Belajar PDE

Example

Partial Differential Equations

Involve Several

Independent Variables

du/dt

du/dx

du/dy du/dz

Dan banyak

lagi

Where to use PDE?

Sound

Heat

Electrostatics

Fluid Flow

Etc

Example PDE

𝜕𝑢

𝜕𝑡−𝜕2𝑢

𝜕𝑥2−𝜕2𝑢

𝜕𝑦2= 8

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

Soalan

• Suatu bar logam yang ditebat seluruh sisinya mempunyai panjang 1m.

• Mula-mula suhu bar logam tersebut adalah 15°𝐶 pada masa 𝑡 = 0.

• Kemudian, kedua hujung bar logam tersebut diletakkan dalam ais pada suhu pada suhu 0°𝐶.

Dapatkan ungkapan bagi suhu di titik 𝑃 yang mempunyai jarak 𝑥 m dari salah satu hujung logam pada masa 𝑡 saat setelah 𝑡 = 0.

Mari belajar persamaan haba PDE

Persamaan haba adalah:

𝜕2𝑢

𝜕𝑥2=1

𝑐2𝜕𝑢

𝜕𝑡

Apa nak kena buat

• Extract Information From Question

“ianya seperti anda amik satu bar logam dan anda letakkan ais pada kedua-dua hujung.”

Suatu bar logam yang ditebat = panjang 1m.

suhu awal bar logam 15°𝐶 pada masa 𝑡 = 0.

Kemudian, kedua hujung bar logam tersebut diletakkan dalam ais pada suhu pada suhu 0°𝐶.

Step Two : Lukis Graf

Berdasarkan graf, anda akan dapati bahawa jika diletakkan ais, maka suhu yang 15 darjah celcius itu akan menurun ke arah yang lebih bawah pada satu jangka masa t. Suhu awal masa = 0 saat, Suhu kemudian masa = t saat

1 meter 15 celcius

Maka Kita ada Nilai Awal

Based on the graph

𝑢 0, 𝑡 = 0, 𝑢 1, 𝑡 = 0, 𝑢 𝑥, 0 = 15

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

Step 1: Let’s Solve PDE

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

Or perhaps 𝑋(𝑥) => 𝑋, 𝑎𝑛𝑑 𝑇(𝑡) => 𝑇

into-> 𝜕2𝑢

𝜕𝑥2=1

𝑐2𝜕𝑢

𝜕𝑡

Thus is become 𝜕2(𝑋𝑇)

𝜕𝑥2=1

𝑐2𝜕(𝑋𝑇)

𝜕𝑡

Step 1: Let’s Solve PDE

𝜕2(𝑋𝑇)

𝜕𝑥2=1

𝑐2𝜕(𝑋𝑇)

𝜕𝑡

But, remember that T and X need to be placed correctly.

Thus become 𝜕2𝑋

𝜕𝑥2𝑇 =1

𝑐2𝜕𝑇

𝜕𝑡𝑋

Step 1: Let’s Solve PDE

𝜕2𝑋

𝜕𝑥2𝑇 =1

𝑐2𝜕𝑇

𝜕𝑡𝑋

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

𝜕2𝑋

𝜕𝑥2 ∗1

𝑋=1

𝑐2𝜕𝑇

𝜕𝑡∗1

𝑇

We are going to simplify this two fellas 𝜕2 𝑇

𝜕𝑡2= 𝑇′ and

𝜕2 𝑋

𝜕𝑡2= 𝑋′′

Thus 𝑇′

𝑐2𝑇=𝑋′′

𝑋

PASTIKAN ANDA DAPAT FAHAMKAN CONCEPT STEP 1 TADI

Step 2: Let’s Solve PDE

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

Step 2: Let’s Solve PDE

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

Step 2: Let’s Solve PDE

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.

