stretch it into length L

0 L


1. Wait, I want to distort it a little bit. Ok aaa

0 L


1. And then, at the some instant, say t = 0, I release this so called elastic string and of course there is a vibration going through it

0 L


1. So, we actually need to find the vibrations of the string. Is to find the deflection at u(x,t) at any point x and at any time t >0 from the equilibrium position.

0 L


1. equilibrium position…?What is that.

0 L

Definitions of equilibrium on the Web: - a stable situation in which forces cancel one another - balance: equality of distribution


1. Is to find the deflection at u(x,t) at any point x and at any time t >0 from the equilibrium position.

0 L 𝑥 𝑥 + ∆𝑥

∆𝑥 is very small yet we still want to calculate it Why? Because that is research Even the thinnest thing we still to research it. That is why during my master, I’ve just improve at least 25% improvement or 11 seconds improvement of Collision detection method

Learn English

Definitions of deflection on the Web:

• In physics deflection is the event where an object collides and bounces against a plane surface

• the amount by which a propagating wave is bent

• the movement of the pointer or pen of a measuring instrument from its zero position


Assumption:-

1. Mass of the string is constant. Perfectly elastic

2. The tension (ketegangan) is very large and thus we neglected gravitational force

3. It also perform a small traverse motions

Wave Equations – Understand it

Assumption:-

1. The string is perfectly flexible and offers no resistance to bending. Thus we apply Newton’s 2nd Law for a small portion of the string between x and x+∆𝑥. The equation is derived

2. Let 𝑇1𝑎𝑛𝑑 𝑇2 be a tension at the end points P and Q of this portion of the string


0 L 𝑥 𝑥 + ∆𝑥

𝑃 𝑄

𝛼

𝛽

𝑇1

𝑇2


Assumption:- 1. The points on the string move only in the vertical

direction, no motion in the horizontal direction. Thus the sum of the forces in the horizontal direction must be zero

Before that, remember this formula? .

A

B

C

tan 𝛾 =𝐴

𝐵

sin 𝛾 =𝐴

𝐶

cos 𝛾 =𝐵

𝐶

𝛾


Since we are getting the horizontal direction, thus

−𝑇1 cos 𝛼 + 𝑇2 cos 𝛽 = 0

0 L 𝑥 𝑥 + ∆𝑥

𝑃 𝑄

𝛼

𝛽

𝑇1

𝑇2

We assume that to the right is force in positive direction, that is why 𝑇1 is negative


But since we know that both cancel is other, so meaning that both share the same T value, thus

𝑇1 cos 𝛼 = 𝑇2 cos 𝛽 = 𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

0 L 𝑥 𝑥 + ∆𝑥

𝑃 𝑄

𝛼

𝛽

𝑇1

𝑇2

𝑊𝐻𝑒𝑛 −𝑇1 cos𝛼 + 𝑇2 cos𝛽 = 0

Let say 𝑇1 cos 𝛼 = 40𝑁 = 𝑇2 cos 𝛽 Then, −40𝑁 + 40𝑁 = 0


Now we need to calculate the vertical components . Take upward direction as a positive

.

0 L 𝑥 𝑥 + ∆𝑥

𝑃 𝑄

𝛼

𝛽

𝑇1

𝑇2

𝐵𝑦 𝑢𝑠𝑖𝑛𝑔 𝑁𝑒𝑤𝑡𝑜𝑛 2𝑛𝑑 𝐿𝑎𝑤 𝐹 = 𝑚𝐴 , 𝑡𝑕𝑢𝑠

𝑇2 sin 𝛽 − 𝑇1 sin 𝛼 = 𝜌∆𝑥 𝜕2𝑢

𝜕𝑡2


Now we need to calculate the vertical components . Take upward direction as a positive

.

