Question

Need help understanding successive approximation method. Please.
x'= f(t,x)
x(t_0)=x_0
then

Answer 1

\[\dfrac{dx}{dt}=f(t,x)\\dx=f(t,x)dt\\
\int_{x_0}^{x} dx=\int_{t_0}^{t} f(s,x(s))ds \]

Answer 2

I don't get it, Let take a concrete example
x'=x
x(0) = 1
then what is f(s,x(s) ) here?

Answer 3

@Michele_Laino

Answer 4

@myininaya

Answer 5

hmmm... I think they did that so the limits wouldn't be the same as the thing we are integrating with respect to but they didn't that for the other side...
\[x'=x \\ \frac{dx}{dt}=x \\ \frac{dx}{x}=dt \\ \int\limits_{x_0}^x \frac{du}{u}=\int\limits_{t_0}^{t} ds\]

Answer 6

I am working in Matlab and trying to understand the code. :)

Answer 7

well that is the only thing that makes sense to me
but I don't why they wouldn't also apply it to the other side
that is write something like
\[\int\limits_{x_0}^x du=\int\limits_{t_0}^{t} f(s,x(s)) ds\]

Answer 8

but they use exactly the same f, right?
Hence f(s,x(s)) must relate to f, but how?

Answer 9

Let see, the solution of x'=x is x=e^t

Answer 10

hence if \(t_0=0\), then \(x(t_0)=x_0=1\)

Answer 11

if t=1, then \(x(t_1)=x(1)=e\)

Answer 12

oops I tried to separate the variables above...

\[dx= x dt \\ \text{ here } f(t,x(t))=x(t) \\ \text{ and } f(s,x(s))=x(s) \\ \int\limits_{x_{0}}^{x} du= \int\limits_{t_0}^{t} f(s,x(s)) ds \\ \int\limits_{x_{0}}^{x} du=\int\limits_{t_0}^t x(s) ds\]

Answer 13

\[\int_{x_0}^{x}dx=\int_{t_0}^{t} f(s,x(s) dx\]

Answer 14

another example
\[\frac{dx}{dt}=x-t \\ dx=(x-t) dt \\ f(t,x)=x-t \\ \text{ so } f(s,x)=x-s \]

Answer 15

\[\int_{1}^{e}dx=\int_0^1ofwhat? ds\]

Answer 16

did you still have any question?

Answer 17

like I think I answered that last one already

Answer 18

No, it is not that.

Answer 19

\[dx= x dt \\ \text{ here } f(t,x(t))=x(t) \\ \text{ and } f(s,x(s))=x(s) \\ \int\limits\limits_{x_{0}}^{x} du= \int\limits\limits_{t_0}^{t} f(s,x(s)) ds \\ \int\limits\limits_{x_{0}}^{x} du=\int\limits\limits_{t_0}^t x(s) ds \]
I just didn't put your limits that you gave in yet

Answer 20

x(s) is x by the way

Answer 21

we are just saying x is a function of s when writing x(s)

Answer 22

another example
\[\text{ if } f(t,x)=x^2t +t \text{ then } f(s,x)=x^2s+s\]

Answer 23

To your example, just replace t by s. The answer from the handout is \(\int_0^t f(s,x(s) ds =t\)

Answer 24

Let I post my handout

Answer 25

k

Answer 26

oh so you are wondering about
\[\phi_k(s) \]
right?

Answer 27

\[\phi_1(t)=x_0+\int\limits_0^t f(s,\phi_0(t)) \\ \text{ where } \phi_0(t)=x_0\]

Answer 28

so our x_0 here is 1

Answer 29

\[\phi_1=1+\int\limits_0^t f(s,1) ds \\ \text{ and } f(s,x)=x \\ \text{ so } f(s,1)=1\]

Answer 30

\[\phi_1=1+\int\limits_0^t 1 st \\ \phi_1=1+t|_0^1=1+t \\ \text{ then } \\ \phi_2=x_0+\int\limits_0^t f(s,\phi_1) ds\]

Answer 31

\[\phi_1=1+t \\ \text{ and remember} f(s,x)=x \\ \text{ so } f(s,1+s)=1+s\]

Answer 32

\[x_0=1 \\ \phi_2=x_0+\int\limits_0^t f(s,\phi_1) ds \\ \phi_2 =1+\int\limits_0^t (1+s) ds\]

Answer 33

different example
\[\text{ say } x'=e^t x \\ x(0)=e \\ \text{ so } \phi_1=x_0+\int\limits_0^t f(s,\phi_0) d s \\ \phi_0=x_0=e \\ \text{ and } f(t,x)=e^t x \\ \text{ so } f(s,\phi_0)=f(s,e)=e^{s}e=e^{s+1} \\ \text{ so } \\ \phi_1=e+\int\limits_0^t e^{s+1} ds\]

Answer 34

if we were to continue with this example...
\[\phi_1=e+\int\limits_0^t e^{s+1} ds \\ =e+e^{s+1}|_0^t=e+(e^{t+1}-e^{0+1})=e+e^{t+1}-e^{1} \\ \text{ so } f(s,\phi_1)=f(s,e+e^{s+1}-e^1)=e^s(e+e^{s+1}-e^1)=e^{s+1}+e^{2s+1}-e^{s+1}\]


so on to phi2
\[\phi_2=e+\int\limits_0^t f(s,\phi_1) ds \\ \text{ while } \phi_1(s)=e^{s+1}+e^{2s+1}-e^{s+1} \\ f(t,x)=e^t x \\ \text{ so } f(s,\phi_1)=e^s(e^{s+1}+e^{2s+1}-e^{s+1}) \]

Answer 35

and you keep going til you get phi_n...