THIS!!! 𝑢 0, 𝑡 = 0, 𝑢 1, 𝑡 = 0

Step 2: Let’s Solve PDE

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 1, 𝑡 = 𝑋 1 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(1)=0

Step 2: Let’s Solve PDE

In this PDE, we consider (REMEMBER THIS) 𝑘 = −𝑕2

Then the solution for it becomes:-

𝑋′′ + 𝑕2𝑋 = 0

Step 2: Let’s Solve PDE

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)

Step 2: Let’s Solve PDE

𝑋 𝑥 = 𝐴𝑐𝑜𝑠 𝑕𝑥 + 𝐵𝑠𝑖𝑛 (𝑕𝑥)

Right now we have:- 𝑋 0 = 0 𝑎𝑛𝑑 𝑋 1 = 0

So, 𝑋 0 = 0 = 𝐴𝑐𝑜𝑠 0 + 𝐵𝑠𝑖𝑛 0 = 𝐴

And 𝑋 1 = 0 = 𝐴𝑐𝑜𝑠 𝑕 + 𝐵𝑠𝑖𝑛 𝑕 = 0

Step 2: Let’s Solve PDE

But given that 𝐴 = 0, Thus 𝑋 1 = 0 = 𝐵𝑠𝑖𝑛 𝑕 = 0

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

𝑠𝑖𝑛 𝑕 = 0

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

for n = 1,2,3, … .

Step 2: Let’s Solve PDE

We now can obtain infinitely many solutions 𝑋 𝑥 = sin 𝑛𝜋𝑥 𝑛 = 1,2, … . .

Above equation is born from

𝑋 𝑥 = 𝐵𝑠𝑖𝑛 (𝑕𝑥), putting h =𝑛𝜋.

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 1, 𝑡 = 0

Step 2: Let’s Solve PDE

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

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

And from we have :- 𝑕 = 𝑛𝜋

Thus:- 𝑘 = − 𝑛𝜋 2

Step 2: Let’s Solve PDE

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

where 𝑘 = − 𝑛𝜋 2

Thus 𝑇′ + 𝑛𝜋 2𝑐2𝑇 = 𝑇′ + 𝑐𝑛𝜋 2𝑇

Let say 𝑐𝑛𝜋 2 = 𝑠

Then

Step 2: Let’s Solve PDE

Integrate the equation (both side) 𝑇′ + 𝑠𝑇 = 0

𝑇′

𝑇𝑑𝑡 = −𝑠 𝑑𝑡

Thus ln 𝑇 = −𝑠𝑡 + 𝑐

So, 𝑇 = 𝑒−𝑠𝑡+𝑐 = 𝑒−𝑠𝑡. 𝑒𝑐 = 𝑟𝑒−𝑠𝑡

Where 𝑟 = 𝑒𝑐 (assuming)

Step 2: Let’s Solve PDE

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 = 𝐵𝑛 sin 𝑛𝜋𝑥 𝑟𝑒−𝑐𝑛𝜋𝑡

∞

𝑛=1

= 𝐴𝑛 𝑒− 𝑐𝑛𝜋 2𝑡 sin 𝑛𝜋𝑥

∞

𝑛=1

And FYI 𝑤𝑕𝑒𝑟𝑒 𝐴𝑛 = 𝐵𝑛 ∗ r

Step 3 : Solving

• The final initial condition given was that at t = 0, u = 15

15 = 𝐴𝑛 sin 𝑛𝜋𝑥

∞

𝑛=1

(Remember) that 𝐴𝑛 = 2 × 𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 15 sin 𝑛𝜋𝑥

Step 3 : Solving

𝐴𝑛 = 2 × 15 sin 𝑛𝜋𝑥 = 30 −cos 𝑛𝜋𝑥

𝑛𝜋1…0

𝐴𝑛 = −30

𝑛𝜋cos 𝑛𝜋 − 1 =

30

𝑛𝜋1 − cos 𝑛𝜋

Solution

𝑢 𝑥, 𝑡 = 𝐴𝑛 𝑒− 𝑐𝑛𝜋 2𝑡 sin 𝑛𝜋𝑥

∞

𝑛=1

𝑢 𝑥, 𝑡

=30

𝜋 1

𝑛1 − cos 𝑛𝜋 sin 𝑛𝜋𝑥 𝑒− 𝑐𝑛𝜋

2𝑡

∞

𝑛=1