0 L 𝑥 𝑥 + ∆𝑥

𝑃 𝑄

𝛼

𝛽

𝑇1

𝑇2

𝑇2 sin 𝛽 − 𝑇1 sin 𝛼 Is because we want to get the force for that region between P and Q. So the upward direction T2 is minus with T1 in order to get the force between P and Q


0 L 𝑥 𝑥 + ∆𝑥

𝑃 𝑄

𝛼

𝛽

𝑇1

𝑇2

𝐵𝑦 𝑢𝑠𝑖𝑛𝑔 𝑁𝑒𝑤𝑡𝑜𝑛 2𝑛𝑑 𝐿𝑎𝑤 𝐹 = 𝑚𝐴 , 𝑡𝑕𝑢𝑠

𝑇2 sin 𝛽 − 𝑇1 sin 𝛼 = 𝑚∆𝑥 𝜕2𝑢

𝜕𝑡2

𝑚 actually is a each point of mass during that so called equilibrium state at ∆𝑥 where the 𝑚 is actually a small density during that ∆𝑥

𝑚𝑎𝑠𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝜌 =𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑎𝑐𝑕 𝑝𝑜𝑖𝑛𝑡

∆𝑥

𝑆𝑜, 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑎𝑐𝑕 𝑝𝑜𝑖𝑛𝑡 = 𝜌∆𝑥

Definitions of density on the Web: the amount per unit size concentration: the spatial property of being crowded together

Wave Equations – Get the PDE

So, given that we have now:- 𝑇1 cos 𝛼 = 𝑇2 cos𝛽 = 𝑇

And

𝑇2 sin 𝛽 − 𝑇1 sin 𝛼 = 𝜌∆𝑥 ∗ ∆𝑥𝜕2𝑢

𝜕𝑡2

Next, we are going to divide it

𝑇2 sin 𝛽

𝑇2 cos 𝛽−

𝑇1 sin 𝛼

𝑇1 cos𝛼=

𝜌∆𝑥2

𝑇

𝜕2𝑢

𝜕𝑡2

Why we divide it? because we want to get the 𝑇1 𝑎𝑛𝑑 𝑇2 of the slopes. By simplify it

tan𝛽 − tan 𝛼 =𝜌∆𝑥2

𝑇

𝜕2𝑢

𝜕𝑡2


tan 𝛽 − tan 𝛼 =𝜌∆𝑥2

𝑇

𝜕2𝑢

𝜕𝑡2

And now, we are going to replace both tangent with their partial derivatives.

tan 𝛽 =𝜕𝑢(𝑥, 𝑡)

𝜕𝑡

And

tan 𝛼 =𝜕𝑢(𝑥 + ∆𝑥, 𝑡)

𝜕𝑡

Where did I get that? - Remember the tangent rules?


Before that, remember this formula?

.

A

B

C

tan 𝛾 =𝐴

𝐵

sin 𝛾 =𝐴

𝐶

cos 𝛾 =𝐵

𝐶

𝛾


tan 𝛾 =𝐴

𝐵

Where A is the vertical slope And B is the horizontal slope At location P. The equation actually is still u(x,t) But, it is just portion of it. A small value from u(x,t). Just a piece of it. So, we will have a differentiation of u(x,t) at the length x On P slope,

𝜕

𝜕𝑡𝑢 𝑥, 𝑡 =

𝜕𝑢 𝑥, 𝑡

𝜕𝑡

And on slope Q 𝜕

𝜕𝑡𝑢 𝑥 + ∆𝑥, 𝑡 =

𝜕𝑢 𝑥 + ∆𝑥, 𝑡

𝜕𝑡


𝑆𝑜, 𝜕𝑢 𝑥 + ∆𝑥, 𝑡

𝜕𝑡−

𝜕𝑢 𝑥, 𝑡

𝜕𝑡=

𝜌∆𝑥2

𝑇

𝜕2𝑢

𝜕𝑡2

Now, we are taking the limit as ∆𝑥 → 0

lim∆𝑥→0

1

∆𝑥2

𝜕𝑢

𝜕𝑥 𝑥+∆𝑥

−𝜕𝑢

𝜕𝑥 𝑥

=𝜕2𝑢

𝜕𝑥2

Because we must have both equation that is continuous function of two variables, x and t


But the discrete formulation of the equation of state with a finite number of mass point is just the suitable one for a numerical propagation of the string motion. The boundary conditions that we know is:-

𝑢 0, 𝑡 = 0, 𝑢 𝐿, 𝑡 = 0 Where L is the length of the string takes in the discrete formulation the for form that for the outermost points , the equation of motion is:-

𝜕2𝑢

𝜕𝑡2 =𝑐

∆𝑥

2

𝑢 𝑥 + ∆𝑥, 𝑡 − 𝑢 𝑥, 𝑡

Where

𝑐 =𝑇

𝜌


𝑆𝑜, By using the limit we can obtain finally the equation

𝜕2𝑢

𝜕𝑥2=

𝜌

𝑇

𝜕2𝑢

𝜕𝑡2

Arrange it back:- 𝜕2𝑢

𝜕𝑡2=

𝑇

𝜌

𝜕2𝑢

𝜕𝑥2

Thus 𝜕2𝑢

𝜕𝑡2= 𝑐2

𝜕2𝑢

𝜕𝑥2

Where 𝑐 =𝑇

𝜌

This is one-dimensional wave equation, which is 2nd order, homogeneous and hyperbolic type.

Let’s solve PDE

Given the equation of:- 𝜕2𝑢

𝜕𝑡2= 𝑐2

𝜕2𝑢

𝜕𝑥2

From above, we can clearly say that

“It has two independents”, So the function is:- 𝑢(𝑥, 𝑡)

Let’s solve PDE

By looking at the graph, we have a string fixed at x = 0, and x = L. Remember, t is still equal to zero.

Thus:- 𝑢 0, 𝑡 = 0, 𝑢 𝐿, 𝑡 = 0

For all t

X is zero and L, so the function is actually equal to zero because we still don’t release it yet

Let’s solve PDE

However, we will have:- initial deflection (at t = 0) initial velocity Now we need to denote them by giving them some notation:- Where:-

𝑢 𝑥, 0 = 𝑓 𝑥 And

𝜕𝑢

𝜕𝑡 𝑡=0

= 𝑔 𝑥

g(x) is a velocity (initial velocity) And f(x) is the deflection

Let’s solve PDE

Step 1:

We now proceed with the very first step (Separation of variables) to obtain two ordinary differential equations

Step 2:

Determine solutions of those two ODE that satisfy the boundary conditions

Step 1: Let’s Solve PDE

We have what? A function 𝑢(𝑥, 𝑡)

What we are going to do with it?

It is stated that we are going to make it separable:-

𝑢 𝑥, 𝑡 = 𝑋 𝑥 𝑇 𝑡

And this is how PDE define “Make It Separable”

Remember that


Now we need to substitute the: 𝑢 𝑥, 𝑡 = 𝑋 𝑥 𝑇 𝑡

Or perhaps 𝑋(𝑥) => 𝑋, 𝑎𝑛𝑑 𝑇(𝑡) => 𝑇

into-> 𝜕2𝑢

𝜕𝑡2= 𝑐2

𝜕2𝑢

𝜕𝑥2

Thus is become 𝜕2 𝑋𝑇

𝜕𝑡2= 𝑐2

𝜕2 𝑋𝑇

𝜕𝑥2


𝜕2 𝑋𝑇

𝜕𝑡2= 𝑐2

𝜕2 𝑋𝑇

𝜕𝑥2

But, remember that T is always want to “stick” with their friends. So, put aside X from the equation at the LHS equation and Put aside T from the RHS equation. Thus become

𝜕2 𝑇

𝜕𝑡2𝑋 = 𝑐2

𝜕2 𝑋

𝜕𝑥2𝑇


𝜕2 𝑇

𝜕𝑡2𝑋 = 𝑐2

𝜕2 𝑋

𝜕𝑥2𝑇

Right now each want to go to their “hometown” So, we let them:-

𝜕2 𝑇

𝜕𝑡2∗

1

𝑇= 𝑐2

𝜕2 𝑋

𝜕𝑥2∗

1

𝑋

We are going to simplify this two fellas 𝜕2 𝑇

𝜕𝑡2 = 𝑇′′ and 𝜕2 𝑋

𝜕𝑡2 = 𝑋′′

Thus 𝑇′′

𝑇=

𝑐2𝑋′′

𝑋


𝑇′′

𝑇=

𝑐2𝑋′′

𝑋

Right now, that c power of two want to go the left. We let them go

𝑇′′

𝑐2𝑇=

𝑋′′

𝑋


Both side right now at their own “house”.

But wait, both equation want to declare themselves as equal to some constant, they always like a letter “k’ because that is why they want it. Again, we let them.

𝑇′′

𝑐2𝑇=

𝑋′′

𝑋= 𝑘

However, they want to become separate and then…


𝑇′′

𝑐2𝑇=

𝑋′′

𝑋= 𝑘

However, they want to become separate and then…

𝑇′′

𝑐2𝑇= 𝒌

𝑇′′ = 𝑘𝑐2𝑇 𝑇′′ − 𝑘𝑐2𝑇 = 0

Wow, now I’m an ODE. I’m No.1

𝑋′′

𝑋= 𝑘

𝑋′′ = 𝑘𝑋 𝑋′′ − 𝑘𝑋 = 0

Wow, now I’m also an ODE. I’m no.2


Since we already convert their way of life to X(x) and T(t). So we must make sure that u=XT satisfies the boundary condition. Err..do you know what I’m talking about? THIS!!!

𝑢 0, 𝑡 = 0, 𝑢 𝐿, 𝑡 = 0 So, That means, Remember this one?

𝑢 𝑥, 𝑡 = 𝑋 𝑥 𝑇 𝑡 Just change it to:-

0 = 𝑢 0, 𝑡 = 𝑋 0 𝑇 𝑡 , 𝑤𝑕𝑒𝑟𝑒 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡 𝑋 0 = 0 0 = 𝑢 𝐿, 𝑡 = 𝑋 𝐿 𝑇 𝑡 , 𝑤𝑕𝑒𝑟𝑒 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑔𝑒𝑡 𝑋 𝐿 = 0

Because?See the red circles.No matter what the T(t), we will get X(0)=X(L)=0


It can be shown that for 𝑘 ≥ 0, only trivial solutions exists.

Trivial Solutions? What is that stuff?

Well, let’s go to the next slide


• A solution or example that is ridiculously simple and of little interest. Often, solutions or examples involving the number 0 are considered trivial. Nonzero solutions or examples are considered nontrivial.

• For example, the equation x + 5y = 0 has the trivial solution x = 0, y = 0. Nontrivial solutions include x = 5, y = –1 and x = –2, y = 0.4.

http://www.mathwords.com/s/solution.htm

http://www.mathwords.com/n/nonzero.htm

http://www.mathwords.com/n/nontrivial.htm


So, got the idea of trivial solution?We just don’t want it the solution to become absolute zero,why? Because that is mathematician want. So, back again

𝑋′′ − 𝑘𝑋 = 0

So, for non trivial solutions, in this PDE, we consider 𝑘 = −𝑕2 < 0

Then the solution for it becomes:-

𝑋′′ + 𝑕2𝑋 = 0


Solve using ODE 𝑋′′ + 𝑕2𝑋 = 0 𝑚2 + 𝑕2 = 0

Thus 𝑚2 = − 𝑕2

𝑚 = ±𝑕𝑖 From the table we have

𝑋 𝑥 = 𝐴𝑐𝑜𝑠 𝑕𝑥 + 𝐵𝑠𝑖𝑛 (𝑕𝑥) Remember that:- If m is complex number, then from the result,

𝛼 = 0, 𝛽 = 𝑕 Thus

𝑋 𝑥 = 𝑒0 𝐴𝑐𝑜𝑠 𝑕𝑥 + 𝐵𝑠𝑖𝑛 𝑕𝑥 = 𝐴𝑐𝑜𝑠 𝑕𝑥 + 𝐵𝑠𝑖𝑛 (𝑕𝑥)


𝑋 𝑥 = 𝐴𝑐𝑜𝑠 𝑕𝑥 + 𝐵𝑠𝑖𝑛 (𝑕𝑥) Right now we have:-

𝑋 0 = 0 𝑎𝑛𝑑 𝑋 𝐿 = 0 So,

𝑋 0 = 0 = 𝐴𝑐𝑜𝑠 0 + 𝐵𝑠𝑖𝑛 0 = 𝐴 And

𝑋 𝐿 = 0 = 𝐴𝑐𝑜𝑠 𝑕𝐿 + 𝐵𝑠𝑖𝑛 𝑕𝐿 = 0 But given that 𝐴 = 0, Thus

𝑋 𝐿 = 0 = 𝐵𝑠𝑖𝑛 𝑕𝐿 = 0 And B is never equal to zero and we need some value here just like the X(0). So,

𝑠𝑖𝑛 𝑕𝐿 = 0 And we know that 𝑠𝑖𝑛 𝑛𝜋 = 0, thus

𝑕𝐿 = 𝑛𝜋


But given that 𝐴 = 0, Thus 𝑋 𝐿 = 0 = 𝐵𝑠𝑖𝑛 𝑕𝐿 = 0

And B is never equal to zero and we need some value here just like the X(0). So,

𝑠𝑖𝑛 𝑕𝐿 = 0 Because 𝐵 × 0 = 0. 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝐵 𝑚𝑢𝑠𝑡 𝑕𝑎𝑣𝑒 𝑠𝑜𝑚𝑒 𝑣𝑎𝑙𝑢𝑒 And we know that 𝑠𝑖𝑛 𝑛𝜋 = 0, thus

𝑕𝐿 = 𝑛𝜋 Conclude:

𝑕 =𝑛𝜋

L, for n = 1,2,3, … .


We now can obtain infinitely many solutions

𝑋 𝑥 = sin𝑛𝜋𝑥

𝐿 𝑛 = 1,2,… . .

Above equation is born from

𝑋 𝑥 = 𝐵𝑠𝑖𝑛 (𝑕𝑥), putting h = 𝑛𝜋

L.

And for any x, A is zero. And for any B, is still equal to zero as long as 𝐵 > 0

And we have boundary condition 𝑢 0, 𝑡 = 0, 𝑢 𝐿, 𝑡 = 0


Now we are going to solve 𝑇′′ − 𝑘𝑐2𝑇 = 0

Right now we already choose 𝑘 = −𝑕2 < 0

And from we have :-

𝑕 =𝑛𝜋

L

Thus:-

𝑘 = −𝑛𝜋

L

2


Now we are going to solve 𝑇′′ − 𝑘𝑐2𝑇 = 0

where

𝑘 = −𝑛𝜋

L

2

Thus

𝑇′′ +𝑛𝜋

L

2

𝑐2𝑇 = 𝑇′′ +𝑐𝑛𝜋

L

2

𝑇

Let say 𝑐𝑛𝜋

L

2

= 𝑠

Then


Solve using ODE 𝑇′′ + 𝑠𝑇 = 0 𝑚2 + 𝑠 = 0

Thus 𝑚2 = − 𝑠 𝑚 = ± 𝑠𝑖

From the table we have 𝑇 𝑡 = 𝐴𝑐𝑜𝑠 𝑠𝑡 + 𝐵𝑠𝑖𝑛 ( 𝑠𝑡)


𝑇 𝑡 = 𝐴𝑐𝑜𝑠 𝑠𝑡 + 𝐵𝑠𝑖𝑛 ( 𝑠𝑡) But wait, we have only T(t) data So,

𝑇 𝑡 = 𝐴𝑐𝑜𝑠 𝑐𝑛𝜋

L

2

𝑡 + 𝐵𝑠𝑖𝑛 𝑐𝑛𝜋

L

2

𝑡

SImplify,

𝑇 𝑡 = 𝐴𝑐𝑜𝑠 𝑐𝑛𝜋

L𝑡 + 𝐵𝑠𝑖𝑛

𝑐𝑛𝜋

L𝑡

𝑎𝑛𝑑 𝑤𝑒 𝑐𝑎𝑛 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡𝑕𝑎𝑡 𝑖𝑡 𝑎𝑙𝑠𝑜 𝑡𝑕𝑒 𝑠𝑎𝑚𝑒 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑛

𝑇𝑛 𝑡 = 𝐴𝑛𝑐𝑜𝑠 𝑐𝑛𝜋

L𝑡 + 𝐵𝑛𝑠𝑖𝑛

𝑐𝑛𝜋

L𝑡


𝑎𝑛𝑑 𝑤𝑒 𝑐𝑎𝑛 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡𝑕𝑎𝑡 𝑖𝑡 𝑎𝑙𝑠𝑜 𝑡𝑕𝑒 𝑠𝑎𝑚𝑒 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑛

𝑇𝑛 𝑡 = 𝐴𝑛𝑐𝑜𝑠 𝑐𝑛𝜋


𝑐𝑛𝜋

L𝑡

And right now, we already know that 𝑢 𝑥, 𝑡 = 𝑋 𝑥 𝑇 𝑡

Just convert it to support for any ‘n’ 𝑢𝑛 𝑥, 𝑡 = 𝑋𝑛 𝑥 𝑇𝑛 𝑡

So, for any n we can combine X(x) with T(t)

𝑢𝑛 𝑥, 𝑡 = sin𝑛𝜋𝑥

𝐿𝐴𝑛𝑐𝑜𝑠

𝑐𝑛𝜋


𝑐𝑛𝜋

L𝑡

∞

𝑛=1

And FYI 𝐴𝑛 𝑎𝑛𝑑𝐵𝑛 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑢𝑠𝑖𝑛𝑔 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

Step 3: Getting into answer – Where the initial displacement is given

(deflection t = 0) 𝑢𝑛 𝑥, 𝑡 = sin

𝑛𝜋𝑥


𝑐𝑛𝜋


𝑐𝑛𝜋

L𝑡

∞

𝑛=1

And FYI 𝐴𝑛 𝑎𝑛𝑑𝐵𝑛 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑢𝑠𝑖𝑛𝑔 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

Remember that:- 𝑢 𝑥, 0 = 𝑓 𝑥

So, we going to put t=0

𝑢𝑛 𝑥, 0 = 𝑓 𝑥 = sin𝑛𝜋𝑥

𝐿𝐴𝑛

∞

𝑛=1

= 𝐴𝑛sin𝑛𝜋𝑥

𝐿

∞

𝑛=1

Wow, a fourier series. Guess what? It is Half Range Fourier Series. Where the x is from 0 until L. Remember the graph? And

𝐴𝑛 =2

𝐿 𝑓(𝑥)

𝐿

0

sin𝑛𝜋𝑥

𝐿𝑑𝑥, 𝑤𝑕𝑒𝑟𝑒 𝑛 = 1,2,3, … .

Step 3: Getting into answer – Where initial velocity is given

Remember that:- 𝜕𝑢

𝜕𝑡 𝑡=0

= 𝑔 𝑥

So, we going to differentiate



𝑐𝑛𝜋


𝑐𝑛𝜋

L𝑡

∞

𝑛=1

Hence

𝜕𝑢

𝜕𝑡= sin

𝑛𝜋𝑥

𝐿−𝐴𝑛

𝑐𝑛𝜋

L𝑠𝑖𝑛

𝑐𝑛𝜋

L𝑡 + 𝐵𝑛

𝑐𝑛𝜋

L𝑐𝑜𝑠

𝑐𝑛𝜋

L𝑡

∞

𝑛=1

Put t = 0

𝑔 𝑥 =𝜕𝑢

𝜕𝑡 𝑡=0

= sin𝑛𝜋𝑥

𝐿𝐵𝑛

𝑐𝑛𝜋

L

∞

𝑛=1

= 𝐵𝑛

𝑐𝑛𝜋

Lsin

𝑛𝜋𝑥

𝐿

∞

𝑛=1


Simplify it Put t = 0

𝑔 𝑥 =𝜕𝑢

𝜕𝑡 𝑡=0

= sin𝑛𝜋𝑥

𝐿𝐵𝑛

𝑐𝑛𝜋

L

∞

𝑛=1

= 𝐵𝑛

𝑐𝑛𝜋

Lsin

𝑛𝜋𝑥

𝐿

∞

𝑛=1

Then

𝐵𝑛

𝑐𝑛𝜋

L= 𝑙𝑒𝑡 𝑠𝑎𝑦 𝐽𝑛

And it is also a Half Range Fourier Series


Because in fourier We suppose to have

𝐵𝑛 =2

𝐿 𝑓(𝑥)

𝐿

0

sin𝑛𝜋𝑥

𝐿𝑑𝑥

SO, right no, we replacing 𝐵𝑛 = 𝐽𝑛

And 𝑓 𝑥 = 𝑔 𝑥

So, what we have is

𝐽𝑛 =2

𝐿 𝑔(𝑥)

𝐿

0

sin𝑛𝜋𝑥

𝐿𝑑𝑥

And replacing back the 𝐽𝑛, we will have

𝐵𝑛

𝑐𝑛𝜋

L=

2

𝐿 𝑔(𝑥)

𝐿

0

sin𝑛𝜋𝑥

𝐿𝑑𝑥


Simplify it

𝐵𝑛

𝑐𝑛𝜋

L=

2

𝐿 𝑔(𝑥)

𝐿

0

sin𝑛𝜋𝑥

𝐿

𝐵𝑛 =2

𝑐𝑛𝜋 𝑔(𝑥)

𝐿

0

sin𝑛𝜋𝑥

𝐿𝑑𝑥

Step 3: Getting into answer

We going to recalled back the equation



𝑐𝑛𝜋


𝑐𝑛𝜋

L𝑡

∞

𝑛=1

Where

𝐴𝑛 =2

𝐿 𝑓(𝑥)

𝐿

0

sin𝑛𝜋𝑥

𝐿𝑑𝑥, 𝑤𝑕𝑒𝑟𝑒 𝑛 = 1,2,3, … .

And

𝐵𝑛 =2

𝑐𝑛𝜋 𝑔(𝑥)

𝐿

0

sin𝑛𝜋𝑥

𝐿𝑑𝑥, 𝑤𝑕𝑒𝑟𝑒 𝑛 = 1,2,3, … .

THE TOTAL SOLUTION EXISTS WHEN

We know (or assume) that 1: When: 𝑢 𝑥, 0 = 𝑓 𝑥 ≠ 0

And 𝜕𝑢

𝜕𝑡 𝑡=0

= 𝑔 𝑥 = 0

Thus



𝑐𝑛𝜋

L𝑡

∞

𝑛=1

Where we need to find just 𝐴𝑛

THE TOTAL SOLUTION EXISTS WHEN

We know (or assume) that 2: When: 𝑢 𝑥, 0 = 𝑓 𝑥 = 0

And 𝜕𝑢

𝜕𝑡 𝑡=0

= 𝑔 𝑥 ≠ 0

Thus


𝐿𝐵𝑛𝑠𝑖𝑛

𝑐𝑛𝜋

L𝑡

∞

𝑛=1

Where we need to find just 𝐵𝑛

What is Partial Differential Equations?

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